Exam
Name___________________________________
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Answer the question.
1)
Cougar Telemarketing is considering establishing a call center. The initial cost will be
$2,750,000 with a $27,500 market value any time within a 13–year period. The fixed cost of
the center will be $830,000 per year with an average variable cost of $3.00 per call. Cougar
expects to generate revenue of $5.25 per call with a capacity of 110,000 calls for the first
year. The company also expects to increase the capacity uniformly each year. At an
interest rate of 2% per year, determine the uniform amount the capacity must increase each
year so that the company can recover its investment in 3 years.
1)
Answer:
Uniform amount of increase per year =687,934 units
Explanation:
Let G = uniform amount of increase per year
AWRevenue = $(5.25 –3.00)[110,000 + G(A/G, 2%, 3)]
=$247,500 + $(2.22)G
AWCost =$2,750,000(A/P, 2%, 3) +$830,000 –$27,500(A/F, 2%, 3)
=$2,750,000(0.3468) +$830,000 –$27,500(0.3268)
=$1,774,713.00
Set Revenue = Cost:
G = [$1,774,713.00 –$247,500]/$(2.22)
=687,934 units
2)
Two machines are under consideration for a new production line. Machine X costs $50,000
and is expected to have a salvage value of $6500 at the end of its useful life of 5 years. It
will have a fixed cost of $16,000 per year and a variable cost of $55 per unit per year. On
the other hand, machine Y costs $55,000 and is expected to have a salvage value of $7000 at
the end of its useful life of 7 years. It will have a fixed cost of $14,500 per year and a
variable cost of $58 per unit per year. Determine the quantity that must be produced for
the two machines to break even at an interest rate of 3% per year.
2)
Answer:
1094 units per year
Explanation:
Let Q = number of breakeven units per year
AWX= –$50,000(A/P, 3%, 5) –$16,000 +$6500(A/F, 3%, 5) –$55Q
= –$50,000(0.2184) –$16,000 +$6500(0.1884) –$55Q
= –25,695.40–$55Q
AWY= –$55,000(A/P, 3%, 7) –$14,500 +$7000(A/F, 3%, 7) –$58Q
= –$55,000(0.1605) –$14,500 +$7000(0.1305) –$58Q
= –$22,414.00 –$58Q
Set AWX=AWY:
$(58–55)Q =$3281.40
Q=1094 units per year
1
3)
Ginger has agreed to a lawsuit settlement of $600,000 with a certain pharmaceutical
company. The company has offered options to pay her the awarded money. After
discussing the terms with the company, she expects that the company should be able to
pay her back within 3–5 years. Ginger has developed the following estimates. Which
option should she select, if her personal MARR is 13% per year?
Option Delay period Cash Flow Estimates, $
A –$600,000 now
B – Pessimistic 5 years $131,000 per year for every year the payment is delayed.
C – Most likely 4 years $161,200 per year for every year the payment is delayed.
D – Optimistic 3 years $211,500 per year for every year the payment is delayed.
3)
Answer:
PWA(13%) =$600,000
PWB(13%) =$460,753.20
PWC(13%) =$479,489.40
PWD(13%) =$499,393.80
PWA(13%) >PWD(13%) >PWC(13%) >PWB(13%); therefore, Ginger should select
option A.
Explanation:
Option A: PW(13%)=$600,000
Option B: PW(13%) = $(131,000)(P/A, 13%, 5)
=$460,753.20
Option C: PW(13%) = $(161,200)(P/A, 13%, 4)
=$479,489.40
Option D: PW(13%)= $(211,500)(P/A, 13%, 3)
=$499,393.80
PWA(13%) >PWD(13%) >PWC(13%) >PWB(13%); therefore, Ginger should
select option A.
4)
A manufacturer of an inspecting and profiling web controller has a fixed cost of $83,000
per year and variable costs of $60 per unit produced. If the product is sold at $90 per unit,
determine the breakeven quantity per year for the company.
4)
Answer:
Q =2767 units
Explanation:
Q =$83,000/ ($90–$60) =2767 units
5)
The estimated cash flows of an investment project are shown below.
Item Estimated Cash Flows Sensitivity Range
Initial investment, $ 55,000 ±5%
Annual revenue, $ 7000 ±10%
Annual expense, $ 4500 ±10%
Market value, $ 1800 ±15%
Project life, years 8±5%
Using an interest rate of 2% per year, analyze the sensitivity of the PW to changes in initial
investment and annual revenue, and determine the breakeven percentage changes of these
two factors.
