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CHAPTER 10
DETERMINING HOW COSTS BEHAVE
10-1
What two assumptions are frequently made when estimating a cost function?
The two assumptions are
1.
Variations in the level of a single activity (the cost driver) explain the variations in the
related total costs.
2.
Cost behavior is approximated by a linear cost function within the relevant range. A
linear cost function is a cost function where, within the relevant range, the graph of total
costs versus the level of a single activity forms a straight line.
10-2
Describe three alternative linear cost functions.
Three alternative linear cost functions are
1.
Variable cost function––a cost function in which total costs change in proportion to the
changes in the level of activity in the relevant range.
2.
Fixed cost function––a cost function in which total costs do not change with changes in
the level of activity in the relevant range.
3.
Mixed cost function––a cost function that has both variable and fixed elements. Total
costs change but not in proportion to the changes in the level of activity in the relevant
range.
10-3 What is the difference between a linear and a nonlinear cost function? Give an example
of each type of cost function.
A linear cost function is a cost function where, within the relevant range, the graph of total costs
versus the level of a single activity related to that cost is a straight line. An example of a linear
cost function is a cost function for use of a videoconferencing line where the terms are a fixed
charge of $10,000 per year plus a $2 per minute charge for line use. A nonlinear cost function is
a cost function where, within the relevant range, the graph of total costs versus the level of a
single activity related to that cost is not a straight line. Examples include economies of scale in
advertising where an agency can double the number of advertisements for less than twice the
costs, step-cost functions, and learning-curve-based costs.
10-4 ―High correlation between two variables means that one is the cause and the other is the
effect.‖ Do you agree? Explain.
No. High correlation merely indicates that the two variables move together in the data examined.
It is essential also to consider economic plausibility before making inferences about cause and
effect. Without any economic plausibility for a relationship, it is less likely that a high level of
correlation observed in one set of data will be similarly found in other sets of data.
10-5
Name four approaches to estimating a cost function.
Four approaches to estimating a cost function are
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2.
3.
4.
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Industrial engineering method.
Conference method.
Account analysis method.
Quantitative analysis of current or past cost relationships.
10-6 Describe the conference method for estimating a cost function. What are two advantages
of this method?
The conference method estimates cost functions on the basis of analysis and opinions about costs
and their drivers gathered from various departments of a company (purchasing, process
engineering, manufacturing, employee relations, etc.). Advantages of the conference method
include
1.
The speed with which cost estimates can be developed.
2.
The pooling of knowledge from experts across functional areas.
3.
The improved credibility of the cost function to all personnel.
10-7
Describe the account analysis method for estimating a cost function.
The account analysis method estimates cost functions by classifying cost accounts in the
subsidiary ledger as variable, fixed, or mixed with respect to the identified level of activity.
Typically, managers use qualitative, rather than quantitative, analysis when making these costclassification decisions.
10-8 List the six steps in estimating a cost function on the basis of an analysis of a past cost
relationship. Which step is typically the most difficult for the cost analyst?
The six steps are
1.
Choose the dependent variable (the variable to be predicted, which is some type of cost).
2.
Identify the independent variable or cost driver.
3.
Collect data on the dependent variable and the cost driver.
4.
Plot the data.
5.
Estimate the cost function.
6.
Evaluate the cost driver of the estimated cost function.
Step 3 typically is the most difficult for a cost analyst.
10-9 When using the high-low method, should you base the high and low observations on the
dependent variable or on the cost driver?
Causality in a cost function runs from the cost driver to the dependent variable. Thus, choosing
the highest observation and the lowest observation of the cost driver is appropriate in the highlow method.
10-10 Describe three criteria for evaluating cost functions and choosing cost drivers.
Three criteria important when choosing among alternative cost functions are
1.
Economic plausibility.
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3.
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Goodness of fit.
Slope of the regression line.
10-11 Define learning curve. Outline two models that can be used when incorporating learning
into the estimation of cost functions.
A learning curve is a function that measures how labor-hours per unit decline as units of
production increase because workers are learning and becoming better at their jobs. Two models
used to capture different forms of learning are
1.
Cumulative average-time learning model. The cumulative average time per unit declines
by a constant percentage each time the cumulative quantity of units produced doubles.
2.
Incremental unit-time learning model. The incremental time needed to produce the last
unit declines by a constant percentage each time the cumulative quantity of units
produced doubles.
10-12 Discuss four frequently encountered problems when collecting cost data on variables
included in a cost function.
Frequently encountered problems when collecting cost data on variables included in a cost
function are
1.
The time period used to measure the dependent variable is not properly matched with the
time period used to measure the cost driver(s).
2.
Fixed costs are allocated as if they are variable.
3.
Data are either not available for all observations or are not uniformly reliable.
4.
Extreme values of observations occur.
5.
A homogeneous relationship between the individual cost items in the dependent variable
cost pool and the cost driver(s) does not exist.
6.
The relationship between the cost and the cost driver is not stationary.
7.
Inflation has occurred in a dependent variable, a cost driver, or both.
10-13 What are the four key assumptions examined in specification analysis in the case of
simple regression?
Four key assumptions examined in specification analysis are
1.
Linearity of relationship between the dependent variable and the independent variable
within the relevant range.
2.
Constant variance of residuals for all values of the independent variable.
3.
Independence of residuals.
4.
Normal distribution of residuals.
10-14 ―All the independent variables in a cost function estimated with regression analysis are
cost drivers.‖ Do you agree? Explain.
No. A cost driver is any factor whose change causes a change in the total cost of a related cost
object. A cause-and-effect relationship underlies selection of a cost driver. Some users of
regression analysis include numerous independent variables in a regression model in an attempt
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to maximize goodness of fit, irrespective of the economic plausibility of the independent
variables included. Some of the independent variables included may not be cost drivers.
10-15 ―Multicollinearity exists when the dependent variable and the independent variable are
highly correlated.‖ Do you agree? Explain.
No. Multicollinearity exists when two or more independent variables are highly correlated with
each other.
10-16 HL Co. uses the high-low method to derive a total cost formula. Using a range of units
produced from 1,500 to 7,500, and a range of total costs from $21,000 to $45,000, producing
2,000 units will cost HL:
a. $8,000
c. $23,000
b. $12,000
d. $29,000
SOLUTION
Choice "c" is correct. The high-low method is used to estimate both fixed and variable costs, and
can then be applied to determine a total cost formula that is used to estimate total costs for any
level of production.
The difference between the total costs ($45,000 − $21,000) is divided by the difference in units
(7,500 – 1,500) to derive a variable cost per unit of $4 ($24,000 / 6,000). Using either end of the
range, fixed costs can then be estimated. Using total costs of $45,000 for 7,500 units, with
variable costs at $4 per unit, $45,000 − 7,500($4) = $15,000 of fixed costs.
The total cost formula for HL will be equal to: $15,000 + [$4.00 × # units]. 2,000 units will
produce a total cost of: $15,000 + [$4.00 × 2,000] = $23,000.
Choice "a" is incorrect. This calculation fails to account for the fixed costs of $15,000.
Choice "b" is incorrect. This calculation incorrectly assumes that because 7,500 units cost
$45,000 (or $6 overall per unit), that 2,000 units would cost $12,000 ($6 per unit).
Choice "d" is incorrect. This calculation incorrectly applies a variable cost of $7 per unit rather
than $4.
10-17 A firm uses simple linear regression to forecast the costs for its main product line. If
fixed costs are equal to $235,000 and variable costs are $10 per unit, how many units does it
need to sell at $15 per unit to make a $300,000 profit?
a. 21,400
c. 60,000
b. 47,000
d. 107,000
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SOLUTION
Choice "d" is correct. The regression equation set up will be: y (total costs) = $235,000 + $10x,
with x representing volume. In order to make a $300,000 profit, sales ($15x) − costs must equal
$300,000. So the full set up will be: $15x − ($235,000 + $10x) = $300,000. Solving for x, $5x =
$535,000, or 107,000 units. At 107,000 units, sales will total $1,605,000 and costs will total
$1,305,000 for a profit of $300,000.
Choice "a" is incorrect. This choice represents a calculation error where the $15 sale price and
the $10 variable cost are added together and divided into $535,000.
Choice "b" is incorrect. 47,000 is the number of units that is required in order to breakeven.
Choice "c" is incorrect. These are the number of units above breakeven that the company must
sell in order to make a $300,000 profit.
10-18 In regression analysis, the coefficient of determination:
a. Is used to determine the proportion of the total variation in the dependent variable (y)
explained by the independent variable (X).
b. Ranges between negative one and positive one.
c. Is used to determine the expected value of the net income based on the regression line.
d. Becomes smaller as the fit of the regression line improves.
SOLUTION
Choice "a" is correct. This is the definition of the coefficient of determination. It is the square of
the coefficient of correlation. The higher the coefficient of determination, the greater the
proportion of the total variation in y that is explained by the variation in x. The higher it is, the
better is the fit of the regression line.
Choice "b" is incorrect. It ranges between 0 and 1. Remember, the coefficient of determination is
the square of the coefficient of correlation. Because it is a number squared, it will be positive.
Choice "c" is incorrect. This is not a use of the coefficient of determination.
Choice "d" is incorrect. It becomes larger as the fit of the regression line improves.
10-19 A regression equation is set up, where the dependent variable is total costs and the
independent variable is production. A correlation coefficient of 0.70 implies that:
a.
b.
c.
d.
The coefficient of determination is negative.
The level of production explains 49% of the variation in total costs
There is a slightly inverse relationship between production and total costs.
A correlation coefficient of 1.30 would produce a regression line with better fit to the data.
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SOLUTION
Choice "b" is correct. A correlation coefficient (used to measure the strength in the linear
relationship between independent and dependent variables) of 0.70 implies that the coefficient of
determination is 0.49. A coefficient of determination of 0.49 equates to the independent variable
(level of production) explaining 49 percent of the variation in the dependent variable (total
costs).
Choice "a" is incorrect. The coefficient of determination will always be a number between 0 and
1.
Choice "c" is incorrect. A positive correlation coefficient implies a direct relationship between
the two variables.
Choice "d" is incorrect. The correlation coefficient can only be between -1 and 1.
10-20 What would be the approximate value of the coefficient of correlation between
advertising and sales where a company advertises aggressively as an alternative to temporary
worker layoffs and cuts off advertising when incoming jobs are on backorder?
a. 1.0
c. –1.0
b. 0
d. –100
SOLUTION
Choice "c" is correct. The coefficient of correlation measures the strength and direction of the
relationship between two variables. Since the company increases advertising when sales are low
and decreases advertising when sales are high, the movement is in directly opposite directions
and the coefficient would be close to - 1.0.
Choice "a" is incorrect. A coefficient of correlation of 1.0 would imply that both variables move
in the same direction at approximately the same rate. An increase in advertising when sales are
increasing would be characteristic of a correlation of coefficient of 1.0.
Choice "b" is incorrect. A coefficient of correlation of 0 would imply that there is no relationship
between advertising and sales. There is an inverse relationship between advertising and sales.
Choice "d" is incorrect. A relationship exists between advertising and sales. According to the
facts of the question, the relationship is an inverse relationship. The coefficient of correlation is
expressed as a range between 1.0 and +1.0.
10-21 Estimating a cost function. The controller of the Javier Company is preparing the budget
for 2018 and needs to estimate a cost function for delivery costs. Information regarding delivery
costs incurred in the prior two months are:
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Month
Miles Driven
Delivery Costs
August
12,000
$10,000
September
17,000
$13,000
Required:
1. Estimate the cost function for delivery.
2. Can the constant in the cost function be used as an estimate of fixed delivery cost per month?
Explain.
SOLUTION
(10 min.)
1.
Estimating a cost function.
Slope coefficient =
Difference in delivery costs
Difference in miles driven
=
$13, 000  $10, 000
17, 000  12, 000
=
$3,000
= $0.60 per mile
5,000
Constant = Total cost – (Slope coefficient  Quantity of cost driver)
= $10,000 – ($0.60  12,000) = $2,800
= $13,000 – ($0.60  17,000) = $2,800
The cost function based on the two observations is
Delivery costs = $2,800 + $0.60  Miles driven
2.
The cost function in requirement 1 is an estimate of how costs behave within the relevant
range, not at cost levels outside the relevant range. If there are no months with zero miles driven
represented in the delivery cost account in the general ledger, data in that account cannot be used
to estimate the fixed costs at the zero miles driven level. Rather, the constant component of the
cost function provides the best available starting point for a straight line that approximates how a
cost behaves within the relevant range.
10-22 Identifying variable-, fixed-, and mixed-cost functions. The Sunrise Corporation
operates car rental agencies at more than 20 airports. Customers can choose from one of three
contracts for car rentals of one day or less:
 Contract 1: $45 for the day
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 Contract 2: $25 for the day plus $0.30 per mile traveled
 Contract 3: $1.50 per mile traveled
Required:
1. Plot separate graphs for each of the three contracts, with costs on the vertical axis and miles
traveled on the horizontal axis.
2. Express each contract as a linear cost function of the form y  a  bX .
3. Identify each contract as a variable-, fixed-, or mixed-cost function.
SOLUTION
(15 min.)
Identifying variable-, fixed-, and mixed-cost functions.
1.
See Solution Exhibit 10-22.
2.
Contract 1: y = $45
Contract 2: y = $25 + $0.30X
Contract 3: y = $1.50X
where X is the number of miles traveled in the day.
3.
Contract
1
2
3
Cost Function
Fixed
Mixed
Variable
SOLUTION EXHIBIT 10-22
Plots of Car Rental Contracts Offered by Sunrise Corp.
Contract 1: Fixed Costs
$100
Car Rental Costs
$80
$60
$40
$20
$0
0
10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
Miles Traveled Per Day
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Contract 2: Mixed Costs
$100
Car Rental Costs
$80
$60
$40
$20
$0
0
10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
Miles Traveled Per Day
Contract 3: Variable Costs
$250
Car Rental Costs
$200
$150
$100
$50
$0
0
10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
Miles Traveled Per Day
10-23 Various cost-behavior patterns. (CPA, adapted).
The vertical axes of the graphs below represent total cost, and the horizontal axes represent units
produced during a calendar year. In each case, the zero point of dollars and production is at the
intersection of the two axes.
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Required:
Select the graph that matches the numbered manufacturing cost data (requirements 1–9). Indicate
by letter which graph best fits the situation or item described. The graphs may be used more than
once.
1. Annual depreciation of equipment, where the amount of depreciation charged is computed by
the machine-hours method.
2. Electricity bill—a flat fixed charge, plus a variable cost after a certain number of kilowatthours are used, in which the quantity of kilowatt-hours used varies proportionately with
quantity of units produced.
3. City water bill, which is computed as follows:
First 1,000,000 gallons or less
$1,000 flat fee
Next 10,000 gallons
$0.003 per gallon used
Next 10,000 gallons
$0.006 per gallon used
Next 10,000 gallons
$0.009 per gallon used
and so on
and so on
The gallons of water used vary proportionately with the quantity of production output.
4. Cost of direct materials, where direct material cost per unit produced decreases with each
pound of material used (for example, if 1 pound is used, the cost is $10; if 2 pounds are used,
the cost is $19.98; if 3 pounds are used, the cost is $29.94), with a minimum cost per unit of
$9.20.
