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SIMPLIFYING ACTIVITYBASED COSTING By ANNEMARIE TERESA LELKES Bachelor of Arts in Accounting Cameron University Lawton, OK 2002 Master of Science in Accounting Oklahoma State University Stillwater, OK 2004 Submitted to the Faculty of the Graduate College of the Oklahoma State University in partial fulfillment of the requirements for the Degree of DOCTOR OF PHILOSOPHY December, 2009 ii SIMPLIFYING ACTIVITYBASED COSTING Dissertation Approved: Dr. Don R. Hansen Dissertation Adviser Dr. Kevin M. Currier Dr. Robert M. Cornell Dr. Thomas S. Wetzel Dr. A. Gordon Emslie Dean of the Graduate College iii ACKNOWLEDGMENTS I wish to express sincere gratitude to my dissertation chair, Dr. Don R. Hansen, for providing much guidance on every detail of the dissertation to make sure I stayed on the right track. Additionally, I am grateful to Dr. Kevin M. Currier for his suggestions and assistance on some of the analytics of the dissertation. Appreciation is also expressed to Dr. Robert M. Cornell and Dr. Thomas S. Wetzel for their suggestions and proofreading. I am grateful to my parents, Tony and Rose, for their care, support, and encouragement throughout my life and especially through the Ph.D. program. Most importantly, I express deep thankfulness to God for helping, guiding, and giving me the ability to do the work for a Ph.D. iv TABLE OF CONTENTS Chapter Page 1. INTRODUCTION .....................................................................................................1 2. LITERATURE REVIEW ..........................................................................................5 Section 2.1. Development of ABC ..........................................................................5 Section 2.2. Success of ABC ...................................................................................7 Section 2.3. Implementation Issues and Limitations of ABC ..................................7 Section 2.4. AftertheFact Simplification ...............................................................8 Section 2.5. BeforetheFact Simplification ..........................................................10 Section 2.6. Motivation ..........................................................................................11 3. EQUIVALENCY ANALYSIS OF TDABC ...........................................................13 Section 3.1. Model Definitions ..............................................................................13 Section 3.2. Assumptions .......................................................................................16 Section 3.3. Equivalency Analysis.........................................................................16 Section 3.4. Time Equations and Unused Capacity ...............................................26 Section 3.5. Resource Diversity .............................................................................29 Section 3.6. Implications........................................................................................33 4. STAGE 2 SIMPLIFICATION .................................................................................34 Section 4.1. IABC Applied to Stage 2 and Model Definition ...............................34 Section 4.2. TDABC Applied to Stage 2 and Model Definition ...........................37 Section 4.3. Equivalency Analysis.........................................................................39 Section 4.4. Unused Capacity in TDABC2 ...........................................................47 Section 4.5. Implications........................................................................................48 v Chapter Page 5. STAGE 1 ERROR ANALYSIS...............................................................................50 Section 5.1. Analysis of the Maximum Error of TDABC .....................................50 Section 5.2. Examples Demonstrating Proposition V and Its Corollaries .............59 Section 5.3. Implications........................................................................................74 6. STAGE 2 ERROR ANALYSIS...............................................................................75 Section 6.1. Analysis of the Maximum Error of TDABC2 ...................................75 Section 6.2. Examples Demonstrating Proposition VI and Its Corollaries ............81 Section 6.3. Implications......................................................................................102 7. CASE ANALYSES OF THE EQUIVALENCY CONDITIONS ........................103 Section 7.1. Introduction to the Case Analyses and Assumptions .......................103 Section 7.2. Case Study 1 ....................................................................................112 Section 7.3. Case Study 2 ....................................................................................120 Section 7.4. Case Study 3 ....................................................................................127 Section 7.5. Case Study 4 ....................................................................................133 Section 7.6. Case Study 5 ....................................................................................140 Section 7.7. Case Study 6 ....................................................................................143 Section 7.8. Case Study 7 ....................................................................................146 Section 7.9. Discussion of Case Study Results ....................................................149 8. SUMMARY OF FINDINGS .................................................................................152 REFERENCES ..........................................................................................................157 vi LIST OF TABLES Table Page 1. Example Illustrating Proposition I ........................................................................19 2. Example Illustrating Corollary IIa ........................................................................23 3. Example Illustrating Corollary IIb ........................................................................24 4. Illustration Not Satisfying Proposition II ..............................................................25 5. Corollary IIa with Unused Capacity .....................................................................28 6. Resource Diversity I .............................................................................................31 7. Resource Diversity II ............................................................................................32 8. The Accuracy of the IABC2 System ....................................................................36 9. Product Example Illustrating Corollary IVa .........................................................42 10. Customer Example Illustrating Corollary IVa ......................................................44 11. Example Illustrating Corollary IVb ......................................................................45 12. Illustration Not Satisfying Proposition IV ............................................................46 13. Corollary IVa with Unused Capacity ....................................................................48 14. Corollary Va for m = n ..........................................................................................61 15. Corollary Va for m < n ..........................................................................................63 16. Corollary Va for m > n ..........................................................................................65 17. Corollary Vb for m = n..........................................................................................68 18. Corollary Vb for m < n..........................................................................................70 19. Corollary Vb for m > n..........................................................................................72 20. Corollary VIa for k = m .........................................................................................83 21. Corollary VIa for k < m .........................................................................................86 22. Corollary VIa for k > m .........................................................................................89 23. Corollary VIb for k = m ........................................................................................93 24. Corollary VIb for k < m ........................................................................................96 25. Corollary VIb for k > m ........................................................................................99 26. Names of Resources, Activities, and Cost Objects for Each Case Study ...........105 27. Case Study 1, Stage 1..........................................................................................115 28. Case Study 1, Stage 2..........................................................................................118 29. Case Study 2, Stage 1..........................................................................................122 30. Case Study 2, Stage 2..........................................................................................125 31. Case Study 3, Stage 1..........................................................................................128 32. Case Study 3, Stage 2..........................................................................................131 33. Case Study 4, Stage 1..........................................................................................135 34. Case Study 4, Stage 2..........................................................................................138 35. Case Study 5 .......................................................................................................141 36. Case Study 6 .......................................................................................................144 vii Table Page 37. Case Study 7 .......................................................................................................147 38. Comparison of Average Absolute Percentage Errors ................................................149 1 CHAPTER 1 INTRODUCTION According to Argyris and Kaplan (1994), activitybased costing (ABC) is a costing model created in the mid1980s that provides more accurate information to managers about the cost and profitability of their business processes, products, services, and customers. ABC provides more accurate cost information by exploiting causal relationships. This is made possible by recognizing that activities consume resources while cost objects (products, customers, etc.) consume activities. Thus, the cost of resources must be first assigned to activities (Stage 1 cost assignment), and then the cost of activities is assigned to cost objects (Stage 2 cost assignment). While ABC is simple in concept, it is complex and costly to implement and operate. An organization must identify and find information for all resources, activities, and their associated drivers, which can number into the hundreds. Consequently, although ABC provides greater accuracy, ABC systems are not as widely adopted as might be expected because of their size, complexity, and cost (Krumwiede 1998a, 1998b; Kaplan and Anderson 2007a). Early attempts to simplify ABC focused on reducing the number of activities and drivers used while attempting to minimize the loss in accuracy (Babad and Balachandran 1993; Homburg 2001). In effect, size and some complexity issues were reduced at the expense of accuracy. These simplified systems also considered 2 the costs to gather information for each activity/driver. However, these attempts required a full implementation of ABC before the simplification could occur. This meant that all activities and drivers had to be identified before the simplification could be done (afterthe fact simplification). If a full implementation must take place, the value of the simplification is questionable. The next major simplification effort is more recent and is a beforethefact simplification. Kaplan and Anderson (2004, 2007a) detail the complexities and costs of ABC. In general, they observe that ABC systems are expensive to build, complex to sustain, and difficult to modify or update. Specifically, they identify the following problems associated with ABC: (1) a timeconsuming and costly interviewing and surveying process is required to identify activities and the resource drivers needed to assign resource costs; (2) since subjectivity is involved in assessing the time spent on various activities, it is difficult to validate the Stage 1 cost assignments; (3) data are expensive to store, process, and report; (4) it is difficult to update the ABC model to accommodate changing circumstances; and (5) the ABC model ignores the potential for unused activity capacity. To address these problems, Kaplan and Anderson (2004, 2007a, 2007b) developed a simplified ABC system called TimeDriven ABC (TDABC). TDABC simplifies Stage 1 by devising a simpler and less timeconsuming approach to assigning resource costs to activities. TDABC provides an easy way to update the ABC model as circumstances change and only assigns the cost of used activity capacity to cost objects. Moreover, it allows an integrated view and approach to cost determination. Thus, 3 TDABC offers a number of significant advantages. However, an examination of its disadvantages and limitations has not been formally addressed. Although the usage of process time equations may reduce the number of activities relative to a fullyimplemented ABC system, TDABC ignores Stage 2 simplification. TDABC calculates activity costs and assigns these costs to cost objects similarly to that of ABC. Since TDABC does not simplify the Stage 2 cost assignment, the size and complexity of TDABC remains considerable because managing the costs and consumption ratios of hundreds of activities is cumbersome for product costing. Hence, under TDABC, Stage 1 is simplified whereas Stage 2 remains complex. Moreover, the accuracy loss of TDABC is another issue that needs to be explored. It is unlikely that TDABC can preserve the same level of accuracy of ABC in all circumstances. The purpose of this study is to extend and expand the beforethefact simplification of ABC. Additional simplification, while overcoming identified limitations of TDABC, should enhance the viability of ABC systems and, thus, represent a significant contribution to ABC literature and actual practice. Hence, the study will first explore the accuracy of TDABC relative to an ideally implemented durationbased ABC system (the benchmark). This will be shown in Chapter 3. Second, as will also be shown in Chapter 3, the study will attempt to specify the conditions that must exist for TDABC to match the ABC assignments (equivalency conditions). Assuming accuracy loss is potentially a significant problem, ways or means of modifying Stage 1 simplification to reduce the accuracy loss will be investigated in Chapter 4. Any such modifications will attempt to preserve the resolution of the problems mentioned by Kaplan and Anderson (2004, 2007a) referred to above. Third, as will be shown also in 4 Chapter 4, the study will provide a new simplified system along with the conditions of equivalency between the new system and ABC to reduce the complexity and, therefore, the cost of Stage 2. Reducing the overall cost and complexity of ABC systems should increase the likelihood of adoption. Fourth, the maximum absolute dollar error between TDABC and ABC systems will be assessed in Chapter 5, with the maximum absolute dollar error between the new simplified system, TDABC2, and ABC in Chapter 6. Finally, in Chapter 7, case studies will be used to explore the validity of the equivalency conditions using a particular company’s data. The next chapter reviews the literature regarding ABC and TDABC, which provides the background for the motivation of this study. 5 CHAPTER 2 LITERATURE REVIEW 2.1. Development of ABC Kaplan (1994) stated that in the early years of ABC, the description of ABC systems was based on an “inner logic” that claims that ABC systems are more accurate than the functionalbased (or, traditional) systems. However, this “inner logic” was not enough to cause a breakthrough for ABC. The academicians, especially Kaplan and Cooper, tried to increase the acceptance of ABC by developing two theories concerning (1) the cost (and activity) hierarchy of factory costs (indirect and support expenses) and (2) what type of resource cost ABC measures. Cooper developed the first ABC theory concerning the cost/activity hierarchy (Cooper 1990). A taxonomy (activity hierarchy) for the activity cost drivers was developed in which activities are classified as (from lowest to highest) unitlevel, batchlevel, productlevel, or facilitysustaininglevel based on the cause and effect relationships between the organizational expense and the level of the organization. Kaplan (1994) states this cost/activity hierarchy provides four advantages. First, all organizational expenses can be mapped to a particular organizational level where cause and effect relationships can be established. Second, the cost/activity hierarchy has provided “a much richer set of drivers of cost variability” (Kaplan 1994, 251). Third, 6 there is a connection between activity levels (unit, batch, product, and facility) and modern developments in operations management. Finally, the activity hierarchy is beneficial for continuous improvement and lean production. Kaplan (1994) states that this activity hierarchy theory helps managers analyze each component of overhead costs to help reduce those costs. Kaplan (1994) developed the second theory in which not all organizational expenses should be assigned to cost objects. ABC systems measure the costs of using resources, not the cost of supplying resources that financial systems measure. The cost of unused capacity is the difference between the cost of resources used and the cost of resources supplied. Once the cost of resources used is found using the ABC system, the cost of unused capacity can be determined. Thus, ABC systems do not directly measure the cost of unused capacity. Additionally, for ABC to provide relevant data, Noreen (1991) found that the cost system must be wellspecified in which the underlying cost function must satisfy three necessary and sufficient conditions. The first condition states that the total overhead cost can be partitioned into cost pools, with each cost pool depending on one activity. The second condition states that there must be a linear relationship between the cost in each cost pool and the level of activity in that cost pool. The third condition eliminates any dependency between products and eliminates joint processes, which means that the production of a product is not dependent on the production of another product. Because of these conditions and the basic intuition behind ABC, there has been some success in implementing ABC as the next section discusses. 7 2.2. Success of ABC The main reason for the success of ABC systems in the firms that adopted and implemented them is the widespread support for ABC within the firm, adequate training, and managers who understand and know ABC information (AlOmiri and Drury 2007). Additionally, research has found that ABC is adopted if 1) there is a current significant risk of cost distortions within the firm, 2) the firm is large, 3) the firm has continuous manufacturing processes as opposed to job shops, and 4) there is product diversity (Krumwiede 1998b). Furthermore, if there is a significant top management support of ABC, then ABC will most likely become integrated within the firm (Krumwiede 1998b). However, the adoption and implementation rates for ABC are low. For instance, one research study stated that the adoption rate is 29 percent (AlOmiri and Drury 2007). Another study stated that the rate is 24 percent (Krumwiede 1998b). Additionally, Gosselin (1997) gave a more informative study and divided the implementation rate from the adoption rate. He found that the adoption rate is 47.8 percent but the implementation rate is only 30.4 percent. Shields (1995) found that 75 percent of the firms that used ABC received a financial benefit. Finally, 85 percent of firms who routinely use ABC feel that it is worth it, whereas 15 percent do not think it is worth the cost (Krumwiede 1998b). The next section discusses the implementation issues and problems of ABC. 2.3. Implementation Issues and Limitations of ABC The last section discussed what drives successful ABC implementation. However, there are reasons that ABC is not successfully adopted. For instance, Krumwiede (1998b) found a strong IT system can prevent ABC adoption or the continuation of implementing it. The reason is that firms with strong IT perceive that 8 they already have enough information for decision making; thus, ABC is not worth the cost to implement it. Additionally, he found that weak top management support and insufficient training in ABC hinders implementation. Insufficient training causes employees to not understand and respect the benefits of an ABC system. Finally, some firms do not have enough patience to wait for the full benefits of implementation and that small firm size and job shops hinder ABC implementation (Krumwiede 1998a, 1998b). Along with these implementation issues, ABC poses some limitations within the system. One limitation of ABC is that the linear approach of activitybased costing provides poor estimates of actual expenditures when there is a nonlinear or discontinuous relation between the demand for and provision of resources (e.g. the resources are provided on a joint and indivisible basis) (Maher and Marais 1998). A second limitation is that an ABC system is expensive, complex, and difficult to modify/update (Krumwiede 1998a; Kaplan and Anderson 2007a). A third limitation is that ABC systems also ignore unused capacity. A fourth limitation is that workers give subjective estimates of their time spent on various activities for Stage 1 cost assignments (Kaplan and Anderson 2007a). In spite of these limitations, the main reason that firms do not implement ABC is that they feel that the perceived benefits do not outweigh the implementation costs and that ABC will not enhance the control of costs (AlOmiri and Drury 2007). Consequently, there is a tradeoff between cost and accuracy. The next two sections focus on the published research that alleviates some of these limitations. 2.4. AftertheFact Simplification Simplification research that focuses on Stage 2 simplification (activity/driver reduction) includes the research by Babad and Balachandran (1993) and Homburg 9 (2001). Babad and Balachandran (1993) developed a model to identify an optimal subset of drivers from the fully specified ABC system that takes into consideration information costs of production and accuracy. Their model allows the decision maker to specify, as a constraint, the maximum number of drivers allowed in the simplified system. This approach combines the costs of the activities corresponding to the eliminated drivers with the activity costs associated with the selected drivers, defining a new, aggregated cost pool for each selected driver. In building more aggregate cost pools, all of the associated activity costs of an eliminated driver are given to the cost pool of a corresponding selected driver. Homburg (2001) extends the Babad and Balachandran (1993) model by allowing the activity costs of the eliminated drivers to be allocated to multiple selected drivers, rather than one corresponding driver. The optimal subset of drivers is selected that minimizes accuracy loss with information costs expressed as a constraint in the model (drivers are selected that do not exceed a prespecified level of information costs). The cost pool for a selected driver is the cost of the selected driver’s associated activity plus a share of the costs of the eliminated activities. He then shows that his approach creates a simplified system with the same level of complexity as the Babad and Balachandran approach but with more accurate product costs compared to a benchmark system. The fact that Homburg’s model produces a more accurate system with no greater information cost illustrates that the Babad and Balachandran model did not identify the optimal simplified system. However, both models assume that a simplified system must sacrifice accuracy. 10 If the system has to be fully specified before it can be simplified, then there is no benefit of simplification since the firm already has a fully specified ABC system. Additionally, whenever the system has to be updated, the fullyspecified system must be updated and then simplified, which seems to be more costly and time consuming in the long run. The next section discusses some research providing a better approach: beforethe fact simplification. 2.5. BeforetheFact Simplification Kaplan and Anderson (2007a) identified a new system called TimeDriven ABC (TDABC) to alleviate some of the complexity of ABC. TDABC skips the stage of driving resource costs to activities and introduces process time equations to take care of diverse and complex transactions (Kaplan and Anderson 2007b). These time equations summarize the time it takes to perform each activity within a process. Hence, TDABC focuses on processes instead of activities, which makes the system more manageable. Kaplan and Anderson (2007a) state The TDABC model simulates the actual processes used to perform work throughout an enterprise. It can therefore capture far more variation and complexity than a conventional ABC model, without creating an exploding demand for data estimates, storage, or processing capabilities. Using TDABC, a company can embrace complexity rather than being forced to use simplified, inaccurate ABC models… (p. 8). Anderson, et al. (2007) claim that TDABC is more accurate since actual transaction data are used instead of estimates. In addition, when the process time equations are built, it is easy to determine which step within the process time equation is consuming too much time. Kaplan and Anderson (2007a) provide other benefits of TDABC over ABC. First, employees do not need to be interviewed or surveyed to allocate resource costs to activities. Second, Stage 1 cost assignment is reduced because 11 resource costs are assigned to the activities using two sets of estimates: 1) the cost of supplying resource capacity for the department (capacity cost rate) and 2) the demand for resource capacity (capacity usage rate, typically time) by each transaction processed in the department. These rates are used to allocate resource costs to activities. Third, TDABC simulates the actual processes, thus capturing more variation and complexity than does ABC without creating greater need for data estimating, storage, or processing capabilities. Fourth, the TDABC model can be updated easier. In contrast, Kaplan and Anderson (2007a, 12) mention that “ABC requires a geometric expansion to capture the increase in complexity.” Additionally, when a new activity is identified, the unit time required only needs to be estimated. The system is updated based on events instead of the calendar. Fifth, it takes only a couple of days instead of weeks to load, calculate, validate, and report findings. Finally, research has found that TDABC can incorporate unused capacity within the TDABC system (Kaplan and Anderson 2007a). Previously, researchers did not understand that unused capacity is vital in ABC systems. However, there are disadvantages. Although TDABC is simpler and cheaper than ABC, TDABC does not reduce the number of activities/drivers that a company has to keep track of for the Stage 2 cost assignments. Additionally, TDABC will not work if the time to perform the activities cannot be reliably clocked or if the activities are not performed in a repetitive manner (Sherratt 2005). 2.6. Motivation In conclusion, TDABC is a better simplification approach as opposed to the afterthe fact simplification models. In TDABC, Stage 1 cost assignment is simplified, but 12 Stage 2 remains complex and similar to the ABC system since all activity costs and their corresponding consumption ratios have to be known. The contribution that this paper will make is to prove that there is a way to simplify the ABC system considerably while maintaining accuracy when compared to the benchmark ABC system. With the simplification method, Stage 1 cost assignment is eliminated with the additional fact that the individual activity costs do not have to be known. If the individual activity costs do not have to be known, then Stage 2 cost assignment is somewhat simplified. To simplify Stage 2 further, TDABC will be modified and applied to Stage 2 as shown in Chapter 4. This simplification will eliminate the need to know the individual activity consumption ratios. The main purpose of this study is to show the limitations of TDABC and provide a simpler and cheaper beforethefact simplified system. It is possible that there are more limitations to TDABC since research has not shown the conditions in which TDABC matches a fullyspecified benchmark ABC system (the benchmark). This study will mathematically analyze those conditions in the next chapter. 