25)
HealthRx, a major manufacturer of non–invasive breast and cervical cancer detection
products is planning to expand its market in Asia. Three single–patient–use disposable
devices based on its proprietary technology to identify cancers and precancers painlessly
and noninvasively by scanning the cervix with light are being considered. The costs to
manufacture these devices are the following:
Alternative Soft Touch Light Touch Gentle Touch
Initial costs $500,000 $520,000 $535,000
Annual operating costs $15,000 $14,500 $14,000
Salvage Value $70,000 $80,000 $85,000
Sell Price per unit $3500 $3600 $3650
If the company sells 1000 units per year, which device should be selected based on the
present worth method? Assume the company uses a MARR of 11% and wants to recover
its investment in 7 years.
25)
Answer:
PW(ST) =$15,955,736.00
PW(LT) =$16,414,129.10
PW(GT) =$16,639,503.70
Therefore, Gentle Touch should be selected.
Explanation:
PW(ST) = – 500,000 –15,000(P/A, 11%, 7) + (3500)(1000)(P/A, 11%, 7) +
70,000(P/F, 11%, 7)
= – 500,000 + (–15,000 +3,500,000)(4.7122) +70,000(0.4817)
=15,955,736.00
PW(LT) = – 520,000 –14,500(P/A, 11%, 7) + (3600)(1000)(P/A, 11%, 7) +
80,000(P/F, 11%, 7)
= – 520,000 + (–14,500 +3,600,000)(4.7122) +80,000(0.4817)
=16,414,129.10
PW(GT) = – 535,000 –14,000(P/A, 11%, 7) + (3650)(1000)(P/A, 11%, 7) +
85,000(P/F, 11%, 7)
= – 535,000 + (–14,000 +3,650,000)(4.7122) +85,000(0.4817)
=16,639,503.70
PW(ST) < PW (LT) < PW (GT); therefore, Gentle Touch should be selected.
21
26)
Two structural designs for a large public monument in Indonesia are under evaluation.
Use the repeatability assumption and the CW method to determine which design should
be selected if the service period of the monument is indefinite and the interest rate is 2%
per year.
Alternative A B
Initial costs $310,000 $325,000
Annual maintenance costs $23,000 $24,500
Usual life, years 811
26)
Answer:
CW (A) = – $57,715.00
CW (B) = – $2,885,750.00
Design B should be selected.
Explanation:
AW(A) = – 310,000(A/P, 2%, 8) –23,000
= – 310,000(0.1365) –23,000
= – 65,315.00
CW (A) = AW/0.02
= (–65,315.00)/(0.02)
= – 3,265,750.00
AW(B) = – 325,000(A/P, 2%, 11) –24,500
= – 325,000(0.1022) –24,500
= – 57,715.00
CW (B) = AW/0.02
= (–57,715.00)/(0.02)
= – 2,885,750.00
CW(B) > CW(A); therefore, the design B should be selected.
22
27)
A large textile company is trying to decide among three alternatives of sludge dewatering
processes. The costs associated with these alternatives are shown below. Alternative Y
will need an upgrade of $9700 at the end of year 2. At the end of year 2, alternative Z
would be replaced with another alternative Z having the same installed and operating
costs. If the MARR is 14% per year, which alternative should be chosen?
Alternative X Y Z
Installed costs $68,500 $48,500 $33,500
Annual operating costs $6000 $4000 $5000
Overhaul cost in year 2–$9700 –
Salvage value $33,250 $28,250 $15,750
Useful life, years 8 4 2
27)
Answer:
AW (X) = – $18,254.90
AW(Y) = – $17,466.50
AW (Z) = – $17,984.57
Select the alternative with the least negative annual worth.
Explanation:
AW(X) = – 68,500(A/P, 14%, 8) –6000 +33,250(A/F, 14%, 8)
= – 68,500(0.2156) –6000 +33,250(0.0756)
= – 18,254.90
AW(Y) = – 48,500(A/P, 14%, 4) –4000 –9700(P/F, 14%, 2)(A/P, 14%, 4) +
28,250(A/F,14%, 4)
= – 48,500(0.3432) –4000 –9700(0.7695)(0.3432) +28,250(0.2032)
= – 17,466.50
AW(Z) = – 33,500(A/P, 14%, 2) –5000 +15,750(A/F, 14%, 2)
= – 33,500(0.6073) –5000 +15,750(0.4673)
= – 17,984.57
Select the alternative with the least negative annual worth.
28)
A major defense supplier is planning for the manufacturing of its Desert Hawk VII
Unmanned Aerial Systems, a hand–launched air vehicle that provides intelligence,
reconnaissance, and surveillance capabilities. Three manufacturing facilities are under
consideration. Each site provides the same manufacturing capability but has different
costs to manufacture and transport the system to the customers. The estimated costs are
provided in the table below.
