Engineering Economic Analysis
Chapter 13  Replacement Analysis
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… from the Deacon/s Masterpiece by Oliver Wendell
Holmes
…
Now in building of chaises, I tell you what,There is always somewhere a weakest spot,—In hub, tire, felloe, in spring or thill,In panel, or crossbar, or floor, or sill,In screw, bolt, thoroughbrace,—lurking still,Find it somewhere, you must and will,—Above or below, or within or without,—And that's the reason, beyond a doubt,A chaise breaks down, but doesn't wear out.
You see, of course, if you 're not a dunce,
How it went to pieces all at once,-All at once, and nothing first,-Just as bubbles do when they burst.
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Reasons for Replacement Analysis
Physical Impairment (Deterioration)
Altered Requirements (Obsolescence)
Technology (Obsolescence
Financing (taxes, lease versus buy)
Replacement study is designed to decide to retain or to
replace now.
If to replace, then done. If not, then revisit annually.
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Types of Lives
Economic life – resulting in minimum equivalent uniform
annual cost (EUAC)
Ownership life – date of acquisition and date of
abandonment or replacement
Physical life – original acquisition to disposal
Useful life – time an asset is kept in productive services
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Replacement Factors
Recognition and acceptance of past errors
unable to see future, bad estimates
Sunk costs (forget the past, and the book value)
Use existing asset value and outsider viewpoint
Economic life of best challenge
Remaining economic life of defender
Income tax considerations
(sunk cost relevant here)
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Replacement Assumptions
Infinite Horizon (repeatability)
No technological change
Do Nothing option is not feasible
(or Keep defender is the do nothing option)
Economic life of asset is the age at which costs are
minimized.
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Defender
Available
Identify Participants
of Comparison
Defender
Marginal
Cost Data
Best Challenger
Not available
Defender
Marginal Cost
Increasing
no
Find lowest EUAC
for Defender
Find EUAC
over Given Life
yes
AT# 1 Compare
next year marginal
cost with C-EUAC
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AT# 2 Compare
lowest D-EUAC with
C-EUAC at min cost
life
rd
Lacking defender data,
compare D-EUAC over
life with C-EUAC at
minimum cost life
7
Annualize the Costs
Find the annual cost for the cash flow below.
n
0
1
2
3
cf
-15M -500K -825K 7.68M & -1.36M
$7.68M
$500K
825K
$15M
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1.36M
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Economic Life
You bought a car for $3K. Determine economic life at 12%.
Year
1
2
3
4
Annual Expenses
950 1050 1150 1550
Market Value
2250 1800 1450 1160
Loss in Market Value 750 450 350 290
Loss in interest
360 270 216 174
Total Marginal costs 2060 1770 1716 2014
Find EUAC of Total Marginal Costs.
(EUAC '(2060 1770 1716 2014) 12)
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Problem 13-12
First cost = $1050K; Salvage value = $225K at any time;
O&M = $235K with $75K gradient. MARR = 10%. Find economic
service life.
n
CR
O&M
CR+O&M
EUAC
1
930
235
1165
1165
2
497.86
310
807.86
994.93
3
354.24
385
739.24
917.68
4
282.76
460
742.76
880.00
5
240.13
535
775.13
862.82
6
211.93
610
821.93
857.52 ***
7
191.96
685
876.96
859.57
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Example
Defender has AOC of $60K per year, a life of 5 years with
zero salvage. Present market value is $30K. Challenger
costs $120K with AOC of $30K and a salvage after 5 years
of $50K. Use before tax MARR of 20%. AT#3
PWD(20%) = -30K – 60K(P/A, 20%, 5) = -$209,436.73
PWC(20%) = -120K - 30K(P/A, 20%, 5) + 50K(P/F, 20%, 5)
= -$189,624.49
Choose the challenger.
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Example
Defender bought 3 years ago for $120K with expected life
of 10 years and $25K salvage value with AOC of $30K.
Current book value is $80K and current life is 3 years.
Challenger available for $100K with trade-in of $70K for
defender. Useful life is 10 years, salvage value is $20K
and ASOC is $20K. Market value for defender is $70K.
