Transcript CHAPTER V
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ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 11
Replacement and Retention
Decisions
Blank & Tarquin: 5th edition. Ch.5 Authored by Dr. Don Smith, Texas A&M University
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ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 11
Learning Objectives
Blank & Tarquin: 5th edition. Ch.5 Authored by Dr. Don Smith, Texas A&M University
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Learning Objectives
1.
2.
3.
4.
5.
Basics of Replacement Study
Economic Service Life
Performing a Replacement Study
Additional Considerations
Replacement Study over a Specified
Study Period
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ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 11
1. Basics of Replacement Study
Blank & Tarquin: 5th edition. Ch.5 Authored by Dr. Don Smith, Texas A&M University
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11.1 Why Replace Assets
Reduced Performance:
Wear and tear;
Decreasing reliability and productivity;
Increasing operating and maintenance costs.
Altered Requirements:
New production needs, accuracy, speed, etc.
Obsolescence:
Current assets may be less productive;
Not meet competition.
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11.1 Terminology
Defender Asset:
Currently installed asset;
Challenger Asset:
The potential replacement or “challenging”
asset;
Under consideration to replace the defender
asset.
Together, the Defender and Challenger:
Constitute mutually exclusive alternatives;
Select one and reject the other.
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11.1 Terminology
Annual Worth Values:
EUAC – since costs tend to dominate
the study (-) cash flows;
Salvage values – if any – are also part
of the analysis (+) value.
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11.1 Terminology
Economic Service Life (ESL)
Number of years for an alternative for
which the AW is minimum;
Computing the AW for 1 year; then 2
years; …until a minimum cost time
period is found;
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11.1 Terminology
Two investment costs are critical:
1. defender first cost;
2. challenger first cost.
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11.1 Defender first cost
The current market value (MV) is the
correct estimate to use for P for the
defender.
Both trade-in value and book value
are not correct.
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11.1 Challenger First Cost
This is the total investment required to
replace the current defender.
Always equal to P.
What if a trade in value is offered for
the defender to apply to the challenger?
At times, a “high” trade-in value may be offered
for the defender by the vendor compared to its
current fair market value.
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11.1 Trade-In Issues
If a trade-in is offered, what should be
the proper investment cost in the
challenger?
Let PC = the cash sale price with no trade-in
Let TIV = the trade-in value for the defender
Let MVD = the Fair Market Value of current
defender
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11.1 Trade-In Issues
Investment in the challenger
PC – (TIV – MVD)
Cash Price
with
no trade-in
The market value
Trade in value
This represents the true investment in the
challenger to the firm!
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11.1 Example
Bought a system 3 years ago for
$120,000 (Defender);
A fair market value of the current
defender is $70,000 right now;
A challenger can be purchased for cash
for $100,000 now!
The vendor selling the challenger offers
a trade-in of $80,000 on the current
defender.
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11.1 Trade-In Issues – Example
What should be the proper investment
cost for the challenger to the firm if the
defender is traded?
PC = $100,000
TIV = $80,000
MVD = $70,000
InvestmentChallenger now = $100,000 –
($80,000 - $70,000) = $90,000.
This represents the “true” investment in the
challenger with the trade.
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11.1 Assumptions
The traditional approach to
conducting a replacement analysis
is:
The Annual Cost or Annual Worth
approach!
With an assumed interest rate;
Assumed lives for each alternative.
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ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 11
2. Economic Service Life
Blank & Tarquin: 5th edition. Ch.5 Authored by Dr. Don Smith, Texas A&M University
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11.2 Economic Service Life (ESL)
The ESL for a given asset is:
The number of years where the AW
of the future costs is minimum;
Termed “The minimum cost life”
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11.2 ESL – General Format
Compute:
Total AW =-Capital Recovery-AW of AOC
Observe the min. cost “n” value.
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11.2 Capital Recovery Formula
Sn
0
1
2
////
...
n-1
n
$P
CRC(i%) = -P(A/P,i%,n) + S(A/F,i%,n)
CRC(i%) is the annual cost of “owing” an
asset over “n” time periods.
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11.2 Typical ESL Plot
Min. Total AW of costs life
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11.2 AW over “k” Years
Notation:
P = initial investment in the asset;
Sk = estimated salvage value after “k”
years;
AOCj = annual operating costs for
year j (j = 1 to k)
“k” the number of years for the
analysis.