13 CHAPTER 3 EQUIVALENCY ANALYSIS OF TDABC 3.1. Model Definitions In this section, the mathematical models for the ABC and TDABC are shown and used to compare the differences in cost assignments. The original models of Kaplan and Anderson are used to explore potential accuracy differences. In this study, the Stage 1 and Stage 2 models for ABC will incorporate duration drivers (timebased drivers) for easier comparison with TDABC. Assuming m activities and n resources, the Stage 1 cost assignment for ABC is modeled as follows: Σ= = n j j j aj a C t t C 1 α j n j ajC Σ= = 1 ρ , a = 1,…, m, (1) Where α a C = cost assigned to activity a under ABC; aj t = activity a’s consumption of time for resource j; j t = total time used to supply resource j (Σ= m a aj t 1 ); aj ρ = relative frequency of use of resource j by activity a (the resource consumption ratio); and Cj = total cost of resource j. 14 Equation 1 states that the total cost of an activity under the ABC system is the sum of the resource consumption ratios, aj ρ , multiplied by the corresponding resource costs, Cj. Assuming k cost objects and m activities, the model for ABC for Stage 2 cost assignment is as follows: Σ= ℑ ℑ = m a a a ia i D C 1 α α =Σ= m a ia a C 1 υ α , i = 1,…, k, (2) Where α i D = cost assigned to cost object i under ABC; α a C = total cost of activity a; ia ℑ = volume or actual absolute frequency of use of activity a by cost object i; a ℑ = total usage of activity a (Σ= ℑ k i ia 1 ); and ia υ = relative frequency of use of activity a by cost object i. Equation 2 states that the total cost of a cost object under the ABC is the sum of the activity consumption ratios, ia υ , multiplied by the corresponding activity costs, α a C . The model for TDABC Stage 1 cost assignment is given below (for simplicity only one resource pool is assumed1): Σ= = n j a aj C c t 1 τ a = cℑ , a = 1,…, m, (3) Where τ a C = cost of activity a under TDABC; 1 The analysis can be easily generalized to more than one resource pool. 15 c = cost per unit of resource time; and a ℑ = total resource time for activity a. Equation 3 states that the total cost assigned to activity a is the sum of the total resource time used by this activity multiplied by the cost per unit of time. The cost per unit of time, c, is simply the total resource cost for the pool divided by the total resource time used by all activities: T T n j j n j j t C t C c = = Σ Σ = = 1 1 , (4) Where j t = total time used to supply resource j; T C = total cost of resources; and T t = total resource time (Σ= n j j t 1 ). The model for TDABC for Stage 2 cost assignment is ia m a i a Dτ Cτυ Σ= = 1 , i = 1,…, k (5) Equation 5 states that the total cost of cost object i ( τ i D ) under TDABC is the sum of each activity cost τ a C multiplied by the corresponding activity consumption ratio ia υ . Equations (2) and (5) for the Stage 2 model for both ABC and TDABC are identical. Any differences in cost assignment between the two models are attributable to differences between α a C and τ a C . Thus, any potential accuracy loss must occur in Stage 1. Before 16 any equivalency analysis is shown, the assumptions behind the analysis are first discussed in the next section. 3.2. Assumptions Two major assumptions are needed to perform the equivalency analysis to find the necessary conditions for equivalency between TDABC and the fullyspecified, benchmark ABC. The first assumption requires a linear relationship between the cost in each cost pool and the level of activity in that cost pool (Noreen 1991). Although Maher and Marais (1998) found that a linear relationship is a limitation of ABC due to poor estimates when there is a nonlinear or discontinuous relation between the demand for and provision of resources, this assumption is fundamental to ABC and will be used for the analysis. TDABC assumes that resources are time driven; thus, the second assumption initially requires that all resources in the ABC system are assigned using duration drivers (timebased drivers). This assumption facilitates the equivalency analysis between TDABC and the benchmark ABC for Stage 1. This assumption is relaxed in Section 3.5 so that the effect of resource diversity on the equivalency conditions can be assessed. 3.3. Equivalency Analysis Differences between α a C and τ a C are highlighted by differences in the information required to calculate each value. The information set for calculating α a C is { } aj j t ,C . Detailed individual resource driver information and resource cost information are needed. Much effort and cost must be expended to gather this information through surveys, interviews, and unbundling the general ledger. The information set needed for calculating τ a C is { } T a T t ,ℑ ,C . Total time and total resource cost are readily available 17 within an existing traditional cost system. TDABC avoids the need to collect detailed information for a ℑ by 1) determining the time to perform one unit of activity; 2) determining the number of times the activity will be performed (usually defined by practical capacity); and 3) multiplying the time to perform one unit of activity by the number of times the activity will be performed. Thus, TDABC allows activity costs to be calculated without knowing individual resource drivers or individual resource costs (only total resource time and total resource cost are needed). Whether the activity cost determined by TDABC is the same as that of ABC is a critical question. It is initially assumed that all resources are time driven. Later this assumption is relaxed. First, an intermediate ABC (IABC) costing system is developed and analyzed that requires knowledge of total resource cost and individual resource drivers. Accordingly, the information set is { } aj T t ,C . The development of the IABC system helps identify the conditions required for equivalency between ABC and TDABC. In the IABC system, an activity’s cost is calculated by multiplying the activity’s average resource consumption ratio by the total resource cost: a T I a C = ρ C , (6) Where I a C = cost of activity a for the IABC system; and a ρ = n n j aj Σ= 1 ρ , the average resource consumption ratio. 18 Equivalency between the Stage 1 cost assignments of ABC and IABC is established by the following reasoning. If a resource costs more (less), it does not mean that an activity has to consume a higher (lower) proportion of that resource’s time. If this state of no linear correlation between resource consumption ratios and individual resource costs exists for every activity, then ABC and IABC are equivalent.2 This equivalency is stated by the following proposition: Proposition I: I a a Cα = C , a = 1,…, m, if and only if there is no correlation between aj ρ and j C , j = 1,…, n. Proof: First, assume there is no correlation between aj ρ and j C for each activity a (a = 1,…, m). The correlation between aj ρ and j C , cρ r , is defined as c n j j n j aj j n j aj c C n C r σ σ ρ ρ ρ ρ Σ Σ Σ = = = − = 1 1 1 , where Σ Σ = = = − n j n j aj aj 1 n 2 2 1 ρ σ ρ ρ and Σ Σ = = = − n j n j j c j n C C 1 2 σ 2 1 . If cρ r = 0, then Σ Σ Σ = = = = n j j n j aj j n j aj C n C 1 1 1 ρ ρ , which implies that I a a Cα = C . 2 Based on the linearity assumption from Section 3.2, all correlations discussed in this dissertation are linear. 19 Next, assume that I a a Cα = C . Since Σ= = n j a aj j C C 1 α ρ and Σ Σ = = = n j j n j aj I a C n C 1 1 ρ , then from the definition of cρ r , this immediately implies that cρ r = 0. QED Table 1 provides a simple illustrative example of Proposition I, using two activities. Note that when the correlation between aj ρ and j C is zero, multiplying the average consumption ratios by the total cost produces the ABC cost assignments. As shown in Table 1, for Activity 1 (A1) and Activity (A2), IABC Stage 1 cost assignments are identical to those under ABC ( α 1 C = CI 1 = $615 and = CI 2 = $585). Hence, under IABC, there is no need to know the individual resource costs. TABLE 1 Example Illustrating Proposition I Resource R1 R2 R3 R4 ABC Cost Assignmenta a ρ b IABC Cost Assignmentc rcρ A1 0.60 0.20 0.30 0.95 $615 0.513 $615 0 A2 0.40 0.80 0.70 0.05 $585 0.488 $585 0 $95 $335 $370 $400 $1,200 $1,200 a Σ= = n j j j aj a C t t C 1 α b a ρ = 4 1 Σ= n j aj ρ c a T I a C = ρ C The information set for IABC is { } aj T t ,C . IABC eliminates the need to know the individual resource costs Cj required for ABC; however, the detailed resource consumption ratios must be known. The correlation between aj ρ and j C is exploited to α 2 C 20 reduce the fineness of the ABC information set. This suggests the possibility of exploiting correlation relationships to establish equivalency between ABC and TDABC. Note that the information set for TDABC is { } T a T t ,ℑ ,C , which eliminates the need to know both Cj and taj of ABC. For TDABC, ρ t r (the correlation between aj ρ and tj) and cρ r are both needed as shown by the following proposition. Proposition II: α τ a a C = C , a = 1,…, m, if and only if t t c c cr σ r σ ρ ρ = . Proof: First assume that t t c c cr σ r σ ρ ρ = . By definition, t n j j a j n j aj t t t r σ σ ρ ρ ρ ρ Σ Σ = = − = 1 1 , where Σ Σ = = = − n j n j j t j n t t 1 2 σ 2 1 . Substitute ρ t r , cρ r , and c with their corresponding formulas and simplify to obtain j n j j aj n j n aj j j n j j t C t C Σ Σ Σ Σ = = = = = 1 1 1 1 ρ ρ . Note that j n j ajt Σ= 1 ρ is equivalent to Σ= n j aj t 1 . Hence, j n j aj n j n aj j j n j j t C t C Σ Σ Σ Σ = = = = = 1 1 1 1 ρ ⇒ τ α a a C = C . Next, assume that α τ a a C = C . From the definition of ρ t r , t t n j j a j n j aj ρ t ρ t r σ σ ρ ρ Σ − Σ = =1 =1 21 ⇒ t t n j a j n j aj t ρ t r σ σ ρ ρ Σ − Σ = =1 =1 . Since Σ Σ Σ = = = = n j n aj j j n j j a t t C C 1 1 τ 1 , then τ n a j j n j n j j aj C C t t Σ Σ Σ = = = = 1 1 1 . Substituting for Σ= n j aj t 1 provides t t n j n a a j j j n j j C t r C t ρ σ σ ρ ρ τ − Σ = Σ Σ = = = 1 1 1 ⇒ Σ Σ Σ = = = = + n j j n j j t t n j a a j t C C C r 1 1 1 ρ σ σ ρ ρ τ ⇒ t t n j a a j C ρ C cr σ σ ρ ρ τ + = Σ= 1 . From the cρ r equation, Σ Σ = = − = n j j a j c c n j aj C C r 1 1 ρ ρ σ σ ρ ρ ⇒ Σ= − = n j a a j c c C C r 1 ρ σ σ ρ ρ α ⇒ Σ= = + n j a a j c c C C r 1 ρ σ σ ρ ρ α . Accordingly, [ ] a a c c t t C C σ r σ cr σ ρ ρ ρ α − τ = − . Thus, if α τ a a C = C , then t t c c cr σ r σ ρ ρ = . QED In the proof of the above proposition, it is shown that [ ] a a c c t t C C σ r σ cr σ ρ ρ ρ α − τ = − . Interestingly, since the dollar value of the error between the two systems equals α τ a a C −C , then the dollar value of the error can be expressed as follows: a ε = [ ] c c t t σ r σ cr σ ρ ρ ρ − , a = 1,…, m (7) 22 When the error for each activity a is zero, there is equivalency, which implies that − = 0 c c t t r σ cr σ ρ ρ . This expression implies that if cρ r = 0, then ρ t r = 0. Thus, the following corollary to the Proposition II has been proved: Corollary IIa: If cρ r = 0 and ρ t r = 0, then α τ a a C = C , a = 1,…, m. In the event that both cρ r and ρ t r are nonzero, then it is also possible to establish an equivalency condition based on a required value for c. When [ ] a c c t t ε σ r σ cr σ ρ ρ ρ = − = 0, solving for c and simplifying yields the following equivalency condition: Σ Σ Σ Σ Σ Σ = = = = = = − − = − − = n j j a j n j aj I a a n j j a j n j aj n j j a j n j aj t t C C t t C C c 1 1 1 1 1 1 ρ ρ ρ ρ ρ ρ α This then establishes a second corollary: Corollary IIb: If T T n j j a j n j aj I a a t C t t C C c = − − = Σ Σ =1 =1 ρ ρ α , then α τ a a C = C , a = 1,…, m. According to Corollary IIb, if the rationale for zero correlation is not valid, it is still possible to obtain equivalency. However, a very special relationship must exist. The numerator I a a Cα −C is the dollar error between ABC and IABC for activity a. The 23 denominator Σ Σ = = − n j j a j n j aj t t 1 1 ρ ρ is the unit time error between time allocated to activity a using ABC and the time allocated to activity a using IABC. Consequently, ε α ρ ρ c t t C C n j j a j n j aj I a a = − − Σ Σ =1 =1 represents the absolute dollar error per unit of error time. Note that ε c must be written in absolute form since the IABC cost assignment for activity a could be greater than that of ABC. Table 2 provides an example illustrating Corollary IIa of Proposition II. Table 2 compares TDABC and ABC when there is no correlation between tj and aj ρ and between aj ρ and j C for each activity a. When cρ r = 0 and ρ t r = 0 for each activity, the activity costs under TDABC ( τ 1 C = $615 and τ 2 C = $585) are equal to those under ABC. TABLE 2 Example Illustrating Corollary IIa ABC Resource R1 R2 R3 R4 ABC Cost Assignmenta A1 0.60 0.20 0.30 0.95 $615 A2 0.40 0.80 0.70 0.05 $585 Cost $95 $335 $370 $400 $1,200 Time 101 300 321 350 1,072 TDABC Unit Time Total Units of Activity a ℑ c TDABC Cost Assignmentb rcρ rtρ A1 36.63 15 549 $1.12 $615 0 0 A2 26.15 20 523 $585 0 0 Total Time 1,072 $1,200 a Σ= = n j j j aj a C t t C 1 α b τ a C a = cℑ 24 Corollary IIb of Proposition II is demonstrated in the following example shown in Table 3. According to Table 3, cρ r and ρ t r are nonzero and the cost per unit of time, c, and dollar error per unit of time, ε c , are both equal to $4. This satisfies Corollary IIb so that α τ a a C = C , where τ 1 C = α 1 C = $216 and τ 2 C = α 2 C = $264. TABLE 3 Example Illustrating Corollary IIb ABC Resource R1 R2 R3 R4 ABC Cost Assignmenta a ρ IABC Cost Assignmentb A1 0.20 0.40 0.60 0.80 $216 0.50 $240 A2 0.80 0.60 0.40 0.20 $264 0.50 $240 Cost $180 $60 $180 $60 $480 $480 Time 35 40 25 20 120 TDABC Unit Time Total Units of Activity a ℑ ε c c c TDABC Cost Assignmentd rcρ rtρ A1 9 6 54 $4 $4 $216 0.45 0.85 A2 3 22 66 $264 0.45 0.85 Total Time 120 $480 a Σ= = n j j j aj a C t t C 1 α b a T I a C = ρ C c T T n j j a j n j aj I a a t C c t t C C c = = − − = Σ Σ =1 =1 ρ ρ α ε d τ a C a = cℑ If Proposition II does not hold, then there is a difference in the cost assigned to TDABC relative to that of ABC. Table 4 shows that when there is perfect correlation between tj and aj ρ and between aj ρ and j C (where t t c c cr σ r σ ρ ρ ≠ ), the activity costs 25 under TDABC ( τ 1 C = $151 and τ 2 C = $384) are not equal to those under ABC ( α 1 C = $249 and α 2 C = $286). The average absolute percentage error of TDABC is 36.7 percent, with dollar error ( a ε ) is $97.93 for A1 and $97.93 for A2 (c = $0.4977 and ε c = $0.488). TABLE 4 Illustration Not Satisfying Proposition II ABC Resource R1 R2 R3 R4 ABC Cost Assignmenta a ρ IABC Cost Assignmentb A1 0.80 0.25 0.30 0.15 $249 0.375 $201 A2 0.20 0.75 0.70 0.85 $286 0.625 $334 Cost $215 $110 $120 $90 $535 $535 Time 101 318 299 357 1,075 TDABC Unit Time Total Units of Activity a ℑ c TDABC Cost Assignmentc rcρ rtρ A1 19 16 304 $0.4977 $151 1 1 A2 25.7 30 771 $384 1 1 Total Time 1,075 $535 ABC TDABC a ε % a ε d ε c e A1 $249 $151 $98 39.24% $0.488 A2 $286 $384 ($98) 34.16% Avg a %ε f 36.7% a Σ= = n j j j aj a C t t C 1 α e cε = T T n j j a j n j aj I a a t C c t t C C ≠ = − − Σ Σ =1 =1 ρ ρ α b a T I a C = ρ C f 2 39.24% + − 34.16% c τ a C a = cℑ d% a ε = α α τ a a a C C − C 26 If cρ r and ρ t r are nonzero and TDABC and ABC are not equivalent, then the cost per unit of time would not be equal to the dollar error per unit of time. Therefore, a ε > 0 for each a. From Equation 7, it is possible to analyze the effects of various variables on the magnitude of the error. For instance, the error will be larger in absolute magnitude if cρ r and ρ t r are opposite in sign, which makes the two terms on the right hand side of Equation 7 additive. The magnitude of the error is also affected by variability in ρaj, tj, and Cj. Additional analysis of the dollar error is needed in which the maximum absolute dollar error is identified and will be shown in Chapter 5. 3.4. Time Equations and Unused Capacity Kaplan and Anderson (2007a, 2007b) stated that a process can be expressed in a process time equation that consists of all of the individual activities that make up the process. Time equations summarize the TDABC time information. Using time equations is a way of obtaining granularity (the level of detail) without having a separate activity for each event. If TDABC and ABC have the same granularity, then Proposition II holds. Time equations are based on the unit time for each activity and the number of times it is actually performed (or, actual activity used). The difference between the time equation based on practical activity and the time equation based on actual activity used is unused capacity. When unused capacity exists and the equivalency conditions are satisfied, the cost of activity a under ABC is equal to the cost of activity a under TDABC plus the cost of unused capacity for activity a. This only means that the cost of unused capacity for activity a is separated from the actual cost of the activity a used. Thus, there is no significant effect on the equivalency conditions. The next table is similar to Table 2 but 27 has been modified to incorporate unused capacity. Time equations are then developed to illustrate the summarization of the TDABC time information. Table 5 shows the same illustration as in Table 2 except that unused capacity now exists. For TDABC, the activity used represents the number of times the activity is performed. Practical activity represents the number of times the activity should be performed under normal operating conditions. Notice that, for equivalency, the ABC cost for an activity must equal the TDABC cost for an activity plus the cost of unused portion of the activity. Therefore, Table 5 illustrates that unused capacity has no significant effect on the conditions for equivalency. To develop the process time equations, assume that the illustration in Table 5 concerns an ordering department that has two activities: number of repeat orders (A1) and the number of new orders (A2). The time equation that represents the total order processing time based on actual activity used is Actual time used = 36.60(# of repeat orders used) + 26.15(# of new orders used) = 36.60(10) + 26.15(15) = 758 minutes The time equation that represents the total order processing time based on practical activity is Practical time = 36.60(# of repeat orders) + 26.15(# of new orders) = 36.60(15) + 26.15(20) = 1,072 minutes The unused capacity time is the difference between the practical time and the actual time used, which is 314 minutes (1,072 minus 758). The total cost of unused capacity is $351 (314 x $1.12). To find the cost of unused capacity for each activity, the activities would 28 have to be separated out of the time equation, and the results would be identical to those displayed in Table 5. TABLE 5 Corollary IIa with Unused Capacity Resource R1 R2 R3 R4 ABC Cost rcρ rtρ c A1 0.60 0.20 0.30 0.95 $615 0 0 $1.12 A2 0.40 0.80 0.70 0.05 $585 0 0 Cost $95 $335 $370 $400 $1,200 Time 101 300 321 350 1,072 Unit Time Activity Used Time Used Practical Activity a ℑ TDABC Cost Unused Cost Total Cost A1 36.60 10 366 15 549 $410 $205 $615 A2 26.15 15 392 20 523 $439 $146 $585 758 1,072 $849 $351 $1,200 Only Corollary IIa is shown for this analysis because if unused capacity is applied to Corollary IIb, the results are similar to the illustration in Table 3 of Corollary IIb and follow the same process as above for unused capacity. Consequently, it has been illustrated that time equations summarize the information of the TDABC system and have no bearing on the equivalency conditions since they are developed after the TDABC system has been implemented. Therefore, both unused capacity and time equations do not affect the equivalency conditions. In Section 3.3, the conditions for equivalency between ABC and TDABC assume that all resources are timedriven. However, there are, in general, resources that are not time driven (e.g. some forms of capital, materials, and some forms of energy). In TDABC, the costs of these nontimedriven resources are pooled with the costs of resources that are time driven. This resource diversity can produce inaccurate activity 29 costs. This inaccuracy can pose a major problem for TDABC if the costs of the nontime driven resources are significant. When nontimedriven resources are significant, pooling can cause inaccurate cost assignments since there would be a lack of causal relationships for nontimedriven resources. In Section 3.5, resource diversity is examined and examples are used to illustrate this problem. 3.5. Resource Diversity Resource diversity exists when there are a significant proportion of nontimedriven resources that are consumed in a different pattern from timedriven resources. In Stage 1 cost assignment, TDABC assigns the cost of all resources to the activities using timebased drivers, which means that timebased drivers are used to assign the costs of both timebased and nontimebased resources to activities. Let the set of all resources be R = {1,…, n}. Next, partition R into a set of timedriven resources, TD = {1,…, l}, and a set of nontimedriven resources, NTD = {l+1,…, n}, where R = TD∪NTD. If, on average, activities consume nontimedriven resources in the same pattern as timedriven resources, then equivalence between ABC and TDABC remains possible. As a result, n t l t n j aj a l j aj a Σ Σ = = = = = 1 1 ρ τ ρ , where ρ τa is the average consumption ratio for timedriven resources for activity a, a = 1,..., m. When ρ τa = a ρ , there is no resource diversity and Proposition II applies. However, if ρ τa ≠ a ρ , then resource diversity (RD) exists and can be measured as follows: a a RD = ρ τ − ρ , a = 1,…, m (8) 30 This suggests the possibility that as RD increases, then the potential difference between ABC and TDABC also increases. Table 6 provides an illustration of resource diversity that shows the potential inaccuracy of TDABC. There are two activities, four timedriven resources (j = 1,…, l), and four nontimedriven resources (j = l+1,…, n). In the example, the timedriven resources are the labor resources (L1 – L4), and the nontime driven resources are the materials resources (M1 and M2) and energy resources (E1 and E2). Additionally, cρ r = 0 and ρ t r = 0 so that . Although the conditions for α τ a a C = C are satisfied, α τ a a C ≠ C because of the effect of nontimebased resources. t t c c cr σ r σ ρ ρ = 31 TABLE 6 Resource Diversity I ABC Resources L1 L2 L3 L4 M1 M2 E1 E2 A1 0.25 0.70 0.15 0.10 0.80 0.40 0.30 0.50 A2 0.75 0.30 0.85 0.90 0.20 0.60 0.70 0.50 Cost $3,565 $3,400 $2,900 $1,000 $1,500 $7,000 $3,000 $2,400 Time 3,800 1,500 1,400 1,000 ABC Costa a ρ rcρ rtρ A1 $9,906 0.40 0 0 A2 $14,859 0.60 0 0 Cost $24,765 TDABC a ℑ c TDABC Costb ρ τa A1 2,310 $3.22 $7,430 0.30 A2 5,390 $17,336 0.70 7,700 $24,765 ABC TDABC a ε % a ε c A1 $9,906 $7,430 $2,477 25.0% A2 $14,859 $17,336 ($2,477) 16.7% Avg a %ε 20.8% a Σ= = n j j j aj a C t t C 1 α b τ a C a = cℑ c% a ε = α α τ a a a C C − C According to Table 6, ρ τa for A1 and A2 are 0.3 and 0.7, respectively, whereas a ρ for A1 and A2 are 0.4 and 0.6, respectively. Thus, RD = 0.1 for A1 and 0.1 for A2. The dollar value of the error a ε is $2,477 for A1 and $2,477 for A2. The average absolute percentage error of TDABC is 20.8 percent. However, if RD increases, then the 32 average absolute percentage error increases as shown in Table 7. Compared to Table 6, Table 7 shows that as RD doubles, the average absolute percentage error almost doubles. Hence, the illustration supports the claim that as RD increases, error increases. TABLE 7 Resource Diversity II ABC Resources L1 L2 L3 L4 M1 M2 E1 E2 A1 0.25 0.70 0.15 0.10 0.60 0.85 0.40 0.95 A2 0.75 0.30 0.85 0.90 0.40 0.15 0.60 0.05 Cost $3,565 $3,400 $2,900 $1,000 $3,000 $1,500 $6,234 $3,166 Time 3,800 1,500 1,400 1,000 ABC Costa a ρ rcρ rtρ A1 $12,383 0.50 0 0 A2 $12,382 0.50 0 0 Cost $24,765 TDABC a ℑ c TDABC Costb ρ τa A1 2,310 $3.22 $7,430 0.30 A2 5,390 $17,336 0.70 Total Time 7,700 $24,765 ABC TDABC a ε % a ε c A1 $12,383 $7,430 $4,953 40.0% A2 $12,382 $17,336 ($4,953) 40.0% Avg a %ε 40.0% a Σ= = n j j j aj a C t t C 1 α b τ a C a = cℑ c% a ε = α α τ a a a C C −C If resource diversity is significant, then TDABC may be significantly less accurate than ABC. One possible resolution to this problem is discussed in Chapter 4. 33 3.6. Implications The major implication that the equivalency conditions for TDABC have on research and practice is to show when the TDABC system will replicate the ABC system. The equivalency holds when the underlying conditions as outlined in Propositions I and II are satisfied and when all resources are time driven. However, when there is resource diversity, the assumption of all resources being time driven is violated. When this one assumption is violated, there is no equivalency although the conditions in Proposition I and Proposition II and its corollaries are met. This issue needs to be resolved. The next chapter provides a resolution by analyzing a Stage 2 simplification procedure in order to eliminate Stage 1 cost assignments while maintaining accurate costing of the cost objects. 34 CHAPTER 4 STAGE 2 SIMPLIFICATION 4.1. IABC Applied to Stage 2 and Model Definition The previous chapter showed the conditions for accuracy for the TDABC system. Because of the potential inaccuracy of the TDABC system when there is resource diversity, this section will look at a way to simplify ABC while resolving the potential inaccuracy issue of TDABC. One resolution method is to extend Proposition I to Stage 2. Recall that Proposition I states that when the resource costs and resource consumption ratios are not linearly correlated, then the cost of a particular activity is basically its average resource consumption ratio multiplied by the total resource cost for all resources. This can be applied to other cost objects as well. The Stage 2 Intermediate system (IABC2) uses the IABC model to resolve the TDABC resource diversity issue and simultaneously offers some simplification for Stage 2. Assuming k cost objects and m activities, the IABC2 model is described as follows: i T m a i a I i C C D υ υ α = = Σ= 1 , i = 1,…, k, (9) Where I i D = cost assigned to cost object i under IABC2; and 35 i υ = m m m a a ia m a ia Σ Σ = = ℑ ℑ 1 = 1 υ , the average activity consumption ratio of cost object i. Equation 9 states that the cost of cost object i is the average activity consumption ratio multiplied by the total cost. Accordingly, the information set for IABC2 is { } ia T ℑ ,C . For IABC2, the individual activity costs do not have to be known; only the total cost needs to be known and the individual activity consumption ratios. Since the individual activity costs do not have to be known, Stage 1 cost allocation is eliminated, which is a significant simplification and the issue of resource diversity is resolved. Equivalency between α i D and I i D is established by the following proposition: Proposition III: I i T i m a i ia a D C C D = = =Σ= α υ α υ 1 , if and only if there is no correlation between ia υ and α a C for each cost object i, a = 1,…, m and i = 1,…, k. The proof of Proposition III parallels that of Proposition I and is, therefore, omitted. Let Cυ r represent the correlation between ia υ and α a C for cost object i. Parallel to the definition of cρ r , C m a a i a m a ia C C C r σ σ υ υ υ α α υ Σ Σ = = − = 1 1 , where Σ( ) Σ = = = − m a m a a C a m C C 1 2 2 1 α σ α and Σ Σ = = = − m a m a ia ia 1 m 2 2 1 υ σ υ υ . Table 8 shows an illustration of Proposition III. The 36 illustration contains two cost objects and four activities. Since Cυ r = 0, the costs assigned to the cost objects under IABC2 are identical to those under ABC. TABLE 8 The Accuracy of the IABC2 System Stage 1 A1 A2 A3 A4 Resource Cost Labor 1 0.16 0.26 0.20 0.38 $300,000 Labor 2 0.38 0.53 0.00 0.09 $650,000 Energy 0.20 0.29 0.10 0.41 $750,000 Materials 0.10 0.20 0.30 0.40 $800,000 Cost $525,000 $800,000 $375,000 $800,000 $2,500,000 Stage 2 A1 A2 A3 A4 ABC Costa Cυ r CO1 0.20 0.30 0.69 0.80 $1,243,750 0 CO2 0.80 0.70 0.31 0.20 $1,256,250 0 Cost $525,000 $800,000 $375,000 $800,000 $2,500,000 IABC2 Costb i υ CO1 $1,243,750 0.4975 CO2 $1,256,250 0.5025 Cost $2,500,000 a Σ= = m a i ia a D C 1 α υ α b i T I i D =υ C TDABC simplifies Stage 1 cost assignment by eliminating the need to know the resource consumption ratios. However, TDABC must calculate the individual activity costs needed for Stage 2 calculations. As shown in Table 8 (which illustrates Proposition III), IABC2 eliminates the need to know activity costs for Stage 2, thus eliminating the potential problem of resource diversity introduced by TDABC. In addition, since IABC2 does not need activity costs, Stage 2 is also simplified. However, the activity 37 consumption ratios have to be found for all activities to calculate the average activity consumption ratios for each cost object. Gathering this information is timeconsuming and costly. Thus, further simplification is desirable. The next section analyzes a more desirable method in which TDABC is applied to Stage 2. 4.2. TDABC Applied to Stage 2 and Model Definition A more desirable method is to develop a simplification of Stage 2 that avoids the need to gather all of the information necessary to calculate the average activity consumption ratios. One approach is to extend TDABC concepts found in Stage 1 to Stage 2. The presence of the IABC2 model suggests the possibility that a TDABC2 model is feasible. The TDABC2 model builds on IABC2 by eliminating the need to know all of the activities and their associated consumption ratios. If TDABC concepts are transferred to Stage 2, then TDABC2 would only require knowledge of the total cost, total time, the unit cycle time, and the number of units of the cost object that will be produced. Thus, TDABC2 is performed by 1) determining the cycle time for one unit of product (e.g. from the time the sales order is received until the finished good goes to the warehouse); 2) determining the number of units that will be produced; and 3) multiplying the cycle time by the number of units that will be produced. The TDABC2 cost assignment model is as follows: i i m a m ia a a m a a i c C D β θ α Σ Σ Σ = Ζ = Ζ = ℑ = ℑ = 1 1 1 , i = 1,…, k, (10) Where Ζ i D = cost of cost object i under TDABC2; ia ℑ = time consumed of activity a by cost object i; 38 a ℑ = total time of activity a; i β = unit cycle time for cost object i; i θ = number of units produced for cost object i at practical capacity; and cΖ = cost per unit of activity time, where Z stands for TDABC2. Equation 10 states that the total cost assigned to cost object i is the unit cycle time multiplied by the number of units produced and then multiplied by the cost per unit of time. The cost per unit of time, c Ζ , is simply the total activity cost for the pool divided by the total activity time used by all cost objects: T T m a a m a a t C C c = ℑ = Σ Σ = Ζ = 1 1 α , (11) Where T C = total overhead cost; and T t = total time in the system (Σ= ℑ m a a 1 ). Additionally, from Equation 10, the cycle time multiplied by the number of units produced is the sum of the time consumed of activity a by cost object i across all a (a = 1,…, m): i i m a ia θ β = ℑ Σ= 1 , i = 1,…, k (12) Notice also from Equation 10 that the cost for one unit of a cost object is the unit cycle time multiplied by c Ζ , or i c Ζβ . Hence, the information set for TDABC2 is { } T T i i C ,t ,β ,θ . The total overhead cost ( T C ), the total time in the system ( T t ), and the 39 number of units produced for cost object i at practical capacity ( i θ ) can be found from the accounting records. The unit cycle time for cost object i ( i β ) is found by clocking how long it takes from the time the sales order is received until the finished good goes to the warehouse. 