5. Annual depreciation of equipment, where the amount is computed by the straight-line
method. When the depreciation schedule was prepared, it was anticipated that the
obsolescence factor would be greater than the wear-and-tear factor.
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6. Rent on a manufacturing plant donated by the city, where the agreement calls for a fixed-fee
payment unless 200,000 labor-hours are worked, in which case no rent is paid.
7. Salaries of repair personnel, where one person is needed for every 1,000 machine-hours or
less (that is, 0 to 1,000 hours requires one person, 1,001 to 2,000 hours requires two people,
and so on).
8. Cost of direct materials used (assume no quantity discounts).
9. Rent on a manufacturing plant donated by the county, where the agreement calls for rent of
$100,000 to be reduced by $1 for each direct manufacturing labor-hour worked in excess of
200,000 hours, but a minimum rental fee of $20,000 must be paid.
SOLUTION
(20 min.)
1.
K
2.
B
3.
G
4.
J
5.
6.
7.
8.
9.
I
L
F
K
C
Various cost-behavior patterns.
Note that A is incorrect because, although the cost per pound eventually equals a
constant at $9.20, the total dollars of cost increases linearly from that point
onward.
The total costs will be the same regardless of the volume level.
This is a classic step-cost function.
10-24 Matching graphs with descriptions of cost and revenue behavior. (D. Green, adapted)
Given here are a number of graphs.
Required:
The horizontal axis of each graph represents the units produced over the year, and the vertical
axis represents total cost or revenues.
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Indicate by number which graph best fits the situation or item described (a–h). Some graphs
may be used more than once; some may not apply to any of the situations.
Direct material costs
Supervisors’ salaries for one shift and two shifts
A cost–volume–profit graph
Mixed costs—for example, car rental fixed charge plus a rate per mile driven
Depreciation of plant, computed on a straight-line basis
Data supporting the use of a variable-cost rate, such as manufacturing labor cost of $14 per unit
produced
g. Incentive bonus plan that pays managers $0.10 for every unit produced above some level of
production
h. Interest expense on $2 million borrowed at a fixed rate of interest
a.
b.
c.
d.
e.
f.
SOLUTION
(30 min.)
a.
b.
c.
d.
e.
f.
(1)
(6)
(9)
(2)
(8)
(10)
g.
h.
(3)
(8)
Matching graphs with descriptions of cost and revenue behavior.
A step-cost function.
It is data plotted on a scatter diagram, showing a linear variable cost function with
constant variance of residuals. The constant variance of residuals implies that
there is a uniform dispersion of the data points about the regression line.
10-25 Account analysis, high-low. Stein Corporation wants to find an equation to estimate
some of their monthly operating costs for the operating budget for 2018. The following cost and
other data were gathered for 2017:
Month
Maintenance Machine
Costs
Hours
Health
Insurance
Number of
Employees
Shipping
Costs
Units
Shipped
January
$4,500
165
$8,600
68
$25,776
7,160
February
$4,452
120
$8,600
75
$29,664
8,240
March
$4,600
230
$8,600
92
$28,674
7,965
April
$4,850
318
$8,600
105
$23,058
6,405
May
$5,166
460
$8,600
89
$21,294
5,915
June
$4,760
280
$8,600
87
$33,282
9,245
July
$4,910
340
$8,600
93
$31,428
8,730
August
$4,960
360
$8,600
88
$30,294
8,415
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September
$5,070
420
$8,600
95
$25,110
6,975
October
$5,250
495
$8,600
102
$25,866
7,185
November
$5,271
510
$8,600
97
$20,124
5,590
December
$4,760
275
$8,600
94
$34,596
9,610
Required:
1. Which of the preceding costs is variable? Fixed? Mixed? Explain.
2. Using the high-low method, determine the cost function for each cost.
3. Combine the preceding information to get a monthly operating cost function for the Stein
Corporation.
4. Next month, Stein expects to use 400 machine hours, have 80 employees, and ship 9,000
units. Estimate the total operating cost for the month.
SOLUTION
(20 min.) Account analysis, high-low
1. The maintenance cost is a mixed cost because the cost neither remains constant in total nor
remains constant per unit.
The health insurance cost is fixed because, although the number of employees varies from month
to month, the total cost remains constant at $8,600.
The shipping cost is variable because, in each month, the cost divided by the number of units
shipped equals a constant $3.60. The definition of a variable cost is one that remains constant per
unit.
2. The month with the highest number of machine hours is November, with 510 machine hours
and $5,271 of cost. The month with the lowest is February, with 120 machine hours and $4,452
in cost. The difference in cost ($5,271 – $4,452), divided by the difference in machine hours
(510 – 120) equals $2.10 per machine hour of variable maintenance cost. Inserted into the cost
formula for November:
$5,271 = a fixed cost + ($2.10 × number of machine hours used)
$5,271 = a + ($2.10 × 510)
$5,271 = a + $1,071
a = $4,200 monthly fixed maintenance cost
Therefore, Stein’s cost formula for monthly maintenance cost is:
y = $4,200 + ($2.10 × number of machine hours used)
3. The shipping rate is $3.60 per unit shipped
The maintenance cost is $4,200 + ($2.10 per machine hour used)
The health insurance cost is $8,600.
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Adding them together we get:
Fixed operating costs = $4,200 (maintenance) + $8,600 (health insurance) = $12,800
Monthly Operating Cost = $12,800 + ($3.60 per unit shipped) + ($2.10 per machine hour used)
4. Estimated operating cost for November:
$12,800 + ($3.60 × 9,000 units shipped) + ($2.10 × 400 machine hours)
= $12,800 + $32,400 + $840
= $46,040
10-26 Account analysis method. Gower, Inc., a manufacturer of plastic products, reports the
following manufacturing costs and account analysis classification for the year ended December
31, 2017.
Account
Classification
Amount
Direct materials
All variable
$300,000
Direct manufacturing labor
All variable
225,000
Power
All variable
37,500
Supervision labor
20% variable
56,250
Materials-handling labor
50% variable
60,000
Maintenance labor
40% variable
75,000
Depreciation
0% variable
95,000
Rent, property taxes, and administration
0% variable
100,000
Gower, Inc., produced 75,000 units of product in 2017. Gower’s management is estimating costs
for 2018 on the basis of 2017 numbers. The following additional information is available for
2018.
a. Direct materials prices in 2018 are expected to increase by 5% compared with 2017.
b. Under the terms of the labor contract, direct manufacturing labor wage rates are expected to
increase by 10% in 2018 compared with 2017.
c. Power rates and wage rates for supervision, materials handling, and maintenance are not
expected to change from 2017 to 2018.
d. Depreciation costs are expected to increase by 5%, and rent, property taxes, and
administration costs are expected to increase by 7%.
e. Gower expects to manufacture and sell 80,000 units in 2018.
Required:
1. Prepare a schedule of variable, fixed, and total manufacturing costs for each account category
in 2018. Estimate total manufacturing costs for 2018.
2. Calculate Gower’s total manufacturing cost per unit in 2017, and estimate total
manufacturing cost per unit in 2018.
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3. How can you obtain better estimates of fixed and variable costs? Why would these better
estimates be useful to Gower?
SOLUTION
(30 min.) Account analysis method.
1.
Manufacturing cost classification for 2017:
Account
Direct materials
Direct manufacturing labor
Power
Supervision labor
Materials-handling labor
Maintenance labor
Depreciation
Rent, property taxes, admin
Total
Total
Costs
(1)
$300,000
225,000
37,500
56,250
60,000
75,000
95,000
100,000
$948,750
% of
Total Costs
That is
Variable
Fixed
Variable
Variable
Costs
Costs
Cost per Unit
(2)
(3) = (1)  (2) (4) = (1) – (3) (5) = (3) ÷ 75,000
100%
100
100
20
50
40
0
0
$300,000
225,000
37,500
11,250
30,000
30,000
0
0
$633,750
$
0
0
0
45,000
30,000
45,000
95,000
100,000
$315,000
$4.00
3.00
0.50
0.15
0.40
0.40
0
0
$8.45
Total manufacturing cost for 2017 = $948,750
Variable costs in 2018:
Account
Direct materials
Direct manufacturing labor
Power
Supervision labor
Materials-handling labor
Maintenance labor
Depreciation
Rent, property taxes, admin.
Total
Unit
Variable
Increase in
Cost per
Variable Variable Cost
Unit for Percentage
Cost
per Unit
2017
Increase
per Unit
for 2018
(6)
(7)
(8) = (6)  (7) (9) = (6) + (8)
$4.00
3.00
0.50
0.15
0.40
0.40
0
0
$8.45
5%
10
0
0
0
0
0
0
$0.20
0.30
0
0
0
0
0
0
$0.50
Total Variable
Costs for 2018
(10) = (9)  80,000
$4.20
3.30
0.50
0.15
0.40
0.40
0
0
$8.95
$336,000
264,000
40,000
12,000
32,000
32,000
0
0
$716,000
Fixed and total costs in 2018:
Account
Fixed
Costs
for 2018
(11)
Percentage
Increase
(12)
10-15
Dollar
Increase in
Fixed Costs
(13) =
(11)  (12)
Fixed Costs
for 2018
(14) =
(11) + (13)
Variable
Costs for
2018
(15)
Total
Costs
(16) =
(14) + (15)
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Direct materials
$
0
Direct manufacturing labor
0
Power
0
Supervision labor
45,000
Materials-handling labor
30,000
Maintenance labor
45,000
Depreciation
95,000
Rent, property taxes, admin. 100,000
Total
$315,000
0%
0
0
0
0
0
5
7
$
0
0
0
0
0
0
4,750
7,000
$11,750
$
0
0
0
45,000
30,000
45,000
99,750
107,000
$326,750
$336,000 $ 336,000
264,000
264,000
40,000
40,000
12,000
57,000
32,000
62,000
32,000
77,000
0
99,750
0
107,000
$716,000 $1,042,750
Total manufacturing costs for 2018 = $1,042,750
2.
Total cost per unit, 2017
Total cost per unit, 2018
$948,750
= $12.65
75,000
$1,042,750
=
= $13.03
80,000
=
3.
Cost classification into variable and fixed costs is based on qualitative, rather than
quantitative, analysis. How good the classifications are depends on the knowledge of individual
managers who classify the costs. Gower may want to undertake quantitative analysis of costs,
using regression analysis on time-series or cross-sectional data to better estimate the fixed and
variable components of costs. Better knowledge of fixed and variable costs will help Gower to
better price his products, to know when he is getting a positive contribution margin, and to better
manage costs.
10-27 Estimating a cost function, high-low method. Reisen Travel offers helicopter service
from suburban towns to John F. Kennedy International Airport in New York City. Each of its 10
helicopters makes between 1,000 and 2,000 round-trips per year. The records indicate that a
helicopter that has made 1,000 round-trips in the year incurs an average operating cost of $350
per round-trip, and one that has made 2,000 round-trips in the year incurs an average operating
cost of $300 per round-trip.
Required:
1. Using the high-low method, estimate the linear relationship y  a  bX , where y is the total
annual operating cost of a helicopter and X is the number of round-trips it makes to JFK
airport during the year.
2. Give examples of costs that would be included in a and in b.
3. If Reisen Travel expects each helicopter to make, on average, 1,200 round-trips in the
coming year, what should its estimated operating budget for the helicopter fleet be?
SOLUTION
(15–20 min.)
Estimating a cost function, high-low method.
10-16
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1.
The key point to note is that the problem provides high-low values of X (annual round
trips made by a helicopter) and Y  X (the operating cost per round trip). We first need to
calculate the annual operating cost Y (as in column (3) below), and then use those values to
estimate the function using the high-low method.
Highest observation of cost driver
Lowest observation of cost driver
Difference
Cost Driver:
Annual RoundTrips (X)
(1)
2,000
1,000
1,000
Operating
Cost per
Round-Trip
(2)
$300
$350
Annual
Operating
Cost (Y)
(3) = (1)  (2)
$600,000
$350,000
$250,000
Slope coefficient = $250,000  1,000 = $250 per round-trip
Constant = $600,000 – ($250  2,000) = $100,000
The estimated relationship is Y = $100,000 + $250 X; where Y is the annual operating cost of a
helicopter and X represents the number of round trips it makes annually.
2.
The constant a (estimated as $100,000) represents the fixed costs of operating a
helicopter, irrespective of the number of round trips it makes. This would include items such as
insurance, registration, depreciation on the aircraft, and any fixed component of pilot and crew
salaries. The coefficient b (estimated as $250 per round-trip) represents the variable cost of each
round trip—costs that are incurred only when a helicopter actually flies a round trip. The
coefficient b may include costs such as landing fees, fuel, refreshments, baggage handling, and
any regulatory fees paid on a per-flight basis.
3.
If each helicopter is, on average, expected to make 1,200 round trips a year, we can use
the estimated relationship to calculate the expected annual operating cost per helicopter:
Y = $100,000 + $250 X
X = 1,200
Y = $100,000 + $250  1,200 = $100,000 + $300,000 = $400,000
With 10 helicopters in its fleet, Reisen’s estimated operating budget is 10  $400,000 = $4,000,000.
10-28 Estimating a cost function, high-low method. Lacy Dallas is examining customerservice costs in the southern region of Camilla Products. Camilla Products has more than 200
separate electrical products that are sold with a 6-month guarantee of full repair or replacement
with a new product. When a product is returned by a customer, a service report is prepared. This
service report includes details of the problem and the time and cost of resolving the problem.
Weekly data for the most recent 8-week period are as follows:
Week
1
Customer-Service Department Costs
$13,300
10-17
Number of Service Reports
185
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2
20,500
285
3
12,000
120
4
18,500
360
5
14,900
275
6
21,600
440
7
16,500
350
8
21,300
315
Required:
1. Plot the relationship between customer-service costs and number of service reports. Is the
relationship economically plausible?
2. Use the high-low method to compute the cost function relating customer-service costs to the
number of service reports.
3. What variables, in addition to number of service reports, might be cost drivers of weekly
customer-service costs of Camilla Products?
SOLUTION
(20 min.)
Estimating a cost function, high-low method.
1.
See Solution Exhibit 10-28. There is a positive relationship between the number of
service reports (a cost driver) and the customer-service department costs. This relationship is
economically plausible.
2.
Highest observation of cost driver
Lowest observation of cost driver
Difference
Number of
Service Reports
440
120
320
Customer-Service
Department Costs
$21,600
12,000
$ 9,600
Customer-service department costs = a + b (number of service reports)
Slope coefficient (b)
Constant (a)
$9,600
= $30 per service report
320
= $21,600 – ($30  440) = $8,400
= $12,000 – ($30  120) = $8,400
=
Customer-service
= $8,400 + $30 (number of service reports)
department costs
3.
Other possible cost drivers of customer-service department costs are:
a.
Number of products replaced with a new product (and the dollar value of the new
products charged to the customer-service department).
b.
Number of products repaired and the time and cost of repairs.
