Dr Azer Önel
Engineering Economy Review Problems-V
2010
Comparing Mutually Exclusive Alternatives
(Note: There may be typographical errors. Check results for all problems!)
Study Period ≠ Useful Life
• Repeatability Assumption
• Study period is a common multiple of the alternatives' lives.
• Compare alternatives over the lowest common multiple of lives. Assume identical
investmet, costs, revenue, etc.
• Compute AW. (Easy!)
• Cotermination Assumption
• Compare alternatives over the same study period.
• Study period > Useful life
Cost alternatives: Assuming repeatability, repeat part of the useful life of the original
alternative, and then use an estimated MV to truncate it at the end of the study period.
Without repeatability, we must purchase/lease the service/asset for the remaining years.
Investment alternatives: Assume all cash flows will be reinvested at the MARR to the end
of the study period (i.e., calculate FW at end of useful life and move this to the end of the
study period using the MARR).
• Study period < Useful life
Truncate (shorten) the alternative at the end of the study period using an estimated market
value.
1) (Prob. 5-24 Sullivan, 12th ed.) Consider the two cost alternatives A & B.
Investment cost
Annual costs
Useful life years
Market value @end of useful life
A
$14,000
14,000
5
8,000
B
$65,000
9,000
20
13,000
a) If the study period = 20 years, which alternative is preferred? The MARR is 15%.
b) If the study period = 10 years and the estimated market value for alternative B = $25,000
@ EOY 10, which alternative is preferred?
Solution:
a) Repeatability assumption; study period = 20 years
Let’s use PW method:
Project A is repeated 4 times, so
PWA = -14,000-14,000(P|A,15%,20) + (-14,000+8000=-6,000)(P|F,15%,5)6,000(P|F,15%,10) -6,000(P|F,15%,15)+8,000(P|F,15%,20)
=-$106,345
PWB = -65,000 - 9,000(P|A, 15, 20) + 13,000(P|F, 15, 20)
= -$120,539
Select A to minimize costs.
1
Also, AWA = -$106,345(A|P, 15%, 20) = -$16,990
AWB = -$120,539(A|P, 15%, 20) = -$19,262
Note: When the study period equals a common multiple of the alternatives' lives, simply
compare AW computed over the respective useful lives (assuming repeatability is valid)!
So: AWA = AW 1-20 = AW 1-5 = AW 6-10 = AW 11-15 = AW 15-20 = -$16,990
b) Cotermination assumption; study period = 10 years
AWA = -$16,990
(Here assume repeatability; so use AW method.)
AWB = -65,000(A|P,15%,10) - 9,000 + 25,000(A|F,15%,10) = -$20,722
Alternative A is still preferred.
2) Select the preferred investment alternative from the mutually exclusive pair shown in the
following table based on
a) the repeatability assumption,
b) the coterminated assumption with a four-year study period and the market value of
Alternative 2 (at the end of year four) determined using the imputed market value technique,
c) the coterminated assumption with an eight-year study period (Alternative 1 would not be
repeated).
The MARR is 10% per year.
End of Year
0
1
2
3
4
5
6
7
8
8 (MV)
Alternative 1
Alternative 2
$40,000
$50,000
12,000
12,000
12,000
36,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
40,000
Solution:
a) Repeatability assumption; study period = 8 years
AW1(10%)
= -$40,000 (A/P,10%,4) + $12,000 + $24,000 (A/F,10%,4)
= $4,552
AW2(10%)
= -$50,000 (A/P,10%,8) + $10,000 + $40,000 (A/F,10%,8)
= $4,126
Select Alternative 1 to maximize annual worth.
b) Coterminated assumption; study period = 4 years
AW1(10%) = $4,552 (from Part a)
2
To calculate AW2, we will use an imputed market (salvage) value estimate at the end of year
4 (MV4) based on the remaining annual capital recovery amount and the market value at end
of year 8.
Methods for estimating market (salvage) value
• Obtain a current estimate from the market place
• Estimate salvage value at time t using straight line depreciation
Investment cost- Annual depreciation expense*t = $I - ($I/useful life)*t
• Market value at time t
(PW at the end of year t of remaining CRC) + PW at the end of year t of original
market value at end of useful life)
MVt = [$I(A/P,i%,n) - $S(A/F,i%,n)] (P/A,i%,t) + S (P/F,i%,t)
Remaining CRC
PW of CRC
PW of S at time t
Imputed market value (MV4) for Alternative 2 using the third method:
Estimated S
0
t=4
Original S
Useful life=8 years
MV4 = [$50,000(A/P,10%,8) - $40,000 (A/F,10%,8)] (P/A,10%,4) + 40,000 (P/F,10%,4)
CRC
PW of CRC
PW of S at time t
= $45,940
AW2(10%) = - $50,000 (A/P,10%,4) + $10,000 + $45,940 (A/F,10%,4)
= $4,125
Select Alternative 1 to maximize annual worth. Notice that AW2 is the same (slight
difference due to rounding) in part (b) using the imputed MV concept as it was in part (a) with
the repeatability assumption.
b) Coterminated assumption; study period = 8 years
Since the investment in Alternative 1 will not be repeated, assume that the positive cash flows
of Alternative 1 are reinvested at the MARR for the 8-year study period.
