5
Cost-Volume-Profit Analysis
Exercise 5.1
Basic Problems
1. a. Variable cost
b. Fixed cost
c. Mixed cost
d. Variable cost
3. a. Fixed costs consist of:
Cell phone
Advertising
Sarah’s wages
Variable costs consist of:
Cleaning supplies
Fuel
Friend’s wages
e.
f.
g.
h.
Fixed cost
Mixed cost
Variable cost
Fixed cost
Business licence
Business liability insurance
Car insurance and licence
Oil changes
Tires
Depreciation
b. Monthly fixed costs consist of:
Cell phone = $80.00/month
Advertising = $75.00/month
Sarah’s wages = $1500.00/month
Business licence = $150/12 = $12.50/month
Car insurance and licence = $1500/12 = $125.00/month
Business liability insurance = $600/12 = $50.00/month
Therefore:
Total monthly fixed costs = $80.00 + $75.00 + $1500.00 + $12.50 + $125.00 + $50.00
= $1842.50/month
Variable costs are:
Cleaning supplies = $10.00/home
Friend’s wages = $30.00/home
$75.00
Fuel =
 10 km  $1.50/home
500 km
$50.00
Oil changes =
 10 km  $0.10/home
5000 km
$500.00
 10 km  $0.05/home
Tires =
100,000 km
$150.00
Depreciation =
 10 km  $1.50/home
1000 km
Therefore:
Total variable cost per home = $10.00 + $30.00 + $1.50 + $0.10 + $0.05 + $1.50
= $43.15/home
Chapter 5: Cost-Volume-Profit Analysis
121
Exercise 5.1 (continued)
Intermediate Problems
5. Last year:
X = 580,000 units
Total costs were $3,710,000, of which $3,335,000 were variable costs.
Since:
Then:
TC = Total Variable Costs + Total Fixed Costs
$3,710,000 = $3,335,000 + FC
FC= $3,710,000 – $3,335,000 = $375,000
And:
Total Variable Costs = VC (X)
$3,335,000 = VC (580,000)
VC = $3,335,000/580,000 = $5.75 per unit
To produce X = 650,000 units next year,
TC = (VC)X + FC
TC = $5.75(650,000) + $375,000
TC = $4,112,500
7. Total variable costs to produce 10,000 units was
($50.4 + $93 + $16 + $14.2) million = $173,600,000
Total fixed costs = ($22.2 + 19.2 + $9.6 + $5) million = $56,000,000
To produce 9700 units,
9700
 $173,600,000 = $168,392,000
Total variable costs =
10,000
and
Total costs = ($56 + $168.392) million = $224,392,000
122
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 5.2
Basic Problems
1. Given: S = $30; VC = $10; FC = $100,000 per year
a. TR = (S)X = $30X
TC = (VC)X + FC = $10X + $100,000
To break even, TR = TC
Then,
$30X = $10X + $100,000
$20X = $100,000
X = 5000 toys per year
b. Break-even revenue = (S)X = $30(5000) = $150,000 per year
3. Given: S = $3.20; VC = $1.20; FC = $250 per month
a. FC = 12($250) = $3000 per year
TR = (S)X = $3.20X
TC = (VC)X + FC = $1.20X + $3000
To break even, TR = TC
Then,
$3.20X = $1.20X + $3000
$2.00X = $3000
X = 1500 jars per year
b. If Ingrid sells 3000 jars per year,
NI = (S – VC)X – FC = ($3.20 – $1.20)3000 – $3000 = $3000
Chapter 5: Cost-Volume-Profit Analysis
123
Exercise 5.2 (continued)
5.Given: Capacity = 250 units per week
S = $20, VC = $12, FC = $1200 per week
a. TR = (S)X = $20X
TC = (VC)X + FC = $12X + $1200
To break even, TR = TC
Therefore, $20X = $12X + $1200
X = 150 units/week
b. NI = TR – TC = $20X – $12X – $1200 = $8X – $1200
(i)
If X = 120 units/week,
then
NI = $8(120) – $1200 = –$240
There will be a loss of $240/week.
(ii)
If X = 250 units/week,
then NI = $8(250) – $1200 = $800/week profit.
c. If NI = $400/week,
then
400 = $8X – $1200
$1600 = $8X
X = 200 units/week
124
Fundamentals of Business Mathematics in Canada, 2/e
Intermediate Problems
7.