10-18
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SOLUTION EXHIBIT 10-28
Plot of Number of Service Reports versus Customer-Service Dept. Costs for Camilla Products
Customer-Service Department Costs
$24,000
$22,000
$20,000
$18,000
$16,000
$14,000
$12,000
$10,000
100
150
200
250
300
350
400
450
500
Number of Service Reports
10-29 Linear cost approximation. Dr. Young, of Young and Associates, LLP, is examining
how overhead costs behave as a function of monthly physician contact hours billed to patients.
The historical data are as follows:
Total Overhead Costs
Physician Contact Hours Billed to Patients
$ 90,000
150
105,000
200
111,000
250
125,000
300
137,000
350
150,000
400
Required:
1. Compute the linear cost function, relating total overhead costs to physician contact hours,
using the representative observations of 200 and 300 hours. Plot the linear cost function.
Does the constant component of the cost function represent the fixed overhead costs of
Young and Associates? Why?
2. What would be the predicted total overhead costs for (a) 150 hours and (b) 400 hours using
the cost function estimated in requirement 1? Plot the predicted costs and actual costs for 150
and 400 hours.
3. Dr. Young had a chance to do some school physicals that would have boosted physician
contact hours billed to patients from 200 to 250 hours. Suppose Dr. Young, guided by the
linear cost function, rejected this job because it would have brought a total increase in
10-19
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contribution margin of $9,000, before deducting the predicted increase in total overhead cost,
$10,000. What is the total contribution margin actually forgone?
SOLUTION
(30–40 min.)
1.
Linear cost approximation.
Slope coefficient (b)
Constant (a)
Cost function
=
Difference in overhead costs
Difference in contact hours billed
=
=
($125,000 - $105,000)/(300 – 200)
$200
= $125,000 – ($200 × 300)
= $65,000
= $65,000 + ($200  contact hours billed)
The linear cost function is plotted in Solution Exhibit 10-29.
No, the constant component of the cost function does not represent the fixed overhead cost of
Young and Associates. The relevant range of physician contact hours billed is from 150 to 400.
The constant component provides the best available starting point for a straight line that
approximates how a cost behaves within the relevant range.
2.
A comparison at various levels of contact hours billed follows. The linear cost function is
based on the formula of $65,000 per month plus $200 per physician contact hours billed.
Total overhead cost behavior:
Month 1 Month 2 Month 3 Month 4 Month 5 Month 6
Professional labor-hours
150
200
250
300
350
400
Actual total overhead costs
$90,000 $105,000 $111,000 $125,000 $137,000 $150,000
Linear approximation
95,000 105,000 115,000 125,000 135,000 145,000
Actual minus linear
Approximation
$(5,000) $
0 $ (4,000) $
0 $ 2,000 $ 5,000
The data are shown in Solution Exhibit 10-29. The linear cost function overstates costs by
$5,000 at the 150-hour level and understates costs by $5,000 at the 400-hour level.
3.
Contribution before deducting incremental overhead
Incremental overhead
Contribution after incremental overhead
The total contribution margin actually forgone is $3,000.
10-20
Based on
Actual
$9,000
6,000
$ 3,000
Based on Linear
Cost Function
$9,000
10,000
$ (1,000)
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SOLUTION EXHIBIT 10-29
Linear Cost Function Plot of Physician Contact Hours Billed
and Total Overhead Costs for Young and Associates
$160,000
Total Overhead Costs
$150,000
$140,000
$130,000
$120,000
$110,000
$100,000
$90,000
$80,000
100
150
200
250
300
350
400
Physician Contact Hours Billed
450
500
10-30 Cost-volume-profit and regression analysis. Relling Corporation manufactures a drink
bottle, model CL24. During 2017, Relling produced 210,000 bottles at a total cost of $808,500.
Kraff Corporation has offered to supply as many bottles as Relling wants at a cost of $3.75 per
bottle. Relling anticipates needing 225,000 bottles each year for the next few years.
Required:
1. a. What is the average cost of manufacturing a drink bottle in 2017? How does it compare to
Kraff’s offer?
b. Can Relling use the answer in requirement 1a to determine the cost of manufacturing
225,000 drink bottles? Explain.
2. Relling’s cost analyst uses annual data from past years to estimate the following regression
equation with total manufacturing costs of the drink bottle as the dependent variable and
drink bottles produced as the independent variable:
y  $445, 000  $1.75 X
During the years used to estimate the regression equation, the production of bottles varied from
200,000 to 235,000. Using this equation, estimate how much it would cost Relling to
manufacture 225,000 drink bottles. How much more or less costly is it to manufacture the
bottles than to acquire them from Kraff?
3. What other information would you need to be confident that the equation in requirement 2
accurately predicts the cost of manufacturing drink bottles?
SOLUTION
(20 min.) Cost-volume-profit and regression analysis.
10-21
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Average cost of manufacturing
=
Total manufacturing costs
Number of drink bottles
=
$808,500
= $3.85 per bottle
210,000
This cost is higher than the $3.75 per bottle that Kraff has quoted.
1b.
Rellings cannot take the average manufacturing cost in 2017 of $3.85 per bottle and
multiply it by 225,000 drink bottles to determine the total cost of manufacturing 225,000 drink
bottles. The reason is that some of the $808,500 (or equivalently the $3.85 cost per bottle) are
fixed costs and some are variable costs. Without distinguishing fixed from variable costs,
Rellings cannot determine the cost of manufacturing 225,000 drink bottles. For example, if all
costs are fixed, the manufacturing costs of 225,000 bottles will continue to be $808,500. If,
however, all costs are variable, the cost of manufacturing 225,000 bottles would be $3.85 
225,000 = $866,250. If some costs are fixed and some are variable, the cost of manufacturing
225,000 bottles will be somewhere between $808,500 and $866,250.
Some students could argue that another reason for not being able to determine the cost of
manufacturing 225,000 drink bottles is that not all costs are output unit-level costs. If some costs
are, for example, batch-level costs, more information would be needed on the number of batches
in which the 225,000 drink bottles would be produced, in order to determine the cost of
manufacturing 225,000 drink bottles.
2.
Expected cost to make
225,000 drink bottles
= $445,000 + ($1.75  225,000)
= $445,000 + $393,750 = $838,750
Purchasing drink bottles from Kraff will cost $3.75  225,000 = $843,750. Hence, it will
cost Rellings $843,750  $838,750 = $5,000 more to purchase the bottles from Kraff rather than
manufacture them in-house.
3.
Rellings would need to consider several factors before being confident that the equation
in requirement 2 accurately predicts the cost of manufacturing drink bottles.
a. Is the relationship between total manufacturing costs and quantity of drink bottles
economically plausible? For example, is the quantity of bottles made the only cost
driver or are there other cost-drivers (for example batch-level costs of setups,
production-orders or material handling) that affect manufacturing costs?
b. How good is the goodness of fit? That is, how well does the estimated line fit the
data?
c. Is the relationship between the number of drink bottles produced and total
manufacturing costs linear?
d. Does the slope of the regression line indicate that a strong relationship exists between
manufacturing costs and the number of drink bottles produced?
10-22
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e. Are there any data problems such as, for example, errors in measuring costs, trends in
prices of materials, labor or overheads that might affect variable or fixed costs over
time, extreme values of observations, or a nonstationary relationship over time
between total manufacturing costs and the quantity of bottles produced?
f. How is inflation expected to affect costs?
g. Will Kraff supply high-quality drink bottles on time?
h. What is the length of the agreement with Kraff? Short-term or Long-term?
10-31 Regression analysis, service company. (CMA, adapted) Linda Olson owns a
professional character business in a large metropolitan area. She hires local college students to
play these characters at children’s parties and other events. Linda provides balloons, cupcakes,
and punch. For a standard party the cost on a per-person basis is as follows:
Balloons, cupcakes, and punch
$ 7
Labor (0.25 hour  $20 per hour)
5
Overhead (0.25 hour  $40 per hour
10
Total cost per person
$22
Linda is quite certain about the estimates of the materials and labor costs, but is not as
comfortable with the overhead estimate. The overhead estimate was based on the actual data for
the past 9 months, which are presented here. These data indicate that overhead costs vary with
the direct labor-hours used. The $40 estimate was determined by dividing total overhead costs
for the 9 months by total labor-hours.
Month
Labor-Hours
Overhead
Costs
April
1,400
$ 65,000
May
1,800
71,000
June
2,100
73,000
July
2,200
76,000
August
1,650
67,000
September
1,725
68,000
October
1,500
66,500
November
1,200
60,000
December
1,900
72,500
15,475
$619,000
Total
Linda has recently become aware of regression analysis. She estimated the following regression
equation with overhead costs as the dependent variable and labor-hours as the independent
variable:
10-23
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y  $43,563  $14.66 X
Required:
1. Plot the relationship between overhead costs and labor-hours. Draw the regression line and
evaluate it using the criteria of economic plausibility, goodness of fit, and slope of the
regression line.
2. Using data from the regression analysis, what is the variable cost per person for a standard
party?
3. Linda Olson has been asked to prepare a bid for a 20-child birthday party to be given next
month. Determine the minimum bid price that Linda would be willing to submit to recoup
variable costs.
SOLUTION
(25 min.) Regression analysis, service company.
1.
Solution Exhibit 10-31 plots the relationship between labor-hours and overhead costs and
shows the regression line.
y = $43,563 + $14.66 X
Economic plausibility. Labor-hours appears to be an economically plausible driver of
overhead costs for the character company. Overhead costs such as scheduling, hiring and training
of workers, and managing the workforce are largely incurred to support labor.
Goodness of fit. The vertical differences between actual and predicted costs are extremely
small, indicating a very good fit. The good fit indicates a strong relationship between the laborhour cost driver and overhead costs.
Slope of regression line. The regression line has a reasonably steep slope from left to
right. Given the small scatter of the observations around the line, the positive slope indicates that,
on average, overhead costs increase as labor-hours increase.
2.
The regression analysis indicates that, within the relevant range of 1,200 to 2,200 laborhours, the variable cost per person for a birthday party equals:
Balloons, cupcakes and punch
Labor (0.25 hrs.  $20 per hour)
Variable overhead (0.25 hrs.  $14.66 per labor-hour)
Total variable cost per person
$ 7.00
5.00
3.67
$15.67
3.
To earn a positive contribution margin, the minimum bid for a 20-child birthday party
would be any amount greater than $313.40. This amount is calculated by multiplying the variable
cost per child of $15.67 by the 20 children. At a price above the variable cost of $313.40, Linda
Olson will be earning a contribution margin toward coverage of her fixed costs.
10-24
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Of course, Linda Olson will consider other factors in developing her bid including (a) an
analysis of the competition––vigorous competition will limit Linda’s ability to obtain a higher
price (b) a determination of whether or not her bid will set a precedent for lower prices––overall,
the prices Linda Olson charges should generate enough contribution to cover fixed costs and earn
a reasonable profit, and (c) a judgment of how representative past historical data (used in the
regression analysis) is about future costs.
SOLUTION EXHIBIT 10-31
Regression Line of Overhead Costs on Labor-Hours for Linda Olson’s Character Business
80,000
Overhead Costs
75,000
y = 14.664x + 43563
R² = 0.9551
70,000
65,000
60,000
55,000
50,000
1,000
1,200
1,400
1,600
1,800
Labor-Hours
2,000
2,200
2,400
10-32 High-low, regression. May Blackwell is the new manager of the materials storeroom for
Clayton Manufacturing. May has been asked to estimate future monthly purchase costs for part
#696, used in two of Clayton’s products. May has purchase cost and quantity data for the past 9
months as follows:
Month
Cost of Purchase
Quantity Purchased
January
$12,675
February
13,000
2,810
March
17,653
4,153
April
15,825
3,756
May
13,125
2,912
June
13,814
3,387
July
15,300
3,622
August
10,233
2,298
September
14,950
3,562
10-25
2,710 parts
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Estimated monthly purchases for this part based on expected demand of the two products for the
rest of the year are as follows:
Month
Purchase Quantity Expected
October
3,340 parts
November
3,710
December
3,040
Required:
1. The computer in May’s office is down, and May has been asked to immediately provide an
equation to estimate the future purchase cost for part #696. May grabs a calculator and uses
the high-low method to estimate a cost equation. What equation does she get?
2. Using the equation from requirement 1, calculate the future expected purchase costs for each
of the last 3 months of the year.
3. After a few hours May’s computer is fixed. May uses the first 9 months of data and
regression analysis to estimate the relationship between the quantity purchased and purchase
costs of part #696. The regression line May obtains is as follows:
y  $2,582.6  3.54 X
Evaluate the regression line using the criteria of economic plausibility, goodness of fit, and
significance of the independent variable. Compare the regression equation to the equation
based on the high-low method. Which is a better fit? Why?
4. Use the regression results to calculate the expected purchase costs for October, November,
and December. Compare the expected purchase costs to the expected purchase costs
calculated using the high-low method in requirement 2. Comment on your results.
SOLUTION
(25 min.) High-low, regression
1.
May will pick the highest point of activity, 4,153 parts (March) at $17,653 of cost, and
the lowest point of activity, 2,298 parts (August) at $10,233.
Highest observation of cost driver
Lowest observation of cost driver
Difference
Cost driver:
Quantity Purchased
4,153
2,298
1,855
Purchase costs = a + b  Quantity purchased
Slope Coefficient = $7,420/1,855 = $4 per part
Constant (a) = $17,653 ─ ($4  4,153) = $1,041
10-26
Cost
$17,653
10,233
$ 7,420
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The equation May gets is:
Purchase costs = $1,041 + ($4  Quantity purchased)
2.
Using the equation above, the expected purchase costs for each month will be:
Purchase
Quantity
Expected
3,340 parts
3,710
3,040
Month
October
November
December
Expected
Formula
cost
y = $1,041 + ($4  3,340) $14,401
y = $1,041 + ($4  3,710) 15,881
y = $1,041 + ($4  3,040) 13,201
3.
Economic Plausibility: Clearly, the cost of purchasing a part is associated with the
quantity purchased.
Goodness of Fit: As seen in Solution Exhibit 10-32, the regression line fits the data well. The
vertical distance between the regression line and observations is small. An r-squared value of
close to 0.96 indicates that almost 98% of the change in cost can be explained by the change in
quantity purchased.
Significance of the Independent Variable: The relatively steep slope of the regression line
suggests that the quantity purchased is correlated with purchasing cost for part #696.
SOLUTION EXHIBIT 10-32
Clayton Manufacturing Purchase Costs for Part #696
$20,000
Cost of Purchase
$18,000
y = 3.5376x + 2582.6
R² = 0.9583
$16,000
$14,000
$12,000
$10,000
$8,000
2,000
2,500
3,000
3,500
Quantity Purchased
10-27
4,000
4,500
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According to the regression, May’s original estimate of fixed cost is too low given all the data
points. The original slope is too steep, but only by 46 cents. So, the variable rate is lower but
the fixed cost is higher for the regression line than for the high-low cost equation.
The regression is the more accurate estimate because it uses all available data (all nine data
points) while the high-low method only relies on two data points and may therefore miss some
important information contained in the other data.
4.