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11.2 Closed Form of AWk
Analytical Form for Total AWk:
Total AWk P ( A / P, i, k ) S k ( A / F , i, k )
k
AOC j ( P / F , i, j ) ( A / P, i, k )
j=1
Procedure: Year-by-year analysis for “k” years – where
“k” is given or assumed.
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11.2 Example 11.2 – Overview
Defender Asset;
3 years old now;
Market value now $13,000;
5-year study period assumed;
Requires estimates of the future
salvage values and annual operating
costs for the 5-year period.
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11.2 Example: Future Market Values
Estimated Future Market Values and
AOC’s:
MktVt
AOCt
t = 1: $9,000
$-2,500
t = 2: $8,000
-2,700
t = 3: $6,000
-3,000
t = 4: $2,000
-3,500
t = 5: $0
-4,500
Mkt. Values are
decreasing:
AOC’s are
increasing:
Assume the interest rate is 10% per year.
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11.2 Example: Find the ESL
Period-by-period analysis
For “k” = 1 year:
S1 =
$9,000
0
1
P=$13,000
AOC1 = -2,500
AW(10%)1 = (-$13,000)(A/P,10%,1) + $9,000(A/F,10%,1) –2,500
= -$7,800 ( for one year!)
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11.2 Example: Find the ESL
Period-by-period analysis
For “k” = 2 years:
S2 = $8,000
0
1
2
AOC1 = -2,500
P=$13,000
AOC2 = -$2,700
AW(10%)2 = (-13,000)(A/P,10%,2) + 8,000(A/F,10%,2)
-[2,500(P/F,10%,1) + 2,700(P/F,10%,2)](A/P,10%,2)
= -$6,276/yr for 2 years.
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11.2 Example: Find the ESL
Period-by-period analysis
For “k” = 3 years:
0
1
2
S3 = $6,000
3
AOC1 = -2,500
P=$13,000
AOC2 = -$2,700
AOC3 = -$3,000
AW(10%)3 = (-13,000)(A/P,10%,3) +6,000(A/F,10%,3)
-[2500(P/F,10%,1) + 2,700(P/F,10%,2) + 3,000(P/F,10%,3](A/P,10%,3) =
-$6,132/yr for 3 years.
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11.2 Example – Continued
A similar analysis for k = 4 and 5 is conducted;
The AW(10)k, K = {1,2,3,4,5} are tabulated as:
Total AWk
k=1:
-7,800
k=2:
-6,276
k=3:
-6,132
k=4:
-6,556
k=5:
-6,579
Min. Cost Year = 3 years
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11.2 Spreadsheet Format
1
2
3
Input Parameters
Interrest Rate (%)
Investment Cost ($)
No of Years to Study
Year
1
2
3
4
5
(3)
Cap. Rec. Costs
-$5,300
-$3,681
-$3,415
-$3,670
-$3,429
$
10.00%
13,000.00
5
(1)
Mkt. Value
$9,000.00
$8,000.00
$6,000.00
$2,000.00
$0.00
(4)
AW of AOC's
-$2,500
-$2,595
-$2,718
-$2,886
-$3,150
(2)
AOC/Yr
-$2,500.00
-$2,700.00
-$3,000.00
-$3,500.00
-$4,500.00
Year
1
2
3
4
5
(5)
Total AW(i%)
-$7,800
-$6,276
-$6,132
-$6,556
-$6,580
Base Input
Parameters
Schedule of
Est. Mkt. Values
and AOC’s/year
Tabulation of
CRS’s, AOC’s and
Total AW(i%)
Min Life
ID
Min Life
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11.2 Plot of Capital Recovery Costs
ESL
CRC
$6,000.00
$5,000.00
$AW
$4,000.00
$3,000.00
$2,000.00
$1,000.00
$0.00
0
1
2
3
4
5
6
Years
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11.2 Plot of the AOC’s
ESL
AOC
$3,500
$3,000
$AW
$2,500
$2,000
$1,500
$1,000
$500
$0
0
1
2
3
4
5
6
Years
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11.2 Combined Plots for Example
ESL
CRC
AOC
AW
$9,000.00
$8,000.00
$7,000.00
$AW
$6,000.00
Min AW
Year!