Given: S = 10¢ per copy; FC = $300 per month;
$5.00 $100.00
+ $0.005 = $0.05 per copy

500
5000
a. TR = (S)X = $0.10X
TC = (VC)X + FC = $0.05X + $300
To break even, TR = TC
Then,
$0.10X = $0.05X + $300
$0.05X = $300
X = 6000 copies per month
b. If X = 6000 + 1000 = 7000 copies per month.
NI = (S – VC)X – FC = ($0.10  $0.05)7000 – $300 = $50 per month
The additional 1000 copies add $50 per month profit.
VC = $0.015 +
9. Given: FC = $450,000 per year; S = $25; VC = $15.
a. At the break-even point,
NI = 0 = (S – VC)X – FC
0 = ($25 – $15) X – $450,000
$450,000
X
 45,000 units per year
$10
As a percentage of capacity, 45,000 units per year is
45,000
 100  75%.
60,000
b. Annual revenue at breakeven = 45,000($25) = $1,125,000 per year
c.
(i)
(ii)
(iii)
NI = (S – VC)X – FC
= ($25 – $15)50,000 – $450,000
= $50,000 profit
Since VC = $$15
25 = 0.60 of S,
NI = TR – TC = $1,000,000 – 0.60($1,000,000) – $450,000 = –$50,000
That is, a $50,000 loss.
90% of capacity represents 0.90 x 60,000 units = 54,000 units
TR = $25 x 54,000 units = $1,350,000
NI = TR – TC = $1,350,000 – $15(54,000) – $450,000 = $90,000
That is, a $90,000 profit.
d. For NI = –$20,000 = (S – VC)X – FC
–$20,000 = ($25 – $15)X –$450,000
$430,000
X 
 43,000 units per year
$10
e. If S and FC are unchanged, but VC increases by $1 to $16,
NI = 0 =($25 – $16)X –$450,000
$450,000
X 
 50,000 units per year at breakeven
$9
Therefore, the break-even volume increases by 50,000 – 45,000 = 5000 units
and the break-even revenue increases by $25(5000) = $125,000.
Chapter 5: Cost-Volume-Profit Analysis
125
$6,000,000
1
 of total revenue,
$18,000,000 3
FC = $10,000,000 per year
$18,000,000
At full capacity, Sales revenue =
= $20,000,000
0.90
a. At the break-even point, TR – TC = 0
That is, TR – 31 TR – $10,000,000 = 0
11. Given:
Total variable costs are
2
3
TR = $10,000,000
TR = $15,000,000
$15,000,000
 $100% = 75% of capacity
$20,000,000
b. At 70% of capacity, TR = 0.70($20,000,000) = $14,000,000
NI = TR – TC = $14,000,000 – 31 ($14,000,000) – $10,000,000 = –$666,667
That is, Beta would have a $666,667 loss.
which is
126
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 5.2 (continued)
13. Given:
S = $180/tonne, VC = $42 + $24 = $66 per tonne,
FC = $400 + $600 + $450 = $1450 per hectare
a. To break even,
FC
$1450
X=
=
= 12.719 tonnes per hectare
$
180
 $66
S  VC
b. (i) NI = (S – VC)X – FC = ($180 – $66)15 – $1450 = $260 per hectare.
(ii) NI = ($180 – $66)10 – $1450 = –$310.00 per hectare.
That is, a loss of $310 per hectare.
c. If S = $190 per tonne instead of $180 per tonne,
$1450
X=
= 11.694 tonnes per hectare
$190  $66
The break-even tonnage will be 11.694 tonnes per hectare.
Advanced Problems
15. Given: S = $270, VC = $220, FC = $1400 for 15 to 36 participants
a. To break even,
FC
$1400
X=
=
= 28 participants
$270  $220
S  VC
b. If X = 36, then
NI = (S – VC)X – FC = ($270 – $220)36 – $1400 = $400
c. If a loss of $400 is incurred,
–$400 = ($270 – $220) X – $1400
X = 20
Hence, the minimum number of participants is 20.
Chapter 5: Cost-Volume-Profit Analysis
127
Exercise 5.2 (continued)
17. Given: S = $37.50, VC = $13.25, FC = $5600/month
FC
$5600
a. Last year’s break-even volume =
=
= 231 units/month
$37.50  $13.25
S  VC
This year, VC = $15, FC = $6000/month.