5)
2
Answer:
Sensitivity to changes in initial investment:
+5%: PW(2%) = –$37,899.95
–5%: PW(2%) = –$32,399.95
Breakeven percent change = –64%
Sensitivity to changes in annual revenue:
+10%: PW(2%) = –$30,022.10
–10%: PW(2%) = –$40,277.80
Breakeven percent change =68.55%
Explanation:
PW(2%) = –$55,000 + ($7000 –$4500)(P/A, 2%, 8) +$1800(P/F, 2%, 8)
= –$55,000 + ($2500)(7.3255) +$1800(0.8535)
= –$35,149.95
Sensitivity to changes in initial investment:
+5%: PW(2%) = –$55,000(1.05) + ($7000 –$4500)(P/A, 2%, 8) +$1800(P/F, 2%, 8)
= –$37,899.95
–5%: PW(2%) = –$55,000(0.95) + ($7000 –$4500)(P/A, 2%, 8) +$1800(P/F, 2%, 8)
= –$32,399.95
Breakeven percent change:
PW(2%) = 0 = –$55,000(1 + X%) + ($7000 –$4500)(P/A, 2%, 8) +$1800(P/F, 2%, 8)
X= –0.64
or X = –64% change in initial investment
Sensitivity to changes in annual revenue:
+10%: PW(2%) = –$55,000 + [$(7000)(1.1)–$4500](P/A, 2%, 8) +$1800(P/F, 2%,
8) = –$30,022.10
–10%: PW(2%) = –$55,000 + [$(7000)(0.9)–$4500](P/A, 2%, 8) +$1800(P/F, 2%,
8) = –$40,277.80
Breakeven percent change:
PW(2%) = 0 = –$55,000 + [$(7000)(1 + X%) –$4500](P/A, 2%, 8) +$1800(P/F, 2%, 8)
X= [35,149.95]/[51,278.50]
=0.6855
Or X =68.55% change in annual revenue
3
6)
Two different machines are under consideration for a reengineering project. Machine X is
expected to have an initial cost of $74,000 and an expected life of 7 years. It will have a
fixed cost of $10,000 per year and a variable cost of $60 per unit per year. Process Y is
expected to have a useful life of 9 years. It will have a fixed cost of $8500 per year and a
variable cost of $57 per unit per year. Determine the amount the company can spend on
Machine Y so the two machines will break even at an interest rate of 11% per year.
Assume the current process capacity of 150 units per year is used for the analysis.
6)
Answer:
Initial cost of machine Y =$97,745.29
Explanation:
Let P = initial cost of machine Y
AWX= –$74,000(A/P, 11%, 7) –$10,000 – $(60)(150)
= –$74,000(0.2122) –$10,000 –$9000
= –$34,702.80
AWY= –$P(A/P, 11%, 9) –$8500 – $(57)(150)
= –$P(0.1806) –$17,050
Set AWX=AWY:
P = [–$34,702.80 +$17,050]/ (–0.1806)
=$97,745.29
7)
Two processes are under consideration for a certain production. Process A requires
acquisition of a new machine that is estimated to have an initial cost of $65,000 and a
salvage value of $52,000 at the end of its useful life of 6 years. In addition, the process
requires a fixed cost of $47,000 per year and a variable cost of $250 per day. Alternatively,
Process B requires the use of human labor. The process will need 6 workers, each earning
$200 per day and will have a fixed cost of $36,000 per year and additional variable costs of
$200 per day. Determine the minimum number of days per year required for the two
processes to break even at an interest rate of 2% per year.
7)
Answer:
12.49 days per year
Explanation:
Let X = number of breakeven days per year
AWA= –$65,000(A/P, 2%, 6) –$47,000 +$52,000(A/F, 2%, 6) –$250X
= –$65,000(0.1785) –$47,000 +$52,000(0.1585) –$250X
= –$50,360.50 –$250X
AWB= –$(6)(200)X –$36,000 –$200X
= –$(1400)X –$36,000
Set AWA=AWB:
–$1400X –$36,000X= –$50,360.50 –$250
–$1150X= –$14,360.50
X= –$14,360.50/ –$1150
=12.49 days per year
8)
A distribution center wants to evaluate an alternative product tracking system. The system
has an initial cost of $500,000 and a salvage value of $80,000 at the end of its useful life of 7
years. The operating cost is estimated to be $550 per metric ton of product moved per day.
The center can handle between 30 and 50 tons per day. Analyze the sensitivity of the PW
to changes in a 10–metric–ton increment of product moved. Use an interest rate of 3% per
year and 200 days of work per year.