4.3. Equivalency Analysis Similar to the analysis of TDABC with the benchmark ABC for Stage 1, the assumptions behind the necessary equivalency conditions are linearity and that the Stage 2 cost assignments are duration based in benchmark ABC system. Conditions needed to establish equivalency between TDABC2 and ABC are derived from extending Proposition II and its corollaries to Stage 2. Like in Proposition II for TDABC, for TDABC2 Cυ r (the linear correlation between ia υ and α a C for cost object i) and ℑυ r (the linear correlation between ia υ and a ℑ for cost object i) are both needed. Parallel to the definition of ρ t r , ℑ = = ℑ Σ ℑ − Σℑ = σ σ υ υ υ υ m a a i a m a ia r 1 1 , where Σ Σ = = ℑ ℑ = ℑ − m a m a a a 1 m 2 σ 2 1 and υ σ is defined as before. The extension of Proposition II to Stage 2 is shown in the following proposition. Proposition IV: = Ζ i i Dα D if and only if ℑ ℑ σ = Ζ σ υ υ r c r C C , i = 1,…, k. The proof is parallel to that of Proposition II and is, therefore, omitted. 40 From the proof of the above proposition, [ ] ℑ ℑ − Ζ =σ σ − Ζ σ υ υ υ Dα D r c r i i C C . Since the dollar value of the error between the two systems equals − Ζ i i Dα D , then the dollar value of the error can be expressed as follows: i ε = [ ] ℑ ℑ σ σ − Ζ σ υ υ υ r c r C C , i = 1,…, k (13) When the error for cost object i is zero, there is equivalency, which implies that − = 0 ℑ ℑ σ Ζ σ υ υ r c r C C . This expression shows that if Cυ r = 0, then ℑυ r = 0. Using the same rationale that establishes equivalency between ABC and TDABC, ℑυ r should also equal zero since this implies that a cost object does not need to consume a higher (lower) proportion of that activity’s time if an activity has more (less) time available. Thus, the following corollary to Proposition IV has been proved: Corollary IVa: If Cυ r = 0 and ℑυ r = 0, then = Ζ i i Dα D , i = 1,…, k. In the event that both Cυ r and ℑυ r are nonzero, then it is also possible to establish an equivalency condition based on a required value for cΖ . When [ ] ℑ ℑ ε =σ σ − Ζ σ υ υ υ r c r i C C = 0, solving for cΖ and simplifying yields the following equivalency condition: Σ Σ Σ Σ Σ Σ = = = = Ζ = = ℑ − ℑ − = ℑ − ℑ − = m a a i a m a ia I i i m a a i a m a ia m a a i a m a ia D D C C c 1 1 1 1 1 1 υ υ υ υ υ υ α α α 41 This then establishes a second corollary: Corollary IVb: If T T m a a i a m a ia m a a i a m a ia t C C C c = ℑ − ℑ − = Σ Σ Σ Σ = = Ζ = = 1 1 1 1 υ υ υ α υ α , then = Ζ i i Dα D , i = 1,…, k. According to Corollary IVb, if the rationale for zero correlation is not valid, it is still possible to obtain equivalency. However, a very special relationship must exist. The numerator I i i Dα − D is the dollar error between Stage 2 ABC and IABC2 for cost object i. The denominator Σ Σ = = ℑ − ℑ m a a i a m a ia 1 1 υ υ is the unit time error between time allocated to cost object i using ABC and the time allocated to cost object i using IABC2. Accordingly, Ζ = = = ℑ − ℑ − Σ Σ ε α υ υ c D D m a a i a m a ia I i i 1 1 represents the dollar error per unit of error time for Stage 2. Again, parallel to ε c in Chapter 3, Ζ ε c must be written in absolute form since IABC2 cost for cost object i could be greater than that of ABC Stage 2. Table 9 provides an illustration of Corollary IVa in which the cost objects are product lines (P1 and P2). Only Stage 2 is shown of ABC. Table 9 compares TDABC2 and ABC when there is no correlation between ia υ and α a C and between ia υ and a ℑ for cost object i. Notice from Table 9 that once cΖ and i β (the unit cycle time for cost object i) are known, the cost per unit of product can be found. Again, i β is an observed value. If the ABC system is durationbased, i β must equal the cycle time calculated from the 42 durationbased benchmark ABC system. The cycle time from the durationbased ABC is calculated by dividing the total hours for cost object i by the number of units produced at practical capacity (e.g. 400/800 = 0.5 unit cycle time for P1). To find the cost of the entire product line, the cost per unit of product is multiplied by the number of units produced ( i θ ). Table 9 shows that the cost of each product under TDABC2 ( Ζ 1 D = $185 and Ζ 2 D = $185) are equal to those under ABC since Cυ r = 0 and ℑυ r = 0. TABLE 9 Product Example Illustrating Corollary IVa ABC Activities A1 A2 A3 A4 Hours P1 20 120 180 80 400 P2 80 180 120 20 400 100 300 300 100 800 A1 A2 A3 A4 ABC Cost Assignmenta i υ IABC2 Cost Assignmentb P1 0.20 0.40 0.60 0.80 $185 0.50 $185 P2 0.80 0.60 0.40 0.20 $185 0.50 $185 Cost $105 $60 $120 $85 $370 $370 TDABC2 Total Cost $370 Total Hours 800 cΖ $0.46 i β Cost per Unit i θ TDABC2 Cost Assignmentc Cυ r ℑυ r P1 0.5 $0.23 800 $185 0 0 P2 2 $0.93 200 $185 0 0 $370 a υ αa m a ia a i C D Σ= = 1 b i T I i D =υ C c i i i DΖ = cΖβ θ 43 To illustrate that TDABC2 can be applied to something other than products, Table 10 provides another illustration of Corollary IVa in which the cost objects are customers. This is the only customer example that will be shown since the next few product illustrations in this chapter can easily be adapted and applied to customers. Here, i β represents the order cycle time for cost object i (the time from which the order is made to the time the payment is received), and i θ represents the number of orders for cost object i. Table 10 shows that the cost of each customer under TDABC2 ( Ζ 1 D = $5,200 and Ζ 2 D = $6,800) are equal to those under ABC since Cυ r = 0 and ℑυ r = 0 for each customer. 44 TABLE 10 Customer Example Illustrating Corollary IVa ABC Activities A1 A2 A3 Days Customer1 210 30 280 520 Customer2 490 120 70 680 700 150 350 1,200 A1 A2 A3 ABC Costa i υ IABC2 Costb Customer1 0.30 0.20 0.80 $5,200 0.43 $5,200 Customer2 0.70 0.80 0.20 $6,800 0.57 $6,800 Cost $7,000 $1,500 $3,500 $12,000 $12,000 TDABC2 Total Cost $12,000 Total Days 1,200 cΖ $10 i β Cost per Order i θ TDABC2 Costc Cυ r ℑυ r Customer1 26 $260 20 $5,200 0 0 Customer2 34 $340 20 $6,800 0 0 $12,000 a υ αa m a ia a i C D Σ= = 1 b i T I i D =υ C c i i i DΖ = cΖβ θ Corollary IVb of Proposition IV is demonstrated in the following example shown in Table 11. According to Table 11, Cυ r and ℑυ r are nonzero and cΖ and the dollar error per unit of time, Ζ ε c , are both equal to $1.17. This satisfies Corollary IVb so that = Ζ i i Dα D , where Ζ 1 D = α 1 D = $640 and Ζ 1 D = α 1 D = $760. However, notice that IABC2 provides inaccurate results because Proposition III is violated. 45 TABLE 11 Example Illustrating Corollary IVb ABC Activities A1 A2 A3 A4 Hours P1 60 183 146 160 549 P2 240 274 97 40 651 300 457 243 200 1,200 A1 A2 A3 A4 ABC Cost Assignmenta i υ IABC2 Cost Assignmentb P1 0.20 0.40 0.60 0.80 $640 0.50 $700 P2 0.80 0.60 0.40 0.20 $760 0.50 $700 Cost $500 $200 $500 $200 $1,400 $1,400 TDABC2 Total Cost $1,400 cΖ $1.17 Total Hours 1,200 Ζ ε c c $1.17 i β Cost per Unit i θ TDABC2 Cost Assignmentd Cυ r ℑυ r P1 0.686 $0.80 800 $640 0.45 0.59 P2 3.257 $3.80 200 $760 0.45 0.59 $1,400 a υ αa m a ia a i C D Σ= = 1 c Σ Σ = = Ζ ℑ − ℑ − = m a a i a m a ia I i i D D c 1 1 υ υ α ε b i T I i D =υ C d i i i DΖ = cΖβ θ If Proposition IV does not hold, then there is a difference in the cost assigned to TDABC2 relative to that of ABC Stage 2. Table 12 shows that when there is a nonzero correlation for Cυ r and ℑυ r (where ℑ ℑ σ ≠ Ζ σ υ υ r c r C C ), the cost of the cost objects under TDABC2 ( Ζ 1 D = $643 and Ζ 2 D = $657) are not equal to those under ABC Stage 2 ( α 1 D = $465 and α 2 D = $835). The average absolute percentage error of TDABC2 for the 46 illustration is 29.76 percent, with dollar error ( i ε ) is $177.78 for P1 and $177.78 for P2 ( cΖ = $0.72 and Ζ ε c = $1.50). TABLE 12 Illustration Not Satisfying Proposition IV ABC Activities A1 A2 A3 A4 Hours P1 75 160 175 480 890 P2 225 240 325 120 910 300 400 500 600 1,800 A1 A2 A3 A4 ABC Cost Assignmenta i υ IABC2 Cost Assignmentb P1 0.25 0.40 0.35 0.80 $465 0.45 $585 P2 0.75 0.60 0.65 0.20 $835 0.55 $715 Cost $500 $300 $400 $100 $1,300 $1,300 TDABC2 Total Cost $1,300 cΖ $0.72 Total Hours 1,800 Ζ ε c c $1.50 i β Cost per Unit i θ TDABC2 Cost Assignmentd Cυ r ℑυ r P1 1.1125 $0.80 800 $643 0.97 0.86 P2 4.55 $3.29 200 $657 0.97 0.86 $1,300 ABC TDABC2 i ε % i ε e P1 $465 $643 $178 38.23% P2 $835 $657 $178 21.29% Avg i %ε f 29.76% a υ αa m a ia a i C D Σ= = 1 d i i i DΖ = cΖβ θ b i T I i D =υ C e% i ε = α α i i i D DΖ − D c Σ Σ = = Ζ ℑ − ℑ − = m a a i a m a ia I i i D D c 1 1 υ υ α ε f 2 38.23%+ − 21.29% 47 If Cυ r and ℑυ r are nonzero and TDABC2 and ABC are not equivalent, then the cost per unit of time would not be equal to the dollar error per unit of time. Therefore, > 0 i ε . The analysis of the effects of the various variables on the magnitude of the error in Equation 13 is similar to that for Equation 7 under Stage 1. Additional analysis of the dollar error in which the maximum error is identified will be shown in Chapter 6. 4.4. Unused Capacity in TDABC2 Parallel to Stage 1 analysis, unused capacity does not affect the necessary equivalency conditions for TDABC2. For equivalency, the cost of cost object i under ABC must equal that under TDABC plus its unused cost. Table 13 shows an example illustrating unused capacity. Table 13 is similar to Table 9 except that unused capacity is included. Notice that Table 13 verifies that the ABC cost for P1 = TDABC cost for P1 + cost of unused capacity ($185 = $173 + $12) and the cost for P2 = TDABC cost for P2 + cost of unused capacity ($185 = $176 + $9). The cost of unused capacity is equal to the cost per unit ( i c Ζβ ) multiplied by the difference between the number of units produced at practical capacity ( i θ ) and the number of units actually produced ( A i θ ). The rationale is that the unit cycle time i β is an observed value, so it must remain constant for each unit that is produced. Hence, the unused time is the difference between the total time available at practical capacity ( i i β θ ) and the total time actually used ( A i i β θ ). Accordingly, the cost of unused capacity is ( ) ( A ) i i i A i i i i c Ζ β θ −β θ = c Ζβ θ −θ . 48 TABLE 13 Corollary IVa with Unused Capacity ABC Activities A1 A2 A3 A4 Hours Cυ r ℑυ r P1 20 120 180 80 400 0 0 P2 80 180 120 20 400 0 0 100 300 300 100 800 A1 A2 A3 A4 ABC Cost Assignment i υ IABC2 Cost Assignment P1 0.20 0.40 0.60 0.80 $185 0.50 $185 P2 0.80 0.60 0.40 0.20 $185 0.50 $185 Cost $105 $60 $120 $85 $370 $370 TDABC2 Total Cost $370 Total Hours 800 cΖ $0.46 i β Cost per Unit i θ TDABC2 Cost Assignment A i θ Unused Cost Total Cost P1 0.5 $0.23 750 $173 800 $12 $185 P2 2 $0.93 190 $176 200 $9 $185 $349 $21 $370 4.5. Implications As shown analytically, under certain conditions, TDABC2 is equivalent to ABC assignments. TDABC2 has the benefit of IABC2 in which Stage 1 is eliminated and, thus, the problem of resource diversity is eliminated. Additionally, the linear relationship limitation due to poor estimates if there is a nonlinear or discontinuous relation between the demand for and provision of resources (Maher and Marais 1998) has also been resolved since Stage 1 has been eliminated. TDABC2 is a feasible system in which only the unit cycle time, total time, total cost, and number of units produced need to be known. Accordingly, TDABC2 is as 49 simple as a functionalbased costing system but with the accuracy of an ABC system. This will have significant practical relevance. However, if the equivalency conditions are not satisfied, then error will exist. The maximum absolute dollar error for TDABC2 relative to ABC must be identified, but before doing so, the maximum absolute dollar error for TDABC relative to ABC must first be identified as shown in the next chapter and then extended to TDABC2 as shown in Chapter 6. 50 CHAPTER 5 STAGE 1 ERROR ANALYSIS 5.1. Analysis of the Maximum Error of TDABC This chapter shows what the maximum absolute dollar error is for TDABC relative to ABC when the conditions in Proposition II and its corollaries are not met. From Chapter 3, the error for activity a, which is derived from Proposition II and shown in Equation 7, is ε a = [ ] c c t t σ r σ cr σ ρ ρ ρ − . If cρ r and ρ t r are substituted in Equation 7 (let j aj aj t t ρ = , T n j j C C = Σ= 1 , and T n j j t t = Σ= 1 ), then − − =Σ − Σ Σ Σ = = = = n j n j j T aj aj n j j T aj aj n j j j a t t n t c t t t n C t t C 1 1 1 1 ε . Simplifying further yields aj n j j j a c t t C Σ= = − 1 ε , a = 1,…, m (14) The Stage 1 error analysis is based on one assumption that in any given instance in time, the total cost of resource j ( j C ) and the total time available for resource j ( j t ) are likely to be fixed, but the consumption of the resources may vary depending on the activity usage. If j C and j t are treated as constants in the system and the consumption of resource j by activity a, aj t , is allowed to vary, then the following proposition shows 51 that the maximum absolute dollar error of the system is Σ Σ = = = − n j j j n j j C ct 1 1 δ , based on the second assumption that all resources are time driven, where j j j δ ≡ C − ct , which is basically the total dollar error contribution of resource j. c t C j j − is the dollar error contribution per unit of time of resource j (“unit dollar error contribution of j”). Proposition V: Given aj n j j j a c t t C Σ= = − 1 ε and j j t C c ≠ , the maximum absolute dollar error for the system is Σ Σ = = = − n j j j n j j C ct 1 1 δ . Proof: The dollar error for activity a is ( ) j aj n j aj j j n j j j a t t c t C ct t C Σ Σ = = − = = − 1 1 ε . Summing over all a yields the dollar error contribution of resource j: ( ) j j m a aj j j j j aj m a j j j t C ct t C ct t t C ct = − − =Σ − = Σ =1 =1 δ . The total dollar error contribution of all resources is the sum of all j δ : ( ) 0 1 1 1 1 1 1 1 1 = − = − = − Σ = Σ Σ Σ Σ Σ Σ Σ = = = = = = = = n j n j j j n j n j j j n j j n j j n j j j n j j t t C δ C ct C c t C . The total dollar error for all resources is zero implying that some resources provide a positive dollar error contribution and others a negative error contribution, such that 52 − > 0 j j C ct and − c > 0 t C j j for j = 1,…, s and − < 0 j j C ct and − c < 0 t C j j for j = s+1,…, n. Hence, the resources can be partitioned into two sets: one to represent the resources that provide the positive dollar error contribution, { } s R R , R ,..., R 1 2 + = , and the other to represent resources that provide a negative dollar error contribution, { } s s n R R , R ,..., R +1 +2 − = , where − > 0 j j C ct ( − c > 0 t C j j ) with j = 1,…, s ∈ R+ and − < 0 j j C ct ( − c < 0 t C j j ) with j = s+1,…, n ∈ − R . Then, Σ= n j j 1 δ can be rewritten as ( ) ( ) 0 1 1 1 Σ =Σ − + Σ − = = = = + n j s j j s j j j n j j δ C ct C ct . Thus, the total maximum absolute dollar error of the system which is essentially the total absolute dollar error contribution from all resources j = 1,…, n is Σ Σ Σ Σ = = = + = = − + − = − n j j j n j s j j j s j j n j j C ct C ct C ct 1 1 1 1 δ ( ) ( ) . QED Notice that the maximum absolute dollar error for the system, Σ Σ = = = − n j j j n j j C ct 1 1 δ , does not depend on the aj t ’s. Maximizing the dollar error based on the aj t ’s involves finding a corner solution that is part of the maximum set of corner solutions. To find the set of aj t ’s that produces the maximum error, the resources must be ordered from largest to smallest c t C j j − , where c t C − 1 1 > c t C − 2 2 > … > c t C n n − in addition to partitioning the resources into R+ and R− sets. The “unit dollar error contribution” is a better measure of magnitude than j j C − ct (total dollar error 53 contribution of resource j) since the amount of time could vary from one resource to the next causing the corresponding j j C − ct to not be of the same magnitude as the corresponding c t C j j − . Next, partition the activities into two sets in which some activities consume resources that provide a positive dollar error contribution ( A+ set) and the rest of the activities consume resources that provide a negative dollar error contribution ( A− set): A+ = {a a = 1,…, q; = − j ∈ R+ t t C ct j aj aj j j ε ( ) , and = ( − ) = 0 j aj aj j j t t ε C ct where = 0 aj t if j ∈ R− }; and A− = {a a = q+1,…, m; = − j ∈ R− t t C ct j aj aj j j ε ( ) , and = ( − ) = 0 j aj aj j j t t ε C ct where = 0 aj t if j ∈ R+ }, where A+ ∪ A− = A. The above expressions for A+ and A− sets state that to maximize the dollar error using aj t ’s, activities in A+ must only consume resources in R+ and activities in A− only consume resources in R− . As a result, the aj t ’s that result in Σ Σ = = = − n j j j n j j C ct 1 1 δ is identified in the following corollary to Proposition V. Corollary Va: The aj t ’s that produce the total maximum absolute dollar error of the system, Σ Σ = = = − n j j j n j j C ct 1 1 δ , is Σ ΣΣ = = = = − m a n j j aj j j m a a t t C ct 1 1 1 ε ( ) . 54 Proof: Choose taj with j = 1,…, s∈ R+ , such that j q a aj t t = Σ= 1 and aj t with j = s+1,…, n∈ R− , such that j m a q aj Σt = t = +1 . As a result, the dollar error from the A+ set is ( ) j aj s j j j s j a aj t t Σ Σ C ct = = + = = − 1 1 ε ε > 0 and from the A− set is ( ) j aj n j s j j n j s a aj t t Σ Σ C ct = + = + − = = − 1 1 ε ε < 0. Hence, 0 1 1 1 Σ =Σ + Σ = = + − = + = m a q a q a a m a a ε ε ε . This means that ( ) ( ) 0 1 1 1 1 Σ =ΣΣ − =Σ − > = = = = + s j j j q a j aj s j j j q a a C ct t t ε C ct using j q a aj t t = Σ= 1 and ( ) ( ) 0 1 1 1 1 Σ = Σ Σ − = Σ − < = + = + = + = + − n j s j j m a q j aj n j s j j m a q a C ct t t ε C ct using j m a q aj Σt = t = +1 . Thus, using these expressions for Σ= + q a a 1 ε and Σ = + − m a q a 1 ε , Σ ΣΣ( ) Σ Σ( ) ΣΣ = = = = + = + = = = − + − = − m a n j j aj j j m a q j aj n j s j j q a j aj s j j j m a a t t C ct t t C ct t t C ct 1 1 1 1 1 1 1 ε ( ) , which can be rewritten as Σ Σ( ) Σ( ) Σ Σ = = = + = = = − + − = − = n j j n j j j n j s j j s j j j m a a C ct C ct C ct 1 1 1 1 1 ε δ . Since there are no cancellation effects of positive resource dollar error contributions with negative resource dollar error contributions for any a, Σ ΣΣ Σ Σ = = = = = = − = − = n j j n j j j m a n j j aj j j m a a C ct t t C ct 1 1 1 1 1 ε ( ) δ . QED 55 Although Σ Σ = = = n j j m a a 1 1 ε δ , a program is identified that provides the maximum percentage error of each activity a that maximizes the average absolute percentage error of the system.3 The percentage error for any given activity a is α ε ε a a a C % = , which is the dollar error for activity a divided by the ABC cost for activity a. To find the maximum average absolute percentage error, one program is used for the positive sets, A+ and R+ , and another program for the negative sets, A− and R− . From Proposition V and Corollary Va, let m+ represent the number of activities in A+ (a = 1,…, q) and n+ represent the number of resources in R+ (j = 1,…, s). After ordering all resources from largest positive to smallest positive c t C j j − and labeling them as R1, R2,…, Rs, where c t C − 1 1 > c t C − 2 2 > … > c t C s s − , the most positive resource, R1, must be the only resource consumed by m+ − (n+ −1) activities. The program for the maximization of the percentage errors for each activity in the A+ set is as follows: Max Σ Σ Σ = = = − q a s j j j aj s j aj j j C t t c t t C 1 1 1 (P1) s.t. γ ,1 1 t t a ≥ , a = 1,…, q1 (P2) γ aj j t ≥ t , a = 1+ m+ − (n+ −1) ,…, q1, j = 2,…, s1 (P3) 3 An optimization software program such as LINGO can be used. 56 All other ≥ 0 aj t , j ∈ R+ and a ∈ A+ (P4) j q a aj t t = Σ= 1 (P5) The objective function in (P1) provides the maximum percentage error in magnitude across all activities in A+ . (P2) is the first constraint that ensures that for R1, at least the amount of γ 1 t is assigned to q1 activities in A+ . The materiality and uniqueness parameter γ (e.g. it can be set as 0.1 to ensure that 10 percent of the time for resource j is assigned to the appropriate activities) is only assigned to q1 activities instead of q to not over restrict the program and allow it to choose the optimal set of aj t ’s. (P3) is the second constraint that ensures uniqueness among resource vectors without overrestricting the program to allow for some aj t in A+ and R+ to be zero or for some activities in A+ to consume all of a single resource in R+ (this is represented by (P4)). (P5) ensures that the sum of the activity consumptions in A+ of a particular resource in R+ is equal to the total time available for that particular resource j. From Proposition V and its Corollary Va, let m− represent the number of activities in A− (a = q+1,…, m) and n− represent the number of resources in R− (j = s+1,…, n). First order all resources from least negative to most negative c t C j j − and label them as Rs+1, Rs+2,…, Rn, where c t C s s − + + 1 1 > c t C s s − + + 2 2 > … > c t C n n − . The most negative Rn must be the only resource consumed by m− − (n− −1) activities. The 57 program for the maximization the magnitude of the percentage errors in the R− set is as follows (this involves minimizing because of dealing with negative values for c t C j j − ): Min Σ Σ Σ = + = + = + − m a q n j s j j aj n j s aj j j C t t c t t C 1 1 1 (P6) s.t. γ a n n t ≥ t , , a = q+2,…, m (P7) γ aj j t ≥ t , a = q+1,…, m − m− − (n− −1) , j = s+2,…, n1 (P8) All other ≥ 0 aj t in j ∈ R− and a ∈ A− (P9) j m a q aj Σt = t = +1 (P10) The objective function in (P6) provides the maximum percentage error in magnitude across all activities in A− . (P7) is the first constraint that ensures that for Rn, at least the amount of γ n t is assigned to a = q+2,…, m activities in A− without over restricting the program and allow it to choose the optimal set of aj t ’s. (P8) is the second constraint that ensures uniqueness among resource vectors without overrestricting the program to allow for some aj t in A− and R− to be zero or for some activities in A− to consume all of a single resource in R− (this is represented by (P9)). (P10) ensures that the sum of the activity consumptions in A− of a particular resource in R− is equal to the total time available for that particular resource j. 58 Typically, an activity will consume resources from both R+ and R− , thus allowing for some cancellation effects from the positive dollar error contribution from resources in R+ and negative dollar error contribution from resources in R− . Since Σ Σ = = + = − = s j j aj j j aj j s j a j t t t t C ct 1 1 ε ( ) δ and Σ ΣΣ ΣΣ Σ = = = = = = + = − = = s j j q a s j j aj j q a j aj j s j j q a a t t t t C ct 1 1 1 1 1 1 ε ( ) δ δ with j j C − ct > 0 ( c t C j j − > 0) and Σ Σ = + = + − = − = n j s j aj j j aj j n j s a j t t t t C ct 1 1 ε ( ) δ and Σ Σ Σ Σ Σ Σ = + = + = + = + = + = + − = − = = n j s j m a q m a q n j s j aj j j aj j n j s j m a q a t t t t C ct 1 1 1 1 1 1 ε ( ) δ δ with j j C − ct < 0 ( c t C j j − < 0), then if any a in A+ (a = 1,…, q) consumes any resource j in R− (j = s+1,…, n), then there is a cancellation effect of negative resource error contributions with positive resource error contributions; thus, Σ Σ = + = < q a a q a a 1 1 ε ε . Likewise, if any a in A− (a = q+1,…, m) consumes any resource j in R+ (j = 1,…, s), then there is a cancellation effect of positive resource error contributions with negative resource error contributions; consequently, Σ Σ = + + = + < m a q a m a q a 1 1 ε ε . As a result, when there are any cancellation effects, Σ Σ = = < n j j m a a 1 1 ε δ , which implies that, j n j j aj n j j aj j n j j aj a j t t t t t t ε Σδ Σδ Σ δ = = = = ≤ = 1 1 1 . Hence, Σ= m a a 1 ε should be less than Σ= n j j 1 δ as shown in the following corollary to Proposition V. Corollary Vb: As already derived, if activities in A+ only consume resources in R+ and activities in A− only consume resources in R− , thenΣ Σ = = = n j j m a a 1 1 ε δ . If any or all 59 activities consume resources from both R+ and R− , then Σ Σ = = < n j j m a a 1 1 ε δ . Therefore, Σ= m a a 1 ε can never exceed Σ= n j j 1 δ , implyingΣ Σ = = ≤ n j j m a a 1 1 ε δ . Proof: By using the triangle inequality m n ε + ε + ...+ ε ≤ δ + δ + ...+ δ 1 2 1 2 , Σ ΣΣ Σ = = = = ≤ = n j j m a j n j j aj m a a t t 1 1 1 1 ε δ δ ⇒ Σ Σ = = ≤ n j j m a a 1 1 ε δ . QED If there are no cancellation effects (activities in A+ only consume resources in R+ and activities in A− only consume resources in R− ), then Σ Σ = = = n j j m a a 1 1 ε δ . Since j j C − ct > 0 ( c t C j j − > 0) for j ∈ R+ and j j C − ct < 0 ( c t C j j − < 0) for j ∈ R− , if any or all activities consume resources from both R+ and R− , then there will be some cancellation effects within each of those a ε ’s, and thus, the actual absolute dollar error across all activities will be less than the maximum absolute dollar error of the system, Σ Σ = = < n j j m a a 1 1 δ ε . Thus, Σ= m a a 1 ε can never exceed Σ= n j j 1 δ . 5.2. Examples Demonstrating Proposition V and Its Corollaries Examples demonstrating Proposition V and its corollaries are shown in this section. Corollary Va will be shown first. Tables 14, 15, and 16 demonstrate examples of Corollary Va for m = n, m < n, and m > n, respectively. All three cases are shown to demonstrate that Proposition V and Corollary Va as well as program (P1) through (P10) 60 are viable in each situation. The resources are ranked from largest to smallest c t C j j − and the resources are partitioned accordingly. For each table, the first three activities (A1 through A3) are in the A+ set, and the rest of the activities are in the A− set. In each table, Σ Σ = = = n j j m a a 1 1 ε δ = $1,700, thus satisfying Corollary Va. Program (P1) through (P10) is used to maximize each a ε % , which maximizes average %Σ= m a a 1 ε . 61 Table 14 Corollary Va for m = n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0 0 0 0 0 $200 0.61 0.61 A2 0.42 0.10 0 0 0 0 $1,069 0.70 0.70 A3 0.48 0.90 1 0 0 0 $5,231 0.70 0.70 A4 0 0 0 1 0.90 0.56 $6,598 0.68 0.68 A5 0 0 0 0 0.10 0.34 $1,302 0.86 0.86 A6 0 0 0 0 0 0.10 $300 0.74 0.74 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 0 0 0 0 0 77.5 $2 $155 j t2 328.96 97.5 0 0 0 0 426.46 $853 j t3 368.5 877.5 1,075 0 0 0 2,321 $4,642 j t4 0 0 0 1,275 1,327.5 986.7 3,589.2 $7,178 j t5 0 0 0 0 147.5 610.8 758.3 $1,517 j t6 0 0 0 0 0 177.5 177.5 $355 $14,700 62 Table 14 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $0 $0 $0 $0 $0 $45 $45 22.50% j j j j C t c t2 ( / − ) $191 $25 $0 $0 $0 $0 $216 $216 20.21% j j j j C t c t 3 ( / − ) $214 $225 $150 $0 $0 $0 $589 $589 11.26% j j j j C t c t 4 ( / − ) $0 $0 $0 $50 $225 $306 $581 $581 8.80% j j j j C t c t 5 ( / − ) $0 $0 $0 $0 $25 $189 $214 $214 16.45% j j j j C t c t 6 ( / − ) $0 $0 $0 $0 $0 $55 $55 $55 18.33% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 16.26% 63 Table 15 Corollary Va for m < n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0 0 0 0 0 $200 0.61 0.61 A2 0.42 0.10 0 0 0 0 $1,069 0.70 0.70 A3 0.48 0.90 1 0 0 0 $5,231 0.70 0.70 A4 0 0 0 1 0.90 0.64 $6,848 0.72 0.72 A5 0 0 0 0 0.10 0.36 $1,352 0.85 0.85 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 0 0 0 0 0 77.5 $2 $155 j t2 328.96 97.5 0 0 0 0 426.46 $853 j t3 368.5 877.5 1,075 0 0 0 2,321 $4,642 j t4 0 0 0 1,275 1,327.5 1,134.96 3,737.46 $7,475 j t5 0 0 0 0 147.5 640 787.5 $1,575 $14,700 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 64 Table 15 (continued) Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $0 $0 $0 $0 $0 $45 $45 22.50% j j j j C t c t2 ( / − ) $191 $25 $0 $0 $0 $0 $216 $216 20.21% j j j j C t c t 3 ( / − ) $214 $225 $150 $0 $0 $0 $589 $589 11.26% j j j j C t c t 4 ( / − ) $0 $0 $0 $50 $225 $352 $627 $627 9.15% j j j j C t c t 5 ( / − ) $0 $0 $0 $0 $25 $198 $223 $223 16.52% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 15.93% 65 Table 16 Corollary Va for m > n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0 0 0 0 0 $200 0.61 0.61 A2 0.42 0.10 0 0 0 0 $1,069 0.70 0.70 A3 0.48 0.90 1 0 0 0 $5,231 0.70 0.70 A4 0 0 0 1 0.90 0.47 $6,347 0.63 0.63 A5 0 0 0 0 0.10 0.33 $1,253 0.86 0.86 A6 0 0 0 0 0 0.10 $300 0.74 0.74 A7 0 0 0 0 0 0.10 $300 0.74 0.74 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 0 0 0 0 0 77.5 $2 $155 j t2 328.96 97.5 0 0 0 0 426.46 $853 j t3 368.5 877.5 1,075 0 0 0 2,321 $4,642 j t4 0 0 0 1,275 1,327.5 838.5 3,441 $6,882 j t5 0 0 0 0 147.5 581.5 729 $1,458 j t6 0 0 0 0 0 177.5 177.5 $355 j t7 0 0 0 0 0 177.5 177.5 $355 $14,700 66 Table 16 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $0 $0 $0 $0 $0 $45 $45 22.50% j j j j C t c t2 ( / − ) $191 $25 $0 $0 $0 $0 $216 $216 20.21% j j j j C t c t 3 ( / − ) $214 $225 $150 $0 $0 $0 $589 $589 11.26% j j j j C t c t 4 ( / − ) $0 $0 $0 $50 $225 $260 $535 $535 8.43% j j j j C t c t 5 ( / − ) $0 $0 $0 $0 $25 $180 $205 $205 16.38% j j j j C t c t6 ( / − ) $0 $0 $0 $0 $0 $55 $55 $55 18.33% j j j j C t c t7 ( / − ) $0 $0 $0 $0 $0 $55 $55 $55 18.33% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 16.49% 67 Illustrations of Corollary Vb are shown in Tables 17, 18, and 19 for m = n, m < n, and m > n, respectively. The resources are ranked from largest to smallest c t C j j − and the resources are partitioned accordingly. As previously, for each table, the first three activities (A1 through A3) are in the A+ set, and the rest of the activities are in the A− set. In each table, all activities share all of the resources so thatΣ Σ = = < n j j m a a 1 1 ε δ = $1,700. Table 17 shows that the average Σ= m a a 1 %ε = 1.90 percent compared to 16.26 percent in Table 14. Compared to Table 15 with an average Σ= m a a 1 %ε = 15.93 percent, Table 18 shows that the average Σ= m a a 1 %ε = 2.18 percent. Table 19 shows that the average Σ= m a a 1 %ε = 1.77 percent compared to 16.49 percent in Table 16. Notice that in each table, each % a ε is less than 5 percent, which demonstrates that the actual maximum error is small since activities will typically consume resources in both R+ and R− sets. 68 Table 17 Corollary Vb for m = n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0.20 0.10 0.30 0.20 0.30 $3,060 0.74 0.74 A2 0.30 0.10 0.10 0.10 0.20 0.10 $2,140 0.43 0.43 A3 0.20 0.30 0.10 0.10 0.10 0.20 $2,410 0.27 0.27 A4 0.10 0.20 0.10 0.10 0.30 0.10 $2,230 0.17 0.17 A5 0.10 0.10 0.40 0.20 0.10 0.10 $2,410 0.18 0.18 A6 0.20 0.10 0.20 0.20 0.10 0.20 $2,450 0.00 0.00 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 195 107.5 382.5 295 532.5 1,590 $2 $3,180 j t2 232.5 97.5 107.5 127.5 295 177.5 1,037.5 $2,075 j t3 155 292.5 107.5 127.5 147.5 355 1,185 $2,370 j t4 77.5 195 107.5 127.5 442.5 177.5 1,127.5 $2,255 j t5 77.5 97.5 430 255 147.5 177.5 1,185 $2,370 j t6 155 97.5 215 255 147.5 355 1,225 $2,450 $14,700 69 Table 17 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $50 $15 $15 $50 $165 $120 $120 3.92% j j j j C t c t2 ( / − ) $135 $25 $15 $5 $50 $55 $65 $65 3.04% j j j j C t c t 3 ( / − ) $90 $75 $15 $5 $25 $110 $40 $40 1.66% j j j j C t c t 4 ( / − ) $45 $50 $15 $5 $75 $55 $25 $25 1.12% j j j j C t c t 5 ( / − ) $45 $25 $60 $10 $25 $55 $40 $40 1.66% j j j j C t c t 6 ( / − ) $90 $25 $30 $10 $25 $110 $0 $0 0.00% $450 $250 $150 $50 $250 $550 $0 $290 AvgΣ= m a a 1 %ε 1.90% 70 Table 18 Corollary Vb for m < n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0.30 0.10 0.30 0.40 0.20 $3,520 0.41 0.41 A2 0.30 0.10 0.10 0.20 0.10 0.40 $3,020 0.35 0.35 A3 0.20 0.30 0.10 0.10 0.30 0.10 $2,650 0.25 0.25 A4 0.10 0.20 0.10 0.30 0.10 0.10 $2,190 0.10 0.10 A5 0.30 0.10 0.60 0.10 0.10 0.20 $3,320 0.31 0.31 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 292.5 107.5 382.5 590 355 1,805 $2 $3,610 j t2 232.5 97.5 107.5 255 147.5 710 1,550 $3,100 j t3 155 292.5 107.5 127.5 442.5 177.5 1,302.5 $2,605 j t4 77.5 195 107.5 382.5 147.5 177.5 1,087.5 $2,175 j t5 232.5 97.5 645 127.5 147.5 355 1,605 $3,210 $14,700 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 71 Table 18 (continued) Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $75 $15 $15 $100 $110 $90 $90 2.56% j j j j C t c t2 ( / − ) $135 $25 $15 $10 $25 $220 $80 $80 2.65% j j j j C t c t 3 ( / − ) $90 $75 $15 $5 $75 $55 $45 $45 1.70% j j j j C t c t 4 ( / − ) $45 $50 $15 $15 $25 $55 $15 $15 0.68% j j j j C t c t 5 ( / − ) $135 $25 $90 $5 $25 $110 $110 $110 3.31% $450 $250 $150 $50 $250 $550 $0 $340 AvgΣ= m a a 1 %ε 2.18% 72 Table 19 Corollary Vb for m > n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0.10 0.40 0.10 0.10 0.10 $2,160 0.20 0.20 A2 0.30 0.10 0.10 0.10 0.10 0.10 $1,870 0.61 0.61 A3 0.20 0.20 0.10 0.10 0.10 0.30 $2,490 0.27 0.27 A4 0.10 0.10 0.10 0.40 0.30 0.10 $2,760 0.27 0.27 A5 0.10 0.20 0.10 0.10 0.20 0.10 $1,960 0.00 0.00 A6 0.10 0.10 0.10 0.10 0.10 0.10 $1,470 0.00 0.00 A7 0.10 0.20 0.10 0.10 0.10 0.20 $1,990 0.32 0.32 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 97.5 430 127.5 147.5 177.5 1,057.5 $2 $2,115 j t2 232.5 97.5 107.5 127.5 147.5 177.5 890 $1,780 j t3 155 195 107.5 127.5 147.5 532.5 1,265 $2,530 j t4 77.5 97.5 107.5 510 442.5 177.5 1,412.5 $2,825 j t5 77.5 195 107.5 127.5 295 177.5 980 $1,960 j t6 77.5 97.5 107.5 127.5 147.5 177.5 730 $1,470 j t7 77.5 195 107.5 127.5 147.5 355 1,010 $2,020 $14,700 73 Table 19 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $25 $60 $5 $25 $55 $45 $45 2.08% j j j j C t c t2 ( / − ) $135 $25 $15 $5 $25 $55 $90 $90 4.81% j j j j C t c t 3 ( / − ) $90 $50 $15 $5 $25 $165 $40 $40 1.61% j j j j C t c t 4 ( / − ) $45 $25 $15 $20 $75 $55 $65 $65 2.36% j j j j C t c t 5 ( / − ) $45 $50 $15 $5 $50 $55 $0 $0 0.00% j j j j C t c t 6 ( / − ) $45 $25 $15 $5 $25 $55 $0 $0 0.00% j j j j C t c t7 ( / − ) $45 $50 $15 $5 $25 $110 $30 $30 1.51% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 1.77% 74 5.3. Implications This chapter analyzes and demonstrates that if activities consume resources from both + R and − R sets (and they most likely will), then the average Σ= m a a 1 %ε is significantly lower than if activities from A+ only consume resources in R+ and activities in A− only consume resources in R− . Additionally, the percentage error for each activity is not significant (e.g. less than 5 percent in Tables 17, 18, and 19). Hence, for Stage 1, TDABC is not significantly different from ABC provided that there is no resource diversity. However, previous discussion has shown that there could be a potentially significant error when resource diversity exists. TDABC2 eliminates this resource diversity issue and significantly reduces the complexity of Stage 2 cost assignments. The error analysis in this chapter is extended to Stage 2 in the next chapter to show the maximum absolute dollar error for TDABC2 relative to Stage 2 of ABC. 75 CHAPTER 6 STAGE 2 ERROR ANALYSIS 6.1. Analysis of the Maximum Error of TDABC2 If the equivalency conditions of Proposition IV and its corollaries are not satisfied, then error is introduced, and it is necessary to determine the maximum error possible. This chapter identifies the maximum absolute dollar error of TDABC2 relative to ABC and all analytics are parallel to those of Chapter 5. From Chapter 4, the error for activity a, which is derived from Proposition IV and shown in Equation 13, is [ ] ℑ ℑ ε =σ σ − Ζ σ υ υ υ r c r i C C . Parallel to the Stage 1 error analysis, if Cυ r and ℑυ r are substituted in Equation 13 (let a ia ia ℑ ℑ υ = , T m a a C C = Σ=1 , and T m a a t = ℑ Σ= 1 ), then simplifying further yields ia m a a a i c C ℑ − ℑ =Σ= Ζ 1 α ε , i = 1,…, k (15) The assumptions for Stage 2 concerning α a C and a ℑ being fixed and that all activities are time driven are similar in rationale to those in the Stage 1 analysis. If α a C and a ℑ are treated as constants in the system and the consumption of activity a by cost object i, ia ℑ , is allowed to vary, then the following proposition shows that the maximum 76 absolute dollar error of the Stage 2 system is Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α , where a a a δ ≡ Cα − c Ζℑ , which is the total dollar error contribution of activity a. − Ζ ℑ c C a a α is the dollar error contribution per unit of time of activity a (“unit dollar error contribution of a”). Proposition VI: Given ia m a a a i c C ℑ − ℑ =Σ= Ζ 1 α ε and a a C c ℑ Ζ ≠ α , the maximum absolute dollar error for the system is Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α . The proof is parallel to that of Proposition V and is, therefore, omitted. Based on the proof from Proposition VI, the activities can be partitioned into two sets: one to represent the activities that provide the positive dollar error contribution, { } q A A , A ,..., A 1 2 + = and the other to represent activities that provide a negative dollar error contribution, { } q q m A A , A ,..., A +1 +2 − = , where a a Cα − cΖℑ > 0 ( − Ζ ℑ c C a a α > 0) with a = 1,…, q ∈ A+ and a a Cα − c Ζℑ < 0 ( − Ζ ℑ c C a a α < 0) with a = q+1,…, m ∈ A− . Parallel to the Stage 1 error analysis, Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α does not depend on the ia ℑ ’s. To find the set of ia ℑ ’s that produces the maximum error, first (along with the partition of activities into A+ and A− sets) the activities are ordered from largest to 77 smallest − Ζ ℑ c C a a α (similar rational to that in Stage 1 error analysis). Second, partition the cost objects into two sets in which some cost objects consume activities that provide a positive dollar error contribution ( I + ) and the rest of the cost objects consume activities that provide a negative dollar error contribution ( I − ): I + = {i i = 1,…, w; Ζ ∈ + ℑ ℑ = C − c ℑ a A a ia ia a a ε ( α ) , and ( ) = 0 ℑ ℑ = − Ζℑ a ia ia a a ε Cα c where ℑ = 0 ia if a ∈ A− }; and I − = {i i = w+1,…, k; Ζ ∈ − ℑ ℑ = C − c ℑ a A a ia ia a a ε ( α ) , and ( ) = 0 ℑ ℑ = − Ζℑ a ia ia a a ε Cα c where ℑ = 0 ia if a ∈ A+ }, where I + ∪I − = I . To maximize the dollar error using ia ℑ ’s, cost objects in I + only consume activities in A+ and cost objects in I − only consume activities in A− . The following corollary to Proposition VI shows the ia ℑ ’s that result in Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α . Corollary VIa: The ia ℑ ’s that produce the total maximum absolute dollar error of the system, Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α , is Σ ΣΣ = = Ζ = ℑ ℑ = − ℑ k i m a a ia a a k i i C c 1 1 1 ε ( α ) . The proof parallels that of Corollary Va and is therefore omitted. 78 Although Σ Σ = = = m a a k i i 1 1 ε δ , a program is identified that provides the maximum percentage error of each cost object i that maximizes the average absolute percentage error of the system for Stage 2.4 The percentage error for any given cost object is α ε ε i i i D % = , which is the dollar error for cost object i divided by the ABC cost for cost object i. Parallel to the Stage 2 error analysis, to find the maximum average absolute percentage error, one program is used for the positive sets, I + and A+ , and another program for the negative sets, I − and A− . Let m+ represent the number of activities in A+ (a = 1,…, q) and k + represent the number of resources in I + (i = 1,…, w). First order all activities from largest positive to smallest positive − Ζ ℑ c C a a α and label them as A1, A2,…, Aq. The most positive activity, A1, must be the only activity consumed by k + − (m+ −1) cost objects. The program for the maximization of the percentage errors for each cost object in the I + set is as follows: Max Σ Σ Σ = = = Ζ ℑ ℑ ℑ − w ℑ i q a a a ia q a ia a a C c C 1 1 1 α α (P11) s.t. γ ,1 1 ℑ ≥ ℑ i , i = 1,…, w1 (P12) γ ia a ℑ ≥ ℑ , i = 1+ k + − (m+ −1) ,…, w1, a = 2,…, q1 (P13) 4 An optimization software program such as LINGO can be used. 79 All other ℑ ≥ 0 ia , a ∈ A+ and i ∈ I + (P14) a w i ia ℑ = ℑ Σ= 1 (P15) The objective function in (P11) provides the maximum percentage error in magnitude across all cost objects in I + . (P12) is the first constraint that ensures that for A1, at least the amount of γ 1 ℑ is assigned to w1 cost objects in I + . The materiality and uniqueness parameter γ (e.g. it can be set as 0.1 to ensure that 10 percent of the time for activity a is assigned to the appropriate cost objects) is only assigned to w1 cost objects instead of w to not over restrict the program and allow it to choose the optimal set of ia ℑ ’s. (P13) is the second constraint that ensures uniqueness among activity vectors without overrestricting the program to allow for some ia ℑ in I + and A+ to be zero or for some cost objects in I + to consume all of a single activity in A+ (this is represented by (P14)). (P15) ensures that the sum of the cost object consumptions in I + of a particular activity in A+ is equal to the total time available for that particular activity a. Now, let k − represent the number of cost objects in I − (i = w+1,…, k) and m− represent the number of activities in A− (a = q+1,…, m). After ordering all activities from least negative to most negative − Ζ ℑ c C a a α and labeling them as Aq+1, Aq+2,…, Am, the most negative Am must be the only activity consumed by k − − (m− −1) cost objects. The program for the maximization the magnitude of the percentage errors in the A− set is as follows: 80 Min Σ Σ Σ = + = + = + Ζ ℑ ℑ ℑ − k ℑ i w m a q a a ia m a q ia a a C c C 1 1 1 α α (P16) s.t. γ i m m ℑ ≥ ℑ , , i = w+2,…, k (P17) γ ia a ℑ ≥ ℑ , i = w+1,…, k − k − − (m− −1) , a = q+2,…, m1 (P18) All other ℑ ≥ 0 ia in a ∈ A− and i ∈ I − (P19) a k i w ia Σℑ = ℑ = +1 (P20) The objective function in (P16) provides the maximum percentage error in magnitude across all cost objects in I − . (P17) is the first constraint that ensures that for Am, at least the amount of γ m ℑ is assigned to i = w+2,…, k cost objects in I − without over restricting the program. (P18) is the second constraint that ensures uniqueness among activity vectors without overrestricting the program to allow for some ia ℑ in I − and A− to be zero or for some cost objects in I − to consume all of a single activity in A− (this is represented by (P19)). (P20) ensures that the sum of the cost object consumptions in I − of a particular activity in A− is equal to the total time available for that particular activity a. However, parallel to that in Stage 1 error analysis, a cost object can consume activities from both A+ and A− , thus allowing for some cancellation effects and a 81 reduced error. Hence, the maximum error cannot exceed the dollar error contribution from all activities (Σ Σ = = ≤ m a a k i i 1 1 ε δ ) and is stated in the following corollary. Corollary VIb: If cost objects in I + only consume activities in A+ and cost objects in I − only consume activities in A− , thenΣ Σ = = = m a a k i i 1 1 ε δ . If any or all cost objects consume activities from both A+ and A− , then Σ Σ = = < m a a k i i 1 1 δ ε . Therefore, Σ= k i i 1 ε can never exceed Σ= m a a 1 δ , implyingΣ Σ = = ≤ m a a k i i 1 1 ε δ . The proof parallels that of Corollary Vb and is, therefore, omitted. In summary, if there are no cancellation effects, then Σ Σ = = = m a a k i i 1 1 ε δ . If any or all cost objects consume activities from both A+ and A− , then there will be some cancellation effects within each of those i ε ’s, and thus, the actual absolute dollar error across all cost objects will be less than the maximum absolute dollar error of the system, Σ Σ = = < m a a k i i 1 1 ε δ . As a result, Σ Σ = = ≤ n j j m a a 1 1 ε δ . 6.2. Examples Demonstrating Proposition VI and Its Corollaries This section provides examples illustrating Proposition VI and its corollaries and using program (P11) through (P20). First, Corollary VIa will be shown in Tables 20, 21, and 22 for k = m, k < m, and k > m, respectively. The activities in each table are ranked 82 from largest to smallest − Ζ ℑ c C a a α and the activities are partitioned accordingly. For each table, the first three cost objects (CO1 through CO3) are in the I + set, with the rest of the cost objects being in the I − set. Table 20 is a continuation of Table 14, in which the activity costs from Stage 1 are assigned to the cost objects in Stage 2. Table 21 is a continuation of Table 15, and Table 22 is a continuation of Table 16. In each of the following tables, Σ Σ = = = m a a k i i 1 1 ε δ = $1,700, thus satisfying Corollary VIa. Program (P1) through (P10) is used to maximize each i ε % , which maximizes average Σ= k i i 1 %ε . 83 Table 20 Corollary VIa for k = m Activities A1 A2 A3 A4 A5 A6 ABC Cost Cυ r ℑυ r CO1 0.10 0 0 0 0 0 $20 0.40 0.40 CO2 0.90 0.10 0 0 0 0 $287 0.44 0.43 CO3 0 0.90 1 0 0 0 $6,193 0.23 0.11 CO4 0 0 0 1 0.90 0 $7,770 0.46 0.56 CO5 0 0 0 0 0.10 0.90 $400 0.41 0.39 CO6 0 0 0 0 0 0.10 $30 0.38 0.36 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $14,700 Time 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 Total 1a ℑ 7.75 0 0 0 0 0 7.75 2a ℑ 69.75 42.65 0 0 0 0 112.40 3a ℑ 0 383.81 2,321.04 0 0 0 2,704.85 4a ℑ 0 0 0 3,589.22 682.45 0 4,271.67 5a ℑ 0 0 0 0 75.83 159.75 235.58 6a ℑ 0 0 0 0 0 17.75 17.75 Total 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 84 Table 20 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 7.750 1 $16 CO2 28.100 4 $225 CO3 54.097 50 $5,410 CO4 85.433 50 $8,543 CO5 2.356 100 $471 CO6 8.875 2 $35 $14,700 Activities A1 A2 A3 A4 A5 A6 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.162 $0.283 $0.310 a a Cα − c Ζℑ $45 $216 $589 $581 $214 $55 Σ= m a a 1 δ $1,700 85 Table 20 (continued) Activities A1 A2 A3 A4 A5 A6 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $0 $0 $0 $0 $0 $4.50 $4.50 22.50% a a a C c 2 ( α − Ζℑ )ℑ $40.50 $22 $0 $0 $0 $0 $61.50 $61.50 21.65% a a a C c 3 ( α − Ζℑ )ℑ $0 $194 $589 $0 $0 $0 $783 $783 12.65% a a a C c 4 ( α − Ζℑ )ℑ $0 $0 $0 $581 $193 $0 $774 $774 9.96% a a a C c 5 ( α − Ζℑ )ℑ $0 $0 $0 $0 $21 $49.50 $70.50 $70.50 17.72% a a a C c 6 ( α − Ζℑ )ℑ $0 $0 $0 $0 $0 $5.50 $5.50 $5.50 18.33% $45 $216 $589 $581 $214 $55 $0 $1,700 AvgΣ= m a a 1 %ε 17.13% 86 Table 21 Corollary VIa for k < m Activities A1 A2 A3 A4 A5 A6 A7 ABC Cost Cυ r ℑυ r CO1 0.10 0 0 0 0 0 0 $20 0.33 0.33 CO2 0.90 0.10 0 0 0 0 0 $287 0.35 0.35 CO3 0 0.90 1 0 0 0 0 $6,193 0.31 0.19 CO4 0 0 0 1 0.90 0.47 0.17 $7,666 0.35 0.45 CO5 0 0 0 0 0.10 0.43 0.73 $474 0.49 0.46 CO6 0 0 0 0 0 0.10 0.10 $60 0.48 0.46 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $300 $14,700 Time 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 A7 Total 1a ℑ 7.75 0 0 0 0 0 0 7.75 2a ℑ 69.75 42.65 0 0 0 0 0 112.40 3a ℑ 0 383.81 2,321.04 0 0 0 0 2,704.85 4a ℑ 0 0 0 3,440.98 656.12 83.16 30.14 4,210.40 5a ℑ 0 0 0 0 72.9 76.59 129.61 279.1 6a ℑ 0 0 0 0 0 17.75 17.75 35.5 Total 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 87 Table 21 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 7.750 1 $16 CO2 28.100 4 $225 CO3 54.097 50 $5,410 CO4 84.208 50 $8,421 CO5 2.791 100 $558 CO6 17.75 2 $71 $14,700 Activities A1 A2 A3 A4 A5 A6 A7 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.155 $0.281 $0.310 $0.310 a a Cα − c Ζℑ $45 $216 $589 $535 $205 $55 $55 Σ= m a a 1 δ $1,700 88 Table 21 (continued) Activities A1 A2 A3 A4 A5 A6 A7 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $0 $0 $0 $0 $0 $0 $4.50 $4.50 22.50% a a a C c 2 ( α − Ζℑ )ℑ $40.50 $22 $0 $0 $0 $0 $0 $61.50 $62 21.65% a a a C c 3 ( α − Ζℑ )ℑ $0 $194 $589 $0 $0 $0 $0 $783 $783 12.65% a a a C c 4 ( α − Ζℑ )ℑ $0 $0 $0 $535 $185 $26 $9 $755 $755 9.84% a a a C c 5 ( α − Ζℑ )ℑ $0 $0 $0 $0 $20 $24 $40 $84 $84 17.82% a a a C c 6 ( α − Ζℑ )ℑ $0 $0 $0 $0 $0 $5 $5 $5 $5 18.34% $45 $216 $589 $535 $205 $55 $55 $0 $1,700 AvgΣ= m a a 1 %ε 17.13% 89 Table 22 Corollary VIa for k > m Activities A1 A2 A3 A4 A5 ABC Cost Cυ r ℑυ r CO1 0.10 0 0 0 0 $20 0.52 0.51 CO2 0.90 0.10 0 0 0 $287 0.58 0.56 CO3 0 0.90 1 0 0 $6,193 0.10 0.03 CO4 0 0 0 1 0.80 $7,770 0.45 0.56 CO5 0 0 0 0 0.10 $400 0.30 0.25 CO6 0 0 0 0 0.10 $30 0.30 0.25 Cost $200 $1,069 $5,231 $6,848 $1,352 $14,700 Time 77.5 426.46 2,321.04 3,737.46 787.54 7,350 Activities A1 A2 A3 A4 A5 Total 1a ℑ 7.75 0 0 0 0 7.75 2a ℑ 69.75 42.65 0 0 0 112.40 3a ℑ 0 383.81 2,321.04 0 0 2,704.85 4a ℑ 0 0 0 3,737.46 630.03 4,367.49 5a ℑ 0 0 0 0 78.75 78.75 6a ℑ 0 0 0 0 78.75 78.75 Total 77.5 426.46 2,321.04 3,737.46 758.28 7,350 90 Table 22 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 7.750 1 $16 CO2 28.100 4 $225 CO3 54.097 50 $5,410 CO4 87.350 50 $8,735 CO5 0.788 100 $158 CO6 39.375 2 $158 $14,700 Activities A1 A2 A3 A4 A5 a a Cα ℑ $2.58 $2.51 $2.25 $1.83 $1.72 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.168 $0.284 a a Cα − c Ζℑ $45 $216 $589 $627 $223 Σ= m a a 1 δ $1,700 91 Table 22 (continued) Activities A1 A2 A3 A4 A5 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $0 $0 $0 $0 $4.50 $4.50 22.50% a a a C c 2 ( α − Ζℑ )ℑ $40.50 $22 $0 $0 $0 $61.50 $61.50 21.65% a a a C c 3 ( α − Ζℑ )ℑ $0 $194 $589 $0 $0 $783 $783 12.65% a a a C c 4 ( α − Ζℑ )ℑ $0 $0 $0 $627 $179 $805 $805 10.16% a a a C c 5 ( α − Ζℑ )ℑ $0 $0 $0 $0 $22 $22 $22 16.52% a a a C c 6 ( α − Ζℑ )ℑ $0 $0 $0 $0 $22 $22 $22 16.52% $45 $216 $589 $627 $223 $0 $1,700 AvgΣ= m a a 1 %ε 16.67% 92 Illustrations of Corollary VIb are shown in Tables 23, 24, and 25 for k = m, k < m, and k > m, respectively. As in the previous three tables, the activities in each table are ranked from largest to smallest − Ζ ℑ c C a a α and the activities are partitioned accordingly. For each table, the first three cost objects (CO1 through CO3) are in the I + set, with the rest of the cost objects being in the I − set. In each table, all cost objects share all of the activities so thatΣ Σ = = < m a a k i i 1 1 δ ε = $1,700. Table 23 shows that the average Σ= k i i 1 %ε = 0.97 percent compared to 17.13 percent in Table 20. Compared to Table 21 with an average Σ= k i i 1 %ε = 17.13 percent, Table 24 shows that the average Σ= k i i 1 %ε = 0.84 percent. Table 25 shows that the average Σ= k i i 1 %ε = 0.42 percent compared to 16.67 percent in Table 20. In each table, notice that each % i ε is less than 5 percent, which demonstrates that the actual maximum error is very small since cost objects will typically consume activities in both A+ and A− sets. 93 Table 23 Corollary VIb for k = m Activities A1 A2 A3 A4 A5 A6 ABC Cost Cυ r ℑυ r CO1 0.10 0.30 0.10 0.10 0.20 0.10 $1,814 0.34 0.35 CO2 0.10 0.20 0.20 0.20 0.10 0.10 $2,760 0.74 0.69 CO3 0.20 0.10 0.20 0.20 0.10 0.10 $2,673 0.62 0.60 CO4 0.30 0.20 0.10 0.10 0.30 0.10 $1,877 0.60 0.58 CO5 0.20 0.10 0.10 0.20 0.10 0.10 $2,150 0.27 0.33 CO6 0.10 0.10 0.30 0.20 0.20 0.50 $3,426 0.002 0.01 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $14,700 Time 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 Total 1a ℑ 7.75 127.94 232.1 358.92 151.66 17.75 896.12 2a ℑ 7.75 85.29 464.21 717.84 75.83 17.75 1,368.67 3a ℑ 15.5 42.65 464.21 717.84 75.83 17.75 1,333.78 4a ℑ 23.25 85.29 232.1 358.92 227.48 17.75 944.8 5a ℑ 15.5 42.65 232.1 717.84 78.83 17.75 1,101.67 6a ℑ 7.75 42.65 696.31 717.84 151.66 88.75 1,704.96 Total 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 94 Table 23 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 896.120 1 $1,792 CO2 342.168 4 $2,737 CO3 26.676 50 $2,668 CO4 18.896 50 $1,890 CO5 11.017 100 $2,203 CO6 852.48 2 $3,410 $14,700 Activities A1 A2 A3 A4 A5 A6 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.162 $0.283 $0.310 a a Cα − c Ζℑ $45 $216 $589 $581 $214 $55 Σ= m a a 1 δ $1,700 95 Table 23 (continued) Activities A1 A2 A3 A4 A5 A6 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $64.80 $58.90 $58.10 $42.90 $5.50 $22 $22 1.20% a a a C c 2 ( α − Ζℑ )ℑ $4.50 $43.20 $117.80 $116.10 $21.40 $5.50 $22 $22 0.81% a a a C c 3 ( α − Ζℑ )ℑ $9 $21.60 $117.80 $116.10 $21.40 $5.50 $5 $5 0.20% a a a C c 4 ( α − Ζℑ )ℑ $13.50 $43.20 $58.90 $58.10 $64.30 $5.50 $12 $12 0.65% a a a C c 5 ( α − Ζℑ )ℑ $9 $21.60 $58.90 $116.10 $21.40 $5.50 $54 $54 2.49% a a a C c 6 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $116.10 $42.90 $27.5 $16 $16 0.48% $45 $216 $589 $581 $214 $55 $0 $132 AvgΣ= m a a 1 %ε 0.97% 96 Table 24 Corollary VIb for k < m Activities A1 A2 A3 A4 A5 A6 A7 ABC Cost Cυ r ℑυ r CO1 0.10 0.30 0.10 0.10 0.20 0.10 0.40 $1,899 0.44 0.44 CO2 0.10 0.20 0.20 0.20 0.10 0.10 0.10 $2,735 0.77 0.72 CO3 0.20 0.10 0.20 0.20 0.10 0.10 0.10 $2,648 0.66 0.64 CO4 0.30 0.20 0.10 0.10 0.30 0.10 0.10 $1,867 0.44 0.43 CO5 0.20 0.10 0.10 0.20 0.10 0.10 0.10 $2,125 0.31 0.37 CO6 0.10 0.10 0.30 0.20 0.20 0.50 0.20 $3,426 0.03 0.02 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $300 $14,700 Time 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 A7 Total 1a ℑ 7.75 127.94 232.1 344.1 145.8 17.75 71 946.44 2a ℑ 7.75 85.29 464.21 688.2 72.9 17.75 17.75 1,353.85 3a ℑ 15.5 42.65 464.21 688.2 72.9 17.75 17.75 1,318.95 4a ℑ 23.25 85.29 232.1 344.1 218.71 17.75 17.75 938.95 5a ℑ 15.5 42.65 232.1 688.2 72.9 17.75 17.75 1,086.85 6a ℑ 7.75 42.65 696.31 688.2 145.8 88.75 35.5 1,704.96 Total 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 97 Table 24 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 946.442 1 $1,893 CO2 338.463 4 $2,708 CO3 26.379 50 $2,638 CO4 18.779 50 $1,878 CO5 10.869 100 $2,174 CO6 852.48 2 $3,410 $14,700 Activities A1 A2 A3 A4 A5 A6 A7 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.155 $0.281 $0.310 $0.310 a a Cα − c Ζℑ $45 $216 $589 $535 $205 $55 $55 Σ= m a a 1 δ $1,700 98 Table 24 (continued) Activities A1 A2 A3 A4 A5 A6 A7 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $64.80 $58.90 $53.48 $41.04 $5.50 $22 $6.20 $6.20 0.33% a a a C c 2 ( α − Ζℑ )ℑ $4.50 $43.20 $117.80 $106.96 $20.52 $5.50 $5.50 $27 $27 0.99% a a a C c 3 ( α − Ζℑ )ℑ $9 $21.60 $117.80 $106.96 $20.52 $5.50 $5.50 $9.90 $9.90 0.37% a a a C c 4 ( α − Ζℑ )ℑ $13.50 $43.20 $58.90 $53.48 $61.56 $5.50 $5.50 $10.40 $10.40 0.56% a a a C c 5 ( α − Ζℑ )ℑ $9 $21.60 $58.90 $106.96 $20.52 $5.50 $5.50 $49 $49 2.31% a a a C c 6 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $106.96 $41.04 $27.50 $11 $16.30 $16.30 0.48% $45 $216 $589 $535 $205 $55 $55 $0 $119 AvgΣ= m a a 1 %ε 0.84% 99 Table 25 Corollary VIb for k > m Activities A1 A2 A3 A4 A5 ABC Cost Cυ r ℑυ r CO1 0.10 0.30 0.10 0.10 0.30 $1,954 0.54 0.52 CO2 0.30 0.30 0.10 0.20 0.10 $2,409 0.46 0.43 CO3 0.10 0.10 0.10 0.10 0.10 $1,470 0.00 0.00 CO4 0.30 0.10 0.10 0.10 0.20 $1,645 0.68 0.63 CO5 0.10 0.10 0.30 0.20 0.20 $3,336 0.71 0.64 CO6 0.10 0.10 0.30 0.30 0.10 $3,886 0.97 0.93 Cost $200 $1,069 $5,231 $6,848 $1,352 $14,700 Time 77.5 426.46 2,321.04 3,737.46 787.54 7,350 Activities A1 A2 A3 A4 A5 Total 1a ℑ 7.75 127.94 232.10 373.75 236.26 977.80 2a ℑ 23.25 127.94 232.10 747.49 78.75 1,209.54 3a ℑ 7.75 42.65 232.10 373.75 78.75 735 4a ℑ 23.25 42.65 232.10 373.75 157.51 829.25 5a ℑ 7.75 42.65 696.31 747.49 157.51 1,651.71 6a ℑ 7.75 42.65 696.31 1,121.24 78.75 1,946.70 Total 77.5 426.46 2,321.04 3,737.46 758.28 7,350 100 Table 25 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 977.800 1 $1,956 CO2 302.385 4 $2,419 CO3 14.700 50 $1,470 CO4 16.585 50 $1,659 CO5 16.517 100 $3,303 CO6 973.350 2 $3,893 $14,700 Activities A1 A2 A3 A4 A5 a a Cα ℑ $2.58 $2.51 $2.25 $1.83 $1.72 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.168 $0.284 a a Cα − c Ζℑ $45 $216 $589 $627 $223 Σ= m a a 1 δ $1,700 101 Table 25 (continued) Activities A1 A2 A3 A4 A5 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $64.80 $58.90 $62.67 $67.00 $1.46 $1.46 0.07% a a a C c 2 ( α − Ζℑ )ℑ $13.50 $64.80 $58.90 $125.34 $22.33 $10.47 $10.47 0.43% a a a C c 3 ( α − Ζℑ )ℑ $4.50 $21.60 $58.90 $62.67 $22.33 $0.00 $0.00 0.00% a a a C c 4 ( α − Ζℑ )ℑ $13.50 $21.60 $58.90 $62.67 $44.66 $13.33 $13.33 0.81% a a a C c 5 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $125.34 $44.66 $32.80 $32.80 0.98% a a a C c 6 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $188.00 $22.33 $7.54 $7.54 0.19% $45 $216 $589 $627 $223 $0 $66 AvgΣ= m a a 1 %ε 0.42% 102 6.3. Implications This chapter demonstrates that cost objects will typically consume activities from both + A and − A sets. As a result, the average Σ= k i i 1 %ε is significantly lower than if cost objects from set I + only consume resources in A+ and activities in I − only consume resources in A− . This chapter shows that TDABC2 is not significantly different from the fullyspecified ABC system (e.g. the percentage errors are less than 2.5 percent in Tables 23, 24, and 25). Hence, TDABC2 should replicate the accuracy of the ABC system with the benefit of eliminating Stage 1 cost assignments and significantly reducing the Stage 2 cost assignments. However, some empirical analyses are needed to determine whether the equivalency conditions that are analytically proven actually hold. As an initial empirical analysis, the next chapter provides case studies based on data from an actual company to explore the validity of the equivalency conditions. 103 CHAPTER 7 CASE ANALYSES OF THE EQUIVALENCY CONDITIONS 7.1. Introduction to the Case Analyses and Assumptions This chapter presents seven case studies using data from a particular company to provide anecdotal evidence that explores the validity of the equivalency conditions in Propositions I through IV.5 Case studies are initially useful to identify and explore the validity of hypothesized relationships and, thus, serve as an important forerunner and input for other types of empirical testing (Lillis and Mundy 2005; Kaplan 1986). Since the case studies in this chapter contain data from only one company, any evidence of the validity of the equivalency conditions is anecdotal, which is, consequently, the limitation of this chapter. Therefore, more empirical analyses using data from a broad range of companies are needed beyond these case studies to verify the equivalency conditions further. The data used in the case analyses are yearly company data. For the first four cases, enough data are available to perform Stage 1 and Stage 2 cost assignments for ABC, IABC, IABC2, and TDABC. Only Stage 1 data to perform ABC, IABC, and TDABC are available for the fifth case study, and only Stage 2 data to perform ABC and IABC2 are available for the sixth and seventh case studies. An overview of the names 5 The name of the company is withheld for reasons of confidentiality. 104 of the resources, activities, and products/services (cost objects) for each case study are given in Table 26. 105 Table 26 Names of Resources, Activities, and Cost Objects for Each Case Study Case 1 R1 Salaries and Benefits A1 Repair and maintain fixed equipment CO1 Unit 1 R2 Travel A2 Repair and maintain rotating equipment CO2 Unit 2 R3 Communication A3 Prepare equipment CO3 Waste water treatment plant R4 Special studies A4 Fabricate piping and welding CO4 Unit 3  Gas Treater R5 Depreciation A5 Repair electrical equipment CO5 Boiler & steam system R6 Materials A6 Receive and inventory materials CO6 Tanks & Pipelines R7 External contract services A7 Dock and sail ships CO7 Docks R8 Outside Contractors A8 Perform instrument calibration/repairs CO8 Loading Racks R9 Parts Inventory A9 Equipment reliability CO9 General Administration R10 Rent A10 Plan and schedule work activities CO10 Dock and sail ships R11 Internal Labor A11 Manage/supervise departments R12 Training A12 Manage internal contractors R13 Procurement cards A13 Perform housekeeping & administrative R14 Other expenses A14 Maintain pipelines and valves A15 Monitor SAP work orders 106 Table 26 (Continued) Case 2 R1 Wages & Salaries A1 Receive Product/Invoices Disputes CO1 National Accounts / Buybacks R2 Labor Burden
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Rating  
Title  Simplifying Activitybased Costing 
Date  20090501 
Author  Lelkes, AnneMarie Teresa 
Keywords  ABC, ActivityBased, Costing, Simplifying, TDABC, TimeDriven 
Department  Accounting 
Document Type  
Full Text Type  Open Access 
Abstract  This study first analyzes the conditions under which TimeDriven ActivityBased Costing (TDABC) is equivalent to ActivityBased Costing (ABC). When these equivalency conditions do not hold, there is error. Accordingly, this study also analyzes the maximum absolute percentage error of TDABC that can occur when the equivalency conditions do not hold. However, even when the equivalency conditions are satisfied, TDABC can provide inaccurate costing results when resource diversity exists. Because of the potential resource diversity issue of TDABC, this study provides a new simplified ABC system (TDABC2) that overcomes the limitations of both ABC and TDABC. The equivalency conditions for TDABC2 relative to ABC are analyzed mathematically as well as the maximum absolute percentage error, which occurs when the equivalency conditions for TDABC2 do not hold. Finally, case studies containing data from a particular company are used as an initial analysis to provide anecdotal evidence of the validity of the equivalency conditions. This study shows analytically that TDABC2 is a viable and simpler alternative to the ABC and TDABC systems currently in practice. The two major benefits of TDABC2 are that Stage 1 has been eliminated and Stage 2 has been greatly simplified. Since only the total cost, total time, the unit cycle time, and the number of units of the cost object that will be produced need to be known, the cost to implement the system should be, obviously, significantly lower than that of ABC, perhaps as low as that of the functionalbased system. Since TDABC2 has been analytically proven to be as accurate as ABC under certain conditions, there should be typically no significant tradeoff between the benefit of accuracy and the cost of the system. Hence, TDABC2 should be of great benefit to practitioners who want a relatively accurate, lowcost, and easy to implement costing system. 