For the break-even volume to remain 231 units/month,
FC
S – VC =
Break - even volume
$6000
S – $15 =
= $25.97
231
S = $15 + $25.97 = $40.97
b. Last year’s NI = (S – VC)X – FC = ($37.50 – $13.25)300 – $5600 = $1675.00.
In order that this year's profit also be $1675/month on sales of 300 units/month,
$1675 = 300(S – $15) – $6000
$7675
S=
+ $15 = $40.58
300
19. Given: NI = –$400,000; TR = $3 million/year at 60% of capacity;
Total variable costs = ⅓TR .
a. First calculate FC.
Since
NI = TR – TC,
then –$400,000 = $3,000,000 – ⅓($3,000,000) – FC
FC = $2,000,000 + $400,000
= $2,400,000
At breakeven, NI = 0 = TR – TC
Hence,
0 = TR – ⅓ TR – $2,400,000
⅔TR = $2,400,000
TR = $3,600,000
Since sales of $3 million/year represented 60% of capacity,
$3,000,000
Full capacity =
= $5,000,000
0.60
$3,600,000
 100%= 72%
% of capacity at breakeven =
$5,000,000
. b. At 80% capacity, TR = 0.80($5,000,000) = $4,000,000
NI = TR – ⅓ TR – $2,400,000
= $4,000,000 – ⅓($4,000,000) – $2,400,000
= $266,667
c. For NI = $700,000,
$700,000 = TR – ⅓TR – $2,400,000
⅔TR = $3,100,000
TR = $4,650,000
d. One-third of each dollar of sales is consumed by variable costs.
Therefore, the increase in NI is $0.67.
e. At breakeven, NI = 0 = TR – ⅓ TR – FC
⅔TR = FC
TR = 3 2 FC
Therefore, if FC increases by $1.00, TR must increase by $1.50 to still break even.
128
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 5.2 (continued)
21. Given: In the first quarter (Q1), TR = $4,500,000; NI = $900,000; X = 60,000 tires.
In the second quarter (Q2), NI = $700,000; X = 50,000 tires.
TR $4,500,000
S

Since TR = (S)X, then
= $75 per tire
X
60,000
Also, TR = $75  50,000 = $3,750,000 in Q2.
Since NI = (S – VC)X – FC
then for Q1,
$900,000 = (S – VC)60,000  FC

and for Q2,
$700,000 = (S – VC)50,000  FC

Subtract:
$200,000 = (S – VC)10,000
VC = S – $20 = $75 – $20 = $55 per tire
Substitute S – VC = $20 into :
$700,000 = ($20)50,000  FC
Therefore,
FC = $1,000,000  $700,000 = $300,000 per quarter
Chapter 5: Cost-Volume-Profit Analysis
129
Exercise 5.3
Intermediate Problems
1. Given: S = $2.50 per package; VC = $1.00 per package; FC = $60,000 per month.
150000
TR = (S)X
= $2.50X
TC = (VC)X + FC
= $1.00X + $60,000
X: 20,000 60,000
TR: $50,000 $150,000
TC: $80,000 $120,000
TR,TC ($)
140000
Total
Revenue, TR
130000
120000
110000
$7500
Total
Cost, TC
100000
90000
a. 40,000 pkgs/month
80000
b. 45,000 pkgs/month
70000
60000
(b)
(a)
X
50000
20000 25000 30000 35000 40000 45000 50000 55000 60000
Packages per Month
130
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 5.3 (continued)
3. Given: S = $10 per bag; VC = $7.50 per bag; FC = $100,000 per year.
TR = (S)X
= $10X
TC = (VC)X + FC
= $7.5X + $100,000
65,000
X: 20,000
$200,000
$587,500
TR:
TC: $250,000 $650,000
a.
b.
c.
d.
40,000 bags/year
$50,000
64,000 bags/year
36,000 bags/year
650000
TR,TC ($)
600000
(b) $50,000 profit
550000
Total Revenue, TR
500000
450000
400000
Total
Cost, TC
350000
300000
(a)
250000
200000
20000
X
25000
30000
35000
40000
45000
50000
55000
60000
65000
Bags per Year
Chapter 5: Cost-Volume-Profit Analysis
131
Exercise 5.3 (continued)
5.