Using the regression equation, the purchase costs for each month will be:
Month
October
November
December
Purchase
Quantity
Expected
3,340 parts
3,710
3,040
Formula
Expected cost
y = $2,582.60 + ($3.54  3,340) $14,406.20
y = $2,582.60 + ($3.54  3,710)
15,716.00
y = $2,582.60 + ($3.54  3,040)
13,344.20
Although the two equations are different in both fixed element and variable rate, within the
relevant range they give similar expected costs. This implies that the high and low points of the
data are a reasonable representation of the total set of points within the relevant range.
10-33 Learning curve, cumulative average-time learning model. Northern Defense
manufactures radar systems. It has just completed the manufacture of its first newly designed
system, RS-32. Manufacturing data for the RS-32 follow:
Required:
Calculate the total variable costs of producing 2, 4, and 8 units.
SOLUTION
(20 min.) Learning curve, cumulative average-time learning model.
The direct manufacturing labor-hours (DMLH) required to produce the first 2, 4, and 8 units
given the assumption of a cumulative average-time learning curve of 85%, is as follows:
10-28
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85% Learning Curve
Cumulative
Number
of Units (X)
(1)
1
2
4
8
Cumulative
Average Time
per Unit (y): Labor Hours
(2)
4,400
3,740
= (4,400  0.85)
3,179
= (3,740  0.85)
2,702
= (3,179  0.85)
Cumulative
Total Time:
Labor-Hours
(3) = (1)  (2)
4,400
7,480
12,716
21,617
Alternatively, to compute the values in column (2) we could use the formula
y = aXb
where a = 4,400, X = 2, 4, or 8, and b = – 0.234465, which gives
when X = 2, y = 4,400 2– 0.234465 = 7,480
when X = 4, y = 4,400 4– 0.234465 = 12,716
when X = 8, y = 4,400 8– 0.234465 = 21,617
Direct materials $84,000  2; 4; 8
Direct manufacturing labor
$27  7,480; 12,716; 21,617
Variable manufacturing overhead
$13  7,480; 12,716; 21,617
Total variable costs
Variable Costs of Producing
2 Units
4 Units
8 Units
$168,000
$336,000
$672,000
201,960
343,332
583,664
97,240
$467,200
165,308
$844,640
281,024
$1,536,688
10-34 Learning curve, incremental unit-time learning model. Assume the same information
for Northern Defense as in Exercise 10-33, except that Northern Defense uses an 85%
incremental unit-time learning model as a basis for predicting direct manufacturing labor-hours.
(An 85% learning curve means b = –0.234465.)
Required:
1. Calculate the total variable costs of producing 2, 3, and 4 units.
2. If you solved Exercise 10-33, compare your cost predictions in the two exercises for 2 and 4
units. Why are the predictions different? How should Northern Defense decide which model
it should use?
SOLUTION
(20 min.) Learning curve, incremental unit-time learning model.
1.
The direct manufacturing labor-hours (DMLH) required to produce the first 2, 3, and 4
units, given the assumption of an incremental unit-time learning curve of 85%, is as follows:
10-29
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Cumulative
Number of Units (X)
(1)
1
2
3
4
85% Learning Curve
Individual Unit Time for Xth
Unit (y): Labor Hours
(2)
4,400
3,740
= (4,400  0.85)
3,401
3,179
= (3,740  0.85)
Cumulative Total Time:
Labor-Hours
(3)
4,400
8,140
11,541
14,720
Values in column (2) are calculated using the formula y = aXb where a = 4,400, X = 2, 3,
or 4, and b = – 0.234465, which gives
when X = 2, y = 4,400  2– 0.234465 = 3,740
when X = 3, y = 4,400  3– 0.234465 = 3,401
when X = 4, y = 4,400  4– 0.234465 = 3,179
Direct materials $84,000  2; 3; 4
Direct manufacturing labor
$27  8,140; 11,541; 14,720
Variable manufacturing overhead
$13  8,140; 11,541; 14,720
Total variable costs
Variable Costs of Producing
2 Units
3 Units
4 Units
$168,000 $ 252,000
$ 336,000
219,780
311,607
397,440
105,820
$493,600
150,033
$713,640
191,360
$924,800
2.
Incremental unit-time learning model (from requirement 1)
Cumulative average-time learning model (from Exercise 10-34)
Difference
Variable Costs of
Producing
2 Units
4 Units
$493,600
$924,800
467,200
844,640
$ 26,400
$ 80,160
Total variable costs for manufacturing 2 and 4 units are lower under the cumulative
average-time learning curve relative to the incremental unit-time learning curve. Direct
manufacturing labor-hours required to make additional units decline more slowly in the
incremental unit-time learning curve relative to the cumulative average-time learning curve when
the same 85% factor is used for both curves. The reason is that, in the incremental unit-time
learning curve, as the number of units double only the last unit produced has a cost of 85% of the
initial cost. In the cumulative average-time learning model, doubling the number of units causes
the average cost of all the units produced (not just the last unit) to be 85% of the initial cost.
10-35 High-low method. Wayne Mueller, financial analyst at CELL Corporation, is examining
the behavior of quarterly utility costs for budgeting purposes. Mueller collects the following data
on machine-hours worked and utility costs for the past 8 quarters:
10-30
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Quarter
Machine-Hours
Utility Costs
1
120,000
$215,000
2
75,000
150,000
3
110,000
200,000
4
150,000
270,000
5
90,000
170,000
6
140,000
250,000
7
130,000
225,000
8
100,000
195,000
Required:
1. Estimate the cost function for the quarterly data using the high-low method.
2. Plot and comment on the estimated cost function.
3. Mueller anticipates that CELL will operate machines for 125,000 hours in quarter 9.
Calculate the predicted utility costs in quarter 9 using the cost function estimated in
requirement 1.
SOLUTION
(25 min.)
High-low method.
1.
Highest observation of cost driver
Lowest observation of cost driver
Difference
Utility costs
Utility Costs
150,000
75,000
75,000
$270,000
150,000
$120,000
= a + b  Machine-hours
Slope coefficient (b) =
Constant (a)
Machine-Hours
$120,000
= $1.60 per machine-hour
75,000
= $270,000 – ($1.60 × 150,000)
= $270,000 – $240,000 = $30,000
or
Constant (a)
= $150,000 – ($1.60 × 75,000)
= $150,000 – $120,000 = $30,000
Utility costs
= $30,000 + ($1.60 × Machine-hours)
10-31
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2.
SOLUTION EXHIBIT 10-35
Plot and High-Low Line of Utility Costs as a Function of Machine-Hours
$280,000
$260,000
Utility Costs
$240,000
$220,000
$200,000
$180,000
$160,000
$140,000
$120,000
$100,000
60,000
80,000
100,000
120,000
140,000
160,000
Machine-Hours
Solution Exhibit 10-35 presents the high-low line.
Economic plausibility. The cost function shows a positive economically plausible relationship
between machine-hours and utility costs. There is a clear-cut engineering relationship of higher
machine-hours and utility costs.
Goodness of fit. The high-low line appears to ―fit‖ the data well. The vertical differences
between the actual and predicted costs appear to be quite small.
Slope of high-low line. The slope of the line appears to be reasonably steep indicating that, on
average, utility costs in a quarter vary with machine-hours used.
3.
Using the cost function estimated in 1, predicted utility costs would be:
$30,000 + ($1.60 × 125,000 hours) = $230,000.
Mueller should budget $230,000 in quarter 9 because the relationship between machinehours and utility costs in Solution Exhibit 10-35 is economically plausible, has an excellent
goodness of fit, and indicates that an increase in machine-hours in a quarter causes utility costs to
increase in the quarter.
10-36 High-low method and regression analysis. Market Thyme, a cooperative of organic
family-owned farms, has recently started a fresh produce club to provide support to the group’s
member farms and to promote the benefits of eating organic, locally produced food. Families pay
a seasonal membership fee of $100 and place their orders a week in advance for a price of $40
per order. In turn, Market Thyme delivers fresh-picked seasonal local produce to several
10-32
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neighborhood distribution points. Five hundred families joined the club for the first season, but
the number of orders varied from week to week.
Tom Diehl has run the produce club for the first season. Tom is now a farmer but remembers
a few things about cost analysis from college. In planning for next year, he wants to know how
many orders will be needed each week for the club to break even, but first he must estimate the
club’s fixed and variable costs. He has collected the following data over the club’s first season of
operation:
Week Number of Orders per Week Weekly Total Costs
1
415
$26,900
2
435
27,200
3
285
24,700
4
325
25,200
5
450
27,995
6
360
25,900
7
420
27,000
8
460
28,315
9
380
26,425
10
350
25,750
Required:
1. Plot the relationship between number of orders per week and weekly total costs.
2. Estimate the cost equation using the high-low method, and draw this line on your graph.
3. Tom uses his computer to calculate the following regression formula:
Weekly total costs = $18,791 + ($19.97  Number of orders per week)
Draw the regression line on your graph. Use your graph to evaluate the regression line using
the criteria of economic plausibility, goodness of fit, and significance of the independent
variable. Is the cost function estimated using the high-low method a close approximation of
the cost function estimated using the regression method? Explain briefly.
4. Did Market Thyme break even this season? Remember that each of the families paid a
seasonal membership fee of $100.
5. Assume that 500 families join the club next year and that prices and costs do not change.
How many orders, on average, must Market Thyme receive each of 10 weeks next season to
break even?
SOLUTION
(30min.) High-low method and regression analysis.
1.
See Solution Exhibit 10-36.
10-33
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SOLUTION EXHIBIT 10-36
$29,000
$28,500
Weekly Total Costs
$28,000
$27,500
High-Low Line
$27,000
$26,500
$26,000
$25,500
Regression Line
$25,000
$24,500
$24,000
250
300
350
400
450
500
Number of Weekly Orders
2.
Number of
Orders per week
Highest observation of cost driver (Week 8)
460
Lowest observation of cost driver (Week 3)
285
Difference
175
Weekly
Total Costs
$28,315
24,700
$ 3,615
Weekly total costs = a + b (number of orders per week)
Slope coefficient (b)
Constant (a)
Weekly total costs
= $3,615/175=$20.66 per order
= $28,315 – ($20.66  460) = $18,811.14
= $24,700 – ($20.66  285) = $18,811.14
= $18,811.14 + $20.66 × (Number of Orders per week)
See high-low line in Solution Exhibit 10-36.
3.
Solution Exhibit 10-36 presents the regression line:
Weekly total costs
= $18,791 + $19.97 × (Number of Orders per week)
Economic Plausibility. The cost function shows a positive economically plausible relationship
between number of orders per week and weekly total costs. Number of orders is a plausible cost
driver of total weekly costs.
10-34
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Goodness of fit. The regression line appears to fit the data well. The vertical differences
between the actual costs and the regression line appear to be quite small.
Significance of independent variable. The regression line has a steep positive slope and
increases by $19.97 for each additional order. Because the slope is not flat, there is a strong
relationship between number of orders and total weekly costs.
The regression line is the more accurate estimate of the relationship between number of orders
and total weekly costs because it uses all available data points while the high-low method relies
only on two data points and may therefore miss some information contained in the other data
points. In addition, because both the low and high data points lie off the regression line, the
high-low method predicts a higher amount of fixed costs as well as a steeper slope (higher
amount of variable cost per order).
4.
Profit =
Total weekly revenues + Total seasonal membership fees – Total weekly costs
(Total number of orders × $40) + (500 × $100)
– $265,385 =
(3,880 × $40) + (500 × $100)
– $265,385 =
$155,200 + $50,000
– $265,385 = ($60,185).
No, the club did not make a profit.
5.
Let the average number of weekly orders be denoted by AWO. We want to find the value
of AWO for which Market Thyme will achieve zero profit.
Using the format in requirement 4, we want:
Profit = [AWO × 10 weeks × $40] + (500 × $100) – [$18,791 + ($19.97 × AWO)] × 10 weeks
$0 =
($400 AWO)
–
+ $50,000
$187,910 – $199.70 AWO
$200.30 AWO = $137,910
AWO = $137,910 ÷ $200.30 = 688.52
So, Market Thyme will have to get at least orders on average each week in order to break even
next year.
10-37 High-low method; regression analysis. (CIMA, adapted) Catherine McCarthy, sales
manager of Baxter Arenas, is checking to see if there is any relationship between promotional
costs and ticket revenues at the sports stadium. She obtains the following data for the past 9
months:
Month
Ticket Revenues
Promotional Costs
April
$200,000
$52,000
May
270,000
65,000
10-35
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June
320,000
80,000
July
480,000
90,000
August
430,000
100,000
September
450,000
110,000
October
540,000
120,000
November
670,000
180,000
December
751,000
197,000
She estimates the following regression equation:
Ticket revenues = $65,583 + ($3.54  Promotional costs)
Required:
1. Plot the relationship between promotional costs and ticket revenues. Also draw the regression
line and evaluate it using the criteria of economic plausibility, goodness of fit, and slope of
the regression line.
2. Use the high-low method to compute the function relating promotional costs and revenues.
3. Using (a) the regression equation and (b) the high-low equation, what is the increase in
revenues for each $10,000 spent on promotional costs within the relevant range? Which
method should Catherine use to predict the effect of promotional costs on ticket revenues?
Explain briefly.
SOLUTION
(3040 min.)
1.
High-low method, regression analysis.
Solution Exhibit 10-37 presents the plots of promotional costs on revenues.
SOLUTION EXHIBIT 10-37
Plot and Regression Line of Promotional Costs on Ticket Revenues
10-36
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840,000
y = 3.542x + 65583
R² = 0.9348
Ticket Revenues
740,000
640,000
540,000
440,000
340,000
240,000
140,000
40,000
50,000 70,000 90,000 110,000 130,000 150,000 170,000 190,000
Promotional Costs
Solution Exhibit 10-37 also shows the regression line of advertising costs on revenues.
We evaluate the estimated regression equation using the criteria of economic plausibility,
goodness of fit, and slope of the regression line.
Economic plausibility. Promotional costs appear to be a plausible cost driver of ticket revenues.
Sports teams and sporting events frequently use promotions to entice numbers of people to
attend.
Goodness of fit. The vertical differences between actual and predicted revenues appears to be
reasonably small. This indicates that promotional costs are related to ticket revenues.
Slope of regression line. The slope of the regression line appears to be relatively steep. Given the
small scatter of the observations around the line, the steep slope indicates that, on average, ticket
revenues increase with spending on promotions.
2.
The high-low method would estimate the cost function as follows:
Promotional
Costs
$ 52,000
197,000
$145,000
Highest observation of revenue driver
Lowest observation of revenue driver
Difference
Revenues
Slope coefficient (b)
= a + (b  promotional costs)
=
$551,000
= 3.80
$145,000
10-37
Ticket
Revenues
$200,000
751,000
$551,000
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Constant (a)
= $200,000  ($52,000  3.80)
= $200,000  $197,600 = $2,400
or
Constant (a)
= $751,000  ($197,000  3.80)
= $751,000  $748,600 = $2,400
Revenues
3.
= $2,400 + (3.80  Promotional costs)
The increase in revenues for each $1,000 spent on advertising within the relevant range is
a. Using the regression equation, 3.542  $10,000 = $35,420
b. Using the high-low equation, 3.80  $10,000 = $38,000
The high-low equation does moderately well in estimating the relationship between
promotional costs and revenues; it overestimates the impact of marginal changes in promotional
expenses on ticket revenues. McCarthy should use the regression equation because it uses
information from all observations. The high-low method, on the other hand, relies only on the
observations that have the highest and lowest values of the cost driver and these observations are
generally not representative of all the data.