$5,000.00
$4,000.00
$3,000.00
$2,000.00
$1,000.00
$0.00
0
1
2
3
4
5
6
Years
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11.2 ESL vs. AW
Traditional AW Analysis:
“n” is fixed or assumed;
First cost at t = 0;
Est. salvage value at t = “n”;
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11.2 ESL vs. AW
ESL Analysis:
“n” varies from t = 1 to t = “k:
analysis using AW(i%)
Table of possible future salvage values
Table of future AOC’s, year by year.
Seeking the min. AW life in an ESL analysis.
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11.2 Marginal Cost Approach
Marginal Costs are year-by-year
estimates of the costs to:
Own the asset, and
Operate the asset
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11.2 Marginal Cost Approach
Three Components of Marginal
Costs:
1. Cost of ownership (loss in Mkt.
Value/yr);
2. Foregone interest of Mkt. Value at
beginning of the year;
3. AOC for each year.
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11.2 Marginal Cost Analysis
Compute the marginal costs per year;
Find their equivalent annual worth;
AW of marginal costs = total AW of costs
ESL analysis and Marginal Cost analysis
can draw the same result!
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11.2 Marginal Cost Format: Example 11.2
Year
1
2
3
4
5
time
1
2
3
4
5
MV
$9,000
$8,000
$6,000
$2,000
$0
Marginal Cost
For the Year
-$7,800.00
-$4,600.00
-$5,800.00
-$8,100.00
-$6,700.00
Loss in MV
for Year
-$4,000.00
-$1,000.00
-$2,000.00
-$4,000.00
-$2,000.00
AW of
Marginal Costs
-$7,800
-$6,276
-$6,132
-$6,556
-$6,580
Lost Interest
on MV for Year
-$1,300.00
-$900.00
-$800.00
-$600.00
-$200.00
Est.
AOC/Year
-$2,500.00
-$2,700.00
-$3,000.00
-$3,500.00
-$4,500.00
Min. Cost Life:
At t = 3.
Same result as
the ESL analysis.
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11.2 Setting Up for a Replacement Analysis
Conduct the ESL for the defender and
the challenger(s);
Form the following alternatives:
Challenger Alternative (C): AWC for nC yrs;
Defender Alternative: (D): AWD for nD yrs.
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ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 11
3. Performing a Replacement
Study
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11.3 Two Ways to Perform the Study
1. Without a specified study period;
2. With a specified study period
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11.3 Overview of the two Methods
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11.3 New Replacement Study
Given: {(C) or (D)};
Apply:
AWD vs. AWC
Select the best alternative
If AWD<AWC, stay with the defender for
nD years;
If AWD>AWC, go with the challenger for
nC years.
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11.3 If AWD<AWC, One-Year-Later Analysis
Is the current year nD?
If “YES,” replace the defender with (C);
If “NO,” retain defender for one more year, and
then re-evaluate.
If cost estimates have changed, then:
Update all estimates
Calculate AWC and AWD
Initiate a new replacement study.
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11.3 Case Problem (Ex. 11.4) – Defender
Defender Data
Current Market Value: $15,000;
Future Mkt. Values will decrease by 20%/yr;
Keep for no more than 3 years;
AOC’s: {$4,000,$8,000,$12,000}
Retrofit next year = $16,000;
AOC’sD := {$20,000, $8,000,$12,000} (costs).
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11.3 Case Problem: Challenger
Challenger Data
First Cost: $50,000
Future Mkt. Values decreasing by 20%/year;
Retain for no more than 5 years;
AOC’sC := {$5,000, $7,000, $9,000, $11,000, $13,000}
Assume the interest rate is set at 10%/year.