FW1(10%) = -$40,000 (F/P,10%,8) + [$12,000(F/A,10%,4) + $24,000](F/P,10%,4)
= $30,933
FW2(10%)= -$50,000 (F/P,10%,8) + $10,000(F/A,10%,8) + $40,000
= $47,179
Select Alternative 2 to maximize future worth.
3
Comparing Independent, Dependent & Mutually Exclusive Alternatives
1) (Prob. 5-35 Sullivan, 12th ed.) Consider the alternatives below, which are compared over
10 years.
A
$30,000
8,000
3,000
Capital investment
AR-AC
Market value
B1
$22,000
6,000
2,000
B2
$70,000
14,000
5,000
C
$82,000
18,000
7,000
Projects B1 and B2 are mutually exclusive. Project C depends on B2 and A depends upon B1.
Capital is limited to $100,000 and MARR is 12%/year. Which bundle (combination) of
projects should be selected?
Solution:
Project
Combination
1. DN
2. B1
3. A, B1
4. B2
5. B2, C
A
B1
B2
C
Investment $
Feasible?
PW
0
0
1
0
0
0
1
1
0
0
0
0
0
1
1
0
0
0
0
1
0
22,000
52,000
70,000
152,000
yes
yes
yes
yes
no
0
$12,545
$28,713
$10,713
Select A and B1 with maximum PW.
2) Consider the following information:
Incremental IRR in % when
compared with alternative
Alternative Initial investment IRR %
A
B
A
- 40,000
29
B
- 75,000
15
1
C
-100,000
16
7
20
D
-200,000
14
10
13
C
12
If the alternatives are independent, which should be selected if the company's MARR is 15%
per year?
a) Only alternative A
c) Alternatives A and C
b) Alternatives A and B
d) Alternatives A, B, and C**
If the alternatives are mutually exclusive and the company's MARR is 11% per year, which
alternative(s) should be selected?
a) A, B, C, and D
c) Only D
b) Only A**
d) Only A and C
4
If the alternatives are mutually exclusive and the company's MARR is 10% per year, which
alternative(s) should be selected?
a) A, B, C, and D
c) Only A
b) Only, A, B, and C
d) Only D**
Depreciation and After-Tax Cash Flows
1) A computer was purchased for $16,000 and $2,000 was spent installing it. It has a life of 4
years. Compute the depreciation deduction (expense) in year 2 and the book value at the end
of year 3 using:
a) straight-line method
b) declining balance method
Solution:
Straight line rate=1/4=25%; declining balance rate=2/4=50%
EOY
0
1
2
3
4
Straight Line
Depr. Exp.
0
4,500
4,500
4,500
4,500
Book Value
18,000
13,500
9,000
4,500
0
Declining Balance
Depr. Exp.
0
9,000
4,500
2,250
2,250
Book Value
18,000
9,000
4,500
2,250
0
2) Compare the following alternative material handling systems, which provide the same
amount of revenue. Which one would you choose?
System X
System Y
First cost $
80,000
200,000
Estimated useful life, years
20
40
Annual cost, $
18,000
6,000
Straight line depreciation is used; tax rate is 30%; After-tax MARR is 6%.
Solution:
System X
A
EOY BTCF
0
-$80,000
1-20 -$18,000 †
B
Depr. Exp.
--$4,000*
C=A+B†
Taxable Income
--$22,000**
D=t*C
Tax Paid (40%)
-$6,600**
E=A-D
ATCF
-$80,000
-$11,400
†
These are annual costs; thus C=A+B.
*Straight line rate=1/20
**Since taxable income is negative (-$22,000), you don’t pay taxes; $22,000*0.4=$6,600 are
a savings in engineering economic terms!
5
System Y
A
EOY BTCF
0
-$200,000
1-20 -$6,000
B
Depr. Exp.
--$5,000
C=A+B
Taxable Income
--$11,000
D=t*C
Tax Paid (40%)
-$3,300
E=A-D
ATCF
-$200,000
-$2,700
AWX= -80,000(A/P,6%,20)-11,400= -$18,360
AWY= -200,000(A/P,6%,40)-2,700= -$16,000. Select Y with minimum cost.