Given: S = $0.10 per copy; FC = $300 per month;
VC = $0.015 +
$5.00 $100.00
+$0.005 = $0.05 per copy

500
5000
TR = (S)X
= $0.10X
TC = (VC)X + FC
= $0.05X + $300
800
4000
$400
$500
650
X:
TR:
TC:
8000
$800
$700
TR,TC ($)
Total
Revenue, TR
750
700
(b) $50 profit
Total
Cost, TC
600
550
a. 6000 copies/month
500
b. $50 per month
450
400
4000
(a)
X
4500
5000
5500
6000
6500
7000
7500
Copies per month
132
Fundamentals of Business Mathematics in Canada, 2/e
8000
Exercise 5.4
Basic Problems
1. Given: S = $30; VC = $10; FC = $100,000 per year
a. Each toy sold contributes
CM = S – VC = $30 – $10 = $20
to the payment of fixed costs. To cover $100,000 of fixed costs
(and break even), Toys-4-U needs to sell
FC $100,000 per year
= 5000 toys per year

CM
$20
b. Break-even revenue = 5000($30) = $150,000 per year
3. Given: S = $3.20; VC = $1.20; FC = $250 per month
a. FC = 12($250) = $3000 per year
Each jar sold contributes
CM = S – VC = $3.20 – $1.20 = $2.00
to the payment of fixed costs. To cover $3000 of fixed costs
(and break even), Ingrid must sell
FC $3000 per year

= 1500 jars per year
CM
$2.00
b. If Ingrid sells 3000 jars per year, this sales level is 1500 jars above break-even. Each
of these jars contributes CM = $2.00 to profit.
Total profit = 1500($2.00) = $3000
5. Given: Capacity = 250 units per week
a. S = $20, VC = $12, FC = $1200 per week
CM = S – VC = $20 – $12 = $8.00
FC
$1200
Break-even volume =
=
= 150 units per week
$8
CM
b. (i) At 30 units per week short of breakeven,
there will be a loss of X(CM) = 30($8) = $240/week.
(ii) At 100 units per week above breakeven,
there will be a profit of 100($8) = $800/week.
c. For a net income of $400/week, sales must be
NI
$400
=
= 50 units above breakeven
$8
CM
Hence, sales must be 150 + 50 = 200 units per week.
Chapter 5: Cost-Volume-Profit Analysis
133
Intermediate Problems
7. Given: S = 10¢ per copy; FC = $300 per month;
$5.00 $100.00
VC = 1.5¢ +
+ 0.5¢ = 5¢ per copy

500
5000
a. CM = S – VC = 10¢ – 5¢ = 5¢ per copy
FC
$300
Break-even volume =
=
= 6000 copies per month
$0.05
CM
b. Each additional 1000 copies per month will add profit of
1000CM = 1000($0.05) = $50 per month
9. Given: FC = $450,000/year, VC = $15, S = $25
a. CM = S – VC = $25 – $15 = $10
FC
$450,000
Break-even volume =
=
= 45,000 units per year
$10
CM
45,000
 100% = 75%
Break-even volume as a percentage of capacity =
60,000
b. From Part (a), we found that break-even volume = 45,000 units per year.
Hence, annual revenue required to break even = 45,000($25) = $1,125,000
c. (i) If sales exceed breakeven by 5000 units,
NI = 5000(CM) = 5000($25 – $15) = $50,000
(ii) The break-even revenue is $1,125,000. In this instance, the actual
$125,000
revenue is $125,000 below break even, or
 5000 units below
$25
breakeven.
When sales are 5000 units below breakeven, then
NI = -5000(CM) = -5000($25 – $15) = -$50,000 = $50,000 loss
(iii) The break-even revenue is $1,125,000. In this instance, the actual
revenue is 0.90 x 60,000 x $25 = $1,350,000, which is $225,000 above break
$225,000
 9000 units more than
even. At this sales level, the company is selling
$25
the breakeven level. The profit will be:
NI = 9000(CM) = 9000($25 – $15) = $90,000
d. If the loss is $20,000, sales will be
$20,000
$20,000
=
= 2000 units below breakeven.
$10
CM
That is, annual sales will be 45,000 – 2000 = 43,000 units.
e. S and FC are unchanged, but VC increases by $1 to $16.00.
The CM drops $1 to $9 and
$450,000
Break-even volume =
= 50,000 units
$9
The break-even unit sales increases by 5000 units
and the break-even revenue by 5000($25) = $125,000.