10-38 Regression, activity-based costing, choosing cost drivers. Sleep Late, a large hotel
chain, has been using activity-based costing to determine the cost of a night’s stay at their hotels.
One of the activities, ―Inspection,‖ occurs after a customer has checked out of a hotel room.
Sleep Late inspects every 10th room and has been using ―number of rooms inspected‖ as the cost
driver for inspection costs. A significant component of inspection costs is the cost of the supplies
used in each inspection.
Mary Adams, the chief inspector, is wondering whether inspection labor-hours might be a
better cost driver for inspection costs. Mary gathers information for weekly inspection costs,
rooms inspected, and inspection labor-hours as follows:
Week
Rooms Inspected
Inspection Labor-Hours
1
254
66
$1,740
2
322
110
2,500
3
335
82
2,250
4
431
123
2,800
5
198
48
1,400
6
239
62
1,690
7
252
108
1,720
8
325
127
2,200
10-38
Inspection Costs
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Mary runs regressions on each of the possible cost drivers and estimates these cost functions:
Inspection Costs = $193.19 + ($6.26  Number of rooms inspected)
Inspection Costs = $944.66 + ($12.04  Inspection labor-hours)
Required:
1. Explain why rooms inspected and inspection labor-hours are plausible cost drivers of
inspection costs.
2. Plot the data and regression line for rooms inspected and inspection costs. Plot the data and
regression line for inspection labor-hours and inspection costs. Which cost driver of
inspection costs would you choose? Explain.
3. Mary expects inspectors to inspect 300 rooms and work for 105 hours next week. Using the
cost driver you chose in requirement 2, what amount of inspection costs should Mary
budget? Explain any implications of Mary choosing the cost driver you did not choose in
requirement 2 to budget inspection costs.
SOLUTION
(30 min.) Regression, activity-based costing, choosing cost drivers.
1.
Both number of rooms inspected and inspection labor-hours are plausible cost drivers for
inspection costs. The number of rooms inspected is likely related to the materials and supplies
that are used in each room, which is a significant component of inspection costs. Inspection
labor-hours are a plausible cost driver if labor hours vary per room inspected, because costs
would be a function of how much time the inspectors spend on each unit. This is particularly
true if the inspectors are paid a wage, and if they spend more time inspecting certain types of
rooms (e.g., suites) if they are larger or need to be more closely inspected because of the higher
prices being charged for them and the corresponding expectations of the guests.
2.
Solution Exhibit 10-38 presents (a) the plots and regression line for number of rooms
inspected versus inspection costs and (b) the plots and regression line for inspection labor-hours
and inspection costs.
SOLUTION EXHIBIT 10-38A
Plot and Regression Line for Rooms Inspected versus Inspection Costs for Sleep Late
10-39
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Sleep Late
$3,500
y = 6.2625x + 193.19
R² = 0.9362
Inspection costs
$3,000
$2,500
$2,000
$1,500
$1,000
$500
$0
200
250
300
350
Rooms inspected
400
450
SOLUTION EXHIBIT 10-38B
Plot and Regression Line for Inspection Labor-Hours and Inspection Costs for Sleep Late
Sleep Late
$3,000
y = 12.042x + 944.66
R² = 0.583
Inspection costs
$2,500
$2,000
$1,500
$1,000
$500
$0
40
60
80
100
Inspection labor-hours
120
Goodness of Fit. As you can see from the two graphs, the regression line based on number of
rooms inspected better fits the data (has smaller vertical distances from the points to the line)
than the regression line based on inspection labor-hours. The activity of inspection appears to be
more closely linearly related to the number of rooms inspected than to inspection labor-hours.
Hence number of rooms inspected is a better cost driver. This is probably because the number of
rooms inspected is closely related to the supplies used for inspection, and the variation due to
more or less time spent inspecting does not affect inspection costs (e.g., the inspectors may be
paid a salary and so higher labor-hours do not translate to higher costs).
10-40
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Significance of independent variable. It is hard to visually compare the slopes because the
graphs are not the same size, but both graphs have steep positive slopes indicating a strong
relationship between number of rooms inspected and inspection costs, and inspection labor-hours
and inspection costs. Indeed, if labor-hours per inspection do not vary much, number of rooms
inspected and inspection labor-hours will be closely related. Overall, it is the significant cost of
supplies that is driven by the number of rooms inspected (not the inspection labor-hours spent on
inspection) that makes rooms inspected the preferred cost driver.
3.
At 105 inspection labor hours and 300 rooms inspected:
Inspection costs using rooms inspected = $193.19 + ($6.26 × 300) = $2,071.19
Inspection costs using inspection labor-hours = $944.86 + ($12.04 × 105) = $2,209.06
If Mary uses inspection-labor-hours she will estimate inspection costs to be $2,209.06, $137.87
($2,209.06 ─ $2,071.19) higher than if she had used number of rooms inspected. If actual costs
equaled, say, $2,150, Mary would conclude that Sleep Late has performed efficiently in its
inspection activity because actual inspection costs would be lower than budgeted amounts. In
fact, based on the more accurate cost function, actual costs of $2,150 exceeded the budgeted
amount of $2,071.19. Mary should find ways to improve inspection efficiency rather than
mistakenly conclude that the inspection activity has been performing well.
10-39 Interpreting regression results. Spirit Freightways is a leader in transporting agricultural
products in the western provinces of Canada. Reese Brown, a financial analyst at Spirit
Freightways, is studying the behavior of transportation costs for budgeting purposes. Transportation
costs at Spirit are of two types: (a) operating costs (such as labor and fuel) and (b) maintenance
costs (primarily overhaul of vehicles).
Brown gathers monthly data on each type of cost, as well as the total freight miles traveled
by Spirit vehicles in each month. The data collected are shown below (all in thousands):
Month
Operating Costs
Maintenance Costs
Freight Miles
January
$ 942
$ 974
1,710
February
1,008
776
2,655
March
1,218
686
2,705
April
1,380
694
4,220
May
1,484
588
4,660
June
1,548
422
4,455
July
1,568
352
4,435
August
1,972
420
4,990
September
1,190
564
2,990
October
1,302
788
2,610
962
762
2,240
November
10-41
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December
772
1,028
1,490
Required:
1. Conduct a regression using the monthly data of operating costs on freight miles. You should
obtain the following result:
Regression: Operating costs = a + (b  Number of freight miles)
Variable
Coefficient
Standard Error
t-Value
Constant
$445.76
$112.97
3.95
Independent variable: No. of freight miles
$
$
7.83
0.26
0.03
2
r = 0.86; Durbin-Watson statistic = 2.18
2. Plot the data and regression line for the above estimation. Evaluate the regression using the
criteria of economic plausibility, goodness of fit, and slope of the regression line.
3. Brown expects Spirit to generate, on average, 3,600 freight miles each month next year. How
much in operating costs should Brown budget for next year?
4. Name three variables, other than freight miles, that Brown might expect to be important cost
drivers for Spirit’s operating costs.
5. Brown next conducts a regression using the monthly data of maintenance costs on freight
miles. Verify that she obtained the following result:
Regression: Maintenance costs = a + (b  Number of freight miles)
Variable
Coefficient
Standard Error
t-Value
Constant
$1,170.57
$91.07
12.85
Independent variable: No. of freight miles
$
–0.15
$ 0.03
–5.83
2
r = 0.77; Durbin-Watson statistic = 1.94
6. Provide a reasoned explanation for the observed sign on the cost driver variable in the
maintenance cost regression. What alternative data or alternative regression specifications
would you like to use to better capture the above relationship?
SOLUTION
(15-20min.)
Interpreting regression results, matching time periods.
1.
Here is the regression data for monthly operating costs as a function of the total freight
miles travelled by Sprit vehicles:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.927299101
10-42
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R Square
Adjusted R
Square
Standard Error
Observations
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0.859883623
0.845871986
132.0816002
12
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
1
10
11
SS
1070620.18
174455.49
1245075.67
Coefficients
445.76
0.26
Standard
Error
112.97
0.03
MS
1070620.18
17445.55
F
61.37
Significance
F
0.00
t Stat
3.95
7.83
P-value
0.00
0.00
Lower 95%
194.04
0.18
Upper
95%
697.48
0.33
2. The chart below presents the data and the estimated regression line for the relationship
between monthly operating costs and freight miles traveled by Spirit Freightways.
Economic
plausibility
A positive relationship between freight miles traveled and monthly
operating costs is economically plausible since increased levels of
economic activity should lead to the consumption of greater amounts of
labor, fuel and other operating expenses.
Goodness of fit
r2 = 86%, Adjusted r2 = 85%
Standard error of regression = 132.08
Excellent fit; there is indisputable evidence of a linear relationship between
the dependent and independent variables. The distances between the
estimated line and the actual data points are small, other than at the highest
10-43
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level of activity recorded during the year.
Significance of
Independent
Variables
The t-value of 7.83 for freight miles traveled output units is significant at
the 0.05 and 0.01 levels.
3. If Brown expects Spirit to generate an average of 3,600 miles each month next year, the best
estimate of operating costs is given by:
Monthly operating costs = $445.76 + ($0.26) × (3,600 miles) = $1,381.76.
Annual operating costs = ($1,381.76) × 12 = $16,581.12.
4. Three variables, other than freight miles, that Brown might expect to be important cost drivers
for Spirit’s operating costs are: input prices (fuel prices and wage rates), mix of agricultural
output carried (weight, volume, value), and route mix and conditions (weather, flat versus
mountainous terrain, short-haul versus long-haul carriage).
5. Here is the regression data for monthly maintenance costs as a function of the total freight
miles travelled by Sprit vehicles:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.87887319
R Square
0.77241808
Adjusted R
Square
0.74965989
Standard Error
106.470794
Observations
12
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
1
10
11
SS
384747.37
113360.30
498107.67
Coefficients
1170.57
-0.15
Standard
Error
91.07
0.03
MS
384747.37
11336.03
F
33.94
Significance
F
0.00
t Stat
12.85
-5.83
P-value
0.00
0.00
Lower 95%
967.66
-0.21
The data and regression estimate are provided in the chart below:
10-44
Upper
95%
1373.48
-0.09
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6. At first glance, the regression result in requirement 5 is surprising and economicallyimplausible. In the regression, the coefficient on freight miles traveled has a negative sign. This
implies that the greater the number of freight miles (i.e., the more activity Spirit carries out), the
smaller are the maintenance costs; specifically, it suggests that each extra freight mile reduces
maintenance costs by $0.14 (recall that all data are in thousands). Clearly, this estimated
relationship is not economically credible. However, one would think that freight miles should
have some impact on fleet maintenance costs.
The logic behind the estimated regression becomes clearer once one realizes that maintenance
costs have a discretionary component to them, especially in terms of timing. Spirit’s peak months
of work transporting agricultural products in western Canada occur in late spring and summer
(the period from April through August). It is likely that Spirit is simply choosing to defer
maintenance to those months when its vehicles are not in use, thereby creating a negative
relationship between monthly activity and maintenance costs. The causality also goes the other
way – if vehicles are in the shop for maintenance, they are clearly not on the road generating
freight miles. A third reason is that vehicles might need to be serviced at greater frequency
during the winter months because of the wear and tear that comes from driving on icy terrain and
in poor weather conditions.
Possible alternative specifications that would better capture the link between Spirit’s activity
levels and the spending on maintenance are to estimate the relationship using annual data over a
period of several years, to look at spending on corrective rather than preventive maintenance, or
to look at the relation using lags (i.e., freight miles traveled in a period against the spending on
maintenance done in a subsequent period in order to service the vehicles).
10-40 Cost estimation, cumulative average-time learning curve. The Pacific Boat Company,
which is under contract to the U.S. Navy, assembles troop deployment boats. As part of its
research program, it completes the assembly of the first of a new model (PT109) of deployment
boats. The Navy is impressed with the PT109. It requests that Pacific Boat submit a proposal on
the cost of producing another six PT109s.
10-45
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alaa.aliasrei@gmail.com
Pacific Boat reports the following cost information for the first PT109 assembled and uses a
90% cumulative average-time learning model as a basis for forecasting direct manufacturing
labor-hours for the next six PT109s. (A 90% learning curve means b = –0.152004.)
Required:
1. Calculate predicted total costs of producing the six PT109s for the Navy. (Pacific Boat will
keep the first deployment boat assembled, costed at $1,477,600, as a demonstration model
for potential customers.)
2. What is the dollar amount of the difference between (a) the predicted total costs for
producing the six PT109s in requirement 1 and (b) the predicted total costs for producing the
six PT109s, assuming that there is no learning curve for direct manufacturing labor? That is,
for (b) assume a linear function for units produced and direct manufacturing labor-hours.
SOLUTION
(30–40 min.) Cost estimation, cumulative average-time learning curve.
1.
Cost to produce the 2nd through the 7th troop deployment boats:
Direct materials, 6  $199,000
Direct manufacturing labor (DML), 61,8521  $42
Variable manufacturing overhead, 61,852  $26
Other manufacturing overhead, 20% of DML costs
Total costs
1
$1,194,000
2,597,784
1,608,152
519,557
$5,919,493
The direct manufacturing labor-hours to produce the second to seventh boats can be calculated in several
ways, given the assumption of a cumulative average-time learning curve of 90%:
10-46
‫عالء محسن شحم‬
alaa.aliasrei@gmail.com
Use of table format:
Cumulative
Number of Units (X)
(1)
1
90% Learning Curve
Cumulative
Average Time per Unit (y): Labor Hours
(2)
14,700
2
13,230
3
4
5
6
7
12,439
11,907
11,509
11,195
10,936
= (14,700  0.90)
= (13,230  0.90)
Cumulative
Total Time:
Labor-Hours
(3) = (1)  (2)
14,700
26,460
37,317
47,628
57,545
67,170
76,552
The direct labor-hours required to produce the second through the seventh boats is 76,552 – 14,700 =
61,852 hours.
Use of formula: y = aXb
where
a = 14,700, X = 7, and b = – 0.152004
y = 14,700  7– 0.152004 = 10,936 hours
The total direct labor-hours for 7 units is 10,936  7 = 76,552 hours
Note: Some students will debate the exclusion of the $279,000 tooling cost. The question
specifies that the tooling ―cost was assigned to the first boat.‖ Although Pacific Boat may well
seek to ensure its total revenue covers the $1,477,600 cost of the first boat, the concern in this
question is only with the cost of producing six more PT109s.
2.
