Solution Manual of Engineering Economy 16th Edition By William G.SALLAVAN
Solutions to Chapter 6 Problems
6-1
According to Descartes’ Rule and Nordstrom’s Rule, a maximum of three interest rates exist for the above
incremental cash flows. In the interval 0%-100%, there is only a single unique internal rate of return,
which equals 26%. This can be seen from the following:
i PW (B-A)
1% 2,918
10% 1,234
20% 326 ] IRR = 26%
30% -164 ]
40% -463
50% -665
60% -809
70% -917
80% -1001
90% -1,067
100% -1,121
EOY Alt.B Alt. A Δ (B-A)
0 -$5,000 -$3,500 -$1,500
1 1,480 1,255 225
2 1,480 1,255 225
3 1,480 1,255 225
4 1,480 -2,245 3,725
5 1,480 1,255 225
6 -3,520 1,255 -4,775
7 1,480 1,255 225
8 1,480 -2,245 3,725
9 1,480 1,255 225
10 1,480 1,255 225
11 1,480 1,255 225
12 1,480 1,255 225
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372
6-2 Present Worth Method, MARR = 10% per year
PWD1 (10%) = −$600,000 − $780,000(P/A,10%,8) = −$4,761,222
PWD2 (10%) = −$760,000 − $728,000(P/A,10%,8) = −$4,643,807
PWD3 (10%) = −$1,240,000 − $630,000(P/A,10%,8) = −$4,600,987
PWD4 (10%) = −$1,600,000 − $574,000(P/A,10%,8) = −$4,662,233
Select Design D3 to minimize the present worth of costs.
Future Worth Method, MARR = 10% per year
FWD1 (10%) = −$600,000(F/P,10%,8) − $780,000(F/A,10%,8) = −$10,206,162
FWD2 (10%) = −$760,000(F/P,10%,8) − $728,000(F/A,10%,8) = −$9,954,471
FWD3 (10%) = −$1,240,000(F/P,10%,8) − $630,000(F/A,10%,8) = −$9,862,681
FWD4 (10%) = −$1,600,000 (F/P,10%,8) – $574,000(F/A,10%,8) = −$9,993,967
Select Design D3 to minimize the future worth of costs.
Annual Worth Method, MARR = 1% per year
AWD1 (10%) = −$600,000 (A/P,10%,8) – $780,000 = −$892,440
AWD2 (10%) = −$760,000 (A/P,10%,8) – $728,000 = −$870,424
AWD3 (10%) = −$1,240,000 (A/P,10%,8) – $630,000 = −$862,376
AWD4 (10%) = −$1,600,000 (A/P,10%,8) – $574,000 = −$873,840
Select Design D3 to minimize the annual worth of costs.
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373
6-3 Don’t be tempted to choose the project that maximizes the IRR! In this problem we should recommend
Project R15 because it has a larger PW (12%) than Project S19.
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374
6-4 Assume all units are produced and sold each year.
AWA(20%) = −$30,000(A/P,20%,10) + 15,000($3.50 − $1.00) − $15,000 = $15,345
AWB(20%) = −$60,000(A/P,20%,10) + 20,000($4.40 − $1.40) − $30,000 + $20,000(A/F,20%,10)
= $16,460
AWC(20%) = −$50,000(A/P,20%,10) + 18,000($4.10 − $1.15) − $25,000 + $15,000(A/F,20%,10)
= $15,753
Select Design B to maximize the annual worth.
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375
6-5 Let’s examine this problem incrementally. The labor savings for the new system are 32 hours per month
(which is 20% of 160 hours per month for the used system) x $40 per hour = $1,280 per month. The
additional investment for the new system is $75,000, and the incremental market value after five years is
$30,000. So we have:
PW(of difference at 1% per month) = $75,000 + $1,280 (P/A, 1%, 60) + $30,000 (P/F, 1%, 60) = $946
The extra investment for the new system is not justified. But the margin in favor of the used system is
quite small, so management may select the new system because of intangible factors (improved
reliability, improved image due to new technology, etc.).
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376
6-6 Wet Tower, Mechanical Draft
AW(12%) = −$3,000,000 (A/P,12%,30)
− 40
0.9
200hp
hp
0.746kw
(8,760 hr/yr) ($0.022/kWh)
− 20
0.9
150hp
hp
0.746kw
(8,760 hr/yr) ($0.022/kWh) − $150,000
= −$372,300 − $1,277,948 − $479,230 − $150,000 = −$2,279,478/yr.
Wet Tower, Natural Draft
AW(12%) = −$8,700,00(A/P,12%,30)
− 20
0.9
150hp
hp
0.746kw
(8,760 hr/yr) ($0.022/kWh) − $100,000
− $1,076,670 − $479,230 − $100,000 = −$1,658,900/yr.
