1. Activity Cost Behavior
CHAPTER
Activity Cost Behavior

2. Objectives
1. Define cost behavior for fixed, variable, and mixed costs.
2. Explain the role of the resource usage model in understanding cost behavior.
3. Separate mixed costs into their fixed and variable components using the high-low method, the scatterplot method, and the method of least squares.
continued

3. Objectives 4. Evaluate the reliability of a cost equation.
5. Discuss the role of multiple regression in assessing cost behavior.
6. Describe the use of managerial judgment in determining cost behavior.

4. A cost that stays the same as output changes is a fixed cost.
Fixed Costs
A cost that stays the same as output changes is a fixed cost.

5. Fixed Costs
Cutting machines are leased for $60,000 per year and have the capacity to produce up to 240,000 units a year.

6. Fixed Costs Lease of Machines Number of Units 60,000 60,000 $1.00
Total Fixed Cost Graph
Total Costs
$120,000
$100,000
$80,000
$60,000
$40,000
$20,000
Units Produced (000)
F = $60,000
Fixed Costs
Lease of Machines
Number of Units
$60,000 0 N/A
60,000 60,000 $1.00
60, ,
60, ,
60, ,
Units Cost

7. Fixed Costs Lease of Machines Number of Units 60,000 60,000 $1.00
Unit Fixed Cost Graph
Cost per Unit
$1.00
$0.50
$0.33
$0.25
Units Produced (000)
Fixed Costs
Lease of Machines
Number of Units
$60,000 0 N/A
60,000 60,000 $1.00
60, ,
60, ,
60, ,
Units Cost

8. A variable cost is a cost that, in total, varies in direct proportion to changes in output.

9. Variable Cost
As the cutting machines cut each unit, they use 0.1 kilowatt-hour at $2.00 per kilowatt hour. Thus, the cost of each unit is $0.20 ($2 x 0.1).

10. Total Variable Cost Graph
Total Costs
Units Produced (000)
$48,000
$36,000
$24,000
$12,000
Yv = .20x
Variable Cost
Cost of Power
Number of Units
$ $ 0
12,000 60,
24, ,
36, ,
48, ,
Units Cost

11. Unit Variable Cost Graph
Units Produced (000)
$0.40
$0.30
$0.20
$0.10
Cost per Unit
Variable Cost
Cost of Power
Number of Units
$ $ 0
12,000 60,
24, ,
36, ,
48, ,
Units Cost

12. A mixed cost is a cost that has both a fixed and a variable component.
Mixed Costs

13. Mixed Costs
Sales representatives often are paid a salary plus a commission on sales.

14. Variable Cost of Selling
Mixed Cost Behavior
$130,000
$110,000
$90,000
$70,000
$50,000
$30,000
Mixed
Costs
Total Costs
Units Sold (000)
Inserts Sold
Variable Cost of Selling
40,000 $ 20,000 $30,000 $ 50,000 $1.25
80,000 40,000 30,000 70,
120,000 60,000 30,000 90,
160,000 80,000 30, ,
200, ,000 30, ,
Total Selling Cost
Fixed Cost of Selling
Selling Cost per Unit

15. Activity Cost Behavior Model
Input:
Materials
Activities
Energy
Activity Output
Labor
Cost Behavior
Changes in Output
Capital
Changes in Input Cost

16. Flexible Resources
Flexible resources are resources acquired as used and needed. Materials and energy are examples.

17. Committed Resources
Committed resources are supplied in advance of usage. Buying or leasing a building is an example of this form of advance resource acquisition.

18. A step cost displays a constant level of cost for a range of output and then jumps to a higher level of cost at some point.
Step-Cost Behavior

19. Step-Cost Behavior 10 20 30 40 50 $500 400 300 200 100 Cost
Activity Output (units)

20. Normal Operating Range (Relevant Range)
Step-Fixed Costs
Cost
$150,000
100,000
50,000
Normal Operating Range (Relevant Range)
2, , ,500
Activity Usage

21. Step-Cost Behavior Three engineers hired at $50,000 each
Each engineer is capable of processing 2,500 change orders
$90,000 was spent on supplies for the engineering activity
There were 6,000 orders processed
The company could process as many as 7,500 orders

22. Step-Cost Behavior Available orders = Orders used + Orders unused
7,500 orders = 6,000 orders + 1,500 orders
Fixed engineering rate = $150,000/7,500
= $20 per change order
Variable engineering rate = $90,000/6,000
= $15 per change order

23. Step-Cost Behavior
The relationship between resources supplied and resources used is expressed by the following equation:
Resources available = Resources used + Unused capacity

24. Step-Cost Behavior
Cost of orders supplied = Cost of orders used Cost of unused orders
= [($20 + $15) x 6,000] + ($ x 1,500)
= $240,000
The $30,000 of excess engineering capacity means that a new product could be introduced without increasing current spending on engineering.
Equal to the $150,000 spent on engineers and the $90,000 spent on supplies.