134
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 5.4 (continued)
11. a. Since the budget projects total variable costs of $6,000,000 for total sales of
$6,000,000
 0.33 or 33.3 cents of
$18,000,000, we know that variable costs are
$18,000,000
every sales dollar. Since we do not know SP, let us use SP = $1.00. Then
VC = $ 0.33 and CM = $1.00 - $ 0.33 =$ 0.66 .
$10,000,000
FC
 15,000,000 “units” per year. Since we set
Break-even volume =
=
CM
$0.66
SP = $1, then this is really $15,000,000 of sales revenue per year.
If revenue of $18,000,000 represents 90% of capacity, then
100%
Revenue at full capacity =
 $18,000,00 0  $20,000,00 0
90%
$15,000,00 0
 100% = 75% of capacity
and break-even revenue represents
$20,000,00 0
b. At 70% of capacity,
Revenue = 0.70($20,000,000) = $14,000,000
In part (a) we determined that variable costs are 33.3 % of every sales dollar.
Therefore, in this instance, Total variable costs = $14,000,000  0.33 = $4,666,667
Net income = $14,000,000 – $10,000,000 – $4,666,667 = –$666,667
That is, Beta would lose $666,667.
13. Given:
S = $180/tonne, VC = $42 + $24 = $66 per tonne,
FC = $400 + $600 + $450 = $1450 per hectare
a. CM = S – VC = $180 – $66 = $114 per tonne
FC
$1450
Break-even yield =
=
= 12.719 tonnes/hectare
$114
CM
b. (i) If X = 15 tonnes/hectare (which is 2.281 tonnes/hectare above breakeven),
NI = 2.281(CM) = 2.281 ($114)  $260 per hectare profit.
(ii) If X = 10 tonnes/hectare (which is 2.719 tonnes/hectare below breakeven),
the loss will be 2.719($114)  $310 per hectare loss.
c. If S = $190 per tonne, then CM = $124 per tonne and
$1450 per hectare
Break-even yield =
= 11.694 tonnes/hectare
$124 per tonne
Chapter 5: Cost-Volume-Profit Analysis
135
Advanced Problems
15. Given: S = $270, VC = $220, FC = $1400 for 15 to 36 participants
a. CM = S – VC = $270 – $220 = $50
FC
$1400
To break even, X =
=
= 28 participants
$50
CM
b. If X = 36 (which is 8 participants above breakeven),
then NI = 8(CM) = 8($50) = $400
c. A loss of $400 will be incurred if there are
NI
 $400
=
= 8 fewer participants than breakeven
$50
CM
Hence, the minimum number of participants is 28 – 8 = 20.
17. Given: S = $37.50, VC = $13.25, FC = $5600/month
a. Last year's CM = S – VC = $37.50 – $13.25 = $24.25
FC
$5600
Last year’s break-even volume =
=
= 231 units/month
$
24 .25
CM
This year, VC = $15, FC = $6000/month. For the break-even
volume to still be 231 units/month,
$6000
FC
CM =
=
= $25.97 and
231
Break - even volume
S = VC + CM = $15 + $25.97 = $40.97
b. Last year’s NI = (CM)X – FC = ($24.25)300 – $5600 = $1675.00. In order that
this year's profit also be $1675/month on sales of 300 units/month,
$1675 = 300CM – $6000
$7675
CM =
= $25.58
300
and S = CM + VC = $25.58 + $15 = $40.58
136
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 5.4 (continued)
19. Given: NI = –$400,000 on sales of $3 million/year;
Utilization = 60% of capacity; Variable costs = 33.3% of sales.
a. Variable costs are 33.3% of sales. If we set SP = $1, VC = $0.33 and
CM = $1 - $0.33  $0.66.
FC = Sales – Variable costs – NI
= $3 million – 0.33 ($3 million) – (– $400,000)
= $2,400,000
$2,400,000
FC
Break-even volume =
=
= 3,600,000 “units”. Since we set SP = $1,
CM
$0.66
then the break-even level of revenue is $3,600,000.
Since sales of $3 million/year represented 60% of capacity,
$3 million
60% of capacity

$3.6 million % of capacity at break even
% of capacity at breakeven =
b. Sales at 80% capacity =
80
60
3.6
3
 60% = 72%
 $3 million = $4,000,000
which is $400,000 above breakeven. If we assume SP = $1, then 80% of capacity
represents 400,000 “units” above the break-even level.