Note  Dissertation 
Rights  © Oklahoma Agricultural and Mechanical Board of Regents 
Transcript  SIMPLIFYING ACTIVITYBASED COSTING By ANNEMARIE TERESA LELKES Bachelor of Arts in Accounting Cameron University Lawton, OK 2002 Master of Science in Accounting Oklahoma State University Stillwater, OK 2004 Submitted to the Faculty of the Graduate College of the Oklahoma State University in partial fulfillment of the requirements for the Degree of DOCTOR OF PHILOSOPHY December, 2009 ii SIMPLIFYING ACTIVITYBASED COSTING Dissertation Approved: Dr. Don R. Hansen Dissertation Adviser Dr. Kevin M. Currier Dr. Robert M. Cornell Dr. Thomas S. Wetzel Dr. A. Gordon Emslie Dean of the Graduate College iii ACKNOWLEDGMENTS I wish to express sincere gratitude to my dissertation chair, Dr. Don R. Hansen, for providing much guidance on every detail of the dissertation to make sure I stayed on the right track. Additionally, I am grateful to Dr. Kevin M. Currier for his suggestions and assistance on some of the analytics of the dissertation. Appreciation is also expressed to Dr. Robert M. Cornell and Dr. Thomas S. Wetzel for their suggestions and proofreading. I am grateful to my parents, Tony and Rose, for their care, support, and encouragement throughout my life and especially through the Ph.D. program. Most importantly, I express deep thankfulness to God for helping, guiding, and giving me the ability to do the work for a Ph.D. iv TABLE OF CONTENTS Chapter Page 1. INTRODUCTION .....................................................................................................1 2. LITERATURE REVIEW ..........................................................................................5 Section 2.1. Development of ABC ..........................................................................5 Section 2.2. Success of ABC ...................................................................................7 Section 2.3. Implementation Issues and Limitations of ABC ..................................7 Section 2.4. AftertheFact Simplification ...............................................................8 Section 2.5. BeforetheFact Simplification ..........................................................10 Section 2.6. Motivation ..........................................................................................11 3. EQUIVALENCY ANALYSIS OF TDABC ...........................................................13 Section 3.1. Model Definitions ..............................................................................13 Section 3.2. Assumptions .......................................................................................16 Section 3.3. Equivalency Analysis.........................................................................16 Section 3.4. Time Equations and Unused Capacity ...............................................26 Section 3.5. Resource Diversity .............................................................................29 Section 3.6. Implications........................................................................................33 4. STAGE 2 SIMPLIFICATION .................................................................................34 Section 4.1. IABC Applied to Stage 2 and Model Definition ...............................34 Section 4.2. TDABC Applied to Stage 2 and Model Definition ...........................37 Section 4.3. Equivalency Analysis.........................................................................39 Section 4.4. Unused Capacity in TDABC2 ...........................................................47 Section 4.5. Implications........................................................................................48 v Chapter Page 5. STAGE 1 ERROR ANALYSIS...............................................................................50 Section 5.1. Analysis of the Maximum Error of TDABC .....................................50 Section 5.2. Examples Demonstrating Proposition V and Its Corollaries .............59 Section 5.3. Implications........................................................................................74 6. STAGE 2 ERROR ANALYSIS...............................................................................75 Section 6.1. Analysis of the Maximum Error of TDABC2 ...................................75 Section 6.2. Examples Demonstrating Proposition VI and Its Corollaries ............81 Section 6.3. Implications......................................................................................102 7. CASE ANALYSES OF THE EQUIVALENCY CONDITIONS ........................103 Section 7.1. Introduction to the Case Analyses and Assumptions .......................103 Section 7.2. Case Study 1 ....................................................................................112 Section 7.3. Case Study 2 ....................................................................................120 Section 7.4. Case Study 3 ....................................................................................127 Section 7.5. Case Study 4 ....................................................................................133 Section 7.6. Case Study 5 ....................................................................................140 Section 7.7. Case Study 6 ....................................................................................143 Section 7.8. Case Study 7 ....................................................................................146 Section 7.9. Discussion of Case Study Results ....................................................149 8. SUMMARY OF FINDINGS .................................................................................152 REFERENCES ..........................................................................................................157 vi LIST OF TABLES Table Page 1. Example Illustrating Proposition I ........................................................................19 2. Example Illustrating Corollary IIa ........................................................................23 3. Example Illustrating Corollary IIb ........................................................................24 4. Illustration Not Satisfying Proposition II ..............................................................25 5. Corollary IIa with Unused Capacity .....................................................................28 6. Resource Diversity I .............................................................................................31 7. Resource Diversity II ............................................................................................32 8. The Accuracy of the IABC2 System ....................................................................36 9. Product Example Illustrating Corollary IVa .........................................................42 10. Customer Example Illustrating Corollary IVa ......................................................44 11. Example Illustrating Corollary IVb ......................................................................45 12. Illustration Not Satisfying Proposition IV ............................................................46 13. Corollary IVa with Unused Capacity ....................................................................48 14. Corollary Va for m = n ..........................................................................................61 15. Corollary Va for m < n ..........................................................................................63 16. Corollary Va for m > n ..........................................................................................65 17. Corollary Vb for m = n..........................................................................................68 18. Corollary Vb for m < n..........................................................................................70 19. Corollary Vb for m > n..........................................................................................72 20. Corollary VIa for k = m .........................................................................................83 21. Corollary VIa for k < m .........................................................................................86 22. Corollary VIa for k > m .........................................................................................89 23. Corollary VIb for k = m ........................................................................................93 24. Corollary VIb for k < m ........................................................................................96 25. Corollary VIb for k > m ........................................................................................99 26. Names of Resources, Activities, and Cost Objects for Each Case Study ...........105 27. Case Study 1, Stage 1..........................................................................................115 28. Case Study 1, Stage 2..........................................................................................118 29. Case Study 2, Stage 1..........................................................................................122 30. Case Study 2, Stage 2..........................................................................................125 31. Case Study 3, Stage 1..........................................................................................128 32. Case Study 3, Stage 2..........................................................................................131 33. Case Study 4, Stage 1..........................................................................................135 34. Case Study 4, Stage 2..........................................................................................138 35. Case Study 5 .......................................................................................................141 36. Case Study 6 .......................................................................................................144 vii Table Page 37. Case Study 7 .......................................................................................................147 38. Comparison of Average Absolute Percentage Errors ................................................149 1 CHAPTER 1 INTRODUCTION According to Argyris and Kaplan (1994), activitybased costing (ABC) is a costing model created in the mid1980s that provides more accurate information to managers about the cost and profitability of their business processes, products, services, and customers. ABC provides more accurate cost information by exploiting causal relationships. This is made possible by recognizing that activities consume resources while cost objects (products, customers, etc.) consume activities. Thus, the cost of resources must be first assigned to activities (Stage 1 cost assignment), and then the cost of activities is assigned to cost objects (Stage 2 cost assignment). While ABC is simple in concept, it is complex and costly to implement and operate. An organization must identify and find information for all resources, activities, and their associated drivers, which can number into the hundreds. Consequently, although ABC provides greater accuracy, ABC systems are not as widely adopted as might be expected because of their size, complexity, and cost (Krumwiede 1998a, 1998b; Kaplan and Anderson 2007a). Early attempts to simplify ABC focused on reducing the number of activities and drivers used while attempting to minimize the loss in accuracy (Babad and Balachandran 1993; Homburg 2001). In effect, size and some complexity issues were reduced at the expense of accuracy. These simplified systems also considered 2 the costs to gather information for each activity/driver. However, these attempts required a full implementation of ABC before the simplification could occur. This meant that all activities and drivers had to be identified before the simplification could be done (afterthe fact simplification). If a full implementation must take place, the value of the simplification is questionable. The next major simplification effort is more recent and is a beforethefact simplification. Kaplan and Anderson (2004, 2007a) detail the complexities and costs of ABC. In general, they observe that ABC systems are expensive to build, complex to sustain, and difficult to modify or update. Specifically, they identify the following problems associated with ABC: (1) a timeconsuming and costly interviewing and surveying process is required to identify activities and the resource drivers needed to assign resource costs; (2) since subjectivity is involved in assessing the time spent on various activities, it is difficult to validate the Stage 1 cost assignments; (3) data are expensive to store, process, and report; (4) it is difficult to update the ABC model to accommodate changing circumstances; and (5) the ABC model ignores the potential for unused activity capacity. To address these problems, Kaplan and Anderson (2004, 2007a, 2007b) developed a simplified ABC system called TimeDriven ABC (TDABC). TDABC simplifies Stage 1 by devising a simpler and less timeconsuming approach to assigning resource costs to activities. TDABC provides an easy way to update the ABC model as circumstances change and only assigns the cost of used activity capacity to cost objects. Moreover, it allows an integrated view and approach to cost determination. Thus, 3 TDABC offers a number of significant advantages. However, an examination of its disadvantages and limitations has not been formally addressed. Although the usage of process time equations may reduce the number of activities relative to a fullyimplemented ABC system, TDABC ignores Stage 2 simplification. TDABC calculates activity costs and assigns these costs to cost objects similarly to that of ABC. Since TDABC does not simplify the Stage 2 cost assignment, the size and complexity of TDABC remains considerable because managing the costs and consumption ratios of hundreds of activities is cumbersome for product costing. Hence, under TDABC, Stage 1 is simplified whereas Stage 2 remains complex. Moreover, the accuracy loss of TDABC is another issue that needs to be explored. It is unlikely that TDABC can preserve the same level of accuracy of ABC in all circumstances. The purpose of this study is to extend and expand the beforethefact simplification of ABC. Additional simplification, while overcoming identified limitations of TDABC, should enhance the viability of ABC systems and, thus, represent a significant contribution to ABC literature and actual practice. Hence, the study will first explore the accuracy of TDABC relative to an ideally implemented durationbased ABC system (the benchmark). This will be shown in Chapter 3. Second, as will also be shown in Chapter 3, the study will attempt to specify the conditions that must exist for TDABC to match the ABC assignments (equivalency conditions). Assuming accuracy loss is potentially a significant problem, ways or means of modifying Stage 1 simplification to reduce the accuracy loss will be investigated in Chapter 4. Any such modifications will attempt to preserve the resolution of the problems mentioned by Kaplan and Anderson (2004, 2007a) referred to above. Third, as will be shown also in 4 Chapter 4, the study will provide a new simplified system along with the conditions of equivalency between the new system and ABC to reduce the complexity and, therefore, the cost of Stage 2. Reducing the overall cost and complexity of ABC systems should increase the likelihood of adoption. Fourth, the maximum absolute dollar error between TDABC and ABC systems will be assessed in Chapter 5, with the maximum absolute dollar error between the new simplified system, TDABC2, and ABC in Chapter 6. Finally, in Chapter 7, case studies will be used to explore the validity of the equivalency conditions using a particular company’s data. The next chapter reviews the literature regarding ABC and TDABC, which provides the background for the motivation of this study. 5 CHAPTER 2 LITERATURE REVIEW 2.1. Development of ABC Kaplan (1994) stated that in the early years of ABC, the description of ABC systems was based on an “inner logic” that claims that ABC systems are more accurate than the functionalbased (or, traditional) systems. However, this “inner logic” was not enough to cause a breakthrough for ABC. The academicians, especially Kaplan and Cooper, tried to increase the acceptance of ABC by developing two theories concerning (1) the cost (and activity) hierarchy of factory costs (indirect and support expenses) and (2) what type of resource cost ABC measures. Cooper developed the first ABC theory concerning the cost/activity hierarchy (Cooper 1990). A taxonomy (activity hierarchy) for the activity cost drivers was developed in which activities are classified as (from lowest to highest) unitlevel, batchlevel, productlevel, or facilitysustaininglevel based on the cause and effect relationships between the organizational expense and the level of the organization. Kaplan (1994) states this cost/activity hierarchy provides four advantages. First, all organizational expenses can be mapped to a particular organizational level where cause and effect relationships can be established. Second, the cost/activity hierarchy has provided “a much richer set of drivers of cost variability” (Kaplan 1994, 251). Third, 6 there is a connection between activity levels (unit, batch, product, and facility) and modern developments in operations management. Finally, the activity hierarchy is beneficial for continuous improvement and lean production. Kaplan (1994) states that this activity hierarchy theory helps managers analyze each component of overhead costs to help reduce those costs. Kaplan (1994) developed the second theory in which not all organizational expenses should be assigned to cost objects. ABC systems measure the costs of using resources, not the cost of supplying resources that financial systems measure. The cost of unused capacity is the difference between the cost of resources used and the cost of resources supplied. Once the cost of resources used is found using the ABC system, the cost of unused capacity can be determined. Thus, ABC systems do not directly measure the cost of unused capacity. Additionally, for ABC to provide relevant data, Noreen (1991) found that the cost system must be wellspecified in which the underlying cost function must satisfy three necessary and sufficient conditions. The first condition states that the total overhead cost can be partitioned into cost pools, with each cost pool depending on one activity. The second condition states that there must be a linear relationship between the cost in each cost pool and the level of activity in that cost pool. The third condition eliminates any dependency between products and eliminates joint processes, which means that the production of a product is not dependent on the production of another product. Because of these conditions and the basic intuition behind ABC, there has been some success in implementing ABC as the next section discusses. 7 2.2. Success of ABC The main reason for the success of ABC systems in the firms that adopted and implemented them is the widespread support for ABC within the firm, adequate training, and managers who understand and know ABC information (AlOmiri and Drury 2007). Additionally, research has found that ABC is adopted if 1) there is a current significant risk of cost distortions within the firm, 2) the firm is large, 3) the firm has continuous manufacturing processes as opposed to job shops, and 4) there is product diversity (Krumwiede 1998b). Furthermore, if there is a significant top management support of ABC, then ABC will most likely become integrated within the firm (Krumwiede 1998b). However, the adoption and implementation rates for ABC are low. For instance, one research study stated that the adoption rate is 29 percent (AlOmiri and Drury 2007). Another study stated that the rate is 24 percent (Krumwiede 1998b). Additionally, Gosselin (1997) gave a more informative study and divided the implementation rate from the adoption rate. He found that the adoption rate is 47.8 percent but the implementation rate is only 30.4 percent. Shields (1995) found that 75 percent of the firms that used ABC received a financial benefit. Finally, 85 percent of firms who routinely use ABC feel that it is worth it, whereas 15 percent do not think it is worth the cost (Krumwiede 1998b). The next section discusses the implementation issues and problems of ABC. 2.3. Implementation Issues and Limitations of ABC The last section discussed what drives successful ABC implementation. However, there are reasons that ABC is not successfully adopted. For instance, Krumwiede (1998b) found a strong IT system can prevent ABC adoption or the continuation of implementing it. The reason is that firms with strong IT perceive that 8 they already have enough information for decision making; thus, ABC is not worth the cost to implement it. Additionally, he found that weak top management support and insufficient training in ABC hinders implementation. Insufficient training causes employees to not understand and respect the benefits of an ABC system. Finally, some firms do not have enough patience to wait for the full benefits of implementation and that small firm size and job shops hinder ABC implementation (Krumwiede 1998a, 1998b). Along with these implementation issues, ABC poses some limitations within the system. One limitation of ABC is that the linear approach of activitybased costing provides poor estimates of actual expenditures when there is a nonlinear or discontinuous relation between the demand for and provision of resources (e.g. the resources are provided on a joint and indivisible basis) (Maher and Marais 1998). A second limitation is that an ABC system is expensive, complex, and difficult to modify/update (Krumwiede 1998a; Kaplan and Anderson 2007a). A third limitation is that ABC systems also ignore unused capacity. A fourth limitation is that workers give subjective estimates of their time spent on various activities for Stage 1 cost assignments (Kaplan and Anderson 2007a). In spite of these limitations, the main reason that firms do not implement ABC is that they feel that the perceived benefits do not outweigh the implementation costs and that ABC will not enhance the control of costs (AlOmiri and Drury 2007). Consequently, there is a tradeoff between cost and accuracy. The next two sections focus on the published research that alleviates some of these limitations. 2.4. AftertheFact Simplification Simplification research that focuses on Stage 2 simplification (activity/driver reduction) includes the research by Babad and Balachandran (1993) and Homburg 9 (2001). Babad and Balachandran (1993) developed a model to identify an optimal subset of drivers from the fully specified ABC system that takes into consideration information costs of production and accuracy. Their model allows the decision maker to specify, as a constraint, the maximum number of drivers allowed in the simplified system. This approach combines the costs of the activities corresponding to the eliminated drivers with the activity costs associated with the selected drivers, defining a new, aggregated cost pool for each selected driver. In building more aggregate cost pools, all of the associated activity costs of an eliminated driver are given to the cost pool of a corresponding selected driver. Homburg (2001) extends the Babad and Balachandran (1993) model by allowing the activity costs of the eliminated drivers to be allocated to multiple selected drivers, rather than one corresponding driver. The optimal subset of drivers is selected that minimizes accuracy loss with information costs expressed as a constraint in the model (drivers are selected that do not exceed a prespecified level of information costs). The cost pool for a selected driver is the cost of the selected driver’s associated activity plus a share of the costs of the eliminated activities. He then shows that his approach creates a simplified system with the same level of complexity as the Babad and Balachandran approach but with more accurate product costs compared to a benchmark system. The fact that Homburg’s model produces a more accurate system with no greater information cost illustrates that the Babad and Balachandran model did not identify the optimal simplified system. However, both models assume that a simplified system must sacrifice accuracy. 10 If the system has to be fully specified before it can be simplified, then there is no benefit of simplification since the firm already has a fully specified ABC system. Additionally, whenever the system has to be updated, the fullyspecified system must be updated and then simplified, which seems to be more costly and time consuming in the long run. The next section discusses some research providing a better approach: beforethe fact simplification. 2.5. BeforetheFact Simplification Kaplan and Anderson (2007a) identified a new system called TimeDriven ABC (TDABC) to alleviate some of the complexity of ABC. TDABC skips the stage of driving resource costs to activities and introduces process time equations to take care of diverse and complex transactions (Kaplan and Anderson 2007b). These time equations summarize the time it takes to perform each activity within a process. Hence, TDABC focuses on processes instead of activities, which makes the system more manageable. Kaplan and Anderson (2007a) state The TDABC model simulates the actual processes used to perform work throughout an enterprise. It can therefore capture far more variation and complexity than a conventional ABC model, without creating an exploding demand for data estimates, storage, or processing capabilities. Using TDABC, a company can embrace complexity rather than being forced to use simplified, inaccurate ABC models… (p. 8). Anderson, et al. (2007) claim that TDABC is more accurate since actual transaction data are used instead of estimates. In addition, when the process time equations are built, it is easy to determine which step within the process time equation is consuming too much time. Kaplan and Anderson (2007a) provide other benefits of TDABC over ABC. First, employees do not need to be interviewed or surveyed to allocate resource costs to activities. Second, Stage 1 cost assignment is reduced because 11 resource costs are assigned to the activities using two sets of estimates: 1) the cost of supplying resource capacity for the department (capacity cost rate) and 2) the demand for resource capacity (capacity usage rate, typically time) by each transaction processed in the department. These rates are used to allocate resource costs to activities. Third, TDABC simulates the actual processes, thus capturing more variation and complexity than does ABC without creating greater need for data estimating, storage, or processing capabilities. Fourth, the TDABC model can be updated easier. In contrast, Kaplan and Anderson (2007a, 12) mention that “ABC requires a geometric expansion to capture the increase in complexity.” Additionally, when a new activity is identified, the unit time required only needs to be estimated. The system is updated based on events instead of the calendar. Fifth, it takes only a couple of days instead of weeks to load, calculate, validate, and report findings. Finally, research has found that TDABC can incorporate unused capacity within the TDABC system (Kaplan and Anderson 2007a). Previously, researchers did not understand that unused capacity is vital in ABC systems. However, there are disadvantages. Although TDABC is simpler and cheaper than ABC, TDABC does not reduce the number of activities/drivers that a company has to keep track of for the Stage 2 cost assignments. Additionally, TDABC will not work if the time to perform the activities cannot be reliably clocked or if the activities are not performed in a repetitive manner (Sherratt 2005). 2.6. Motivation In conclusion, TDABC is a better simplification approach as opposed to the afterthe fact simplification models. In TDABC, Stage 1 cost assignment is simplified, but 12 Stage 2 remains complex and similar to the ABC system since all activity costs and their corresponding consumption ratios have to be known. The contribution that this paper will make is to prove that there is a way to simplify the ABC system considerably while maintaining accuracy when compared to the benchmark ABC system. With the simplification method, Stage 1 cost assignment is eliminated with the additional fact that the individual activity costs do not have to be known. If the individual activity costs do not have to be known, then Stage 2 cost assignment is somewhat simplified. To simplify Stage 2 further, TDABC will be modified and applied to Stage 2 as shown in Chapter 4. This simplification will eliminate the need to know the individual activity consumption ratios. The main purpose of this study is to show the limitations of TDABC and provide a simpler and cheaper beforethefact simplified system. It is possible that there are more limitations to TDABC since research has not shown the conditions in which TDABC matches a fullyspecified benchmark ABC system (the benchmark). This study will mathematically analyze those conditions in the next chapter. 