Dry Tower, Mechanical Draft
AW(12%) = −$5,100,000(A/P,12%,30)
− 20
0.9
200hp
hp
0.746kw
(8,760 hr/yr) ($0.022/kWh)
− 40
0.9
100hp
hp
0.746kw
(8,760 hr/yr) ($0.022/kWh) − $170,000
= −$632,910 − $638,974 − $638,974 − $170,000 −$2,080,858/yr.
Dry Tower, Natural Draft
AW(12%) = −$9,000,000(A/P,12%,30)
− 40
0.9
100hp
hp
0.746kw
(8,760 hr/yr) ($0.022/kWh) − $120,000
= −$1,116,900 − $638,974 − $120,000 = −$1,875,874/yr.
The wet cooling tower with natural draft heat removal from the condenser water is the most economical
(i.e., least costly) alternative.
Non−economic factors include operating considerations and licensing the plant in a given location with
its unique environmental characteristics.
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377
6-7 PWA(20%) = −$28,000 + ($23,000 − $15,000)(P/A,20%,10) + $6,000(P/F,20%,10)
= $6,509
PWB(20%) = −$55,000 + ($28,000 − $13,000)(P/A,20%,10) + $8,000(P/F,20%,10)
= $9,180
PWC(20%) = −$40,000 + ($32,000 − $22,000)(P/A,20%,10) + $10,000(P/F,20%,10)
= $3,540
Select Alternative B to maximize present worth.
Note: If you were to pick the alternative with the highest total IRR, you would have incorrectly
selected Alternative A.
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378
6-8 ILB: Installation cost is $2,000 plus the cost of the bulbs ($500) for an investment cost of $2,500. This is
assumed to occur at the beginning of each year in the 8-year study period. The cost of electricity for
60,000 kWh at $0.12 per kWh is $7,200 per year. Let’s assume that the electricity expense is incurred at
the end of each year for eight years:
PW(12%) = $2,500 $2,500(P/A, 12%, 7) $7,200(P/A, 12%, 8) = $49,677
CFL: Installation cost is $3,000 plus the cost of the 1,000 CFLs for a total investment cost of $5,000.
This is assumed to be incurred at the beginning of the 8-year study period. The cost of electricity for
13,000 kWh at $0.12 per kWh is $1,560 at the end of each year for eight years:
PW(12%) = $5,000 $1,560(P/A, 12%, 8) = $12,749
The boss will be happy to learn that CFLs offer tremendous cost savings over the ILBs. CFLs cost about
26% of the PW of cost of the ILBs over the 8-year study period. A side note: In Europe ILBs will not be
sold in stores beginning in September of 2009. ILBs are simply too energy inefficient and create too
much of a carbon footprint! Let’s become “enlightened” and make the change.
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379
6-9 Jean’s future worth at age 65 will be $1,000 (F/A, 6%, 10) (F/P, 6%, 25) = $56,571. Doug’s future worth
will be $1,000 (F/A, 6%, 25) = $54,865. Jean’s future worth will be greater than Doug’s even though she
stopped making payments into her plan before Doug started making payments into his plan! The moral is
to start saving for retirement at an early age (the earlier the better).
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380
6-10 We can examine the incremental cash flows (R − O) to determine the IRR on this difference, Δ( R − O):
EOY Δ (R−O)
0 -$6,000
1 0
2 0
3 11,718
The IRR on the incremental cash flow is 25%, so recommend the rectangular re-bar. This can be
confirmed by computing the PW (25%) of each alternative: PW of O equals $729 and PW of R equals
$1,510
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381
6-11 Tool B should not be considered further since its IRR < 8%.
PWA = $55,000 + ($18,250 $6,250)(P/A, 8%, 7) + $18,000(P/F, 8%, 7) = $17,980
PWC = $80,000 + ($20,200 $3,200)(P/A, 8%, 7) + $22,000(P/F, 8%, 7) = $21,346
Select Tool C.
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382
6-12 Design A: All components have a 20 year life.
Capital Investment
Concrete pavement: ($90/ft)(5,280 ft/mi) = $475,200 /mile
Paved ditches: 2($3/ft)(5,280 ft/mi) = $31,680 /mile
Box culverts: (3 culverts/mile)($9,000/culvert) = $27,000 /mile
Total Capital Investment = $533,880 /mile
Maintenance
Annual maintenance: $1,800 /mile
Periodic cleaning of culverts*
:
(3 culverts/mile)($450/culvert) = $1,350 /mile every 5 years
AWA(6%) = −$533,880(A/P,6%,20) − $1,800 − $1,350(A/F,6%,5)*
= −$48,594 /mile
PWA(6%) = −$533,880 − [$1,800 + $1,350(A/F,6%,5)](P/A,6%,20) = −$557,273 /mile
* assumes a cleaning also occurs at the end of year 20.