NI = CM(400,000) = ($0.66)(400,000  $266,667
c. In order that NI = $700,000,
$700,000
NI
= 3,600,000 +
= 4,650,000
CM
$0.66
“units”. Since we set SP = $1, sales revenue of $4,650,000 is required to generate a
net income of $700,000.
Sales volume = Break-even sales volume +
d. The CM gives us this figure directly. That is, CM = $ 0.66 means that
net income increases by 66.6 ¢ for each $1 of sales (above break even).
$1
FC

 $1.50 more sales to cover it.
e. Each $1 of FC requires
CM $0.66
Chapter 5: Cost-Volume-Profit Analysis
137
Exercise 5.4 (continued)
21. Given: In the first quarter (Q1), TR = $4,500,000; NI = $900,000; X = 60,000 tires.
In the second quarter (Q2), NI = $700,000; X = 50,000 tires.
TR $4,500,000

Since TR = (S)X, then S 
= $75 per tire selling price
X
60,000
Also, TR = $75  50,000 = $3,750,000 in Q2.
Since NI = (CM)X – FC
then for Q1,
$900,000 = (CM)60,000  FC

and for Q2,
$700,000 = (CM)50,000  FC

Subtract:
$200,000 = (CM)10,000
$200,000
CM 
= $20 per tire
10,000
Substitute CM = $20 into :
$700,000 = ($20)50,000  FC
Therefore,
FC = $1,000,000  $700,000 = $300,000 per quarter
Since
then
CM = S  VC
VC = S  CM = $75  $20 = $55 per tire
Checkpoint Questions (Exercise 5.5)
1. a. False. A fixed cost can, at some point, change. What makes it is a fixed cost,
however, is the fact that the cost is changing for a reason other than a change in sales
volume. For example, rent is a fixed cost. At the renewal of a rental agreement, the
rent may increase. However, rent is still a fixed cost because its total value does not
change based on changes in sales.
b. False. “Total revenue” is the total dollar value of sales for a given period. “Net
income” is the difference between total revenue and total expenses for a given period.
c. True
d. True
e. False. All else being equal, lower fixed costs will result in a lower break-even point.
3. At the break-even point, all fixed costs and all variable costs are covered and NI = 0.
Production beyond the break-even point results in additional variable costs. Therefore,
only the difference between additional revenue ($1) and additional variable costs
[$1 (1 – CR)] contributes to net income.
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Fundamentals of Business Mathematics in Canada, 2/e
Exercise 5.5
Basic Problems
1.
Given: S = $25.00, VC = $5.00, FC = $680/month
a. CM = S – VC = $25.00 – $5.00 = $20.00 per lawn
CR =
CM
$20.00
 100% = 80%
x100% =
$25.00
S
b. Break-even Revenue =
FC $680
=
= $850 per month
0.80
CR
c. Since TR= S(X), and break-even revenue is $850/month, then:
$850 = $25(X)
$850
X=
 34 lawns per month to break even
$25
Intermediate Problems
3. a. Step 1: Calculate Beta Inc.’s anticipated contribution rate, CR.
Total revenue- Total variablecosts
CR =
x100%
Total revenue
CR =
$18,000,000  $6,000,000
x100%
$18,000,000
CR = 66.6%
Step 2: Calculate Beta Inc.’s break-even revenue level of annual revenue.
FC $10,000,000
Break-even Revenue =
=
= $15,000,000 per year
CR
0.66
Step 3: Calculate Beta Inc’s capacity and the percentage of capacity represented by
break-even sales of $15,000,000 per year
Since the forecasted $18,000,000 of sales represents 90% of capacity, then
$18,000,000 = 0.90(Capacity)
$18,000,000
Capacity =
= $20,000,000 of annual revenue
0.90
The break-even level of revenue, $15,000,000 per year, represents
$15,000,000
 100%  75% of capacity
$20,000,000
b. Since Beta’s capacity was calculated to be $20,000,000 of revenue per year, when
Beta operates at 70% of capacity, total revenue would be
0.70($20,000,000) = $14,000,000
Chapter 5: Cost-Volume-Profit Analysis
139
In part a, Beta’s break-even level of revenue was calculated to be $15,000,000 per
year. Therefore, annual sales of $14,000,000 are $1,000,000 below the break-even
level. Since CR is 66.66%, then sales of $1,000,000 below break even will generate a
$1,000,000  $0.66  $666,666.67 loss.
Exercise 5.5 (continued)
Advanced Problems
5. a. Step 1: Calculate Greenwood Corp.’s contribution rate, CR.