8)
Answer:
PW30(3%) = –$20,994,942.00
PW40(3%) = –$27,848,272.00
PW50(3%) = –$34,701,602.00
Explanation:
PW30(3%) = –$500,000 – $(550)(30)(200)(P/A, 3%, 7) +$80,000(P/F, 3%, 7)
= –$500,000 – $(3,300,000)(6.2303) +$80,000(0.8131)
= –$20,994,942.00
PW40(3%) = –$500,000 – $(550)(40)(200)(P/A, 3%, 7) +$80,000(P/F, 3%, 7)
= –$500,000 – $(4,400,000)(6.2303) +$80,000(0.8131)
= –$27,848,272.00
PW50(3%) = –$500,000 – $(550)(50)(200)(P/A, 3%, 7) +$80,000(P/F, 3%, 7)
= –$500,000 – $(5,500,000)(6.2303) +$80,000(0.8131)
= –$34,701,602.00
9)
Wolfpack, Inc., a textile manufacturing company, is considering opening a production and
shipping facility to keep up with demand for its pillows. The facility is expected to require
an initial investment of $190,000 and will have a $36,000 salvage value after 5 years. Net
annual revenue is estimated to be $100,000. Determine how sensitive the decision to invest
in the new facility is to the estimates of initial cost and net annual revenue. Use a MARR of
4% per year and a 5–year study period.
9)
Answer:
If the change in initial cost is greater than 150%, the investment in the new facility
would no longer be acceptable.
If the change in net annual revenue is lower than –64.00%, the investment in the
new facility would no longer be acceptable.
Explanation:
PW(4%) = –$190,000 +$100,000(P/A, 4%, 5) +$36,000(P/F, 4%, 5)
Let X = change in initial cost that would reverse decision
PW(4%) = 0 = –$190,000(1+ X%) +$100,000(P/A, 4%, 5) +$36,000(P/F, 4%, 5)
X=1.50
If the change in initial cost is greater than 150%, the investment in the new
facility would no longer be acceptable.
Let Y = change in net annual revenue that would reverse decision
PW(4%) = 0 = –$190,000 +$100,000(1 + Y%)(P/A, 4%, 5) +$36,000(P/F, 4%, 5)
Y= –0.64
If the change in net annual revenue drops by –64.00%, the investment in the
new facility would no longer be acceptable.
10)
The estimated cash flows of an investment project are shown below.
Item Optimistic Most likely Pessimistic
Initial investment, $ 745,000 750,000 755,000
Net annual revenue, $/year 81,500 80,000 79,500
Market value, $ 39,500 38,000 37,000
Project life, years 7 7 7
Using a MARR of 14% per year, determine the AW for each of the three estimation
conditions.
10)
Answer:
AWo= –$88,552.60
AWm= –$91,358.40
AWp= –$93,117.60
Explanation:
AW(optimistic) = –$745,000(A/P, 14%, 7) +$81,500 +$39,500(A/F, 14%, 7)
= –$745,000(0.2332) +$81,500 +$39,500(0.0932)
= –$88,552.60
AW(most likely) = –$750,000(A/P, 14%, 7) +$80,000 +$38,000(A/F, 14%, 7)
= –$750,000(0.2332) +$80,000 +$38,000(0.0932)
= –$91,358.40
AW(pessimistic) = –$755,000(A/P, 14%, 7) +$79,500 +$37,000(A/F, 14%, 7)
= –$755,000(0.2332) +$79,500 +$37,000(0.0932)
= –$93,117.60
6
Answer Key
Testname: C11
1)
Uniform amount of increase per year =687,934 units
2)
1094 units per year
3)
PWA(13%) =$600,000
PWB(13%) =$460,753.20
PWC(13%) =$479,489.40
PWD(13%) =$499,393.80
PWA(13%) >PWD(13%) >PWC(13%) >PWB(13%); therefore, Ginger should select option A.
4)
Q =2767 units
5)
Sensitivity to changes in initial investment:
+5%: PW(2%) = –$37,899.95
–5%: PW(2%) = –$32,399.95
Breakeven percent change = –64%
Sensitivity to changes in annual revenue:
+10%: PW(2%) = –$30,022.10
–10%: PW(2%) = –$40,277.80
Breakeven percent change =68.55%
6)
Initial cost of machine Y =$97,745.29
7)
12.49 days per year
8)
PW30(3%) = –$20,994,942.00
PW40(3%) = –$27,848,272.00
PW50(3%) = –$34,701,602.00
9)
If the change in initial cost is greater than 150%, the investment in the new facility would no longer be acceptable.
If the change in net annual revenue is lower than –64.00%, the investment in the new facility would no longer be
acceptable.
10)
AWo= –$88,552.60
AWm= –$91,358.40
AWp= –$93,117.60