Cost to produce the 2nd through the 7th boats assuming linear function for direct laborhours and units produced:
Direct materials, 6  $199,000
$1,194,000
Direct manufacturing labor (DML), 6  14,700 hrs.  $42
3,704,400
Variable manufacturing overhead, 6  14,700 hrs.  $26
2,293,200
Other manufacturing overhead, 20% of DML costs
740,880
Total costs
$7,932,480
The difference in predicted costs is:
Predicted cost in requirement 2
(based on linear cost function)
Predicted cost in requirement 1
(based on 90% learning curve)
Difference in favor of learning curve cost function
$7,932,480
5,919,493
$2,012,987
Note that the linear cost function assumption leads to a total cost that is almost 35% higher than
the cost predicted by the learning curve model. Learning curve effects are most prevalent in large
manufacturing industries such as airplanes and boats where costs can run into the millions or
hundreds of millions of dollars, resulting in very large and monetarily significant differences
10-47
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between the two models. In the case of Pacific Boat, if it is in fact easier to produce additional
boats as the firm gains experience, the learning curve model is the right one to use. The firm can
better forecast its future costs and use that information to submit an appropriate cost bid to the
Navy, as well as refine its pricing plans for other potential customers.
10-41 Cost estimation, incremental unit-time learning model. Assume the same information
for the Pacific Boat Company as in Problem 10-40 with one exception. This exception is that
Pacific Boat uses a 90% incremental unit-time learning model as a basis for predicting direct
manufacturing labor-hours in its assembling operations. (A 90% learning curve means
b = –0.152004.)
Required:
1. Prepare a prediction of the total costs for producing the six PT109s for the Navy.
2. If you solved requirement 1 of Problem 10-40, compare your cost prediction there with the
one you made here. Why are the predictions different? How should Pacific Boat decide
which model it should use?
SOLUTION
(20–30 min.)
1.
Cost estimation, incremental unit-time learning model.
Cost to produce the 2nd through the 7th boats:
Direct materials, 6  $199,000
Direct manufacturing labor (DML), 71,2171  $42
Variable manufacturing overhead, 71,217  $26
Other manufacturing overhead, 20% of DML costs
Total costs
$1,194,000
2,991,114
1,851,642
598,223
$6,634,979
1The
direct labor hours to produce the second through the seventh boats can be calculated via a table
format, given the assumption of an incremental unit-time learning curve of 90%:
90% Learning Curve
Cumulative
Cumulative
Number of
Individual Unit Time for Xth
Total Time:
Units (X)
Unit (y)*: Labor Hours
Labor-Hours
(1)
(2)
(3)
1
14,700
2
3
4
5
6
7
13,230
12,439
11,907
11,510
11,195
10,936
= (14,700  0.90)
= (13,230  0.90)
14,700
27,930
40,369
52,276
63,786
74,981
85,917
*Calculated as y = aXb where a = 14,700, b = – 0.152004, and X = 1, 2, 3,. . .7.
10-48
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The direct manufacturing labor-hours to produce the second through the seventh boat is 85,917 –
14,700 = 71,217 hours.
2.
Difference in total costs to manufacture the second through the seventh boat under the
incremental unit-time learning model and the cumulative average-time learning model is
$6,634,979 (calculated in requirement 1 of this problem) – $5,919,493 (from requirement 1 of
Problem 10-41) = $715,486, i.e., the total costs are higher for the incremental unit-time model.
The incremental unit-time learning curve has a slower rate of decline in the time required
to produce successive units than does the cumulative average-time learning curve (see Problem
10-41, requirement 1). Assuming the same 90% factor is used for both curves:
Cumulative
Number of Units
1
2
4
7
Estimated Cumulative Direct Manufacturing Labor-Hours
Cumulative AverageIncremental Unit-Time
Time Learning Model
Learning Model
14,700
14,700
26,460
27,930
47,628
52,276
76,552
85,917
The reason is that, in the incremental unit-time learning model, as the number of units
double, only the last unit produced has a cost of 90% of the initial cost. In the cumulative
average-time learning model, doubling the number of units causes the average cost of all the
units produced (not just the last unit) to be 90% of the initial cost.
Pacific Boat should examine its own internal records on past jobs and seek information from
engineers, plant managers, and workers when deciding which learning curve better describes the
behavior of direct manufacturing labor-hours on the production of the PT109 boats.
10-42 Regression; choosing among models. Apollo Hospital specializes in outpatient surgeries
for relatively minor procedures. Apollo is a nonprofit institution and places great emphasis on
controlling costs in order to provide services to the community in an efficient manner.
Apollo’s CFO, Julie Chen, has been concerned of late about the hospital’s consumption of
medical supplies. To better understand the behavior of this cost, Julie consults with Rhett Bratt,
the person responsible for Apollo’s cost system. After some discussion, Julie and Rhett conclude
that there are two potential cost drivers for the hospital’s medical supplies costs. The first driver
is the total number of procedures performed. The second is the number of patient-hours
generated by Apollo. Julie and Rhett view the latter as a potentially better cost driver because the
hospital does perform a variety of procedures, some more complex than others.
Rhett provides the following data relating to the past year to Julie.
10-49
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Required:
1. Estimate the regression equation for (a) medical supplies costs and number of procedures and
(b) medical supplies costs and number of patient-hours. You should obtain the following
results:
Regression 1: Medical supplies costs = a + (b  Number of procedures)
Variable
Coefficient
Standard
Error
t-Value
Constant
$36,939.77
$56,404.86
0.65
Independent variable: No. of procedures
$
$
2.37
361.91
152.93
2
r = 0.36; Durbin-Watson statistic = 2.48
Regression 2: Medical supplies costs = a + (b  Number of patient-hours)
Variable
Coefficient
Standard Error
t-Value
Constant
$3,654.86
$23,569.51
0.16
Independent variable: No. of patient-hours
$
$
7.25
56.76
7.82
r2 = 0.84; Durbin-Watson statistic = 1.91
2. On different graphs plot the data and the regression lines for each of the following cost
functions:
a. Medical supplies costs = a + (b  Number of procedures)
b. Medical supplies costs = a + (b  Number of patient-hours)
3. Evaluate the regression models for ―Number of procedures‖ and ―Number of patient-hours‖
as the cost driver according to the format of Exhibit 10-18 (page 406).
10-50
‫عالء محسن شحم‬
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4. Based on your analysis, which cost driver should Julie Chen adopt for Apollo Hospital?
Explain your answer.
SOLUTION
(30 min.)
Regression; choosing among models.
1. See Solution Exhibit 10-42A below.
SOLUTION EXHIBIT 10-42A
(a) Regression Output for Medical Supplies Costs and Number of Procedures
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.599152481
R Square
0.358983696
Adjusted R
Square
0.294882065
Standard Error
52998.71699
Observations
12
ANOVA
df
1
10
11
SS
15730276644
28088640022
43818916667
Coefficients
36939.77
361.91
Standard
Error
56404.86
152.93
Regression
Residual
Total
Intercept
X Variable 1
MS
1.57E+10
2.81E+09
t Stat
0.65
2.37
5.60
Significance
F
0.04
P-value
0.53
0.04
Lower 95%
-88738.09
21.16
F
(b) Regression Output for Medical Supplies Costs and Number of Patient-Hours
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.91669199
R Square
0.84032421
Adjusted R
Square
0.82435663
Standard Error
26451.5032
Observations
12
ANOVA
10-51
Upper
95%
162617.63
702.66
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df
1
10
11
SS
36822096457
6996820210
43818916667
Coefficients
3654.86
56.76
Standard
Error
23569.51
7.82
Regression
Residual
Total
Intercept
X Variable 1
MS
3.68E+10
7E+08
F
52.63
Significance
F
0.00
t Stat
0.16
7.25
P-value
0.88
0.00
Lower 95%
-48861.29
39.33
Upper
95%
56171.00
74.19
2. See Solution Exhibit 10-42B below.
SOLUTION EXHIBIT 10-42B
Plots and Regression Lines for (a) Medical Supplies Costs and Number of Procedures and (b)
Medical Supplies Costs and Number of Patient-Hours
(a)
Apollo Hospitals
Medical supplies costs
250,000
200,000
y = 361.91x + 36940
R² = 0.359
150,000
100,000
50,000
100
200
300
400
Number of procedures
10-52
500
600
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(b)
Apollo Hospitals
Medical supplies costs
250,000
200,000
y = 56.759x + 3654.9
R² = 0.8403
150,000
100,000
50,000
1,000
1,500
2,000
2,500
3,000
3,500
Number of patient-hours
4,000
4,500
3.
Economic
plausibility
Number of Setups
A positive relationship
between medical supplies costs
and the number of procedures
is economically plausible.
Number of Setup Hours
A positive relationship between
medical supplies costs and the number
of patient-hours is also economically
plausible, especially since the time
taken to serve patients is not uniform.
Patient-hours is more likely to capture
the true level of activity in the hospital
since it accounts for the mix of
procedures performed.
Goodness of fit
r2 = 36%
r2 = 84%
Standard error of regression = $52,999 Standard error of regression = $26,452
Reasonable goodness of fit.
Excellent goodness of fit.
Significance of
Independent
Variables
The t-value of 2.37 is significant at the The t-value of 7.25 is highly
0.05 level. It is not significant at the
significant at the 0.05 and 0.01 levels.
0.01 level.
Specification
analysis of
estimation
assumptions
Based on a plot of the data, the
linearity assumption holds, but there is
some possibility that the constant
variance assumption does not hold.
The Durbin-Watson statistic of 2.48
10-53
Based on a plot of the data, the
assumptions of linearity, constant
variance, independence of residuals
(Durbin-Watson = 1.91), and
normality of residuals hold. However,
‫عالء محسن شحم‬
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suggests the residuals are independent. inferences drawn from only 12
The normality of residuals assumption observations are not reliable.
appears to hold. However, inferences
drawn from only 12 observations are
not reliable.
4. The regression model using number of patient-hours should be used to estimate medical
supplies costs because the number of patient-hours is a more economically plausible cost driver
of medical supplies costs (compared to the number of procedures performed). The time taken to
prepare medical facilities and to actually deal with patient issues (surgery, post-procedure care,
etc.) is different for different procedures. The more complex the procedure, the more time is
taken with the patient to analyze and manage the problem, and the greater the supplies costs
incurred. As such, patient-hours might serve as a better driver of medical supplies costs. The
regression of number of patient-hours and medical supplies costs also has a better fit, a
substantially significant independent variable, and better satisfies the assumptions of the
estimation technique used.
10-43 Multiple regression (continuation of 10-42). After further discussion, Julie and Rhett
wonder if they should view both the number of procedures and number of patient-hours as cost
drivers in a multiple regression estimation in order to best understand Apollo’s medical supplies
costs.
Required:
1. Conduct a multiple regression to estimate the regression equation for medical supplies costs
using both number of procedures and number of patient-hours as independent variables. You
should obtain the following result:
Regression 3: Medical supplies costs = a + (b1  No. of procedures) + (b2  No. of patienthours)
Variable
Constant
Coefficient
Standard Error
t-Value
–$3,103.76
$30,406.54
–0.10
Independent variable 1: No. of procedures
$
38.24
$
100.76
0.38
Independent variable 2: No. of patient-hours
$
54.37
$
10.33
5.26
2
r = 0.84; Durbin-Watson statistic = 1.96
2. Evaluate the multiple regression output using the criteria of economic plausibility goodness
of fit, significance of independent variables, and specification of estimation assumptions.
3. What potential issues could arise in multiple regression analysis that are not present in simple
regression models? Is there evidence of such difficulties in the multiple regression presented
in this problem? Explain.
4. Which of the regression models from Problems 10-42 and 10-43 would you recommend Julie
Chen use? Explain.
10-54
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SOLUTION
(30 min.) Multiple regression (continuation of 10-42).
1. Solution Exhibit 10-43 presents the regression output for medical supplies costs using both
number of procedures and number of patient-hours as independent variables (cost drivers).
SOLUTION EXHIBIT 10-43
Regression Output for Multiple Regression for Medical Supplies Costs Using Both Number of
Procedures and Number of Patient-Hours as Independent Variables (Cost Drivers)
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.91806327
R Square
0.84284017
Adjusted R Square 0.80791577
Standard Error
27661.7936
Observations
12
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
X Variable 2
2.
Economic
plausibility
2
9
11
SS
36932343254
6886573413
43818916667
Coefficients
-3103.76
38.24
54.37
Standard
Error
30406.54
100.76
10.33
MS
1.85E+10
7.65E+08
F
24.13
Significance
F
0.00
t Stat
-0.10
0.38
5.26
P-value
0.92
0.71
0.00
Lower 95%
-71888.13
-189.68
31.00
Upper
95%
65680.61
266.17
77.73
A positive relationship between medical supplies costs and each of the
independent variables (number of procedures and number of patient-hours)
is economically plausible.
Goodness of fit
r2 = 84%, Adjusted r2 = 81%
Standard error of regression =$27,662
Excellent goodness of fit.
Significance of
Independent
Variables
The t-value of 0.38 for number of procedures is not significant at the 0.05
level. The t-value of 5.26 for number of patient-hours is significant at the
0.05 and 0.01 levels.
10-55
‫عالء محسن شحم‬
Specification
analysis of
estimation
assumptions
alaa.aliasrei@gmail.com
Assuming linearity, constant variance, and normality of residuals, the
Durbin-Watson statistic of 1.96 suggests the residuals are independent.
However, we must be cautious when drawing inferences from only 12
observations.
3. Multicollinearity is an issue that can arise with multiple regression but not simple regression
analysis. Multicollinearity means that the independent variables are highly correlated.
The correlation feature in Excel’s Data Analysis reveals a coefficient of correlation of
0.61 between number of procedures and number of patient-hours. This is close to the threshold
of 0.70 that is usually taken as a sign of multicollinearity problems. As evidence, note the
substantial drop in the t-value for patient-hours from 7.25 to 5.26, despite a fairly small change
in the estimated coefficient (from $56.76 to $54.37).
4. The simple regression model using the number of patient-hours as the independent variable
achieves a comparable r2 to the multiple regression model. However, the multiple regression
model includes an insignificant independent variable, number of procedures. Adding this
variable does not improve Apollo Hospital’s ability to better estimate medical supplies costs and
it also introduces multicollinearity issues. Julie should use the simple regression model with
number of patient-hours as the independent variable to estimate medical supplies costs.
10-44 Cost estimation. Hankuk Electronics started production on a sophisticated new
smartphone running the Android operating system in January 2017. Given the razor-thin margins
in the consumer electronics industry, Hankuk’s success depends heavily on being able to produce
the phone as economically as possible.
At the end of the first year of production, Hankuk’s controller, Inbee Kim, gathered data on
its monthly levels of output, as well as monthly consumption of direct labor-hours (DLH). Inbee
views labor-hours as the key driver of Hankuk’s direct and overhead costs. The information
collected by Inbee is provided below:
10-56
‫عالء محسن شحم‬
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Required:
1. Inbee is keen to examine the relationship between direct labor consumption and output
levels. She decides to estimate this relationship using a simple linear regression based on the
monthly data. Verify that the following is the result obtained by Inbee:
Regression 1: Direct labor-hours = a + (b  Output units)
Variable
Coefficient
Standard Error
t-Value
345.24
589.07
0.59
0.71
0.93
0.76
Constant
Independent variable: Output units
r2 = 0.054; Durbin-Watson statistic = 0.50
2. Plot the data and regression line for the above estimation. Evaluate the regression using the
criteria of economic plausibility, goodness of fit, and slope of the regression line.