13 CHAPTER 3 EQUIVALENCY ANALYSIS OF TDABC 3.1. Model Definitions In this section, the mathematical models for the ABC and TDABC are shown and used to compare the differences in cost assignments. The original models of Kaplan and Anderson are used to explore potential accuracy differences. In this study, the Stage 1 and Stage 2 models for ABC will incorporate duration drivers (timebased drivers) for easier comparison with TDABC. Assuming m activities and n resources, the Stage 1 cost assignment for ABC is modeled as follows: Σ= = n j j j aj a C t t C 1 α j n j ajC Σ= = 1 ρ , a = 1,…, m, (1) Where α a C = cost assigned to activity a under ABC; aj t = activity a’s consumption of time for resource j; j t = total time used to supply resource j (Σ= m a aj t 1 ); aj ρ = relative frequency of use of resource j by activity a (the resource consumption ratio); and Cj = total cost of resource j. 14 Equation 1 states that the total cost of an activity under the ABC system is the sum of the resource consumption ratios, aj ρ , multiplied by the corresponding resource costs, Cj. Assuming k cost objects and m activities, the model for ABC for Stage 2 cost assignment is as follows: Σ= ℑ ℑ = m a a a ia i D C 1 α α =Σ= m a ia a C 1 υ α , i = 1,…, k, (2) Where α i D = cost assigned to cost object i under ABC; α a C = total cost of activity a; ia ℑ = volume or actual absolute frequency of use of activity a by cost object i; a ℑ = total usage of activity a (Σ= ℑ k i ia 1 ); and ia υ = relative frequency of use of activity a by cost object i. Equation 2 states that the total cost of a cost object under the ABC is the sum of the activity consumption ratios, ia υ , multiplied by the corresponding activity costs, α a C . The model for TDABC Stage 1 cost assignment is given below (for simplicity only one resource pool is assumed1): Σ= = n j a aj C c t 1 τ a = cℑ , a = 1,…, m, (3) Where τ a C = cost of activity a under TDABC; 1 The analysis can be easily generalized to more than one resource pool. 15 c = cost per unit of resource time; and a ℑ = total resource time for activity a. Equation 3 states that the total cost assigned to activity a is the sum of the total resource time used by this activity multiplied by the cost per unit of time. The cost per unit of time, c, is simply the total resource cost for the pool divided by the total resource time used by all activities: T T n j j n j j t C t C c = = Σ Σ = = 1 1 , (4) Where j t = total time used to supply resource j; T C = total cost of resources; and T t = total resource time (Σ= n j j t 1 ). The model for TDABC for Stage 2 cost assignment is ia m a i a Dτ Cτυ Σ= = 1 , i = 1,…, k (5) Equation 5 states that the total cost of cost object i ( τ i D ) under TDABC is the sum of each activity cost τ a C multiplied by the corresponding activity consumption ratio ia υ . Equations (2) and (5) for the Stage 2 model for both ABC and TDABC are identical. Any differences in cost assignment between the two models are attributable to differences between α a C and τ a C . Thus, any potential accuracy loss must occur in Stage 1. Before 16 any equivalency analysis is shown, the assumptions behind the analysis are first discussed in the next section. 3.2. Assumptions Two major assumptions are needed to perform the equivalency analysis to find the necessary conditions for equivalency between TDABC and the fullyspecified, benchmark ABC. The first assumption requires a linear relationship between the cost in each cost pool and the level of activity in that cost pool (Noreen 1991). Although Maher and Marais (1998) found that a linear relationship is a limitation of ABC due to poor estimates when there is a nonlinear or discontinuous relation between the demand for and provision of resources, this assumption is fundamental to ABC and will be used for the analysis. TDABC assumes that resources are time driven; thus, the second assumption initially requires that all resources in the ABC system are assigned using duration drivers (timebased drivers). This assumption facilitates the equivalency analysis between TDABC and the benchmark ABC for Stage 1. This assumption is relaxed in Section 3.5 so that the effect of resource diversity on the equivalency conditions can be assessed. 3.3. Equivalency Analysis Differences between α a C and τ a C are highlighted by differences in the information required to calculate each value. The information set for calculating α a C is { } aj j t ,C . Detailed individual resource driver information and resource cost information are needed. Much effort and cost must be expended to gather this information through surveys, interviews, and unbundling the general ledger. The information set needed for calculating τ a C is { } T a T t ,ℑ ,C . Total time and total resource cost are readily available 17 within an existing traditional cost system. TDABC avoids the need to collect detailed information for a ℑ by 1) determining the time to perform one unit of activity; 2) determining the number of times the activity will be performed (usually defined by practical capacity); and 3) multiplying the time to perform one unit of activity by the number of times the activity will be performed. Thus, TDABC allows activity costs to be calculated without knowing individual resource drivers or individual resource costs (only total resource time and total resource cost are needed). Whether the activity cost determined by TDABC is the same as that of ABC is a critical question. It is initially assumed that all resources are time driven. Later this assumption is relaxed. First, an intermediate ABC (IABC) costing system is developed and analyzed that requires knowledge of total resource cost and individual resource drivers. Accordingly, the information set is { } aj T t ,C . The development of the IABC system helps identify the conditions required for equivalency between ABC and TDABC. In the IABC system, an activity’s cost is calculated by multiplying the activity’s average resource consumption ratio by the total resource cost: a T I a C = ρ C , (6) Where I a C = cost of activity a for the IABC system; and a ρ = n n j aj Σ= 1 ρ , the average resource consumption ratio. 18 Equivalency between the Stage 1 cost assignments of ABC and IABC is established by the following reasoning. If a resource costs more (less), it does not mean that an activity has to consume a higher (lower) proportion of that resource’s time. If this state of no linear correlation between resource consumption ratios and individual resource costs exists for every activity, then ABC and IABC are equivalent.2 This equivalency is stated by the following proposition: Proposition I: I a a Cα = C , a = 1,…, m, if and only if there is no correlation between aj ρ and j C , j = 1,…, n. Proof: First, assume there is no correlation between aj ρ and j C for each activity a (a = 1,…, m). The correlation between aj ρ and j C , cρ r , is defined as c n j j n j aj j n j aj c C n C r σ σ ρ ρ ρ ρ Σ Σ Σ = = = − = 1 1 1 , where Σ Σ = = = − n j n j aj aj 1 n 2 2 1 ρ σ ρ ρ and Σ Σ = = = − n j n j j c j n C C 1 2 σ 2 1 . If cρ r = 0, then Σ Σ Σ = = = = n j j n j aj j n j aj C n C 1 1 1 ρ ρ , which implies that I a a Cα = C . 2 Based on the linearity assumption from Section 3.2, all correlations discussed in this dissertation are linear. 19 Next, assume that I a a Cα = C . Since Σ= = n j a aj j C C 1 α ρ and Σ Σ = = = n j j n j aj I a C n C 1 1 ρ , then from the definition of cρ r , this immediately implies that cρ r = 0. QED Table 1 provides a simple illustrative example of Proposition I, using two activities. Note that when the correlation between aj ρ and j C is zero, multiplying the average consumption ratios by the total cost produces the ABC cost assignments. As shown in Table 1, for Activity 1 (A1) and Activity (A2), IABC Stage 1 cost assignments are identical to those under ABC ( α 1 C = CI 1 = $615 and = CI 2 = $585). Hence, under IABC, there is no need to know the individual resource costs. TABLE 1 Example Illustrating Proposition I Resource R1 R2 R3 R4 ABC Cost Assignmenta a ρ b IABC Cost Assignmentc rcρ A1 0.60 0.20 0.30 0.95 $615 0.513 $615 0 A2 0.40 0.80 0.70 0.05 $585 0.488 $585 0 $95 $335 $370 $400 $1,200 $1,200 a Σ= = n j j j aj a C t t C 1 α b a ρ = 4 1 Σ= n j aj ρ c a T I a C = ρ C The information set for IABC is { } aj T t ,C . IABC eliminates the need to know the individual resource costs Cj required for ABC; however, the detailed resource consumption ratios must be known. The correlation between aj ρ and j C is exploited to α 2 C 20 reduce the fineness of the ABC information set. This suggests the possibility of exploiting correlation relationships to establish equivalency between ABC and TDABC. Note that the information set for TDABC is { } T a T t ,ℑ ,C , which eliminates the need to know both Cj and taj of ABC. For TDABC, ρ t r (the correlation between aj ρ and tj) and cρ r are both needed as shown by the following proposition. Proposition II: α τ a a C = C , a = 1,…, m, if and only if t t c c cr σ r σ ρ ρ = . Proof: First assume that t t c c cr σ r σ ρ ρ = . By definition, t n j j a j n j aj t t t r σ σ ρ ρ ρ ρ Σ Σ = = − = 1 1 , where Σ Σ = = = − n j n j j t j n t t 1 2 σ 2 1 . Substitute ρ t r , cρ r , and c with their corresponding formulas and simplify to obtain j n j j aj n j n aj j j n j j t C t C Σ Σ Σ Σ = = = = = 1 1 1 1 ρ ρ . Note that j n j ajt Σ= 1 ρ is equivalent to Σ= n j aj t 1 . Hence, j n j aj n j n aj j j n j j t C t C Σ Σ Σ Σ = = = = = 1 1 1 1 ρ ⇒ τ α a a C = C . Next, assume that α τ a a C = C . From the definition of ρ t r , t t n j j a j n j aj ρ t ρ t r σ σ ρ ρ Σ − Σ = =1 =1 21 ⇒ t t n j a j n j aj t ρ t r σ σ ρ ρ Σ − Σ = =1 =1 . Since Σ Σ Σ = = = = n j n aj j j n j j a t t C C 1 1 τ 1 , then τ n a j j n j n j j aj C C t t Σ Σ Σ = = = = 1 1 1 . Substituting for Σ= n j aj t 1 provides t t n j n a a j j j n j j C t r C t ρ σ σ ρ ρ τ − Σ = Σ Σ = = = 1 1 1 ⇒ Σ Σ Σ = = = = + n j j n j j t t n j a a j t C C C r 1 1 1 ρ σ σ ρ ρ τ ⇒ t t n j a a j C ρ C cr σ σ ρ ρ τ + = Σ= 1 . From the cρ r equation, Σ Σ = = − = n j j a j c c n j aj C C r 1 1 ρ ρ σ σ ρ ρ ⇒ Σ= − = n j a a j c c C C r 1 ρ σ σ ρ ρ α ⇒ Σ= = + n j a a j c c C C r 1 ρ σ σ ρ ρ α . Accordingly, [ ] a a c c t t C C σ r σ cr σ ρ ρ ρ α − τ = − . Thus, if α τ a a C = C , then t t c c cr σ r σ ρ ρ = . QED In the proof of the above proposition, it is shown that [ ] a a c c t t C C σ r σ cr σ ρ ρ ρ α − τ = − . Interestingly, since the dollar value of the error between the two systems equals α τ a a C −C , then the dollar value of the error can be expressed as follows: a ε = [ ] c c t t σ r σ cr σ ρ ρ ρ − , a = 1,…, m (7) 22 When the error for each activity a is zero, there is equivalency, which implies that − = 0 c c t t r σ cr σ ρ ρ . This expression implies that if cρ r = 0, then ρ t r = 0. Thus, the following corollary to the Proposition II has been proved: Corollary IIa: If cρ r = 0 and ρ t r = 0, then α τ a a C = C , a = 1,…, m. In the event that both cρ r and ρ t r are nonzero, then it is also possible to establish an equivalency condition based on a required value for c. When [ ] a c c t t ε σ r σ cr σ ρ ρ ρ = − = 0, solving for c and simplifying yields the following equivalency condition: Σ Σ Σ Σ Σ Σ = = = = = = − − = − − = n j j a j n j aj I a a n j j a j n j aj n j j a j n j aj t t C C t t C C c 1 1 1 1 1 1 ρ ρ ρ ρ ρ ρ α This then establishes a second corollary: Corollary IIb: If T T n j j a j n j aj I a a t C t t C C c = − − = Σ Σ =1 =1 ρ ρ α , then α τ a a C = C , a = 1,…, m. According to Corollary IIb, if the rationale for zero correlation is not valid, it is still possible to obtain equivalency. However, a very special relationship must exist. The numerator I a a Cα −C is the dollar error between ABC and IABC for activity a. The 23 denominator Σ Σ = = − n j j a j n j aj t t 1 1 ρ ρ is the unit time error between time allocated to activity a using ABC and the time allocated to activity a using IABC. Consequently, ε α ρ ρ c t t C C n j j a j n j aj I a a = − − Σ Σ =1 =1 represents the absolute dollar error per unit of error time. Note that ε c must be written in absolute form since the IABC cost assignment for activity a could be greater than that of ABC. Table 2 provides an example illustrating Corollary IIa of Proposition II. Table 2 compares TDABC and ABC when there is no correlation between tj and aj ρ and between aj ρ and j C for each activity a. When cρ r = 0 and ρ t r = 0 for each activity, the activity costs under TDABC ( τ 1 C = $615 and τ 2 C = $585) are equal to those under ABC. TABLE 2 Example Illustrating Corollary IIa ABC Resource R1 R2 R3 R4 ABC Cost Assignmenta A1 0.60 0.20 0.30 0.95 $615 A2 0.40 0.80 0.70 0.05 $585 Cost $95 $335 $370 $400 $1,200 Time 101 300 321 350 1,072 TDABC Unit Time Total Units of Activity a ℑ c TDABC Cost Assignmentb rcρ rtρ A1 36.63 15 549 $1.12 $615 0 0 A2 26.15 20 523 $585 0 0 Total Time 1,072 $1,200 a Σ= = n j j j aj a C t t C 1 α b τ a C a = cℑ 24 Corollary IIb of Proposition II is demonstrated in the following example shown in Table 3. According to Table 3, cρ r and ρ t r are nonzero and the cost per unit of time, c, and dollar error per unit of time, ε c , are both equal to $4. This satisfies Corollary IIb so that α τ a a C = C , where τ 1 C = α 1 C = $216 and τ 2 C = α 2 C = $264. TABLE 3 Example Illustrating Corollary IIb ABC Resource R1 R2 R3 R4 ABC Cost Assignmenta a ρ IABC Cost Assignmentb A1 0.20 0.40 0.60 0.80 $216 0.50 $240 A2 0.80 0.60 0.40 0.20 $264 0.50 $240 Cost $180 $60 $180 $60 $480 $480 Time 35 40 25 20 120 TDABC Unit Time Total Units of Activity a ℑ ε c c c TDABC Cost Assignmentd rcρ rtρ A1 9 6 54 $4 $4 $216 0.45 0.85 A2 3 22 66 $264 0.45 0.85 Total Time 120 $480 a Σ= = n j j j aj a C t t C 1 α b a T I a C = ρ C c T T n j j a j n j aj I a a t C c t t C C c = = − − = Σ Σ =1 =1 ρ ρ α ε d τ a C a = cℑ If Proposition II does not hold, then there is a difference in the cost assigned to TDABC relative to that of ABC. Table 4 shows that when there is perfect correlation between tj and aj ρ and between aj ρ and j C (where t t c c cr σ r σ ρ ρ ≠ ), the activity costs 25 under TDABC ( τ 1 C = $151 and τ 2 C = $384) are not equal to those under ABC ( α 1 C = $249 and α 2 C = $286). The average absolute percentage error of TDABC is 36.7 percent, with dollar error ( a ε ) is $97.93 for A1 and $97.93 for A2 (c = $0.4977 and ε c = $0.488). TABLE 4 Illustration Not Satisfying Proposition II ABC Resource R1 R2 R3 R4 ABC Cost Assignmenta a ρ IABC Cost Assignmentb A1 0.80 0.25 0.30 0.15 $249 0.375 $201 A2 0.20 0.75 0.70 0.85 $286 0.625 $334 Cost $215 $110 $120 $90 $535 $535 Time 101 318 299 357 1,075 TDABC Unit Time Total Units of Activity a ℑ c TDABC Cost Assignmentc rcρ rtρ A1 19 16 304 $0.4977 $151 1 1 A2 25.7 30 771 $384 1 1 Total Time 1,075 $535 ABC TDABC a ε % a ε d ε c e A1 $249 $151 $98 39.24% $0.488 A2 $286 $384 ($98) 34.16% Avg a %ε f 36.7% a Σ= = n j j j aj a C t t C 1 α e cε = T T n j j a j n j aj I a a t C c t t C C ≠ = − − Σ Σ =1 =1 ρ ρ α b a T I a C = ρ C f 2 39.24% + − 34.16% c τ a C a = cℑ d% a ε = α α τ a a a C C − C 26 If cρ r and ρ t r are nonzero and TDABC and ABC are not equivalent, then the cost per unit of time would not be equal to the dollar error per unit of time. Therefore, a ε > 0 for each a. From Equation 7, it is possible to analyze the effects of various variables on the magnitude of the error. For instance, the error will be larger in absolute magnitude if cρ r and ρ t r are opposite in sign, which makes the two terms on the right hand side of Equation 7 additive. The magnitude of the error is also affected by variability in ρaj, tj, and Cj. Additional analysis of the dollar error is needed in which the maximum absolute dollar error is identified and will be shown in Chapter 5. 3.4. Time Equations and Unused Capacity Kaplan and Anderson (2007a, 2007b) stated that a process can be expressed in a process time equation that consists of all of the individual activities that make up the process. Time equations summarize the TDABC time information. Using time equations is a way of obtaining granularity (the level of detail) without having a separate activity for each event. If TDABC and ABC have the same granularity, then Proposition II holds. Time equations are based on the unit time for each activity and the number of times it is actually performed (or, actual activity used). The difference between the time equation based on practical activity and the time equation based on actual activity used is unused capacity. When unused capacity exists and the equivalency conditions are satisfied, the cost of activity a under ABC is equal to the cost of activity a under TDABC plus the cost of unused capacity for activity a. This only means that the cost of unused capacity for activity a is separated from the actual cost of the activity a used. Thus, there is no significant effect on the equivalency conditions. The next table is similar to Table 2 but 27 has been modified to incorporate unused capacity. Time equations are then developed to illustrate the summarization of the TDABC time information. Table 5 shows the same illustration as in Table 2 except that unused capacity now exists. For TDABC, the activity used represents the number of times the activity is performed. Practical activity represents the number of times the activity should be performed under normal operating conditions. Notice that, for equivalency, the ABC cost for an activity must equal the TDABC cost for an activity plus the cost of unused portion of the activity. Therefore, Table 5 illustrates that unused capacity has no significant effect on the conditions for equivalency. To develop the process time equations, assume that the illustration in Table 5 concerns an ordering department that has two activities: number of repeat orders (A1) and the number of new orders (A2). The time equation that represents the total order processing time based on actual activity used is Actual time used = 36.60(# of repeat orders used) + 26.15(# of new orders used) = 36.60(10) + 26.15(15) = 758 minutes The time equation that represents the total order processing time based on practical activity is Practical time = 36.60(# of repeat orders) + 26.15(# of new orders) = 36.60(15) + 26.15(20) = 1,072 minutes The unused capacity time is the difference between the practical time and the actual time used, which is 314 minutes (1,072 minus 758). The total cost of unused capacity is $351 (314 x $1.12). To find the cost of unused capacity for each activity, the activities would 28 have to be separated out of the time equation, and the results would be identical to those displayed in Table 5. TABLE 5 Corollary IIa with Unused Capacity Resource R1 R2 R3 R4 ABC Cost rcρ rtρ c A1 0.60 0.20 0.30 0.95 $615 0 0 $1.12 A2 0.40 0.80 0.70 0.05 $585 0 0 Cost $95 $335 $370 $400 $1,200 Time 101 300 321 350 1,072 Unit Time Activity Used Time Used Practical Activity a ℑ TDABC Cost Unused Cost Total Cost A1 36.60 10 366 15 549 $410 $205 $615 A2 26.15 15 392 20 523 $439 $146 $585 758 1,072 $849 $351 $1,200 Only Corollary IIa is shown for this analysis because if unused capacity is applied to Corollary IIb, the results are similar to the illustration in Table 3 of Corollary IIb and follow the same process as above for unused capacity. Consequently, it has been illustrated that time equations summarize the information of the TDABC system and have no bearing on the equivalency conditions since they are developed after the TDABC system has been implemented. Therefore, both unused capacity and time equations do not affect the equivalency conditions. In Section 3.3, the conditions for equivalency between ABC and TDABC assume that all resources are timedriven. However, there are, in general, resources that are not time driven (e.g. some forms of capital, materials, and some forms of energy). In TDABC, the costs of these nontimedriven resources are pooled with the costs of resources that are time driven. This resource diversity can produce inaccurate activity 29 costs. This inaccuracy can pose a major problem for TDABC if the costs of the nontime driven resources are significant. When nontimedriven resources are significant, pooling can cause inaccurate cost assignments since there would be a lack of causal relationships for nontimedriven resources. In Section 3.5, resource diversity is examined and examples are used to illustrate this problem. 3.5. Resource Diversity Resource diversity exists when there are a significant proportion of nontimedriven resources that are consumed in a different pattern from timedriven resources. In Stage 1 cost assignment, TDABC assigns the cost of all resources to the activities using timebased drivers, which means that timebased drivers are used to assign the costs of both timebased and nontimebased resources to activities. Let the set of all resources be R = {1,…, n}. Next, partition R into a set of timedriven resources, TD = {1,…, l}, and a set of nontimedriven resources, NTD = {l+1,…, n}, where R = TD∪NTD. If, on average, activities consume nontimedriven resources in the same pattern as timedriven resources, then equivalence between ABC and TDABC remains possible. As a result, n t l t n j aj a l j aj a Σ Σ = = = = = 1 1 ρ τ ρ , where ρ τa is the average consumption ratio for timedriven resources for activity a, a = 1,..., m. When ρ τa = a ρ , there is no resource diversity and Proposition II applies. However, if ρ τa ≠ a ρ , then resource diversity (RD) exists and can be measured as follows: a a RD = ρ τ − ρ , a = 1,…, m (8) 30 This suggests the possibility that as RD increases, then the potential difference between ABC and TDABC also increases. Table 6 provides an illustration of resource diversity that shows the potential inaccuracy of TDABC. There are two activities, four timedriven resources (j = 1,…, l), and four nontimedriven resources (j = l+1,…, n). In the example, the timedriven resources are the labor resources (L1 – L4), and the nontime driven resources are the materials resources (M1 and M2) and energy resources (E1 and E2). Additionally, cρ r = 0 and ρ t r = 0 so that . Although the conditions for α τ a a C = C are satisfied, α τ a a C ≠ C because of the effect of nontimebased resources. t t c c cr σ r σ ρ ρ = 31 TABLE 6 Resource Diversity I ABC Resources L1 L2 L3 L4 M1 M2 E1 E2 A1 0.25 0.70 0.15 0.10 0.80 0.40 0.30 0.50 A2 0.75 0.30 0.85 0.90 0.20 0.60 0.70 0.50 Cost $3,565 $3,400 $2,900 $1,000 $1,500 $7,000 $3,000 $2,400 Time 3,800 1,500 1,400 1,000 ABC Costa a ρ rcρ rtρ A1 $9,906 0.40 0 0 A2 $14,859 0.60 0 0 Cost $24,765 TDABC a ℑ c TDABC Costb ρ τa A1 2,310 $3.22 $7,430 0.30 A2 5,390 $17,336 0.70 7,700 $24,765 ABC TDABC a ε % a ε c A1 $9,906 $7,430 $2,477 25.0% A2 $14,859 $17,336 ($2,477) 16.7% Avg a %ε 20.8% a Σ= = n j j j aj a C t t C 1 α b τ a C a = cℑ c% a ε = α α τ a a a C C − C According to Table 6, ρ τa for A1 and A2 are 0.3 and 0.7, respectively, whereas a ρ for A1 and A2 are 0.4 and 0.6, respectively. Thus, RD = 0.1 for A1 and 0.1 for A2. The dollar value of the error a ε is $2,477 for A1 and $2,477 for A2. The average absolute percentage error of TDABC is 20.8 percent. However, if RD increases, then the 32 average absolute percentage error increases as shown in Table 7. Compared to Table 6, Table 7 shows that as RD doubles, the average absolute percentage error almost doubles. Hence, the illustration supports the claim that as RD increases, error increases. TABLE 7 Resource Diversity II ABC Resources L1 L2 L3 L4 M1 M2 E1 E2 A1 0.25 0.70 0.15 0.10 0.60 0.85 0.40 0.95 A2 0.75 0.30 0.85 0.90 0.40 0.15 0.60 0.05 Cost $3,565 $3,400 $2,900 $1,000 $3,000 $1,500 $6,234 $3,166 Time 3,800 1,500 1,400 1,000 ABC Costa a ρ rcρ rtρ A1 $12,383 0.50 0 0 A2 $12,382 0.50 0 0 Cost $24,765 TDABC a ℑ c TDABC Costb ρ τa A1 2,310 $3.22 $7,430 0.30 A2 5,390 $17,336 0.70 Total Time 7,700 $24,765 ABC TDABC a ε % a ε c A1 $12,383 $7,430 $4,953 40.0% A2 $12,382 $17,336 ($4,953) 40.0% Avg a %ε 40.0% a Σ= = n j j j aj a C t t C 1 α b τ a C a = cℑ c% a ε = α α τ a a a C C −C If resource diversity is significant, then TDABC may be significantly less accurate than ABC. One possible resolution to this problem is discussed in Chapter 4. 33 3.6. Implications The major implication that the equivalency conditions for TDABC have on research and practice is to show when the TDABC system will replicate the ABC system. The equivalency holds when the underlying conditions as outlined in Propositions I and II are satisfied and when all resources are time driven. However, when there is resource diversity, the assumption of all resources being time driven is violated. When this one assumption is violated, there is no equivalency although the conditions in Proposition I and Proposition II and its corollaries are met. This issue needs to be resolved. The next chapter provides a resolution by analyzing a Stage 2 simplification procedure in order to eliminate Stage 1 cost assignments while maintaining accurate costing of the cost objects. 34 CHAPTER 4 STAGE 2 SIMPLIFICATION 4.1. IABC Applied to Stage 2 and Model Definition The previous chapter showed the conditions for accuracy for the TDABC system. Because of the potential inaccuracy of the TDABC system when there is resource diversity, this section will look at a way to simplify ABC while resolving the potential inaccuracy issue of TDABC. One resolution method is to extend Proposition I to Stage 2. Recall that Proposition I states that when the resource costs and resource consumption ratios are not linearly correlated, then the cost of a particular activity is basically its average resource consumption ratio multiplied by the total resource cost for all resources. This can be applied to other cost objects as well. The Stage 2 Intermediate system (IABC2) uses the IABC model to resolve the TDABC resource diversity issue and simultaneously offers some simplification for Stage 2. Assuming k cost objects and m activities, the IABC2 model is described as follows: i T m a i a I i C C D υ υ α = = Σ= 1 , i = 1,…, k, (9) Where I i D = cost assigned to cost object i under IABC2; and 35 i υ = m m m a a ia m a ia Σ Σ = = ℑ ℑ 1 = 1 υ , the average activity consumption ratio of cost object i. Equation 9 states that the cost of cost object i is the average activity consumption ratio multiplied by the total cost. Accordingly, the information set for IABC2 is { } ia T ℑ ,C . For IABC2, the individual activity costs do not have to be known; only the total cost needs to be known and the individual activity consumption ratios. Since the individual activity costs do not have to be known, Stage 1 cost allocation is eliminated, which is a significant simplification and the issue of resource diversity is resolved. Equivalency between α i D and I i D is established by the following proposition: Proposition III: I i T i m a i ia a D C C D = = =Σ= α υ α υ 1 , if and only if there is no correlation between ia υ and α a C for each cost object i, a = 1,…, m and i = 1,…, k. The proof of Proposition III parallels that of Proposition I and is, therefore, omitted. Let Cυ r represent the correlation between ia υ and α a C for cost object i. Parallel to the definition of cρ r , C m a a i a m a ia C C C r σ σ υ υ υ α α υ Σ Σ = = − = 1 1 , where Σ( ) Σ = = = − m a m a a C a m C C 1 2 2 1 α σ α and Σ Σ = = = − m a m a ia ia 1 m 2 2 1 υ σ υ υ . Table 8 shows an illustration of Proposition III. The 36 illustration contains two cost objects and four activities. Since Cυ r = 0, the costs assigned to the cost objects under IABC2 are identical to those under ABC. TABLE 8 The Accuracy of the IABC2 System Stage 1 A1 A2 A3 A4 Resource Cost Labor 1 0.16 0.26 0.20 0.38 $300,000 Labor 2 0.38 0.53 0.00 0.09 $650,000 Energy 0.20 0.29 0.10 0.41 $750,000 Materials 0.10 0.20 0.30 0.40 $800,000 Cost $525,000 $800,000 $375,000 $800,000 $2,500,000 Stage 2 A1 A2 A3 A4 ABC Costa Cυ r CO1 0.20 0.30 0.69 0.80 $1,243,750 0 CO2 0.80 0.70 0.31 0.20 $1,256,250 0 Cost $525,000 $800,000 $375,000 $800,000 $2,500,000 IABC2 Costb i υ CO1 $1,243,750 0.4975 CO2 $1,256,250 0.5025 Cost $2,500,000 a Σ= = m a i ia a D C 1 α υ α b i T I i D =υ C TDABC simplifies Stage 1 cost assignment by eliminating the need to know the resource consumption ratios. However, TDABC must calculate the individual activity costs needed for Stage 2 calculations. As shown in Table 8 (which illustrates Proposition III), IABC2 eliminates the need to know activity costs for Stage 2, thus eliminating the potential problem of resource diversity introduced by TDABC. In addition, since IABC2 does not need activity costs, Stage 2 is also simplified. However, the activity 37 consumption ratios have to be found for all activities to calculate the average activity consumption ratios for each cost object. Gathering this information is timeconsuming and costly. Thus, further simplification is desirable. The next section analyzes a more desirable method in which TDABC is applied to Stage 2. 4.2. TDABC Applied to Stage 2 and Model Definition A more desirable method is to develop a simplification of Stage 2 that avoids the need to gather all of the information necessary to calculate the average activity consumption ratios. One approach is to extend TDABC concepts found in Stage 1 to Stage 2. The presence of the IABC2 model suggests the possibility that a TDABC2 model is feasible. The TDABC2 model builds on IABC2 by eliminating the need to know all of the activities and their associated consumption ratios. If TDABC concepts are transferred to Stage 2, then TDABC2 would only require knowledge of the total cost, total time, the unit cycle time, and the number of units of the cost object that will be produced. Thus, TDABC2 is performed by 1) determining the cycle time for one unit of product (e.g. from the time the sales order is received until the finished good goes to the warehouse); 2) determining the number of units that will be produced; and 3) multiplying the cycle time by the number of units that will be produced. The TDABC2 cost assignment model is as follows: i i m a m ia a a m a a i c C D β θ α Σ Σ Σ = Ζ = Ζ = ℑ = ℑ = 1 1 1 , i = 1,…, k, (10) Where Ζ i D = cost of cost object i under TDABC2; ia ℑ = time consumed of activity a by cost object i; 38 a ℑ = total time of activity a; i β = unit cycle time for cost object i; i θ = number of units produced for cost object i at practical capacity; and cΖ = cost per unit of activity time, where Z stands for TDABC2. Equation 10 states that the total cost assigned to cost object i is the unit cycle time multiplied by the number of units produced and then multiplied by the cost per unit of time. The cost per unit of time, c Ζ , is simply the total activity cost for the pool divided by the total activity time used by all cost objects: T T m a a m a a t C C c = ℑ = Σ Σ = Ζ = 1 1 α , (11) Where T C = total overhead cost; and T t = total time in the system (Σ= ℑ m a a 1 ). Additionally, from Equation 10, the cycle time multiplied by the number of units produced is the sum of the time consumed of activity a by cost object i across all a (a = 1,…, m): i i m a ia θ β = ℑ Σ= 1 , i = 1,…, k (12) Notice also from Equation 10 that the cost for one unit of a cost object is the unit cycle time multiplied by c Ζ , or i c Ζβ . Hence, the information set for TDABC2 is { } T T i i C ,t ,β ,θ . The total overhead cost ( T C ), the total time in the system ( T t ), and the 39 number of units produced for cost object i at practical capacity ( i θ ) can be found from the accounting records. The unit cycle time for cost object i ( i β ) is found by clocking how long it takes from the time the sales order is received until the finished good goes to the warehouse. 