Alternative Texas Plant Utah Plant Minnesota Plant
Set up costs $720,000 $770,000 $755,000
Annual operating costs $27,500 $22,500 $24,500
Salvage Value, at end of year 8$420,000 $430,000 $445,000
Transportation costs per unit $2000 $2250 $2300
Useful lives, years 8 9 10
If the company has to deliver 300 units per year for 8 years, which facility should be
selected based on the present worth method? Assume the company uses a MARR of 5%
and a study period of 8 years.
28)
23
Answer:
PW(TX) = – $4,491,402.00
PW(UT) = – $4,987,058.00
PW(MN) = – $5,071,780.40
Therefore, the Texas plant should be selected.
Explanation:
Compare alternatives over a study period of 8 years.
PW(TX) = – 720,000 –27,500(P/A, 5%, 8) – (2000)(300)(P/A, 5%, 8) +420,000(P/F,
5%, 8)
= – 720,000 – (27,500 +600,000)(6.4632) +420,000(0.6768)
= – 4,491,402.00
PW(UT) = – 770,000 –22,500(P/A, 5%, 8) – (2250)(300)(P/A, 5%, 8) +430,000(P/F,
5%, 8)
= – 770,000 – (22,500 +675,000)(6.4632) +430,000(0.6768)
= – 4,987,058.00
PW(MN) = – 755,000 –24,500(P/A, 5%, 8) – (2300)(300)(P/A, 5%, 8) +
445,000(P/F, 5%, 8)
= – 755,000 – (24,500 +690,000)(6.4632) +445,000(0.6768)
= – 5,071,780.40
PW(TX) > PW (UT) > PW (MN); therefore, the Texas plant should be selected.
24
29)
Compare the alternatives shown below on the basis of their future worth, using an interest
rate of 18% per year. Which alternative should be selected?
Alternative L M N
Initial costs $238,000 $213,000 $293,000
Annual revenues $69,000 in year 1,
increasing by $190
each year
$114,000 $69,000
in years 1 to 5,
$69,570
in years 6 to 16
Annual expenses $20,000 $25,000 $20,000
Salvage value $6000 $9000 $6000
Life, years 16 816
29)
Answer:
FW(L) =$277,411.15
FW(M) =$2,724,278.40
FW(N) = – $543,414.65
The alternative with the largest positive future worth should be selected.
Explanation:
FW(L) = – 238,000(F/P, 18%, 16)+[69,000 –20,000](F/A, 18%, 16) +190(P/G, 18%,
16)(F/P, 18%, 16) +6000
= – 238,000(14.129) + [69,000 –20,000](72.939) +190(22.3885)(14.129) +
6000
=277,411.15
FW(M) = – 213,000(F/P, 18%, 16)+[114,000 –25,000](F/A, 18%, 16)+ [9000 –
213,000](F/P, 18%, 8) +9000
= – 213,000(14.129) + [89,000](72.939) + [–204,000](3.7589) +9000
=2,724,278.40
FW(N) = – 293,000(F/P, 18%, 16) +69,000(F/A, 18%, 5)(F/P, 18%, 11) +69,570(F/A,
18%, 11)
–20,000(F/A, 18%, 16) +6000
= – 293,000(14.129) +69,000(7.1542)(6.1759) +69,570(28.7551) –20,000(
72.9390) +6000
= – 543,414.65
The alternative with the largest positive future worth should be selected.
25
30)
An insulated shipping container supplier is considering expanding its product line. Three
materials with different insulation properties are under consideration. The following
information is prepared for the economic evaluation of the best material. If the company’s
MARR is 4% per year and the study period is 5 years, use an AW–based incremental rate
of return equation to determine which alternative is preferred. Assume the salvage value is
negligible.
Material Q R S
First costs $46,000 $58,000 $61,000
Net annual revenue $10,000 in year 1,
increasing by
$100 per year
thereafter
$11,900,
increasing by
$450 per year
thereafter
$13,450
IRR (%) 5.93 5.15 5.14
Incremental IRR (%)
Q –
R 2.50% –
S 2.79% 5.09% –
30)
Answer:
AW (R–Q) = 0 = – 12,000(A/P, i%, 5) + 1900 + 350 (A/G, i%, 5)
IRR (R–Q) = 2.50% <4%; therefore, discard material R.
AW(S–Q) = 0 = – 15,000(A/P, i%, 5) + 3450 – 100 (A/G, i%, 5)
IRR (S–Q) = 2.79% <4%; therefore, discard material S.
Select material Q.
Explanation:
Rank Q, R and S
AW (R–Q) = 0 = [–58,000 – (–46,000)](A/P, i%, 5) + 1900 + (450–100) (A/G, i%,
5)
= – 12,000(A/P, i%, 5) + 1900 + 350 (A/G, i%, 5)
IRR (R–Q) = 2.50% <4%; therefore, discard material R.