3. Inbee estimates that Hankuk has a variable cost of $17.50 per direct labor-hour. She expects
that Hankuk will produce 650 units in the next month, January 2018. What should she budget
as the expected variable cost? How confident is she of her estimate?
SOLUTION
(30 min.) Cost estimation.
1. Here is the summary output for the monthly regression of Direct Labor Hours on Output
Units for Hankuk Electronics:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.2333602
R Square
0.054457
Adjusted R
Square
-0.0400973
Standard Error
206.18345
Observations
12
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
1
10
11
SS
24483.86
425116.1
449600
Coefficients
345.24
0.71
Standard
Error
589.07
0.93
MS
24483.86
42511.61
F
0.575933
Significance
F
0.465422344
t Stat
0.59
0.76
P-value
0.57
0.47
Lower 95%
-967.29
-1.37
10-57
Upper
95%
1657.77
2.79
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2. The plot and regression line for monthly direct labor hours on monthly output for Hankuk
Electronics are given below:
Hankuk Electronics
Direct Labor Hours
1,600
1,400
y = 0.7091x + 345.24
R² = 0.0545
1,200
1,000
800
600
400
450
500
550
600
650
Output (Units)
700
750
Economic
plausibility
A positive relationship between direct labor hours and monthly output is
economically plausible since increased levels of production should lead to
the consumption of greater amounts of direct labor.
Goodness of fit
r2 = 5.45%, Adjusted r2 = - 4%
Standard error of regression = 206.18
Terrible fit; in fact, there is no evidence of a linear relationship between the
dependent and independent variables. At least one data point represents a
significant outlier.
Significance of
Independent
Variables
The t-value of 0.76 for output units is not significant at the 0.05 level.
3. Given Inbee’s expectation that Hankuk will produce 650 units in January 2018, her best
estimate given the linear regression above is that Hankuk will use:
345.24 + (0.71 × 650 units) = 806.74 direct labor hours.
At an estimated variable cost of $17.50 per direct labor-hour, this implies that Inbee should
budget
806.74 × $17.50 = $14,118
for direct labor costs for January 2018.
10-58
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Note that 650 units is in the range of output values that were used to find the regression equation,
and therefore falls in the range of predictability for this model. However, there is substantial
uncertainty around the cost estimate of $14,118. In particular, this predicted value relies on the
regression point estimate of 0.71 for the marginal impact of output on labor hours. But, the 95%
confidence interval for the slope of the regression ranges all the way from -1.37 to 2.79, and the
predicted cost would vary accordingly. One cannot reject the null hypothesis that output levels
have no impact on labor consumption, leaving the budgeted cost estimate a highly speculative
one!
10-45 Cost estimation, learning curves (continuation of 10-44). Inbee is concerned that she
still does not understand the relationship between output and labor consumption. She consults
with Jim Park, the head of engineering, and shares the results of her regression estimation. Jim
indicates that the production of new smartphone models exhibits significant learning effects—as
Hankuk gains experience with production, it can produce additional units using less time. He
suggests that it is more appropriate to specify the following relationship:
y = axb
where x is cumulative production in units, y is the cumulative average direct labor-hours per unit
(i.e., cumulative DLH divided by cumulative production), and a and b are parameters of the
learning effect.
To estimate this, Inbee and Jim use the original data to calculate the cumulative output and
cumulative average labor-hours per unit for each month. They then take natural logarithms of
these variables in order to be able to estimate a regression equation. Here is the transformed data:
10-59
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Required:
1. Estimate the relationship between the cumulative average direct labor-hours per unit and
cumulative output (both in logarithms). Verify that the following is the result obtained by
Inbee and Jim:
Regression 1: Ln (Cumulative avg DLH per unit) = a + [b  Ln (Cumulative Output)]
Variable
Coefficient
Standard Error
t-Value
2.087
0.024
85.44
–0.208
0.003
–69.046
Constant
Independent variable: Ln (Cum Output)
2
r = 0.998; Durbin-Watson statistic = 2.66
2. Plot the data and regression line for the above estimation. Evaluate the regression using the
criteria of economic plausibility, goodness of fit, and slope of the regression line.
3. Verify that the estimated slope coefficient corresponds to an 86.6% cumulative average-time
learning curve.
4. Based on this new estimation, how will Inbee revise her budget for Hankuk’s variable cost
for the expected output of 650 units in January 2018? How confident is she of this new cost
estimate?
SOLUTION
(30 min.) Cost estimation, learning curves (continuation of 10-44).
1. Here is the summary output for the monthly regression of the natural log of Cumulative
Average Direct Labor-Hours per Unit on the natural logarithm of Cumulative Output:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.9989528
R Square
0.9979068
Adjusted R
Square
0.9976975
Standard Error
0.0074326
Observations
12
ANOVA
df
Regression
Residual
Total
1
10
11
SS
MS
F
0.263368
0.000552
0.26392
0.263368
5.52E-05
4767.34
10-60
Significance
F
9.89803E15
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Coefficients
2.09
-0.21
Intercept
X Variable 1
Standard
Error
0.02
0.00
t Stat
85.44
-69.05
P-value
0.00
0.00
Lower 95%
2.03
-0.21
Upper
95%
2.14
-0.20
2. The plot of the data and the regression line estimated above are provided next.
Log of Cumulative Average DLH per unit
Hankuk Electronics
0.800
y = -0.2079x + 2.0876
R² = 0.9979
0.700
0.600
0.500
0.400
0.300
0.200
6.000
6.500
7.000
7.500
8.000
Log of Cumulative Output
8.500
9.000
Economic
plausibility
A negative relationship between cumulative average direct-labor hours per
unit and cumulative output (in natural logarithms) is economically
plausible and reflects the presence of learning effects. Specifically, as the
firm gains experience via production, it becomes more efficient and is able
to use fewer direct labor hours to make each unit of product.
Goodness of fit
r2 = 99.8%, Adjusted r2 = 99.8%
Standard error of regression = 0.007
Unparalleled goodness of fit. Virtually perfect linear fit in logarithms.
Significance of
Independent
Variables
The t-value of -69.05 for the logarithm of cumulative output is significant
at all conventional levels. The t-value for the intercept (85.44) is highly
significant as well.
3. The original learning curve specification, y = axb is mathematically identical to the following
log-linear specification:
Ln y = Ln a + b × Ln x
The regression equation we have estimated,
Ln (Cumulative avg DLH per unit) = a + (b × Ln (Cumulative Output))
10-61
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is precisely the above specification, and in particular the slope coefficient directly yields the ―b‖
from the learning curve equation. We know therefore that for Hankuk electronics, b = -0.208. As
explained in Exhibit 10-10, this value is related to the learning curve percentage as follows:
b = Ln (learning-curve % in decimal form)/Ln 2, or
-0.208 = Ln (learning-curve % in decimal form)/0.693, or
Ln (learning-curve % in decimal form) = -0.208 × 0.693 = -0.144.
As the exponent of -0.144 is 0.8659, this implies that Hankuk is experiencing an 86.6%
cumulative average-time learning curve.
4. With an additional 650 units in January 2018, Hankuk’s cumulative output will go from 7,527
at the end of December 2016 to 8,177 (7,527 + 650). As Ln (8,177) = 9.0091, the cumulative
average direct-labor hours in logarithmic terms are given by:
2.0876 – 0.2079 × 9.0091 = 0.2146.
The cumulative direct-labor hours per unit therefore equals Exp (0.2146) = 1.2394. This implies
a total direct labor hours of 1.2394 × 8,177 = 10,134 by the end of January. As Hankuk has
used a total of 9,480 direct labor hours at the end of December 2017, the incremental hours
needed in January therefore are 654 (10,134 – 9,480). At $17.50 per labor hour, this suggest that
Inbee should budget
654 × $17.50 = $11,445
for direct labor costs for January 2018.
While 9.0091 is outside the range of cumulative output values (measured in logarithms) used to
find the regression equation, unless there has been a structural break in the experience curve
Hankuk is facing, it is highly likely that its January costs will be in the neighborhood of $11,445.
The reason is that the estimated regression line is close to perfect and has a standard error close
to zero. There is virtually no uncertainty around the coefficient estimates. The slope coefficient,
for example, has a point estimate of -0.2079, and a narrow 95% confidence interval between 0.2146 and -0.2012. Using either of those estimates would make barely any difference to the
predicted cost for the month of January 2018.
10-46 Interpreting regression results, matching time periods. Nandita Summers works at
Modus, a store that caters to fashion for young adults. Nandita is responsible for the store’s
online advertising and promotion budget. For the past year, she has studied search engine
optimization and has been purchasing keywords and display advertising on Google, Facebook,
and Twitter. In order to analyze the effectiveness of her efforts and to decide whether to continue
online advertising or move her advertising dollars back to traditional print media, Nandita
collects the following data:
10-62
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Required:
1. Nandita performs a regression analysis, comparing each month’s online advertising expense
with that month’s revenue. Verify that she obtains the following result:
Revenue = $51,999.64 – (0.98  Online advertising expense)
Variable
Coefficient
Constant
Independent variable: Online advertising expense
Standard Error t-Value
$51,999.64
7,988.68
6.51
–0.98
1.99
–0.49
r2 = 0.02; Durbin-Watson statistic = 2.14
2. Plot the preceding data on a graph and draw the regression line. What does the cost formula
indicate about the relationship between monthly online advertising expense and monthly
revenues? Is the relationship economically plausible?
3. After further thought, Nandita realizes there may have been a flaw in her approach. In
particular, there may be a lag between the time customers click through to the Modus website
and peruse its social media content (which is when the online ad expense is incurred) and the
time they actually shop in the physical store. Nandita modifies her analysis by comparing
each month’s sales revenue to the advertising expense in the prior month. After discarding
September revenue and August advertising expense, show that the modified regression yields
the following:
Revenue = $28,361.37 + (5.38  Online advertising expense)
10-63
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Variable
Constant
Coefficient
Standard Error
t-Value
$28,361.37
5,428.69
5.22
5.38
1.31
4.12
Independent variable: Previous month’s online
advertising expense r2 = 0.65; Durbin-Watson
statistic = 1.71
4. What does the revised formula indicate? Plot the revised data on a graph. Is this relationship
economically plausible?
5. Can Nandita conclude that there is a cause-and-effect relationship between online advertising
expense and sales revenue? Why or why not?
SOLUTION
(25 min.)
Interpreting regression results, matching time periods
1.
Here is the summary output for the monthly regression of Sales Revenue on Online
Advertising Expense:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.15
R Square
0.02
Adjusted R
Square
-0.07
Standard Error
11837.30
Observations
12.00
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
1
10
11
SS
33972689.79
1401216525
1435189215
Coefficients
51999.64
-0.98
Standard
Error
7988.68
1.99
MS
33972690
1.4E+08
F
0.242451
Significance
F
0.633072
t Stat
6.51
-0.49
P-value
0.00
0.63
Lower 95%
34199.74
-5.41
Upper
95%
69799.54
3.45
2.
SOLUTION EXHIBIT 10-46A presents the data plot for the initial analysis. The formula
of Sales Revenue = $52,000 – (0.98 × Online advertising expense) indicates that there is a fixed
amount of revenue each month of $52,000, which is reduced by 0.98 times that month’s online
advertising expense. This relationship is not economically plausible, as advertising would not
10-64
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reduce revenue. The data points do not appear linear, and the r-square of 0.02 indicates a very
weak goodness of fit (in fact, almost no fit at all).
SOLUTION EXHIBIT 10-46 A
Plot and Regression Line for Sales Revenue and Online Advertising Expense
$70,000
$65,000
Sales Revenue
$60,000
$55,000
$50,000
$45,000
y = -0.9789x + 52000
R² = 0.0237
$40,000
$35,000
$30,000
$25,000
$20,000
$0
$1,000 $2,000 $3,000 $4,000 $5,000 $6,000 $7,000
Online Advertising Expense
3.
Here is the summary output for the regression of monthly Sales Revenue on the prior
month’s Online Advertising Expense:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.808588
R Square
0.653815
Adjusted R
Square
0.61535
Standard Error
7393.922
Observations
11
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
1
9
10
SS
9.29E+08
4.92E+08
1.42E+09
Coefficients
28361.37
5.381665
Standard
Error
5428.687
1.305336
MS
929262059
54670085
F
16.99763
Significance
F
0.002587
t Stat
5.2243522
4.1228186
P-value
0.000546
0.002587
Lower 95%
16080.83
2.428789
10-65
Upper
95%
40641.91
8.33454
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3.
SOLUTION EXHIBIT 10-46 B presents the data plot for the revised analysis. The
formula of Sales Revenue = $28,361 + (5.38 × Online Advertising Expense) indicates that there
is a fixed amount of revenue each month of $28,361, which increases by 5.38 times the prior
month’s advertising expense in the online channel. This relationship is economically plausible.
One would expect a positive correlation between advertising expense and (future) sales revenue.
The slope coefficient of 5.38 has a t stat of 4.12 indicating that it is statistically significant at the
5% level. In the revised analysis, there is improved linearity in the data points, and the r-square
of 0.65 indicates a much stronger goodness of fit.
SOLUTION EXHIBIT 10-46B
Plot and Regression Line for Sales Revenue and Previous Month Online Advertising
70,000
65,000
Sales Revenue
60,000
55,000
y = 5.3817x + 28361
R² = 0.6538
50,000
45,000
40,000
35,000
30,000
0
1,000 2,000 3,000 4,000 5,000 6,000
Online Advertising Expense (Prior Month)
7,000
4.
Nandita must be very careful about making conclusions regarding cause and effect. Even
a strong goodness of fit does not prove a cause and effect relationship. The independent and
dependent variables could both be caused by a third factor, or the correlation could be simply
coincidental. However, there is enough of a correlation in the revised analysis for Nandita to
make a meaningful presentation to the store’s owner.
10-47 Purchasing department cost drivers, activity-based costing, simple regression
analysis. Perfect Fit operates a chain of 10 retail department stores. Each department store makes
its own purchasing decisions. Carl Hart, assistant to the president of Perfect Fit, is interested in
better understanding the drivers of purchasing department costs. For many years, Perfect Fit has
allocated purchasing department costs to products on the basis of the dollar value of merchandise
purchased. A $100 item is allocated 10 times as many overhead costs associated with the
purchasing department as a $10 item.
Hart recently attended a seminar titled ―Cost Drivers in the Retail Industry.‖ In a presentation
at the seminar, Kaliko Fabrics, a leading competitor that has implemented activity-based costing,
reported number of purchase orders and number of suppliers to be the two most important cost
drivers of purchasing department costs. The dollar value of merchandise purchased in each
purchase order was not found to be a significant cost driver. Hart interviewed several members
10-66
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of the purchasing department at the Perfect Fit store in Miami. They believed that Kaliko
Fabrics’ conclusions also applied to their purchasing department.