4.3. Equivalency Analysis Similar to the analysis of TDABC with the benchmark ABC for Stage 1, the assumptions behind the necessary equivalency conditions are linearity and that the Stage 2 cost assignments are duration based in benchmark ABC system. Conditions needed to establish equivalency between TDABC2 and ABC are derived from extending Proposition II and its corollaries to Stage 2. Like in Proposition II for TDABC, for TDABC2 Cυ r (the linear correlation between ia υ and α a C for cost object i) and ℑυ r (the linear correlation between ia υ and a ℑ for cost object i) are both needed. Parallel to the definition of ρ t r , ℑ = = ℑ Σ ℑ − Σℑ = σ σ υ υ υ υ m a a i a m a ia r 1 1 , where Σ Σ = = ℑ ℑ = ℑ − m a m a a a 1 m 2 σ 2 1 and υ σ is defined as before. The extension of Proposition II to Stage 2 is shown in the following proposition. Proposition IV: = Ζ i i Dα D if and only if ℑ ℑ σ = Ζ σ υ υ r c r C C , i = 1,…, k. The proof is parallel to that of Proposition II and is, therefore, omitted. 40 From the proof of the above proposition, [ ] ℑ ℑ − Ζ =σ σ − Ζ σ υ υ υ Dα D r c r i i C C . Since the dollar value of the error between the two systems equals − Ζ i i Dα D , then the dollar value of the error can be expressed as follows: i ε = [ ] ℑ ℑ σ σ − Ζ σ υ υ υ r c r C C , i = 1,…, k (13) When the error for cost object i is zero, there is equivalency, which implies that − = 0 ℑ ℑ σ Ζ σ υ υ r c r C C . This expression shows that if Cυ r = 0, then ℑυ r = 0. Using the same rationale that establishes equivalency between ABC and TDABC, ℑυ r should also equal zero since this implies that a cost object does not need to consume a higher (lower) proportion of that activity’s time if an activity has more (less) time available. Thus, the following corollary to Proposition IV has been proved: Corollary IVa: If Cυ r = 0 and ℑυ r = 0, then = Ζ i i Dα D , i = 1,…, k. In the event that both Cυ r and ℑυ r are nonzero, then it is also possible to establish an equivalency condition based on a required value for cΖ . When [ ] ℑ ℑ ε =σ σ − Ζ σ υ υ υ r c r i C C = 0, solving for cΖ and simplifying yields the following equivalency condition: Σ Σ Σ Σ Σ Σ = = = = Ζ = = ℑ − ℑ − = ℑ − ℑ − = m a a i a m a ia I i i m a a i a m a ia m a a i a m a ia D D C C c 1 1 1 1 1 1 υ υ υ υ υ υ α α α 41 This then establishes a second corollary: Corollary IVb: If T T m a a i a m a ia m a a i a m a ia t C C C c = ℑ − ℑ − = Σ Σ Σ Σ = = Ζ = = 1 1 1 1 υ υ υ α υ α , then = Ζ i i Dα D , i = 1,…, k. According to Corollary IVb, if the rationale for zero correlation is not valid, it is still possible to obtain equivalency. However, a very special relationship must exist. The numerator I i i Dα − D is the dollar error between Stage 2 ABC and IABC2 for cost object i. The denominator Σ Σ = = ℑ − ℑ m a a i a m a ia 1 1 υ υ is the unit time error between time allocated to cost object i using ABC and the time allocated to cost object i using IABC2. Accordingly, Ζ = = = ℑ − ℑ − Σ Σ ε α υ υ c D D m a a i a m a ia I i i 1 1 represents the dollar error per unit of error time for Stage 2. Again, parallel to ε c in Chapter 3, Ζ ε c must be written in absolute form since IABC2 cost for cost object i could be greater than that of ABC Stage 2. Table 9 provides an illustration of Corollary IVa in which the cost objects are product lines (P1 and P2). Only Stage 2 is shown of ABC. Table 9 compares TDABC2 and ABC when there is no correlation between ia υ and α a C and between ia υ and a ℑ for cost object i. Notice from Table 9 that once cΖ and i β (the unit cycle time for cost object i) are known, the cost per unit of product can be found. Again, i β is an observed value. If the ABC system is durationbased, i β must equal the cycle time calculated from the 42 durationbased benchmark ABC system. The cycle time from the durationbased ABC is calculated by dividing the total hours for cost object i by the number of units produced at practical capacity (e.g. 400/800 = 0.5 unit cycle time for P1). To find the cost of the entire product line, the cost per unit of product is multiplied by the number of units produced ( i θ ). Table 9 shows that the cost of each product under TDABC2 ( Ζ 1 D = $185 and Ζ 2 D = $185) are equal to those under ABC since Cυ r = 0 and ℑυ r = 0. TABLE 9 Product Example Illustrating Corollary IVa ABC Activities A1 A2 A3 A4 Hours P1 20 120 180 80 400 P2 80 180 120 20 400 100 300 300 100 800 A1 A2 A3 A4 ABC Cost Assignmenta i υ IABC2 Cost Assignmentb P1 0.20 0.40 0.60 0.80 $185 0.50 $185 P2 0.80 0.60 0.40 0.20 $185 0.50 $185 Cost $105 $60 $120 $85 $370 $370 TDABC2 Total Cost $370 Total Hours 800 cΖ $0.46 i β Cost per Unit i θ TDABC2 Cost Assignmentc Cυ r ℑυ r P1 0.5 $0.23 800 $185 0 0 P2 2 $0.93 200 $185 0 0 $370 a υ αa m a ia a i C D Σ= = 1 b i T I i D =υ C c i i i DΖ = cΖβ θ 43 To illustrate that TDABC2 can be applied to something other than products, Table 10 provides another illustration of Corollary IVa in which the cost objects are customers. This is the only customer example that will be shown since the next few product illustrations in this chapter can easily be adapted and applied to customers. Here, i β represents the order cycle time for cost object i (the time from which the order is made to the time the payment is received), and i θ represents the number of orders for cost object i. Table 10 shows that the cost of each customer under TDABC2 ( Ζ 1 D = $5,200 and Ζ 2 D = $6,800) are equal to those under ABC since Cυ r = 0 and ℑυ r = 0 for each customer. 44 TABLE 10 Customer Example Illustrating Corollary IVa ABC Activities A1 A2 A3 Days Customer1 210 30 280 520 Customer2 490 120 70 680 700 150 350 1,200 A1 A2 A3 ABC Costa i υ IABC2 Costb Customer1 0.30 0.20 0.80 $5,200 0.43 $5,200 Customer2 0.70 0.80 0.20 $6,800 0.57 $6,800 Cost $7,000 $1,500 $3,500 $12,000 $12,000 TDABC2 Total Cost $12,000 Total Days 1,200 cΖ $10 i β Cost per Order i θ TDABC2 Costc Cυ r ℑυ r Customer1 26 $260 20 $5,200 0 0 Customer2 34 $340 20 $6,800 0 0 $12,000 a υ αa m a ia a i C D Σ= = 1 b i T I i D =υ C c i i i DΖ = cΖβ θ Corollary IVb of Proposition IV is demonstrated in the following example shown in Table 11. According to Table 11, Cυ r and ℑυ r are nonzero and cΖ and the dollar error per unit of time, Ζ ε c , are both equal to $1.17. This satisfies Corollary IVb so that = Ζ i i Dα D , where Ζ 1 D = α 1 D = $640 and Ζ 1 D = α 1 D = $760. However, notice that IABC2 provides inaccurate results because Proposition III is violated. 45 TABLE 11 Example Illustrating Corollary IVb ABC Activities A1 A2 A3 A4 Hours P1 60 183 146 160 549 P2 240 274 97 40 651 300 457 243 200 1,200 A1 A2 A3 A4 ABC Cost Assignmenta i υ IABC2 Cost Assignmentb P1 0.20 0.40 0.60 0.80 $640 0.50 $700 P2 0.80 0.60 0.40 0.20 $760 0.50 $700 Cost $500 $200 $500 $200 $1,400 $1,400 TDABC2 Total Cost $1,400 cΖ $1.17 Total Hours 1,200 Ζ ε c c $1.17 i β Cost per Unit i θ TDABC2 Cost Assignmentd Cυ r ℑυ r P1 0.686 $0.80 800 $640 0.45 0.59 P2 3.257 $3.80 200 $760 0.45 0.59 $1,400 a υ αa m a ia a i C D Σ= = 1 c Σ Σ = = Ζ ℑ − ℑ − = m a a i a m a ia I i i D D c 1 1 υ υ α ε b i T I i D =υ C d i i i DΖ = cΖβ θ If Proposition IV does not hold, then there is a difference in the cost assigned to TDABC2 relative to that of ABC Stage 2. Table 12 shows that when there is a nonzero correlation for Cυ r and ℑυ r (where ℑ ℑ σ ≠ Ζ σ υ υ r c r C C ), the cost of the cost objects under TDABC2 ( Ζ 1 D = $643 and Ζ 2 D = $657) are not equal to those under ABC Stage 2 ( α 1 D = $465 and α 2 D = $835). The average absolute percentage error of TDABC2 for the 46 illustration is 29.76 percent, with dollar error ( i ε ) is $177.78 for P1 and $177.78 for P2 ( cΖ = $0.72 and Ζ ε c = $1.50). TABLE 12 Illustration Not Satisfying Proposition IV ABC Activities A1 A2 A3 A4 Hours P1 75 160 175 480 890 P2 225 240 325 120 910 300 400 500 600 1,800 A1 A2 A3 A4 ABC Cost Assignmenta i υ IABC2 Cost Assignmentb P1 0.25 0.40 0.35 0.80 $465 0.45 $585 P2 0.75 0.60 0.65 0.20 $835 0.55 $715 Cost $500 $300 $400 $100 $1,300 $1,300 TDABC2 Total Cost $1,300 cΖ $0.72 Total Hours 1,800 Ζ ε c c $1.50 i β Cost per Unit i θ TDABC2 Cost Assignmentd Cυ r ℑυ r P1 1.1125 $0.80 800 $643 0.97 0.86 P2 4.55 $3.29 200 $657 0.97 0.86 $1,300 ABC TDABC2 i ε % i ε e P1 $465 $643 $178 38.23% P2 $835 $657 $178 21.29% Avg i %ε f 29.76% a υ αa m a ia a i C D Σ= = 1 d i i i DΖ = cΖβ θ b i T I i D =υ C e% i ε = α α i i i D DΖ − D c Σ Σ = = Ζ ℑ − ℑ − = m a a i a m a ia I i i D D c 1 1 υ υ α ε f 2 38.23%+ − 21.29% 47 If Cυ r and ℑυ r are nonzero and TDABC2 and ABC are not equivalent, then the cost per unit of time would not be equal to the dollar error per unit of time. Therefore, > 0 i ε . The analysis of the effects of the various variables on the magnitude of the error in Equation 13 is similar to that for Equation 7 under Stage 1. Additional analysis of the dollar error in which the maximum error is identified will be shown in Chapter 6. 4.4. Unused Capacity in TDABC2 Parallel to Stage 1 analysis, unused capacity does not affect the necessary equivalency conditions for TDABC2. For equivalency, the cost of cost object i under ABC must equal that under TDABC plus its unused cost. Table 13 shows an example illustrating unused capacity. Table 13 is similar to Table 9 except that unused capacity is included. Notice that Table 13 verifies that the ABC cost for P1 = TDABC cost for P1 + cost of unused capacity ($185 = $173 + $12) and the cost for P2 = TDABC cost for P2 + cost of unused capacity ($185 = $176 + $9). The cost of unused capacity is equal to the cost per unit ( i c Ζβ ) multiplied by the difference between the number of units produced at practical capacity ( i θ ) and the number of units actually produced ( A i θ ). The rationale is that the unit cycle time i β is an observed value, so it must remain constant for each unit that is produced. Hence, the unused time is the difference between the total time available at practical capacity ( i i β θ ) and the total time actually used ( A i i β θ ). Accordingly, the cost of unused capacity is ( ) ( A ) i i i A i i i i c Ζ β θ −β θ = c Ζβ θ −θ . 48 TABLE 13 Corollary IVa with Unused Capacity ABC Activities A1 A2 A3 A4 Hours Cυ r ℑυ r P1 20 120 180 80 400 0 0 P2 80 180 120 20 400 0 0 100 300 300 100 800 A1 A2 A3 A4 ABC Cost Assignment i υ IABC2 Cost Assignment P1 0.20 0.40 0.60 0.80 $185 0.50 $185 P2 0.80 0.60 0.40 0.20 $185 0.50 $185 Cost $105 $60 $120 $85 $370 $370 TDABC2 Total Cost $370 Total Hours 800 cΖ $0.46 i β Cost per Unit i θ TDABC2 Cost Assignment A i θ Unused Cost Total Cost P1 0.5 $0.23 750 $173 800 $12 $185 P2 2 $0.93 190 $176 200 $9 $185 $349 $21 $370 4.5. Implications As shown analytically, under certain conditions, TDABC2 is equivalent to ABC assignments. TDABC2 has the benefit of IABC2 in which Stage 1 is eliminated and, thus, the problem of resource diversity is eliminated. Additionally, the linear relationship limitation due to poor estimates if there is a nonlinear or discontinuous relation between the demand for and provision of resources (Maher and Marais 1998) has also been resolved since Stage 1 has been eliminated. TDABC2 is a feasible system in which only the unit cycle time, total time, total cost, and number of units produced need to be known. Accordingly, TDABC2 is as 49 simple as a functionalbased costing system but with the accuracy of an ABC system. This will have significant practical relevance. However, if the equivalency conditions are not satisfied, then error will exist. The maximum absolute dollar error for TDABC2 relative to ABC must be identified, but before doing so, the maximum absolute dollar error for TDABC relative to ABC must first be identified as shown in the next chapter and then extended to TDABC2 as shown in Chapter 6. 50 CHAPTER 5 STAGE 1 ERROR ANALYSIS 5.1. Analysis of the Maximum Error of TDABC This chapter shows what the maximum absolute dollar error is for TDABC relative to ABC when the conditions in Proposition II and its corollaries are not met. From Chapter 3, the error for activity a, which is derived from Proposition II and shown in Equation 7, is ε a = [ ] c c t t σ r σ cr σ ρ ρ ρ − . If cρ r and ρ t r are substituted in Equation 7 (let j aj aj t t ρ = , T n j j C C = Σ= 1 , and T n j j t t = Σ= 1 ), then − − =Σ − Σ Σ Σ = = = = n j n j j T aj aj n j j T aj aj n j j j a t t n t c t t t n C t t C 1 1 1 1 ε . Simplifying further yields aj n j j j a c t t C Σ= = − 1 ε , a = 1,…, m (14) The Stage 1 error analysis is based on one assumption that in any given instance in time, the total cost of resource j ( j C ) and the total time available for resource j ( j t ) are likely to be fixed, but the consumption of the resources may vary depending on the activity usage. If j C and j t are treated as constants in the system and the consumption of resource j by activity a, aj t , is allowed to vary, then the following proposition shows 51 that the maximum absolute dollar error of the system is Σ Σ = = = − n j j j n j j C ct 1 1 δ , based on the second assumption that all resources are time driven, where j j j δ ≡ C − ct , which is basically the total dollar error contribution of resource j. c t C j j − is the dollar error contribution per unit of time of resource j (“unit dollar error contribution of j”). Proposition V: Given aj n j j j a c t t C Σ= = − 1 ε and j j t C c ≠ , the maximum absolute dollar error for the system is Σ Σ = = = − n j j j n j j C ct 1 1 δ . Proof: The dollar error for activity a is ( ) j aj n j aj j j n j j j a t t c t C ct t C Σ Σ = = − = = − 1 1 ε . Summing over all a yields the dollar error contribution of resource j: ( ) j j m a aj j j j j aj m a j j j t C ct t C ct t t C ct = − − =Σ − = Σ =1 =1 δ . The total dollar error contribution of all resources is the sum of all j δ : ( ) 0 1 1 1 1 1 1 1 1 = − = − = − Σ = Σ Σ Σ Σ Σ Σ Σ = = = = = = = = n j n j j j n j n j j j n j j n j j n j j j n j j t t C δ C ct C c t C . The total dollar error for all resources is zero implying that some resources provide a positive dollar error contribution and others a negative error contribution, such that 52 − > 0 j j C ct and − c > 0 t C j j for j = 1,…, s and − < 0 j j C ct and − c < 0 t C j j for j = s+1,…, n. Hence, the resources can be partitioned into two sets: one to represent the resources that provide the positive dollar error contribution, { } s R R , R ,..., R 1 2 + = , and the other to represent resources that provide a negative dollar error contribution, { } s s n R R , R ,..., R +1 +2 − = , where − > 0 j j C ct ( − c > 0 t C j j ) with j = 1,…, s ∈ R+ and − < 0 j j C ct ( − c < 0 t C j j ) with j = s+1,…, n ∈ − R . Then, Σ= n j j 1 δ can be rewritten as ( ) ( ) 0 1 1 1 Σ =Σ − + Σ − = = = = + n j s j j s j j j n j j δ C ct C ct . Thus, the total maximum absolute dollar error of the system which is essentially the total absolute dollar error contribution from all resources j = 1,…, n is Σ Σ Σ Σ = = = + = = − + − = − n j j j n j s j j j s j j n j j C ct C ct C ct 1 1 1 1 δ ( ) ( ) . QED Notice that the maximum absolute dollar error for the system, Σ Σ = = = − n j j j n j j C ct 1 1 δ , does not depend on the aj t ’s. Maximizing the dollar error based on the aj t ’s involves finding a corner solution that is part of the maximum set of corner solutions. To find the set of aj t ’s that produces the maximum error, the resources must be ordered from largest to smallest c t C j j − , where c t C − 1 1 > c t C − 2 2 > … > c t C n n − in addition to partitioning the resources into R+ and R− sets. The “unit dollar error contribution” is a better measure of magnitude than j j C − ct (total dollar error 53 contribution of resource j) since the amount of time could vary from one resource to the next causing the corresponding j j C − ct to not be of the same magnitude as the corresponding c t C j j − . Next, partition the activities into two sets in which some activities consume resources that provide a positive dollar error contribution ( A+ set) and the rest of the activities consume resources that provide a negative dollar error contribution ( A− set): A+ = {a a = 1,…, q; = − j ∈ R+ t t C ct j aj aj j j ε ( ) , and = ( − ) = 0 j aj aj j j t t ε C ct where = 0 aj t if j ∈ R− }; and A− = {a a = q+1,…, m; = − j ∈ R− t t C ct j aj aj j j ε ( ) , and = ( − ) = 0 j aj aj j j t t ε C ct where = 0 aj t if j ∈ R+ }, where A+ ∪ A− = A. The above expressions for A+ and A− sets state that to maximize the dollar error using aj t ’s, activities in A+ must only consume resources in R+ and activities in A− only consume resources in R− . As a result, the aj t ’s that result in Σ Σ = = = − n j j j n j j C ct 1 1 δ is identified in the following corollary to Proposition V. Corollary Va: The aj t ’s that produce the total maximum absolute dollar error of the system, Σ Σ = = = − n j j j n j j C ct 1 1 δ , is Σ ΣΣ = = = = − m a n j j aj j j m a a t t C ct 1 1 1 ε ( ) . 54 Proof: Choose taj with j = 1,…, s∈ R+ , such that j q a aj t t = Σ= 1 and aj t with j = s+1,…, n∈ R− , such that j m a q aj Σt = t = +1 . As a result, the dollar error from the A+ set is ( ) j aj s j j j s j a aj t t Σ Σ C ct = = + = = − 1 1 ε ε > 0 and from the A− set is ( ) j aj n j s j j n j s a aj t t Σ Σ C ct = + = + − = = − 1 1 ε ε < 0. Hence, 0 1 1 1 Σ =Σ + Σ = = + − = + = m a q a q a a m a a ε ε ε . This means that ( ) ( ) 0 1 1 1 1 Σ =ΣΣ − =Σ − > = = = = + s j j j q a j aj s j j j q a a C ct t t ε C ct using j q a aj t t = Σ= 1 and ( ) ( ) 0 1 1 1 1 Σ = Σ Σ − = Σ − < = + = + = + = + − n j s j j m a q j aj n j s j j m a q a C ct t t ε C ct using j m a q aj Σt = t = +1 . Thus, using these expressions for Σ= + q a a 1 ε and Σ = + − m a q a 1 ε , Σ ΣΣ( ) Σ Σ( ) ΣΣ = = = = + = + = = = − + − = − m a n j j aj j j m a q j aj n j s j j q a j aj s j j j m a a t t C ct t t C ct t t C ct 1 1 1 1 1 1 1 ε ( ) , which can be rewritten as Σ Σ( ) Σ( ) Σ Σ = = = + = = = − + − = − = n j j n j j j n j s j j s j j j m a a C ct C ct C ct 1 1 1 1 1 ε δ . Since there are no cancellation effects of positive resource dollar error contributions with negative resource dollar error contributions for any a, Σ ΣΣ Σ Σ = = = = = = − = − = n j j n j j j m a n j j aj j j m a a C ct t t C ct 1 1 1 1 1 ε ( ) δ . QED 55 Although Σ Σ = = = n j j m a a 1 1 ε δ , a program is identified that provides the maximum percentage error of each activity a that maximizes the average absolute percentage error of the system.3 The percentage error for any given activity a is α ε ε a a a C % = , which is the dollar error for activity a divided by the ABC cost for activity a. To find the maximum average absolute percentage error, one program is used for the positive sets, A+ and R+ , and another program for the negative sets, A− and R− . From Proposition V and Corollary Va, let m+ represent the number of activities in A+ (a = 1,…, q) and n+ represent the number of resources in R+ (j = 1,…, s). After ordering all resources from largest positive to smallest positive c t C j j − and labeling them as R1, R2,…, Rs, where c t C − 1 1 > c t C − 2 2 > … > c t C s s − , the most positive resource, R1, must be the only resource consumed by m+ − (n+ −1) activities. The program for the maximization of the percentage errors for each activity in the A+ set is as follows: Max Σ Σ Σ = = = − q a s j j j aj s j aj j j C t t c t t C 1 1 1 (P1) s.t. γ ,1 1 t t a ≥ , a = 1,…, q1 (P2) γ aj j t ≥ t , a = 1+ m+ − (n+ −1) ,…, q1, j = 2,…, s1 (P3) 3 An optimization software program such as LINGO can be used. 56 All other ≥ 0 aj t , j ∈ R+ and a ∈ A+ (P4) j q a aj t t = Σ= 1 (P5) The objective function in (P1) provides the maximum percentage error in magnitude across all activities in A+ . (P2) is the first constraint that ensures that for R1, at least the amount of γ 1 t is assigned to q1 activities in A+ . The materiality and uniqueness parameter γ (e.g. it can be set as 0.1 to ensure that 10 percent of the time for resource j is assigned to the appropriate activities) is only assigned to q1 activities instead of q to not over restrict the program and allow it to choose the optimal set of aj t ’s. (P3) is the second constraint that ensures uniqueness among resource vectors without overrestricting the program to allow for some aj t in A+ and R+ to be zero or for some activities in A+ to consume all of a single resource in R+ (this is represented by (P4)). (P5) ensures that the sum of the activity consumptions in A+ of a particular resource in R+ is equal to the total time available for that particular resource j. From Proposition V and its Corollary Va, let m− represent the number of activities in A− (a = q+1,…, m) and n− represent the number of resources in R− (j = s+1,…, n). First order all resources from least negative to most negative c t C j j − and label them as Rs+1, Rs+2,…, Rn, where c t C s s − + + 1 1 > c t C s s − + + 2 2 > … > c t C n n − . The most negative Rn must be the only resource consumed by m− − (n− −1) activities. The 57 program for the maximization the magnitude of the percentage errors in the R− set is as follows (this involves minimizing because of dealing with negative values for c t C j j − ): Min Σ Σ Σ = + = + = + − m a q n j s j j aj n j s aj j j C t t c t t C 1 1 1 (P6) s.t. γ a n n t ≥ t , , a = q+2,…, m (P7) γ aj j t ≥ t , a = q+1,…, m − m− − (n− −1) , j = s+2,…, n1 (P8) All other ≥ 0 aj t in j ∈ R− and a ∈ A− (P9) j m a q aj Σt = t = +1 (P10) The objective function in (P6) provides the maximum percentage error in magnitude across all activities in A− . (P7) is the first constraint that ensures that for Rn, at least the amount of γ n t is assigned to a = q+2,…, m activities in A− without over restricting the program and allow it to choose the optimal set of aj t ’s. (P8) is the second constraint that ensures uniqueness among resource vectors without overrestricting the program to allow for some aj t in A− and R− to be zero or for some activities in A− to consume all of a single resource in R− (this is represented by (P9)). (P10) ensures that the sum of the activity consumptions in A− of a particular resource in R− is equal to the total time available for that particular resource j. 58 Typically, an activity will consume resources from both R+ and R− , thus allowing for some cancellation effects from the positive dollar error contribution from resources in R+ and negative dollar error contribution from resources in R− . Since Σ Σ = = + = − = s j j aj j j aj j s j a j t t t t C ct 1 1 ε ( ) δ and Σ ΣΣ ΣΣ Σ = = = = = = + = − = = s j j q a s j j aj j q a j aj j s j j q a a t t t t C ct 1 1 1 1 1 1 ε ( ) δ δ with j j C − ct > 0 ( c t C j j − > 0) and Σ Σ = + = + − = − = n j s j aj j j aj j n j s a j t t t t C ct 1 1 ε ( ) δ and Σ Σ Σ Σ Σ Σ = + = + = + = + = + = + − = − = = n j s j m a q m a q n j s j aj j j aj j n j s j m a q a t t t t C ct 1 1 1 1 1 1 ε ( ) δ δ with j j C − ct < 0 ( c t C j j − < 0), then if any a in A+ (a = 1,…, q) consumes any resource j in R− (j = s+1,…, n), then there is a cancellation effect of negative resource error contributions with positive resource error contributions; thus, Σ Σ = + = < q a a q a a 1 1 ε ε . Likewise, if any a in A− (a = q+1,…, m) consumes any resource j in R+ (j = 1,…, s), then there is a cancellation effect of positive resource error contributions with negative resource error contributions; consequently, Σ Σ = + + = + < m a q a m a q a 1 1 ε ε . As a result, when there are any cancellation effects, Σ Σ = = < n j j m a a 1 1 ε δ , which implies that, j n j j aj n j j aj j n j j aj a j t t t t t t ε Σδ Σδ Σ δ = = = = ≤ = 1 1 1 . Hence, Σ= m a a 1 ε should be less than Σ= n j j 1 δ as shown in the following corollary to Proposition V. Corollary Vb: As already derived, if activities in A+ only consume resources in R+ and activities in A− only consume resources in R− , thenΣ Σ = = = n j j m a a 1 1 ε δ . If any or all 59 activities consume resources from both R+ and R− , then Σ Σ = = < n j j m a a 1 1 ε δ . Therefore, Σ= m a a 1 ε can never exceed Σ= n j j 1 δ , implyingΣ Σ = = ≤ n j j m a a 1 1 ε δ . Proof: By using the triangle inequality m n ε + ε + ...+ ε ≤ δ + δ + ...+ δ 1 2 1 2 , Σ ΣΣ Σ = = = = ≤ = n j j m a j n j j aj m a a t t 1 1 1 1 ε δ δ ⇒ Σ Σ = = ≤ n j j m a a 1 1 ε δ . QED If there are no cancellation effects (activities in A+ only consume resources in R+ and activities in A− only consume resources in R− ), then Σ Σ = = = n j j m a a 1 1 ε δ . Since j j C − ct > 0 ( c t C j j − > 0) for j ∈ R+ and j j C − ct < 0 ( c t C j j − < 0) for j ∈ R− , if any or all activities consume resources from both R+ and R− , then there will be some cancellation effects within each of those a ε ’s, and thus, the actual absolute dollar error across all activities will be less than the maximum absolute dollar error of the system, Σ Σ = = < n j j m a a 1 1 δ ε . Thus, Σ= m a a 1 ε can never exceed Σ= n j j 1 δ . 5.2. Examples Demonstrating Proposition V and Its Corollaries Examples demonstrating Proposition V and its corollaries are shown in this section. Corollary Va will be shown first. Tables 14, 15, and 16 demonstrate examples of Corollary Va for m = n, m < n, and m > n, respectively. All three cases are shown to demonstrate that Proposition V and Corollary Va as well as program (P1) through (P10) 60 are viable in each situation. The resources are ranked from largest to smallest c t C j j − and the resources are partitioned accordingly. For each table, the first three activities (A1 through A3) are in the A+ set, and the rest of the activities are in the A− set. In each table, Σ Σ = = = n j j m a a 1 1 ε δ = $1,700, thus satisfying Corollary Va. Program (P1) through (P10) is used to maximize each a ε % , which maximizes average %Σ= m a a 1 ε . 