25. Methods for Separating Mixed Costs
The High-Low Method
The Scatterplot Method
The Method of Least Squares
Variable Component
Fixed Component

26. The linearity assumption assumes that variable costs increase in direct proportion to the number of units produced (or activity units used).

27. Methods for Separating Mixed Costs
Y = a + bx
Variable Cost per Unit
Number of Units
Total Fixed Cost
Total Cost

28. Step 1: Solve for variable cost (b)
The High-Low Method
Month
Setup Costs
Setup Hours
January
$1,000
100
February
1,250
200
March
2,250
300
April
2,500
400
May
3,750
500
Step 1: Solve for variable cost (b)

29. The High-Low Method High Cost – Low Cost High Units – Low Units b =
Month Setup Costs Setup Hours
January $1,
February ,
March ,
April ,
May ,
b =
High Cost – Low Cost
High Units – Low Units

30. The High-Low Method b = $3,750 – Low Cost 500 – Low Units b =
Month Setup Costs Setup Hours
January $1,
February ,
March ,
April ,
May ,
b =
$3,750 – Low Cost
– Low Units
b =
High Cost – Low Cost
High Units – Low Units

31. The High-Low Method b = $3,750 – Low Cost 500 – Low Units b =
Month Setup Costs Setup Hours
January $1,
February ,
March ,
April ,
May ,
b =
$3, – Low Cost
– Low Units
b =
$3, – $1,000 –

32. The High-Low Method $3,750 – $1,000 500 – 100 b = b = $6.875
$3, – $1,000 –
b = $6.875
Step 2: Using either the high cost or low cost, solve for the total fixed cost (a).

33. The High-Low Method Y = a + b (x) $3,750 = a + $6.875(500) $312.50 = a
High End
$3,750 = a $6.875(500)
$ = a
Y = a b (x)
Low End
$1,000 = a $6.875(100)
$ = a
The cost formula using the high-low method is:
Total cost = $ ($6.875 x Setup hours)

34. The Scatterplot Method

35. The Scatterplot Method
Nonlinear Relationship
Activity
Cost * * * * *
Activity Output

36. The Scatterplot Method
Upward Shift in Cost Relationship
Activity
Cost * * * * * *
Activity Output

37. The Scatterplot Method
Presence of Outliers
Activity
Cost * * * *
Estimated regression line * *
Estimated fixed cost
Activity Output

38. LEAST SQUARE METHOD Y = a + bX ∑XY = a ∑X + b ∑X^2 (1)
∑Y = na + b ∑X (2)
Ekstrapolasi rumus (1) dan (2)

39. The Method of Least Squares
Spreadsheet Data for Larson Company

40. The Method of Least Squares
Regression Output for Larson Company

41. The Method of Least Squares
The results give rise to the following equation:
Setup costs = $125 + ($6.75 x Setup hours)
R2 = .944, or 94.4 percent of the variation in setup costs is explained by the number of setup hours variable.

42. Coefficient of Correlation
Positive Correlation
r approaches +1
Machine Hours
Utilities Costs
Machine Hours
Utilities Costs

43. Coefficient of Correlation
Negative Correlation
r approaches -1
Hours of Safety Training
Industrial Accidents
Hours of Safety Training
Industrial Accidents

44. Coefficient of Correlation
No Correlation
r ~ 0
Hair Length
Accounting Grade
Hair Length
Accounting Grade

45. Multiple Regression TC = b0 + ( b1X1) + (b2X2) + . . .
b0 = the fixed cost or intercept
b1 = the variable rate for the first independent variable
X1 = the first independent variable
b2 = the variable rate for the second independent variable
X2 = the second independent variable

46. Data for Phoenix Factory Utilities Cost Regression
Multiple Regression
Data for Phoenix Factory Utilities Cost Regression

47. Multiple Regression for Phoenix Factory Utilities Cost

48. Multiple Regression The results gives rise to the following equation:
Utilities cost = $ $1.097(Machine hours) + ($ x Summer)
R2 = .967, or 96.7 percent of the variation in utilities cost is explained by the machine hours and summer variables.

49. Managerial Judgment
Managerial judgment is critically important in determining cost behavior, and it is by far the most widely used method in practice.

50. Chapter Three
The End