Hart collects the following data for the most recent year for Perfect Fit’s 10 retail department
stores:
Hart decides to use simple regression analysis to examine whether one or more of three variables
(the last three columns in the table) are cost drivers of purchasing department costs. Summary
results for these regressions are as follows:
Regression 1: PDC = a + (b  MP$)
Variable
Constant
Coefficient
$1,041,421
Independent variable 1: MP$
Standard Error
$346,709
0.0031
t-Value
3.00
0.0038
0.83
r2 = 0.08; Durbin-Watson statistic = 2.41
Regression 2: PDC = a + (b  No. of POs)
Variable
Coefficient
Standard Error
t-Value
Constant
$722,538
$265,835
2.72
Independent variable 1: No. of POs
$
$
2.46
2
r = 0.43; Durbin-Watson statistic = 1.97
Regression 3: PDC = a + (b  No. of Ss)
10-67
159.48
64.84
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Variable
Coefficient
Standard Error
t-Value
Constant
$828,814
$246,571
3.36
Independent variable 1: No. of Ss
$
$
2.25
3,816
1,698
2
r = 0.39; Durbin-Watson statistic = 2.01
1. Compare and evaluate the three simple regression models estimated by Hart. Graph each one.
Also, use the format employed in Exhibit 10-18 (page 406) to evaluate the information.
2. Do the regression results support the Kaliko Fabrics’ presentation about the purchasing
department’s cost drivers? Which of these cost drivers would you recommend in designing
an ABC system?
3. How might Hart gain additional evidence on drivers of purchasing department costs at each
of Perfect Fit’s stores?
SOLUTION
(40–50 min.)
Purchasing Department cost drivers, activity-based costing, simple
regression analysis.
The problem reports the exact t-values from the computer runs of the data. Because the
coefficients and standard errors given in the problem are rounded to three decimal places,
dividing the coefficient by the standard error may yield slightly different t-values.
1.
Plots of the data used in Regressions 1 to 3 are in Solution Exhibit 10-47A. See Solution
Exhibit 10-47B for a comparison of the three regression models.
2.
Both Regressions 2 and 3 are well-specified regression models. The slope coefficients on
their respective independent variables are significantly different from zero. These results support
the Kaliko Fabrics’ presentation in which the number of purchase orders and the number of
suppliers were reported to be drivers of purchasing department costs.
In designing an activity-based cost system, Perfect Fit should use number of purchase
orders and number of suppliers as cost drivers of purchasing department costs. As the chapter
appendix describes, Perfect Fit can either (a) estimate a multiple regression equation for
purchasing department costs with number of purchase orders and number of suppliers as cost
drivers, or (b) divide purchasing department costs into two separate cost pools, one for costs
related to purchase orders and another for costs related to suppliers, and estimate a separate
relationship for each cost pool.
3.
Guidelines presented in the chapter could be used to gain additional evidence on cost
drivers of purchasing department costs.
1. Use physical relationships or engineering relationships to establish cause-and-effect
links. Hart could observe the purchasing department operations to gain insight into
how costs are driven.
10-68
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2. Use knowledge of operations. Hart could interview operating personnel in the purchasing
department to obtain their insight on cost drivers.
SOLUTION EXHIBIT 10-47A
Regression Lines of Various Cost Drivers for Purchasing Dept. Costs for Perfect Fit
Purchasing Department Costs
$2,500,000
y = 0.0031x + 1E+06
R² = 0.0798
$2,000,000
$1,500,000
$1,000,000
$500,000
$0
0
50,000,000
100,000,000
Dollar Value of Merchandise Purchased
150,000,000
Purchasing Department Costs
$2,500,000
y = 159.48x + 722538
R² = 0.4306
$2,000,000
$1,500,000
$1,000,000
$500,000
$0
0
1,000
2,000
3,000
4,000
5,000
6,000
Number of Purchase Orders
10-69
7,000
8,000
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Purchasing Department Costs
$2,500,000
$2,000,000
y = 3815.7x + 828814
R² = 0.3868
$1,500,000
$1,000,000
$500,000
$0
0
50
100
150
Number of Suppliers
200
250
SOLUTION EXHIBIT 10-47B
Comparison of Alternative Cost Functions for Purchasing Department
Costs Estimated with Simple Regression for Perfect Fit
Regression 2
PDC = a + (b  # of POs)
Economically plausible.
The higher the number of
purchase orders, the more
tasks undertaken.
Regression 3
PDC = a + (b  # of Ss)
Economically plausible.
Increasing the number of
suppliers increases the
costs of certifying
vendors and managing
the Perfect Fit-supplier
relationship.
2. Goodness of fit r2 = 0.08. Poor
goodness of fit.
r2 = 0.43. Reasonable
goodness of fit.
r2 = 0.39. Reasonable
goodness of fit.
3. Significance of
Independent
Variables
t-value on # of POs of 2.46 t-value on # of Ss of 2.25
is significant.
is significant.
Criterion
1. Economic
Plausibility
Regression 1
PDC = a + (b  MP$)
Result presented at
seminar by Kaliko
Fabrics found little
support for MP$ as a
driver. Purchasing
personnel at the
Miami store believe
MP$ is not a
significant cost driver.
t-value on MP$ of
0.83 is insignificant.
10-70
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4. Specification
Analysis
A. Linearity
within the
relevant range
Appears questionable Appears reasonable.
but no strong evidence
against linearity.
Appears reasonable.
B. Constant
variance of
residuals
Appears questionable, Appears reasonable.
but no strong evidence
against constant
variance.
Appears reasonable.
C. Independence
of residuals
Durbin-Watson
Statistic = 2.41
Assumption of
independence is not
rejected.
Durbin-Watson
Statistic = 2.01
Assumption of
independence is not
rejected.
D. Normality of
residuals
Data base too small to Data base too small to
make reliable
make reliable inferences.
inferences.
Durbin-Watson
Statistic = 1.97
Assumption of
independence is not
rejected.
Data base too small to
make reliable inferences.
10-48 Purchasing department cost drivers, multiple regression analysis (continuation of
10-47). Carl Hart decides that the simple regression analysis used in Problem 10-47 could be
extended to a multiple regression analysis. He finds the following results for two multiple
regression analyses:
Regression 4: PDC = a + (b1  No. of POs) + (b2  No. of Ss)
Variable
Coefficient
Standard Error
t-Value
Constant
$484,522
$256,684
1.89
Independent variable 1: No. of POs
$
$
2.19
Independent variable 2: No. of Ss
$
126.66
2,903
$
57.80
1,459
1.99
2
r = 0.64; Durbin-Watson statistic = 1.91
Regression 5: PDC = a + (b1  No. of POs) + (b2  No. of Ss) + (b3  MP$)
Variable
Constant
Coefficient
Standard Error
t-Value
$483,560
$312,554
1.55
$
1.99
$
Independent variable 1: No. of POs
Independent variable 2: No. of Ss
$
10-71
126.58
2,901
$
63.75
1,622
1.79
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Independent variable 3: MP$
0.00002
0.0029
0.01
2
r = 0.64; Durbin-Watson statistic = 1.91
The coefficients of correlation between combinations of pairs of the variables are as follows:
PDC
MP$
MP$
0.28
No. of POs
0.66
0.27
No. of Ss
0.62
0.30
No. of POs
0.29
Required
1. Evaluate regression 4 using the criteria of economic plausibility, goodness of fit,
significance of independent variables, and specification analysis. Compare regression 4
with regressions 2 and 3 in Problem 10-47. Which one of these models would you
recommend that Hart use? Why?
2. Compare regression 5 with regression 4. Which one of these models would you recommend
that Hart use? Why?
3. Hart estimates the following data for the Baltimore store for next year: dollar value of
merchandise purchased, $78,500,000; number of purchase orders, 4,100; number of
suppliers, 110. How much should Hart budget for purchasing department costs for the
Baltimore store for next year?
4. What difficulties do not arise in simple regression analysis that may arise in multiple
regression analysis? Is there evidence of such difficulties in either of the multiple regressions
presented in this problem? Explain.
5. Give two examples of decisions in which the regression results reported here (and in Problem
10-47) could be informative.
SOLUTION
(30–40 min.)
Purchasing Department cost drivers, multiple regression analysis
(continuation of 10-47).
The problem reports the exact t-values from the computer runs of the data. Because the
coefficients and standard errors given in the problem are rounded to three decimal places,
dividing the coefficient by the standard error may yield slightly different t-values.
1.
Regression 4 is a well-specified regression model:
Economic plausibility: Both independent variables are plausible and are supported by the
findings of the Kaliko Fabrics study.
Goodness of fit: The r2 of 0.64 indicates an excellent goodness of fit.
10-72
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Significance of independent variables: The t-value on # of POs is 2.19 while the t-value on # of
Ss is 1.99. These t-values are either significant or border on significance.
Specification analysis: Results are available to examine the independence of residuals
assumption. The Durbin-Watson statistic of 1.91 indicates that the assumption of independence
is not rejected.
Regression 4 is consistent with the findings in Problem 10-47 that both the number of
purchase orders and the number of suppliers are drivers of purchasing department costs.
Regressions 2, 3, and 4 all satisfy the four criteria outlined in the text. Regression 4 has the best
goodness of fit (0.64 for Regression 4 compared to 0.43 and 0.39 for Regressions 2 and 3,
respectively). Most importantly, it is economically plausible that both the number of purchase
orders and the number of suppliers drive purchasing department costs. We would recommend
that Hart use Regression 4 over Regressions 2 and 3.
2.
Regression 5 adds an additional independent variable (MP$) to the two independent
variables in Regression 4. This additional variable (MP$) has a t-value of -0.01, implying its
slope coefficient is insignificantly different from zero. The r2 in Regression 5 (0.64) is the same
as that in Regression 4 (0.64), implying the addition of this third independent variable adds close
to zero explanatory power. In summary, Regression 5 adds very little to Regression 4. We would
recommend that Hart use Regression 4 over Regression 5.
3.
Budgeted purchasing department costs for the Baltimore store next year are
$484,522 + ($126.66  4,100) + ($2,903  110) = $1,323,158
4.
Multicollinearity is a frequently encountered problem in cost accounting; it does not arise
in simple regression because there is only one independent variable in a simple regression. One
consequence of multicollinearity is an increase in the standard errors of the coefficients of the
individual variables. This frequently shows up in reduced t-values for the independent variables
in the multiple regression relative to their t-values in the simple regression:
Variables
Regression 4:
# of POs
# of Ss
Regression 5:
# of POs
# of Ss
MP$
t-value in
Multiple Regression
t-value from
Simple Regressions
in Problem 10-47
2.19
1.99
2.46
2.25
1.99
1.79
-0.01
2.46
2.25
0.83
10-73
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The decline in the t-values in the multiple regressions is consistent with some (but not very high)
collinearity among the independent variables. Pairwise correlations between the independent
variables are:
# of POs and # of Ss
# of POs and MP$
# of Ss and MP$
Correlation
0.29
0.27
0.30
There is no evidence of difficulties due to multicollinearity in Regressions 4 and 5.
5.
are
Decisions in which the regression results in Problems 10-47 and 10-48 could be useful
Cost management decisions: Perfect Fit could restructure relationships with the suppliers so that
fewer separate purchase orders are made. Alternatively, it may aggressively reduce the number
of existing suppliers.
Purchasing policy decisions: Perfect Fit could set up an internal charge system for individual
retail departments within each store. Separate charges to each department could be made for each
purchase order and each new supplier added to the existing ones. These internal charges would
signal to each department ways in which their own decisions affect the total costs of Perfect Fit.
Accounting system design decisions: Perfect Fit may want to discontinue allocating purchasing
department costs on the basis of the dollar value of merchandise purchased. Allocation bases
that better capture cause-and-effect relations at Perfect Fit are the number of purchase orders and
the number of suppliers.
Try It! 10-1
a. y = $1.70X
b. y = $8,000
c. y = $80 + $2.00X
d. y = $1,000 + $12X
Try It! 10-2
The highest and lowest observations of the cost driver correspond to 5,850 hours and 3,000
hours, respectively. Using those data points:
a. Slope = ($69,850 − $38,500) / (5,850 − 3,000)
= $31,350/2,850
= $11
10-74
‫عالء محسن شحم‬
alaa.aliasrei@gmail.com
b. $69,850 = Constant + ($11 × 5,850)
Constant = $5,500
c. y = $5,500 + ($11 × Hours)
d. y = $5,500 + ($11 × 3,100) = $39,600
Try It! 10-3
a. Unit
1
2
Hours Cumulative Cumulative Average
6
6
6
4.8
10.8
5.4
Learning percentage = 5.4/6.0 = 0.90
b.
b
y =a×X
− 0.1520
=6×8
= 4.37 hours
(or) Cumulative average time for:
1 unit = 6 hours
2 units = 6 × 0.9 = 5.4 hours
4 units = 5.4 × 0.9 = 4.86 hours
8 units = 4.86 × 0.9 = 4.37 hours
Therefore, total time to build 8 units = 8 × 4.37 = 34.96 hours.
c. Total time = 34.96 hours
Manufacturing overhead charge = 34.96 × $25 = $624
d. As production doubles from 1 to 2 units, the incremental time of the second unit relative to
the first is 4.8 hours/6 hours = 0.8.
Therefore, this represents an 80% learning curve under the incremental unit-time learning model.
b
e. y = a × X
− 0.3219
= 6 × 16
= 2.458 hours
(or) Time for:
1st unit = 6 hours
2nd unit = 6 × 0.8 = 4.8 hours
4th unit = 4.8 × 0.8 = 3.84 hours
8th unit = 3.84 × 0.8 = 3.072 hours
16th unit = 3.072 × .8 = 2.458 hours
10-75
‫عالء محسن شحم‬
alaa.aliasrei@gmail.com
Try It! 10-4
a, b and c.
Plot and Regression Line of Sales Revenues on Promotional Costs
850,000
800,000
y = 14.184x + 397163
R² = 0.6271
Sales Revenues
750,000
700,000
650,000
600,000
550,000
500,000
450,000
400,000
5,000
10,000
15,000
Promotional Costs
20,000
25,000
The above plot includes the regression line of sales revenues on promotional costs. Here are the
details, from carrying out a regression analysis in Excel:
Sales Revenues  $397,163  (14.18  Promotional Costs)
Variable
Coefficient
Constant
$397,163.12
Independent variable:
Promotional costs
r2  0.63; Durbin-Watson statistic  2.55
14.18
Standard Error
t-Value
59,169.06
6.71
3.87
3.67
We evaluate the estimated regression equation using the criteria of economic plausibility,
goodness of fit, and slope of the regression line.
Economic plausibility. Promotional costs would appear to be a plausible cost driver of sales
revenues. Restaurants frequently use promotional activities such as advertising, sponsoring of
local events, etc. to engender interest among potential clients and increase their patronage.
Goodness of fit. As seen in the plot, the regression line fits the data well. The vertical differences
between actual and predicted revenues are reasonably small. This indicates that promotional
10-76
‫عالء محسن شحم‬
alaa.aliasrei@gmail.com
costs are related to restaurant revenues. An r-squared value of 0.627 indicates that almost 63%
of the change in revenues can be explained by the change in promotional costs.
Slope of regression line. The slope of the regression line is relatively steep. Given the small
scatter of the observations around the line, this indicates that, on average, restaurant revenues
increase with newspaper advertising at a slope of 14.18. The t-value of 3.67 is statistically
significant at the 0.05 and 0.01 levels.
d.
The increase in sales revenues for each $1,000 spent on promotion, within the relevant
range, is:
$14.184  $1,000 = $14,184.
10-77