Defender
Challenger
First cost
$70K
$100K
AOC
30K
20K
Salvage
10K
20K
Life (years)
3
10
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Example
First cost (sunk 5 years)
Depreciation
Market Value
Useful Life
Annual Operating Cost
Annual Benefit
Market Value after life
Combined tax rate
$12,500
Straight Line (salvage = $2,500)
$8,000
10 years
$3,000
$4,500
$1,500
34%
n
0
0
5
TI
500
-500
1500
BTCF
8K
-8K
1500
Dep
7500
-
Tax Rate (34%) ATCF
-170
$7830 (sell)
170
-$7830 (keep)
-510
990
Cost Basis = 12,500 - 5 * (12,500 –2500) / 10 = $7500
$8000 – 7500 = $500 (depreciation recapture)
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Example 13-1
$7500(A/P) $900(A/G) $500+$400(A/G, 8%, n)
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n
CRWR
1
2
3
4
5
6
7
8
9
1
11
12
13
14
15
8100
4206
2910
2264
1878
1622
1440
1305
1200
1117
1050
995
948
909
876
Maint
0
433
854
1264
1661
2048
2425
2789
3142
3484
3816
4136
4466
4746
5035 rd
Operating Total EUAC
500
692
880
1062
1238
1410
1578
1740
1896
2048
2196
2338
2476
2609
2738
8600
5331
4644
4589
4779
5081
5443
5834
6239
6650
7063
7470
7871
8265
8648
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Example 13-1
Minimum Cost Life
10000
9000
8000
Series1
7000
EUAC of Capital Recovery
Cost
6000
5000
EUAC of Mainteneance and
Repair
4000
EUAC of Operating
3000
Total EUAC
2000
1000
0
1
2
3
4
5
6
7
8 9 10 11 12 13 14 15
Year
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Marginal Costs
Challenger First Cost = $25K; Annual O&M = $2K with -$500 gradient;
Annual Risk $5K for 3 yrs; then -1500 Gradient
Market value ($18 13 9 6 4 3 2.5)K from 1 to 7 years; MARR = 15%
n
1
2
3
4
5
6
7
Lost Market Lost Interest
$7K 0.15*25K=$3750
5K
2700
4K
1950
3K
1350
2K
900
1K
600
0.5K
450
O&M
$2.0K
2.5K
3.0K
3.5K
4.0K
4.5K
5.0K
Risk
$5.0K
5.0K
5.0K
6.5K
8.0K
9.5K
11.0K
Total MC
$17,750
15,200
13,950
14,350
14,900
15,600
16,950
EUAC
17,750
16,564
15,811
15,518
15,427 ***
15,447
15,582
(AGP (List-pgf '(0 17750 15200 13950 14350 14900 15600 16950) 15) 15 7) 
$15,582.46
(EUAC cash-flow 15)
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Example 13-3
Defender's Market value= $15K; Annual O&M = $10K with a
$1500 gradient;
Market value ($14 13 12 11 10)K from 1 to 5 years; MARR = 15%
n
1
2
3
4
5
Lost MarketLost Interest
O&M
$1K 0.15*15K=$2250 $10.0K
1K
2100
11.5K
1K
1950
13.0K
1K
1800
14.5K
1K
1650
16K
Total Marginal Cost
$13,250
14,600
15,950 > $15,427 **
17,300
18,650
Note that the marginal costs are increasing each year => AT #1
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Problem 13-4
Maintenance costs are expected to be higher. Fully
depreciated and other factors irrelevant. Cost this year is
expected to be $800.
AT#1 one year,
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Problem 13-12
First cost = $1050K; Salvage value = $225K at any time
O&M = $235K with $75K gradient. MARR = 10%. Find economic
service life.