61 Table 14 Corollary Va for m = n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0 0 0 0 0 $200 0.61 0.61 A2 0.42 0.10 0 0 0 0 $1,069 0.70 0.70 A3 0.48 0.90 1 0 0 0 $5,231 0.70 0.70 A4 0 0 0 1 0.90 0.56 $6,598 0.68 0.68 A5 0 0 0 0 0.10 0.34 $1,302 0.86 0.86 A6 0 0 0 0 0 0.10 $300 0.74 0.74 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 0 0 0 0 0 77.5 $2 $155 j t2 328.96 97.5 0 0 0 0 426.46 $853 j t3 368.5 877.5 1,075 0 0 0 2,321 $4,642 j t4 0 0 0 1,275 1,327.5 986.7 3,589.2 $7,178 j t5 0 0 0 0 147.5 610.8 758.3 $1,517 j t6 0 0 0 0 0 177.5 177.5 $355 $14,700 62 Table 14 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $0 $0 $0 $0 $0 $45 $45 22.50% j j j j C t c t2 ( / − ) $191 $25 $0 $0 $0 $0 $216 $216 20.21% j j j j C t c t 3 ( / − ) $214 $225 $150 $0 $0 $0 $589 $589 11.26% j j j j C t c t 4 ( / − ) $0 $0 $0 $50 $225 $306 $581 $581 8.80% j j j j C t c t 5 ( / − ) $0 $0 $0 $0 $25 $189 $214 $214 16.45% j j j j C t c t 6 ( / − ) $0 $0 $0 $0 $0 $55 $55 $55 18.33% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 16.26% 63 Table 15 Corollary Va for m < n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0 0 0 0 0 $200 0.61 0.61 A2 0.42 0.10 0 0 0 0 $1,069 0.70 0.70 A3 0.48 0.90 1 0 0 0 $5,231 0.70 0.70 A4 0 0 0 1 0.90 0.64 $6,848 0.72 0.72 A5 0 0 0 0 0.10 0.36 $1,352 0.85 0.85 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 0 0 0 0 0 77.5 $2 $155 j t2 328.96 97.5 0 0 0 0 426.46 $853 j t3 368.5 877.5 1,075 0 0 0 2,321 $4,642 j t4 0 0 0 1,275 1,327.5 1,134.96 3,737.46 $7,475 j t5 0 0 0 0 147.5 640 787.5 $1,575 $14,700 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 64 Table 15 (continued) Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $0 $0 $0 $0 $0 $45 $45 22.50% j j j j C t c t2 ( / − ) $191 $25 $0 $0 $0 $0 $216 $216 20.21% j j j j C t c t 3 ( / − ) $214 $225 $150 $0 $0 $0 $589 $589 11.26% j j j j C t c t 4 ( / − ) $0 $0 $0 $50 $225 $352 $627 $627 9.15% j j j j C t c t 5 ( / − ) $0 $0 $0 $0 $25 $198 $223 $223 16.52% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 15.93% 65 Table 16 Corollary Va for m > n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0 0 0 0 0 $200 0.61 0.61 A2 0.42 0.10 0 0 0 0 $1,069 0.70 0.70 A3 0.48 0.90 1 0 0 0 $5,231 0.70 0.70 A4 0 0 0 1 0.90 0.47 $6,347 0.63 0.63 A5 0 0 0 0 0.10 0.33 $1,253 0.86 0.86 A6 0 0 0 0 0 0.10 $300 0.74 0.74 A7 0 0 0 0 0 0.10 $300 0.74 0.74 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 0 0 0 0 0 77.5 $2 $155 j t2 328.96 97.5 0 0 0 0 426.46 $853 j t3 368.5 877.5 1,075 0 0 0 2,321 $4,642 j t4 0 0 0 1,275 1,327.5 838.5 3,441 $6,882 j t5 0 0 0 0 147.5 581.5 729 $1,458 j t6 0 0 0 0 0 177.5 177.5 $355 j t7 0 0 0 0 0 177.5 177.5 $355 $14,700 66 Table 16 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $0 $0 $0 $0 $0 $45 $45 22.50% j j j j C t c t2 ( / − ) $191 $25 $0 $0 $0 $0 $216 $216 20.21% j j j j C t c t 3 ( / − ) $214 $225 $150 $0 $0 $0 $589 $589 11.26% j j j j C t c t 4 ( / − ) $0 $0 $0 $50 $225 $260 $535 $535 8.43% j j j j C t c t 5 ( / − ) $0 $0 $0 $0 $25 $180 $205 $205 16.38% j j j j C t c t6 ( / − ) $0 $0 $0 $0 $0 $55 $55 $55 18.33% j j j j C t c t7 ( / − ) $0 $0 $0 $0 $0 $55 $55 $55 18.33% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 16.49% 67 Illustrations of Corollary Vb are shown in Tables 17, 18, and 19 for m = n, m < n, and m > n, respectively. The resources are ranked from largest to smallest c t C j j − and the resources are partitioned accordingly. As previously, for each table, the first three activities (A1 through A3) are in the A+ set, and the rest of the activities are in the A− set. In each table, all activities share all of the resources so thatΣ Σ = = < n j j m a a 1 1 ε δ = $1,700. Table 17 shows that the average Σ= m a a 1 %ε = 1.90 percent compared to 16.26 percent in Table 14. Compared to Table 15 with an average Σ= m a a 1 %ε = 15.93 percent, Table 18 shows that the average Σ= m a a 1 %ε = 2.18 percent. Table 19 shows that the average Σ= m a a 1 %ε = 1.77 percent compared to 16.49 percent in Table 16. Notice that in each table, each % a ε is less than 5 percent, which demonstrates that the actual maximum error is small since activities will typically consume resources in both R+ and R− sets. 68 Table 17 Corollary Vb for m = n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0.20 0.10 0.30 0.20 0.30 $3,060 0.74 0.74 A2 0.30 0.10 0.10 0.10 0.20 0.10 $2,140 0.43 0.43 A3 0.20 0.30 0.10 0.10 0.10 0.20 $2,410 0.27 0.27 A4 0.10 0.20 0.10 0.10 0.30 0.10 $2,230 0.17 0.17 A5 0.10 0.10 0.40 0.20 0.10 0.10 $2,410 0.18 0.18 A6 0.20 0.10 0.20 0.20 0.10 0.20 $2,450 0.00 0.00 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 195 107.5 382.5 295 532.5 1,590 $2 $3,180 j t2 232.5 97.5 107.5 127.5 295 177.5 1,037.5 $2,075 j t3 155 292.5 107.5 127.5 147.5 355 1,185 $2,370 j t4 77.5 195 107.5 127.5 442.5 177.5 1,127.5 $2,255 j t5 77.5 97.5 430 255 147.5 177.5 1,185 $2,370 j t6 155 97.5 215 255 147.5 355 1,225 $2,450 $14,700 69 Table 17 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $50 $15 $15 $50 $165 $120 $120 3.92% j j j j C t c t2 ( / − ) $135 $25 $15 $5 $50 $55 $65 $65 3.04% j j j j C t c t 3 ( / − ) $90 $75 $15 $5 $25 $110 $40 $40 1.66% j j j j C t c t 4 ( / − ) $45 $50 $15 $5 $75 $55 $25 $25 1.12% j j j j C t c t 5 ( / − ) $45 $25 $60 $10 $25 $55 $40 $40 1.66% j j j j C t c t 6 ( / − ) $90 $25 $30 $10 $25 $110 $0 $0 0.00% $450 $250 $150 $50 $250 $550 $0 $290 AvgΣ= m a a 1 %ε 1.90% 70 Table 18 Corollary Vb for m < n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0.30 0.10 0.30 0.40 0.20 $3,520 0.41 0.41 A2 0.30 0.10 0.10 0.20 0.10 0.40 $3,020 0.35 0.35 A3 0.20 0.30 0.10 0.10 0.30 0.10 $2,650 0.25 0.25 A4 0.10 0.20 0.10 0.30 0.10 0.10 $2,190 0.10 0.10 A5 0.30 0.10 0.60 0.10 0.10 0.20 $3,320 0.31 0.31 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 292.5 107.5 382.5 590 355 1,805 $2 $3,610 j t2 232.5 97.5 107.5 255 147.5 710 1,550 $3,100 j t3 155 292.5 107.5 127.5 442.5 177.5 1,302.5 $2,605 j t4 77.5 195 107.5 382.5 147.5 177.5 1,087.5 $2,175 j t5 232.5 97.5 645 127.5 147.5 355 1,605 $3,210 $14,700 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 71 Table 18 (continued) Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $75 $15 $15 $100 $110 $90 $90 2.56% j j j j C t c t2 ( / − ) $135 $25 $15 $10 $25 $220 $80 $80 2.65% j j j j C t c t 3 ( / − ) $90 $75 $15 $5 $75 $55 $45 $45 1.70% j j j j C t c t 4 ( / − ) $45 $50 $15 $15 $25 $55 $15 $15 0.68% j j j j C t c t 5 ( / − ) $135 $25 $90 $5 $25 $110 $110 $110 3.31% $450 $250 $150 $50 $250 $550 $0 $340 AvgΣ= m a a 1 %ε 2.18% 72 Table 19 Corollary Vb for m > n Resources R1 R2 R3 R4 R5 R6 ABC cρ r ρ t r A1 0.10 0.10 0.40 0.10 0.10 0.10 $2,160 0.20 0.20 A2 0.30 0.10 0.10 0.10 0.10 0.10 $1,870 0.61 0.61 A3 0.20 0.20 0.10 0.10 0.10 0.30 $2,490 0.27 0.27 A4 0.10 0.10 0.10 0.40 0.30 0.10 $2,760 0.27 0.27 A5 0.10 0.20 0.10 0.10 0.20 0.10 $1,960 0.00 0.00 A6 0.10 0.10 0.10 0.10 0.10 0.10 $1,470 0.00 0.00 A7 0.10 0.20 0.10 0.10 0.10 0.20 $1,990 0.32 0.32 Cost $2,000 $2,200 $2,300 $2,500 $2,700 $3,000 $14,700 Time 775 975 1,075 1,275 1,475 1,775 7,350 Resources R1 R2 R3 R4 R5 R6 Total c TDABC j t1 77.5 97.5 430 127.5 147.5 177.5 1,057.5 $2 $2,115 j t2 232.5 97.5 107.5 127.5 147.5 177.5 890 $1,780 j t3 155 195 107.5 127.5 147.5 532.5 1,265 $2,530 j t4 77.5 97.5 107.5 510 442.5 177.5 1,412.5 $2,825 j t5 77.5 195 107.5 127.5 295 177.5 980 $1,960 j t6 77.5 97.5 107.5 127.5 147.5 177.5 730 $1,470 j t7 77.5 195 107.5 127.5 147.5 355 1,010 $2,020 $14,700 73 Table 19 (continued) Resources R1 R2 R3 R4 R5 R6 j j C t $2.58 $2.26 $2.14 $1.96 $1.83 $1.69 C t c j j− $0.581 $0.256 $0.140 $0.039 $0.169 $0.310 j j C − ct $450 $250 $150 $50 $250 $550 Σ= n j j 1 δ $1,700 Resources R1 R2 R3 R4 R5 R6 Total a ε % a ε j j j j C t c t1 ( / − ) $45 $25 $60 $5 $25 $55 $45 $45 2.08% j j j j C t c t2 ( / − ) $135 $25 $15 $5 $25 $55 $90 $90 4.81% j j j j C t c t 3 ( / − ) $90 $50 $15 $5 $25 $165 $40 $40 1.61% j j j j C t c t 4 ( / − ) $45 $25 $15 $20 $75 $55 $65 $65 2.36% j j j j C t c t 5 ( / − ) $45 $50 $15 $5 $50 $55 $0 $0 0.00% j j j j C t c t 6 ( / − ) $45 $25 $15 $5 $25 $55 $0 $0 0.00% j j j j C t c t7 ( / − ) $45 $50 $15 $5 $25 $110 $30 $30 1.51% $450 $250 $150 $50 $250 $550 $0 $1,700 AvgΣ= m a a 1 %ε 1.77% 74 5.3. Implications This chapter analyzes and demonstrates that if activities consume resources from both + R and − R sets (and they most likely will), then the average Σ= m a a 1 %ε is significantly lower than if activities from A+ only consume resources in R+ and activities in A− only consume resources in R− . Additionally, the percentage error for each activity is not significant (e.g. less than 5 percent in Tables 17, 18, and 19). Hence, for Stage 1, TDABC is not significantly different from ABC provided that there is no resource diversity. However, previous discussion has shown that there could be a potentially significant error when resource diversity exists. TDABC2 eliminates this resource diversity issue and significantly reduces the complexity of Stage 2 cost assignments. The error analysis in this chapter is extended to Stage 2 in the next chapter to show the maximum absolute dollar error for TDABC2 relative to Stage 2 of ABC. 75 CHAPTER 6 STAGE 2 ERROR ANALYSIS 6.1. Analysis of the Maximum Error of TDABC2 If the equivalency conditions of Proposition IV and its corollaries are not satisfied, then error is introduced, and it is necessary to determine the maximum error possible. This chapter identifies the maximum absolute dollar error of TDABC2 relative to ABC and all analytics are parallel to those of Chapter 5. From Chapter 4, the error for activity a, which is derived from Proposition IV and shown in Equation 13, is [ ] ℑ ℑ ε =σ σ − Ζ σ υ υ υ r c r i C C . Parallel to the Stage 1 error analysis, if Cυ r and ℑυ r are substituted in Equation 13 (let a ia ia ℑ ℑ υ = , T m a a C C = Σ=1 , and T m a a t = ℑ Σ= 1 ), then simplifying further yields ia m a a a i c C ℑ − ℑ =Σ= Ζ 1 α ε , i = 1,…, k (15) The assumptions for Stage 2 concerning α a C and a ℑ being fixed and that all activities are time driven are similar in rationale to those in the Stage 1 analysis. If α a C and a ℑ are treated as constants in the system and the consumption of activity a by cost object i, ia ℑ , is allowed to vary, then the following proposition shows that the maximum 76 absolute dollar error of the Stage 2 system is Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α , where a a a δ ≡ Cα − c Ζℑ , which is the total dollar error contribution of activity a. − Ζ ℑ c C a a α is the dollar error contribution per unit of time of activity a (“unit dollar error contribution of a”). Proposition VI: Given ia m a a a i c C ℑ − ℑ =Σ= Ζ 1 α ε and a a C c ℑ Ζ ≠ α , the maximum absolute dollar error for the system is Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α . The proof is parallel to that of Proposition V and is, therefore, omitted. Based on the proof from Proposition VI, the activities can be partitioned into two sets: one to represent the activities that provide the positive dollar error contribution, { } q A A , A ,..., A 1 2 + = and the other to represent activities that provide a negative dollar error contribution, { } q q m A A , A ,..., A +1 +2 − = , where a a Cα − cΖℑ > 0 ( − Ζ ℑ c C a a α > 0) with a = 1,…, q ∈ A+ and a a Cα − c Ζℑ < 0 ( − Ζ ℑ c C a a α < 0) with a = q+1,…, m ∈ A− . Parallel to the Stage 1 error analysis, Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α does not depend on the ia ℑ ’s. To find the set of ia ℑ ’s that produces the maximum error, first (along with the partition of activities into A+ and A− sets) the activities are ordered from largest to 77 smallest − Ζ ℑ c C a a α (similar rational to that in Stage 1 error analysis). Second, partition the cost objects into two sets in which some cost objects consume activities that provide a positive dollar error contribution ( I + ) and the rest of the cost objects consume activities that provide a negative dollar error contribution ( I − ): I + = {i i = 1,…, w; Ζ ∈ + ℑ ℑ = C − c ℑ a A a ia ia a a ε ( α ) , and ( ) = 0 ℑ ℑ = − Ζℑ a ia ia a a ε Cα c where ℑ = 0 ia if a ∈ A− }; and I − = {i i = w+1,…, k; Ζ ∈ − ℑ ℑ = C − c ℑ a A a ia ia a a ε ( α ) , and ( ) = 0 ℑ ℑ = − Ζℑ a ia ia a a ε Cα c where ℑ = 0 ia if a ∈ A+ }, where I + ∪I − = I . To maximize the dollar error using ia ℑ ’s, cost objects in I + only consume activities in A+ and cost objects in I − only consume activities in A− . The following corollary to Proposition VI shows the ia ℑ ’s that result in Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α . Corollary VIa: The ia ℑ ’s that produce the total maximum absolute dollar error of the system, Σ Σ = Ζ = = − ℑ m a a a m a a C c 1 1 δ α , is Σ ΣΣ = = Ζ = ℑ ℑ = − ℑ k i m a a ia a a k i i C c 1 1 1 ε ( α ) . The proof parallels that of Corollary Va and is therefore omitted. 78 Although Σ Σ = = = m a a k i i 1 1 ε δ , a program is identified that provides the maximum percentage error of each cost object i that maximizes the average absolute percentage error of the system for Stage 2.4 The percentage error for any given cost object is α ε ε i i i D % = , which is the dollar error for cost object i divided by the ABC cost for cost object i. Parallel to the Stage 2 error analysis, to find the maximum average absolute percentage error, one program is used for the positive sets, I + and A+ , and another program for the negative sets, I − and A− . Let m+ represent the number of activities in A+ (a = 1,…, q) and k + represent the number of resources in I + (i = 1,…, w). First order all activities from largest positive to smallest positive − Ζ ℑ c C a a α and label them as A1, A2,…, Aq. The most positive activity, A1, must be the only activity consumed by k + − (m+ −1) cost objects. The program for the maximization of the percentage errors for each cost object in the I + set is as follows: Max Σ Σ Σ = = = Ζ ℑ ℑ ℑ − w ℑ i q a a a ia q a ia a a C c C 1 1 1 α α (P11) s.t. γ ,1 1 ℑ ≥ ℑ i , i = 1,…, w1 (P12) γ ia a ℑ ≥ ℑ , i = 1+ k + − (m+ −1) ,…, w1, a = 2,…, q1 (P13) 4 An optimization software program such as LINGO can be used. 79 All other ℑ ≥ 0 ia , a ∈ A+ and i ∈ I + (P14) a w i ia ℑ = ℑ Σ= 1 (P15) The objective function in (P11) provides the maximum percentage error in magnitude across all cost objects in I + . (P12) is the first constraint that ensures that for A1, at least the amount of γ 1 ℑ is assigned to w1 cost objects in I + . The materiality and uniqueness parameter γ (e.g. it can be set as 0.1 to ensure that 10 percent of the time for activity a is assigned to the appropriate cost objects) is only assigned to w1 cost objects instead of w to not over restrict the program and allow it to choose the optimal set of ia ℑ ’s. (P13) is the second constraint that ensures uniqueness among activity vectors without overrestricting the program to allow for some ia ℑ in I + and A+ to be zero or for some cost objects in I + to consume all of a single activity in A+ (this is represented by (P14)). (P15) ensures that the sum of the cost object consumptions in I + of a particular activity in A+ is equal to the total time available for that particular activity a. Now, let k − represent the number of cost objects in I − (i = w+1,…, k) and m− represent the number of activities in A− (a = q+1,…, m). After ordering all activities from least negative to most negative − Ζ ℑ c C a a α and labeling them as Aq+1, Aq+2,…, Am, the most negative Am must be the only activity consumed by k − − (m− −1) cost objects. The program for the maximization the magnitude of the percentage errors in the A− set is as follows: 80 Min Σ Σ Σ = + = + = + Ζ ℑ ℑ ℑ − k ℑ i w m a q a a ia m a q ia a a C c C 1 1 1 α α (P16) s.t. γ i m m ℑ ≥ ℑ , , i = w+2,…, k (P17) γ ia a ℑ ≥ ℑ , i = w+1,…, k − k − − (m− −1) , a = q+2,…, m1 (P18) All other ℑ ≥ 0 ia in a ∈ A− and i ∈ I − (P19) a k i w ia Σℑ = ℑ = +1 (P20) The objective function in (P16) provides the maximum percentage error in magnitude across all cost objects in I − . (P17) is the first constraint that ensures that for Am, at least the amount of γ m ℑ is assigned to i = w+2,…, k cost objects in I − without over restricting the program. (P18) is the second constraint that ensures uniqueness among activity vectors without overrestricting the program to allow for some ia ℑ in I − and A− to be zero or for some cost objects in I − to consume all of a single activity in A− (this is represented by (P19)). (P20) ensures that the sum of the cost object consumptions in I − of a particular activity in A− is equal to the total time available for that particular activity a. However, parallel to that in Stage 1 error analysis, a cost object can consume activities from both A+ and A− , thus allowing for some cancellation effects and a 81 reduced error. Hence, the maximum error cannot exceed the dollar error contribution from all activities (Σ Σ = = ≤ m a a k i i 1 1 ε δ ) and is stated in the following corollary. Corollary VIb: If cost objects in I + only consume activities in A+ and cost objects in I − only consume activities in A− , thenΣ Σ = = = m a a k i i 1 1 ε δ . If any or all cost objects consume activities from both A+ and A− , then Σ Σ = = < m a a k i i 1 1 δ ε . Therefore, Σ= k i i 1 ε can never exceed Σ= m a a 1 δ , implyingΣ Σ = = ≤ m a a k i i 1 1 ε δ . The proof parallels that of Corollary Vb and is, therefore, omitted. In summary, if there are no cancellation effects, then Σ Σ = = = m a a k i i 1 1 ε δ . If any or all cost objects consume activities from both A+ and A− , then there will be some cancellation effects within each of those i ε ’s, and thus, the actual absolute dollar error across all cost objects will be less than the maximum absolute dollar error of the system, Σ Σ = = < m a a k i i 1 1 ε δ . As a result, Σ Σ = = ≤ n j j m a a 1 1 ε δ . 6.2. Examples Demonstrating Proposition VI and Its Corollaries This section provides examples illustrating Proposition VI and its corollaries and using program (P11) through (P20). First, Corollary VIa will be shown in Tables 20, 21, and 22 for k = m, k < m, and k > m, respectively. The activities in each table are ranked 82 from largest to smallest − Ζ ℑ c C a a α and the activities are partitioned accordingly. For each table, the first three cost objects (CO1 through CO3) are in the I + set, with the rest of the cost objects being in the I − set. Table 20 is a continuation of Table 14, in which the activity costs from Stage 1 are assigned to the cost objects in Stage 2. Table 21 is a continuation of Table 15, and Table 22 is a continuation of Table 16. In each of the following tables, Σ Σ = = = m a a k i i 1 1 ε δ = $1,700, thus satisfying Corollary VIa. Program (P1) through (P10) is used to maximize each i ε % , which maximizes average Σ= k i i 1 %ε . 83 Table 20 Corollary VIa for k = m Activities A1 A2 A3 A4 A5 A6 ABC Cost Cυ r ℑυ r CO1 0.10 0 0 0 0 0 $20 0.40 0.40 CO2 0.90 0.10 0 0 0 0 $287 0.44 0.43 CO3 0 0.90 1 0 0 0 $6,193 0.23 0.11 CO4 0 0 0 1 0.90 0 $7,770 0.46 0.56 CO5 0 0 0 0 0.10 0.90 $400 0.41 0.39 CO6 0 0 0 0 0 0.10 $30 0.38 0.36 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $14,700 Time 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 Total 1a ℑ 7.75 0 0 0 0 0 7.75 2a ℑ 69.75 42.65 0 0 0 0 112.40 3a ℑ 0 383.81 2,321.04 0 0 0 2,704.85 4a ℑ 0 0 0 3,589.22 682.45 0 4,271.67 5a ℑ 0 0 0 0 75.83 159.75 235.58 6a ℑ 0 0 0 0 0 17.75 17.75 Total 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 84 Table 20 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 7.750 1 $16 CO2 28.100 4 $225 CO3 54.097 50 $5,410 CO4 85.433 50 $8,543 CO5 2.356 100 $471 CO6 8.875 2 $35 $14,700 Activities A1 A2 A3 A4 A5 A6 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.162 $0.283 $0.310 a a Cα − c Ζℑ $45 $216 $589 $581 $214 $55 Σ= m a a 1 δ $1,700 85 Table 20 (continued) Activities A1 A2 A3 A4 A5 A6 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $0 $0 $0 $0 $0 $4.50 $4.50 22.50% a a a C c 2 ( α − Ζℑ )ℑ $40.50 $22 $0 $0 $0 $0 $61.50 $61.50 21.65% a a a C c 3 ( α − Ζℑ )ℑ $0 $194 $589 $0 $0 $0 $783 $783 12.65% a a a C c 4 ( α − Ζℑ )ℑ $0 $0 $0 $581 $193 $0 $774 $774 9.96% a a a C c 5 ( α − Ζℑ )ℑ $0 $0 $0 $0 $21 $49.50 $70.50 $70.50 17.72% a a a C c 6 ( α − Ζℑ )ℑ $0 $0 $0 $0 $0 $5.50 $5.50 $5.50 18.33% $45 $216 $589 $581 $214 $55 $0 $1,700 AvgΣ= m a a 1 %ε 17.13% 86 Table 21 Corollary VIa for k < m Activities A1 A2 A3 A4 A5 A6 A7 ABC Cost Cυ r ℑυ r CO1 0.10 0 0 0 0 0 0 $20 0.33 0.33 CO2 0.90 0.10 0 0 0 0 0 $287 0.35 0.35 CO3 0 0.90 1 0 0 0 0 $6,193 0.31 0.19 CO4 0 0 0 1 0.90 0.47 0.17 $7,666 0.35 0.45 CO5 0 0 0 0 0.10 0.43 0.73 $474 0.49 0.46 CO6 0 0 0 0 0 0.10 0.10 $60 0.48 0.46 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $300 $14,700 Time 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 A7 Total 1a ℑ 7.75 0 0 0 0 0 0 7.75 2a ℑ 69.75 42.65 0 0 0 0 0 112.40 3a ℑ 0 383.81 2,321.04 0 0 0 0 2,704.85 4a ℑ 0 0 0 3,440.98 656.12 83.16 30.14 4,210.40 5a ℑ 0 0 0 0 72.9 76.59 129.61 279.1 6a ℑ 0 0 0 0 0 17.75 17.75 35.5 Total 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 87 Table 21 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 7.750 1 $16 CO2 28.100 4 $225 CO3 54.097 50 $5,410 CO4 84.208 50 $8,421 CO5 2.791 100 $558 CO6 17.75 2 $71 $14,700 Activities A1 A2 A3 A4 A5 A6 A7 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.155 $0.281 $0.310 $0.310 a a Cα − c Ζℑ $45 $216 $589 $535 $205 $55 $55 Σ= m a a 1 δ $1,700 88 Table 21 (continued) Activities A1 A2 A3 A4 A5 A6 A7 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $0 $0 $0 $0 $0 $0 $4.50 $4.50 22.50% a a a C c 2 ( α − Ζℑ )ℑ $40.50 $22 $0 $0 $0 $0 $0 $61.50 $62 21.65% a a a C c 3 ( α − Ζℑ )ℑ $0 $194 $589 $0 $0 $0 $0 $783 $783 12.65% a a a C c 4 ( α − Ζℑ )ℑ $0 $0 $0 $535 $185 $26 $9 $755 $755 9.84% a a a C c 5 ( α − Ζℑ )ℑ $0 $0 $0 $0 $20 $24 $40 $84 $84 17.82% a a a C c 6 ( α − Ζℑ )ℑ $0 $0 $0 $0 $0 $5 $5 $5 $5 18.34% $45 $216 $589 $535 $205 $55 $55 $0 $1,700 AvgΣ= m a a 1 %ε 17.13% 89 Table 22 Corollary VIa for k > m Activities A1 A2 A3 A4 A5 ABC Cost Cυ r ℑυ r CO1 0.10 0 0 0 0 $20 0.52 0.51 CO2 0.90 0.10 0 0 0 $287 0.58 0.56 CO3 0 0.90 1 0 0 $6,193 0.10 0.03 CO4 0 0 0 1 0.80 $7,770 0.45 0.56 CO5 0 0 0 0 0.10 $400 0.30 0.25 CO6 0 0 0 0 0.10 $30 0.30 0.25 Cost $200 $1,069 $5,231 $6,848 $1,352 $14,700 Time 77.5 426.46 2,321.04 3,737.46 787.54 7,350 Activities A1 A2 A3 A4 A5 Total 1a ℑ 7.75 0 0 0 0 7.75 2a ℑ 69.75 42.65 0 0 0 112.40 3a ℑ 0 383.81 2,321.04 0 0 2,704.85 4a ℑ 0 0 0 3,737.46 630.03 4,367.49 5a ℑ 0 0 0 0 78.75 78.75 6a ℑ 0 0 0 0 78.75 78.75 Total 77.5 426.46 2,321.04 3,737.46 758.28 7,350 90 Table 22 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 7.750 1 $16 CO2 28.100 4 $225 CO3 54.097 50 $5,410 CO4 87.350 50 $8,735 CO5 0.788 100 $158 CO6 39.375 2 $158 $14,700 Activities A1 A2 A3 A4 A5 a a Cα ℑ $2.58 $2.51 $2.25 $1.83 $1.72 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.168 $0.284 a a Cα − c Ζℑ $45 $216 $589 $627 $223 Σ= m a a 1 δ $1,700 91 Table 22 (continued) Activities A1 A2 A3 A4 A5 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $0 $0 $0 $0 $4.50 $4.50 22.50% a a a C c 2 ( α − Ζℑ )ℑ $40.50 $22 $0 $0 $0 $61.50 $61.50 21.65% a a a C c 3 ( α − Ζℑ )ℑ $0 $194 $589 $0 $0 $783 $783 12.65% a a a C c 4 ( α − Ζℑ )ℑ $0 $0 $0 $627 $179 $805 $805 10.16% a a a C c 5 ( α − Ζℑ )ℑ $0 $0 $0 $0 $22 $22 $22 16.52% a a a C c 6 ( α − Ζℑ )ℑ $0 $0 $0 $0 $22 $22 $22 16.52% $45 $216 $589 $627 $223 $0 $1,700 AvgΣ= m a a 1 %ε 16.67% 92 Illustrations of Corollary VIb are shown in Tables 23, 24, and 25 for k = m, k < m, and k > m, respectively. As in the previous three tables, the activities in each table are ranked from largest to smallest − Ζ ℑ c C a a α and the activities are partitioned accordingly. For each table, the first three cost objects (CO1 through CO3) are in the I + set, with the rest of the cost objects being in the I − set. In each table, all cost objects share all of the activities so thatΣ Σ = = < m a a k i i 1 1 δ ε = $1,700. Table 23 shows that the average Σ= k i i 1 %ε = 0.97 percent compared to 17.13 percent in Table 20. Compared to Table 21 with an average Σ= k i i 1 %ε = 17.13 percent, Table 24 shows that the average Σ= k i i 1 %ε = 0.84 percent. Table 25 shows that the average Σ= k i i 1 %ε = 0.42 percent compared to 16.67 percent in Table 20. In each table, notice that each % i ε is less than 5 percent, which demonstrates that the actual maximum error is very small since cost objects will typically consume activities in both A+ and A− sets. 93 Table 23 Corollary VIb for k = m Activities A1 A2 A3 A4 A5 A6 ABC Cost Cυ r ℑυ r CO1 0.10 0.30 0.10 0.10 0.20 0.10 $1,814 0.34 0.35 CO2 0.10 0.20 0.20 0.20 0.10 0.10 $2,760 0.74 0.69 CO3 0.20 0.10 0.20 0.20 0.10 0.10 $2,673 0.62 0.60 CO4 0.30 0.20 0.10 0.10 0.30 0.10 $1,877 0.60 0.58 CO5 0.20 0.10 0.10 0.20 0.10 0.10 $2,150 0.27 0.33 CO6 0.10 0.10 0.30 0.20 0.20 0.50 $3,426 0.002 0.01 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $14,700 Time 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 Total 1a ℑ 7.75 127.94 232.1 358.92 151.66 17.75 896.12 2a ℑ 7.75 85.29 464.21 717.84 75.83 17.75 1,368.67 3a ℑ 15.5 42.65 464.21 717.84 75.83 17.75 1,333.78 4a ℑ 23.25 85.29 232.1 358.92 227.48 17.75 944.8 5a ℑ 15.5 42.65 232.1 717.84 78.83 17.75 1,101.67 6a ℑ 7.75 42.65 696.31 717.84 151.66 88.75 1,704.96 Total 77.5 426.46 2,321.04 3,589.22 758.28 177.5 7,350 94 Table 23 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 896.120 1 $1,792 CO2 342.168 4 $2,737 CO3 26.676 50 $2,668 CO4 18.896 50 $1,890 CO5 11.017 100 $2,203 CO6 852.48 2 $3,410 $14,700 Activities A1 A2 A3 A4 A5 A6 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.162 $0.283 $0.310 a a Cα − c Ζℑ $45 $216 $589 $581 $214 $55 Σ= m a a 1 δ $1,700 95 Table 23 (continued) Activities A1 A2 A3 A4 A5 A6 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $64.80 $58.90 $58.10 $42.90 $5.50 $22 $22 1.20% a a a C c 2 ( α − Ζℑ )ℑ $4.50 $43.20 $117.80 $116.10 $21.40 $5.50 $22 $22 0.81% a a a C c 3 ( α − Ζℑ )ℑ $9 $21.60 $117.80 $116.10 $21.40 $5.50 $5 $5 0.20% a a a C c 4 ( α − Ζℑ )ℑ $13.50 $43.20 $58.90 $58.10 $64.30 $5.50 $12 $12 0.65% a a a C c 5 ( α − Ζℑ )ℑ $9 $21.60 $58.90 $116.10 $21.40 $5.50 $54 $54 2.49% a a a C c 6 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $116.10 $42.90 $27.5 $16 $16 0.48% $45 $216 $589 $581 $214 $55 $0 $132 AvgΣ= m a a 1 %ε 0.97% 96 Table 24 Corollary VIb for k < m Activities A1 A2 A3 A4 A5 A6 A7 ABC Cost Cυ r ℑυ r CO1 0.10 0.30 0.10 0.10 0.20 0.10 0.40 $1,899 0.44 0.44 CO2 0.10 0.20 0.20 0.20 0.10 0.10 0.10 $2,735 0.77 0.72 CO3 0.20 0.10 0.20 0.20 0.10 0.10 0.10 $2,648 0.66 0.64 CO4 0.30 0.20 0.10 0.10 0.30 0.10 0.10 $1,867 0.44 0.43 CO5 0.20 0.10 0.10 0.20 0.10 0.10 0.10 $2,125 0.31 0.37 CO6 0.10 0.10 0.30 0.20 0.20 0.50 0.20 $3,426 0.03 0.02 Cost $200 $1,069 $5,231 $6,598 $2,700 $300 $300 $14,700 Time 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 Activities A1 A2 A3 A4 A5 A6 A7 Total 1a ℑ 7.75 127.94 232.1 344.1 145.8 17.75 71 946.44 2a ℑ 7.75 85.29 464.21 688.2 72.9 17.75 17.75 1,353.85 3a ℑ 15.5 42.65 464.21 688.2 72.9 17.75 17.75 1,318.95 4a ℑ 23.25 85.29 232.1 344.1 218.71 17.75 17.75 938.95 5a ℑ 15.5 42.65 232.1 688.2 72.9 17.75 17.75 1,086.85 6a ℑ 7.75 42.65 696.31 688.2 145.8 88.75 35.5 1,704.96 Total 77.5 426.46 2,321.04 3,440.98 729.02 177.5 177.5 7,350 97 Table 24 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 946.442 1 $1,893 CO2 338.463 4 $2,708 CO3 26.379 50 $2,638 CO4 18.779 50 $1,878 CO5 10.869 100 $2,174 CO6 852.48 2 $3,410 $14,700 Activities A1 A2 A3 A4 A5 A6 A7 a a Cα ℑ $2.58 $2.51 $2.25 $1.84 $1.72 $1.69 $1.69 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.155 $0.281 $0.310 $0.310 a a Cα − c Ζℑ $45 $216 $589 $535 $205 $55 $55 Σ= m a a 1 δ $1,700 98 Table 24 (continued) Activities A1 A2 A3 A4 A5 A6 A7 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $64.80 $58.90 $53.48 $41.04 $5.50 $22 $6.20 $6.20 0.33% a a a C c 2 ( α − Ζℑ )ℑ $4.50 $43.20 $117.80 $106.96 $20.52 $5.50 $5.50 $27 $27 0.99% a a a C c 3 ( α − Ζℑ )ℑ $9 $21.60 $117.80 $106.96 $20.52 $5.50 $5.50 $9.90 $9.90 0.37% a a a C c 4 ( α − Ζℑ )ℑ $13.50 $43.20 $58.90 $53.48 $61.56 $5.50 $5.50 $10.40 $10.40 0.56% a a a C c 5 ( α − Ζℑ )ℑ $9 $21.60 $58.90 $106.96 $20.52 $5.50 $5.50 $49 $49 2.31% a a a C c 6 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $106.96 $41.04 $27.50 $11 $16.30 $16.30 0.48% $45 $216 $589 $535 $205 $55 $55 $0 $119 AvgΣ= m a a 1 %ε 0.84% 99 Table 25 Corollary VIb for k > m Activities A1 A2 A3 A4 A5 ABC Cost Cυ r ℑυ r CO1 0.10 0.30 0.10 0.10 0.30 $1,954 0.54 0.52 CO2 0.30 0.30 0.10 0.20 0.10 $2,409 0.46 0.43 CO3 0.10 0.10 0.10 0.10 0.10 $1,470 0.00 0.00 CO4 0.30 0.10 0.10 0.10 0.20 $1,645 0.68 0.63 CO5 0.10 0.10 0.30 0.20 0.20 $3,336 0.71 0.64 CO6 0.10 0.10 0.30 0.30 0.10 $3,886 0.97 0.93 Cost $200 $1,069 $5,231 $6,848 $1,352 $14,700 Time 77.5 426.46 2,321.04 3,737.46 787.54 7,350 Activities A1 A2 A3 A4 A5 Total 1a ℑ 7.75 127.94 232.10 373.75 236.26 977.80 2a ℑ 23.25 127.94 232.10 747.49 78.75 1,209.54 3a ℑ 7.75 42.65 232.10 373.75 78.75 735 4a ℑ 23.25 42.65 232.10 373.75 157.51 829.25 5a ℑ 7.75 42.65 696.31 747.49 157.51 1,651.71 6a ℑ 7.75 42.65 696.31 1,121.24 78.75 1,946.70 Total 77.5 426.46 2,321.04 3,737.46 758.28 7,350 100 Table 25 (continued) c Ζ $2 i β β i θ θ TDABC2 Cost CO1 977.800 1 $1,956 CO2 302.385 4 $2,419 CO3 14.700 50 $1,470 CO4 16.585 50 $1,659 CO5 16.517 100 $3,303 CO6 973.350 2 $3,893 $14,700 Activities A1 A2 A3 A4 A5 a a Cα ℑ $2.58 $2.51 $2.25 $1.83 $1.72 C ℑ − cΖ a a α $0.581 $0.507 $0.254 $0.168 $0.284 a a Cα − c Ζℑ $45 $216 $589 $627 $223 Σ= m a a 1 δ $1,700 101 Table 25 (continued) Activities A1 A2 A3 A4 A5 Total i ε ε % i ε ε a a a C c 1 ( α − Ζℑ )ℑ $4.50 $64.80 $58.90 $62.67 $67.00 $1.46 $1.46 0.07% a a a C c 2 ( α − Ζℑ )ℑ $13.50 $64.80 $58.90 $125.34 $22.33 $10.47 $10.47 0.43% a a a C c 3 ( α − Ζℑ )ℑ $4.50 $21.60 $58.90 $62.67 $22.33 $0.00 $0.00 0.00% a a a C c 4 ( α − Ζℑ )ℑ $13.50 $21.60 $58.90 $62.67 $44.66 $13.33 $13.33 0.81% a a a C c 5 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $125.34 $44.66 $32.80 $32.80 0.98% a a a C c 6 ( α − Ζℑ )ℑ $4.50 $21.60 $176.70 $188.00 $22.33 $7.54 $7.54 0.19% $45 $216 $589 $627 $223 $0 $66 AvgΣ= m a a 1 %ε 0.42% 102 6.3. Implications This chapter demonstrates that cost objects will typically consume activities from both + A and − A sets. As a result, the average Σ= k i i 1 %ε is significantly lower than if cost objects from set I + only consume resources in A+ and activities in I − only consume resources in A− . This chapter shows that TDABC2 is not significantly different from the fullyspecified ABC system (e.g. the percentage errors are less than 2.5 percent in Tables 23, 24, and 25). Hence, TDABC2 should replicate the accuracy of the ABC system with the benefit of eliminating Stage 1 cost assignments and significantly reducing the Stage 2 cost assignments. However, some empirical analyses are needed to determine whether the equivalency conditions that are analytically proven actually hold. As an initial empirical analysis, the next chapter provides case studies based on data from an actual company to explore the validity of the equivalency conditions. 103 CHAPTER 7 CASE ANALYSES OF THE EQUIVALENCY CONDITIONS 7.1. Introduction to the Case Analyses and Assumptions This chapter presents seven case studies using data from a particular company to provide anecdotal evidence that explores the validity of the equivalency conditions in Propositions I through IV.5 Case studies are initially useful to identify and explore the validity of hypothesized relationships and, thus, serve as an important forerunner and input for other types of empirical testing (Lillis and Mundy 2005; Kaplan 1986). Since the case studies in this chapter contain data from only one company, any evidence of the validity of the equivalency conditions is anecdotal, which is, consequently, the limitation of this chapter. Therefore, more empirical analyses using data from a broad range of companies are needed beyond these case studies to verify the equivalency conditions further. The data used in the case analyses are yearly company data. For the first four cases, enough data are available to perform Stage 1 and Stage 2 cost assignments for ABC, IABC, IABC2, and TDABC. Only Stage 1 data to perform ABC, IABC, and TDABC are available for the fifth case study, and only Stage 2 data to perform ABC and IABC2 are available for the sixth and seventh case studies. An overview of the names 5 The name of the company is withheld for reasons of confidentiality. 104 of the resources, activities, and products/services (cost objects) for each case study are given in Table 26. 105 Table 26 Names of Resources, Activities, and Cost Objects for Each Case Study Case 1 R1 Salaries and Benefits A1 Repair and maintain fixed equipment CO1 Unit 1 R2 Travel A2 Repair and maintain rotating equipment CO2 Unit 2 R3 Communication A3 Prepare equipment CO3 Waste water treatment plant R4 Special studies A4 Fabricate piping and welding CO4 Unit 3  Gas Treater R5 Depreciation A5 Repair electrical equipment CO5 Boiler & steam system R6 Materials A6 Receive and inventory materials CO6 Tanks & Pipelines R7 External contract services A7 Dock and sail ships CO7 Docks R8 Outside Contractors A8 Perform instrument calibration/repairs CO8 Loading Racks R9 Parts Inventory A9 Equipment reliability CO9 General Administration R10 Rent A10 Plan and schedule work activities CO10 Dock and sail ships R11 Internal Labor A11 Manage/supervise departments R12 Training A12 Manage internal contractors R13 Procurement cards A13 Perform housekeeping & administrative R14 Other expenses A14 Maintain pipelines and valves A15 Monitor SAP work orders 106 Table 26 (Continued) Case 2 R1 Wages & Salaries A1 Receive Product/Invoices Disputes CO1 National Accounts / Buybacks R2 Labor Burden 



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