Design B: All components have a 10 year life.
Capital Investment (Year 0)
Bituminous pavement: ($45/ft)(5,280 ft/mi) = $237,600 /mile
Sodded ditches: 2($1.50/ft)(5,280 ft/mi) = $15,840 /mile
Pipe culverts: (3 culverts/mile)($2,250/culvert) = $6,750 /mile
Total = $260,190 /mile
Capital Investment (EOY 10)
Bituminous pavement: ($45/ft)(5,280 ft/mi) = $237,600 /mile
Sodded ditches: 2($1.50/ft)(5,280 ft/mi) = $15,840 /mile
Replacement culverts:
(3 culverts/mile)($2,400/culvert) = $7,200 /mile
Total = $260,640 /mile
Maintenance
Annual pavement maintenance: = $2,700 /mile
Annual cleaning of culverts:
(3 culverts/mile)($225/culvert) = $675 /mile
Annual ditch maintenance:
2($1.50/ft)(5280 ft/mi) = $15,840 /mile
Total = $19,215 /mile
AWB(6%) = −[$260,190 + $260,640(P/F,6%,10)](A/P,6%,20) − $19,215 = −$54,595 /mile
PWB(6%) = −$260,190 − $260,640(P/F,6%,10) − $19,215(P/A,6%,20) = −$626,126 /mile
Select Design A (concrete pavement) to minimize costs.
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383
6-13 Method: Incremental PW
Order alternatives by increasing capital investment: ER3, ER1, ER2.
Is ER3 an acceptable base alternative?
PWER3(12%) = −$81,200 + $19,750(P/A,12%,6) = $0.15 0.
Since PW(MARR=12%) 0, ER3 is an acceptable base alternative.
Analyze (ER1 − ER3)
PW(12%)= −($98,600 − $81,200) +
0.12 0.06
$25,800[1 ( / ,12%,6)( / ,6%,6)]
P F F P
− $19,750(P/A,12%,6)
= −$17,400 +
0.06
$25,800(0.2814)
− $19,750(4.1114)
= $22,402 > 0
The additional capital investment earns more than the MARR. Therefore, design ER1 is preferred to
design ER3.
Analyze (ER2 − ER1)
PW(12%) = −($115,000 − $98,600) + $29,000(P/A,12%,6) + $150(P/G,12%,6)
−
0.12 0.06
$25,800[1 ( / ,12%,6)( / ,6%,6)]
P F F P
= −$16,400 + $29,000(4.1114) + $150(8.93) −
0.06
$25,800(0.2814)
= −$16,832 < 0
The additional capital investment required by design ER2 has a negative PW (earns less than the MARR).
Therefore, design ER1 is preferred to design ER2.
Decision: Recommend Design ER1
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384
6-14 PWA = $15,500(P/A, 12%, 10) + $500(P/G, 12%, 10) = $97,705
PWB = $12,000(P/A, 12%, 10) + $2,000(P/G, 12%, 10) = $108,310
Based on property tax assessments, Parcel A is preferred to Parcel B.
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385
6-15 (a) PW1(10%) = −$10,000 + $5,125 (P/A, 10%, 3) = −$10,000 + $5,125 (2.4869) = $2,745
PW2(10%) = −$8,500 + $4,450 (P/A, 10%, 3) = $2,567
PW3(10%) = −$11,000 + $5,400 (P/A, 10%, 3) = $2,429
(b) IRR1 = 25%; IRR2 = 26.5%; IRR3 = 22.2%
(c) Select Project 1 to maximize profitability.
(d) This is because the IRR method assumes reinvestment of cash flows at the IRR whereas the PW
method assumes reinvestment at the MARR.
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386
6-16 (a) Cost of electricity for the 90% efficient motor:
[30 hp/0.90](0.746 kW/hp)($0.10/kWh)(4,000 hr/yr) = $9,946.67 per year
Cost of electricity of the 93% efficient motor:
[30 hp/0.93](0.746 kW/hp)($0.10/kWh)(4,000 hr/yr) = $9,625.81 per year
PW90%(15%) = −$2,200 − $9,946.67(P/A, 15%, 8) = −$46,834
PW93%(15%) = −$3,200 − $9,625.81(P/A, 15%, 8) = −$46,394
The 93% efficient motor is the better choice. Notice that energy expense dominates the analysis.
For instance, the 93% efficient motor has a PW of energy expense that is $43,194 / $3,200, or over
13 times its purchase price. Generally speaking, the potential for savings is enourmous for high
efficiency motors and pumps.
(b) $9,946.67 − $9,625.81 = $320.86 per year savings
PW(15%) of savings = $320.86(P/A, 15%, 8) = $1,440, which exceeds the extra investment cost of
$1,000, so the more efficienct motor is preferred. Thus, the PW of the incremental investment is
$440.
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