AW(S–Q) = 0 = [–61,000 – (–46,000)](A/P, i%, 5) + 3450 – 100 (A/G, i%, 5) = 0
= – 15,000(A/P, i%, 5) + 3450 – 100 (A/G, i%, 5)
IRR (S–Q) = 2.79% <4%; therefore, discard material S.
Select material Q.
26
Answer Key
Testname: C6
1)
AW(Separate Lease) = – 53,550
AW(Combined Lease) = – 63,000
AW(Purchase) = – $38,076.00
The alternative with the least negative annual worth should be selected.
2)
IRR (V) =2.66% <6%; Discard vacuum forming.
Incremental IRR (F–D) = 19.43% >6%; Discard drape forming.
Incremental IRR (Z–M) = 8.30% >6%; Discard free blowing.
Pressure forming should be selected.
3)
AW(X) =$41,438.00
AW(Y) =$84,840.00
AW(Z) =$105,926.00
Select the alternative with the largest positive annual worth.
4)
FW(quarterly payment) =$231,621.33
FW(annual payment) =$220,021.10
Continuing to make quarterly payment offers more money at the time of retirement.
5)
PW(A) = – $119,406.22
PW(B) = – $137,556.60
Therefore, system X should be selected.
6)
AW(A) = – $10,104.55
AW(B) = – $10,275.85
Therefore machine A should be recommended.
7)
MV at the end of year 8=$16,792.91
AW(X) =$4997.36
AW(Y) =$6647.76
Select machine Y.
8)
PW(Purchase) = – $319,020.50
PW(Lease) = – $254,669.80
Therefore, the facility should be leased.
9)
PW(A) =$40,849.49
PW(B) =$41,381.95
Bond B should be purchased.
10)
PW(A) = – $102,140.80
PW(B) = – $83,550.73
PW(C) = – $83,349.30
Therefore, plan C should be selected.
11)
PW(A) =$9799.40
PW(B) =$9355.42
Therefore, alternative A should be selected.
Answer Key
Testname: C6
12)
PW(Delta–T) = $6370.50
PW(Alpha–B) =$10,487.50
Therefore Alpha–B should be recommended.
13)
Incremental IRR (X–Y) =14.83% >4%; discard machine Y
Machine X should be selected.
14)
FW(A) = – $5,053,855.29
FW(B) = – $5,682,668.31
The alternative with the least negative future worth should be selected.
15)
AW (Purchase) = – $101,744.50
AW (Lease) = – $54,188.40
Select the alternative with the least negative annual worth.
16)
AW(M) =$79,645.00
AW(N) =$99,640.00
AW(P) =$85,506.00
The alternative with the largest positive annual worth should be selected.
17)
CW (Fund) = – $487,500,000.00
CW (Building) = – $81,564,500.00
Select the alternative with the least negative capitalized worth.
18)
AW(A) = – $8566.00
AW(B) = – $7195.50
The alternative with the least negative annual worth should be selected.
19)
PW(CM/GC) = – $10,208.40
PW(DB) =$23,411.35
Therefore, the Design–Build method should be selected.
20)
FW(A) =$8,012,971.00
FW(B) =$7,436,476.25
The alternative with the largest positive future worth should be selected.
21)
FW(Electric) = – $176,217.48
FW(Gas) = – $91,517.46
FW(Oil) = – $153,104.10
The alternative with the least negative future worth should be selected.
22)
Incremental IRR (M–D) = 54.22% >9%; Discard Dynasty
Incremental IRR (Z–M) = 44.32% >9%; Discard Mighty–E
Zenn should be selected.
23)
FW(E) = – $13,748,192.81
FW(F) = – $12,232,706.60
The alternative with the least negative future worth should be selected.
28
Answer Key
Testname: C6
24)
ERR (M) =4.11% > 4%
ERR (N–M) =5.55% > 4%
Select project N.
25)
PW(ST) =$15,955,736.00
PW(LT) =$16,414,129.10
PW(GT) =$16,639,503.70
Therefore, Gentle Touch should be selected.
26)
CW (A) = – $57,715.00
CW (B) = – $2,885,750.00
Design B should be selected.
27)
AW (X) = – $18,254.90
AW(Y) = – $17,466.50
AW (Z) = – $17,984.57
Select the alternative with the least negative annual worth.
28)
PW(TX) = – $4,491,402.00
PW(UT) = – $4,987,058.00
PW(MN) = – $5,071,780.40
Therefore, the Texas plant should be selected.
29)
FW(L) =$277,411.15
FW(M) =$2,724,278.40
FW(N) = – $543,414.65
The alternative with the largest positive future worth should be selected.
30)
AW (R–Q) = 0 = – 12,000(A/P, i%, 5) + 1900 + 350 (A/G, i%, 5)
IRR (R–Q) = 2.50% <4%; therefore, discard material R.
AW(S–Q) = 0 = – 15,000(A/P, i%, 5) + 3450 – 100 (A/G, i%, 5)
IRR (S–Q) = 2.79% <4%; therefore, discard material S.
Select material Q.