n
CR
O&M
EO&M
EUAC
1
$930K 235K
235K
1165K
2
497.857K 310K
270.71K
768.567K
3 354.244K 385K
305.24K
659K
4 282.763K 460K
338.59K
621.354K
5 240.133K 535K
370.76K
610.893K ***
6 211.926K 610K
401.77K
613.696K
CR = (P – S)(A/P, i%, n) + Si
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19
Big-J Construction Problem 13-21
MARR = 20%
n Operating
Cost
Maint
MV
Lost Lost
MV int
Total
Marginal
Cost
NPW
EUAC
0
1
120K
15K
9K
85K
35K 24K
83K
69167.67
82999.98
2
15K
10K
65K
20K 17K
62K
112222.22
73454.53
3
17K
12K
50K
15K 13K
57K
145208.33
68934.05
4
20K
18K
40K
10K 10K
58K
173179.01
66897.16
5
25K
20K
35K
5K
8K
58K
196487.91
65701.55
6
30K
25K
30K
5K
7K
67K
218926.08
65832.31
7
35K
30K
25K
5K
6K
76K
240136.28
66619.55
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Problem 13-22
MARR = 10%
Year D-MC
1
$2500
2
2400
3
2300
4
2550
5
2900
6
3400
7
4000
D-EUAC
$2500.00
2452.38
2406.34 ***
2437.30
2513.09
2628.04
2772.65
C-EUAC
$4500
3600
3000
2600 ***
2700
3500
4000
a) What is the lowest EUAC of the defender?
$2406.34
b) What is minimum cost life of challenger?
$2600
c) When should defender be replaced with challenger? Never
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Problem 13-23
n
1
2
3
4
5
6
MCD-D
$3000
3150
3400
3800
4250
4950
EUAC-C
$4500
4000
3300
4100
4400
6000
MARR = 10%
(EUAC '(3000 3150 3400 3800 4250 4950) 10) 
0 3000 3071.43 3170.69 3306.29 3460.87 3653.87
Shouldn't replace Defender at all, but had year 1 for D
> 3300 etc., replace then/now.
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Problem 13-25
Old forklift MARR = 8% AT#1
n
1
2
3
Maint Cost
400
600 800
Year
1
BTCF
-400
Depc
0
4
1000
TI
-400
5-10
1400/year
Tax 40% ATCF
$160
-$240
Challenger
Year BTCF
Depc
TI
Tax 40% ATCF
0
-6500
-$6500
1-10
-50
650
-700 $280
230
EUAC = 6500(A/P, 8%, 10) – 230 = $968.69 – $230
= $738.69 => keep old forklift for another year.
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Example
The economic service life given current market value of asset is $15K
and expected cash flows shown below and MARR at 10% is
Year Salvage Value
O&M Cost
1
$10K
$50K
2
8K
53K
3
5K
60K
4
0K
68K
a) i year
b) 2 years
c) 3 years
d) 4 years
EUAC1 = (15 – 10)K(A/P, 10%, 1) + 10K*0.1 + 50K =
$56.50K
EUAC2 = (15 – 8)K (A/P, 10%, 2) + 8K*0.1 + 51.53K = 56.36K ***
EUAC3 = (15 – 5)K (A/P, 10%, 3) + 5K*0.1 + 54.02K = 58.54K
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Example
When should the defender be replaced?
a) now b) 1 year from now
c) 2 years from now d) 3 years from now
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Year
AW-D Cost
AW-C Cost
1
$24K
$31K
2
25K
28K
3
26K
25K
4
27K
25.9K
5
28K
27.5K
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Economic Life
Asset bought at $8K at 8% cost of money.
n
1
2
3
4
5
Aexpense 3K
3K
3.5K 4K
4.5K
MV
4700 3200 2200 1450 950
Loss in MV 3300 1500 1000 750 500
Loss in Int 640
376 256
176 116
Total Margin 6940 4876 4756 4928 5116
6
5.52K
600
350
76
5946
(EUAC '(6940 4876 4756 4928 5116 5946) 8)
$5381.27 5 years
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PW of ATCF
Present Worth of After Tax data, find economic life of each
and when to replace defender.
Year Defender
Challenger
1
14020
18630
2
28100
34575
3
43075
48130
4
-65320
5
-77910
(mapcar #' AGP '(18630 34575 48130 65320 77910)
(list-of 5 12) (upto 5)) 
(20865.60 20457.96 20038.88 21505.59 21613)
(mapcar #' AGP '(14020 28100 43075) (list-of 3 12)(upto 3))
15702.40 16626.72 17934.23
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