Total revenue- Total variablecosts
CR =
x100%
Total revenue
CR =
$1,200,000  $600,000
x100%
$1,200,000
CR = 50%
Step 2: Calculate Greenwood Corporation’s break-even revenue level of annual
revenue.
FC $400,000
Break-even Revenue =
=
= $800,000 per year
0.50
CR
b. In the past year TR = $1,200,000, FC = $400,000, and TVC = $600,000. Therefore,
net income in the past year was
NI = TR – FC – TVC
NI = $1,200,000 – $400,000 – $600,000
NI = $200,000 in the past year
If sales increase by 15%, the new total revenue will be 1.15($1,200,000) = $1,380,000.
This new sales level is $1,380,000 - $800,000 = $580,000 beyond the break-even
level of revenue.
Since CR = 0.50, 50 cents from every dollar of sales beyond the
break-even level contributes toward net income or profit. Therefore, the new level of
net income will be $580,000 x 0.50 = $290,000, an increase of $290,000 - $200,000 =
$90,000 increase in net income.
c. If fixed costs are 10% lower in the year ahead, then FC = 0.90($400,000) = $360,000.
Since sales remain unchanged at $1,200,000 and variable costs remain at $600,000,
then:
NI = TR – FC – TVC
NI = $1,200,000 – $360,000 – $600,000
NI = $240,000
Since net income in the past year was $200,000, the reduction in fixed costs in the
year ahead would result in a $240,000 - $200,000 = $40,000 increase in net income.
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Fundamentals of Business Mathematics in Canada, 2/e
d. If variable costs are 10% higher in the year ahead, then total variable costs will be
1.1($600,000) = $660,000. Since sales remain unchanged at $1,200,000 and fixed
costs remain unchanged at $400,000, then:
NI = TR – FC – TVC
NI = $1,200,000 – $400,000 – $660,000
NI = $140,000
Since net income in the past year was $200,000, the increase in variable costs in the
year ahead would result in a $200,000 – $140,000 = $60,000 reduction in net income.
Review Problems
Basic Problems
1. a. The amount of sugar used will grow or decline in proportion to sales. Therefore, sugar
is a variable cost.
b. The cost of weekly newspaper advertisement will not change if sales increase or
decrease. Therefore, weekly newspaper advertisement is a fixed cost.
c. The cost of the lease for a delivery vehicle will typically be a set amount per month,
and will not change if sales increase or decrease. Therefore, lease costs for a delivery
vehicle is an example of a fixed cost.
d. The number of cake boxes used will grow or decline in proportion to sales. Therefore,
cake boxes are a variable cost.
3. Given: S = $40; X = 384; TC= $4848; FC = $3600 per year
a.
TC = (VC)X + FC
$4848 = (VC)384 + $3600
$4848 – $3600 =(VC)384
$1248
VC =
= $3.25 per dog
384
b. In a year in which Sabine groomed 500 dogs,
TC = (VC)X + FC = $3.25(500)+$3600 = $5225
Chapter 5: Cost-Volume-Profit Analysis
141
Review Problems (continued)
5. Given: S = $180; VC = $110; FC = $1,260,000 per year
a. TR = (S)X = $180X
TC = (VC)X + FC = $110X + $1,260,000
To break even, TR = TC
Then,
$180X = $110X + $1,260,000
$70X = $1,260,000
X = 18,000 units
b. Break-even sales revenue = (S)X = $180(18,000) = $3,240,000
c. If X = 20,000 units are sold,
NI = (S – VC)X – FC
= ($180 – $110)20,000 – $1,260,000
= $140,000
That is, the company will have a profit of $140,000.
If X = 17,500 units are sold,
NI = (S – VC)X – FC
= ($180 – $110)17,500 – $1,260,000
= –$35,000
That is, the company will suffer a loss of $35,000.
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Fundamentals of Business Mathematics in Canada, 2/e
Review Problems (continued)
7. Given: S = $155; VC = $65; FC = $18,000 per month
a. TR = (S)X = $155X
TC = (VC)X + FC = $65X + $18,000
To break even, TR = TC
Then,
$155X = $65X + $18,000
$90X = $18,000
X = 200 seats per month = 2400 seats per year
b. If X = 2000 seats per year,
NI = (S – VC)X – FC
= ($155 – $65)2000 – 12($18,000)
= $180,000 – $216,000
= – $36,000
That is, ChildCare will have a loss of $36,000.