Benefit/Cost Ratio Method
1) (Prob. 11-5 Sullivan, 12th ed.) There are 2 alternatives for generating power. Which one
should be selected?
Capital investment $
Annual power sales $
Annual O&M costs
Annual flood-control savings $
Annual irrigation benefits $
Annual recreation benefits $
Ability to attract new industry $
Useful life, years
Interest rate: 5% per year
A (Coal powered)
20,000,000
1,000,000
200,000
500,000
50
B (Hydroelectric)
30,000,000
800,000
100,000
600,000
200,000
100,000
400,000
50
Solution:
B/C ratio= (AW of Benefits - AW of Disbenefits)/ (AW of Inv. Cost+Annual O&M costs)
B/C for A: (1,000,000+500,000)/ [20,000,000(A/P,5%,50) +200,000] =1.16>1.0
B/C for B: (800,000+1,300,000)/ [30,000,000(A/P,5%,50)+100,000]=1.20>1.0
Both A and B are acceptable. (Note B/C ratios above are actually ∆ (DN→A) and ∆
(DN→B). Both ratios are greater than 1.0, and so acceptable!)
Use incremental analysis.
Rank order alternatives from low capital investment to high capital investment: A→B.
∆ (A→B):
∆ Capital investment= $10,000,000 ∆ Annual O&M= $100,000
∆ Annual power sales= $200,000
∆ Other benefits= $800,000
∆B/C: (800,000-200,000)/ [10,000,000(A/P,5%,50)-100,000]=1.34>1.0 Select B.
6
Dealing with Uncertainty
1) (Prob. 10-18 Sullivan, 12th ed.) Suppose for an engineering project the optimistic, most
likely, and pessimistic estimates are as shown below.
Capital Investment
Useful Life
Market Value
Net Annual Cash Flow
MARR
Optimistic
-$80,000
12 years
$30,000
$35,000
12%/yr
Most Likely
-$95,000
10 years
$20,000
$30,000
12%/yr
Pessimistic
-$120,000
6 years
0
$20,000
12%/yr
a) What is the AW for each of the three estimation conditions?
b) It is thought the most critical elements are useful life and net annual cash flow. Develop a
table showing the AW for all combinations of estimates for these two factors, assuming that
all other factors remain at their most likely values.
Solution:
a) AWO = -80,000(A/P,12%,12)+30,000(A/F,12%,12)+35,000= $23,330
AWML = -95,000(A/P,12%,10)+20,000(A/F,12%,10)+30,000= $14,325
AWP = -120,000(A/P,12%,6)+20,000= -$9,184
b) Summary results
O
ML
P
1
2
3
Net Annual
Cash Flow
$35,000
$30,000
$20,000
O
(12 years)
20,4951
15,495
5,495
Useful Life
ML
(10 years)
19,325
14,325
4,3252
P
(6 years)
14,360
9,360
-6403
AWO = -95,000(A/P,12%,12)+20,000(A/F,12%,12)+35,000= $20,495
AWML = -95,000(A/P,12%,10)+20,000(A/F,12%,10)+20,000= $4,325
AWP = -95,000(A/P,12%,6)+20,000(A/F,12%,6)+20,000= -$640
2) (Prob. 10-18 Sullivan, 12th ed.) A certain potential investment project is critical to a firm.
The following are “best” or “most likely” estimates:
Investment
$100,000
Life
10 yr
Salvage value
$20,000
Net annual cash flow
$30,000
MARR
10%
It is desired to show the sensitivity of a measure of merit (net annual worth) to variation, over
a range of ±50% of the expected values, in the following elements: (a) life, (b) net annual cash
flow, and (c) interest rate. Graph the results. To which element is the decision most sensitive?
7
Solution:
If n = project life, i = interest rate, and A = net annual cash flow, we have
AW = -$100,000(A/P,i,n) + A + $20,000(A/F,i,n)
%
change
-50%
-40%
-30%
-20%
-10%
0%
10%
20%
30%
40%
50%
Net Annual Worth
Life
Net Annual MARR
Cash Flow
$6,896
$(20)
$18,640
$9,631
$2,980
$17,931
$11,568 $5,980
$17,210
$13,004 $8,980
$16,478
$14,109 $11,980
$15,734
$14,980 $14,980
$14,980
$15,683 $17,980
$14,216
$16,259 $20,980
$13,441
$16,738 $23,980
$12,657
$17,140 $26,980
$11,863
$17,482 $29,980
$11,060
Net Annual Worth is most sensitive to changes in the net annual cash flow. However, the
decision to invest in the project is relatively insensitive to changes in the specified range.
$24,980
CF
$19,980
$14,980
LIFE
MARR
$9,980
$4,980
-$20
-50
-30
-10
10
30
50
8