Intermediate Problems
9. Given: FC = $45,000 + $7000 + $8000 = $60,000;
VC = $8.00 + 0.08($35) = $10.80; S = $35.00
FC
$60,000
a. Break-even volume =
=
= 2479 books
$35.00  $10.80
S  VC
b. NI = (S – VC)X – FC = ($35.00 – $10.80)4800 – $60,000 = $56,160
c. If the price is reduced 10%,
S = $35(1 – 0.10) = $31.50
and if sales are 15% higher,
X = 4800(1 + 0.15) = 5520 units
Note that a reduction in selling price changes VC because the 8% royalty will be a
smaller dollar value. At the new S = $31.50, the new VC = $8.00 + 0.08($31.50) =
$10.52.
Then
NI = ($31.50 – $10.52)5520 – $60,000 = $55,809.60
Select the $35 price because (forecast) net income will be larger (by about $350).
d. If FC = $65,000 and VC = $11.80, then
FC
$65,000
Break-even volume =
=
= 2802 books
$35.00  $11.80
S  VC
That is, the break-even volume would be increased by 2802 – 2479 = 323 books
Chapter 5: Cost-Volume-Profit Analysis
143
11. Given: S = $100, FC = $200,000, VC = $60
Forecast X = 8000 units
a. CM = S – VC = $100 – $60 = $40
FC
$200,000
Break-even volume =
=
= 5000 units
$40
CM
b. Volume = Break-even volume +
NI
$100,000
= 5000 +
= 7500 units
$40
CM
c. If sales are 8000 – 5000 = 3000 units above breakeven, then
NI = 3000(CM) = 3000($40) = $120,000
d. (i)
(ii)
e. (i)
f.
NI = (CM)X – FC = 8000(CM) – FC
For each $1 increase in FC, NI will drop $1.
If FC is 5% (or $10,000) higher, NI will be $10,000 lower.
If FC is 10% (or $20,000) lower, NI will be $20,000 higher.
If VC is 10% (or $6) higher, then
CM = S – VC = $100 – $66 = $34, and
NI = 8000(CM) – FC = 8000($34) – $200,000 = $72,000
That is, NI will be $120,000 – $72,000 = $48,000 lower.
(ii)
If VC is 5% (or $3) lower, then
CM = $100 – $57 = $43 and
NI = 8000($43) – $200,000 = $144,000
That is, NI will be $144,000 – $120,000 = $24,000 higher.
Note the “leverage” effect here. That is, (% change in NI) = – 4(% change in VC)
(i)
If S is 5% (or $5) higher, then
CM = $105 – $60 = $45, and
NI = 8000($45) – $200,000 = $160,000
That is, NI will be $160,000 – $120,000 = $40,000 higher.
(ii)
If S is 10% (or $10) lower, then
CM = $90 – $60 = $30 and
NI = 8000($30) – $200,000 = $40,000
That is, NI will be $120,000 – $40,000 = $80,000 lower.
Note the “leverage” effect here. That is,
% change in NI = 6. 6 (% change in S)
(assuming that the sales volume does not change).
g. If X = 8000, VC = 1.1($60) = $66. S = $100, and
FC = 0.9($200,000) = $180,000, then
CM = $100 – $66 = $34 and
NI = 8000($34) – $180,000 = $92,000
That is, NI will be $120,000 – $92,000 = $28,000 lower.
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Fundamentals of Business Mathematics in Canada, 2/e
Advanced Problems
13.
Given: S = $10 (= Average revenue per person)).
Plan 1: VC = (0.1 + 0.3)S = $4, FC = $15,000
Plan 2: VC = (0.1 + 0.5)S = $6, FC = $10,000
a.
CM = S – VC:
Break-even point =
FC
:
CM
Plan 1
$10 – $4 = $6
$15,000
= 2500
$6
Plan 2
$10 – $6 = $4
$10,000
= 2500
$4
b. (i)
At X = 3000, NI:
500($6) = $3000
500($4) = $2000
(ii)
At X = 2200, NI:
–300($6) = –$1800
(loss of $1800)
–300($4) = –$1200
(loss of $1200)
c. If attendance surpasses the break-even point, the 30% commission rate generates
the higher profit. However, if attendance falls short of the 2500 break-even point,
the 30% commission will produce the larger loss.
Chapter 5: Cost-Volume-Profit Analysis
145