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11.3 Defender Analysis
Input Parameters
Interrest Rate (%)
Investment Cost ($)
No of Years to Study
(1)
Mkt. Value
$12,000.00
$9,600.00
$7,680.00
Year
1
2
3
(3)
Cap. Rec. Costs
-$4,500
-$4,071
-$3,711
$
10.00%
15,000.00
3
(4)
AW of AOC's
-$20,000
-$14,286
-$13,595
Year
1
2
3
(2)
AOC/Yr
-$20,000.00
-$8,000.00
-$12,000.00
(5)
Total AW(i%)
-$24,500
-$18,357
-$17,307
Min Life
ID
Min Life
ESL(Defender) = 3 yrs: AW = -$17,307/year
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11.3 Cost Plots for Defender
ESL
CRC
AOC
AW
$30,000.00
$25,000.00
$AW
$20,000.00
$15,000.00
$10,000.00
$5,000.00
$0.00
0
1
2
3
Years
Min AW Cost Life = 3 years
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11.3 Example 11.4: Challenger
Input Parameters
Interrest Rate (%)
Investment Cost ($)
No of Years to Study
Year
1
2
3
4
5
(3)
Cap. Rec. Costs
-$15,000
-$13,571
-$12,372
-$11,361
-$10,506
$
(1)
Mkt. Value
$40,000.00
$32,000.00
$25,600.00
$20,480.00
$16,384.00
(4)
AW of AOC's
-$5,000
-$5,952
-$6,873
-$7,762
-$8,620
10.00%
50,000.00
5
AWC = -19,123/year
(2)
AOC/Yr
-$5,000.00
-$7,000.00
-$9,000.00
-$11,000.00
-$13,000.00
Year
1
2
3
4
5
Min. Cost
Life
N = 4 Years
(5)
Total AW(i%)
-$20,000
-$19,524
-$19,245
-$19,123
-$19,126
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11.3 Challenger Plots
$AW
AW
$20,100
$20,000
$19,900
$19,800
$19,700
$19,600
$19,500
$19,400
$19,300
$19,200
$19,100
$19,000
ESL: Challenger
Min. Cost Life = 4 years
At -$19,123/year
0
1
2
3
4
5
6
Years
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11.3 Case Problem: Summary
AWD = -$17,307/year;
nD = 3 years;
AWC = -$19,123/year;
nC = 4 years;
Conclusion:
Stay with the Defender for at least one more
year – lowest AW(10%) cost: -$17,307/yr vs. $19,123/yr.
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11.3 Market Value of Defender
What minimum market value of the
defender will make the current
challenger economically attractive?
If a high enough market value (trade-in)
is possible for the defender asset, one
should do so and go with the Challenger
immediately!
Break-even or replacement value (RV)
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11.3 Replacement Value (Defender)
Economic Fact:
If the actual market value (trade-in)
exceeds the breakeven replacement value,
then the challenger is the better
alternative.
If this is the case, replace now with the
challenger!
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ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 11
4. Additional Considerations
in a Replacement Study
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11.4 Three Additional Considerations
1. Future-year replacement decisions;
2. Opportunity-cost vs. cash flow
approaches;
3. Anticipation of improved future
challengers.
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11.4 Future Replacement Decisions
1.
Future-year replacement decisions:
Replace now? One year from now?
Two years from now?
The procedure just presented does
assist with answering this question
provided:
The cost estimates for (C) and (D) do
not change!
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11.4 Future Replacement Decisions
1. Future-year Replacement Decisions:
If cost estimates change for either
(D) or ©, then the analysis should
be initiated again within a
reasonable time frame.
With changing estimates the
decision to replace may be altered!
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11.4 Opportunity Cost vs. Cash Flow
The method just presented applies the
opportunity cost approach:
(D)’s investment now is the fair market
value of (D) now;
Represents the opportunity cost foregone
by not disposing of (D) now;
(C)’s investment now is the investment in
(C) now.
This is a correct approach for a
replacement analysis.
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11.4 Cash Flow Approach
2. Cash Flow Approach:
Assumes that when the challenger is
selected and a cash inflow for the
defender is received, then:
The investment in the challenger is
immediately reduced.
Works only if the lives of the (C) and (D)
are the same!
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11.4 Anticipation of Future Challengers
3. Assumption: At some time in the
future a worthy challenger will appear
and replace the defender.
Always study future trends of challengers;
May be best to augment the defender
until such time that a more worthy
challenger becomes available.
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ENGINEERING ECONOMY Fifth Edition
Blank and Tarquin
Mc
Graw
Hill
CHAPTER 11
5. Replacement Study Over a
Specified Study Period
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11.5 Specified Study Period
If a fixed study period, then…
Determine the AW for both C and D over
the prescribed study period;
No need to perform an ESL analysis;
Assumption is that the services of C
and D are not needed beyond the study
period.
Use the AW approach.
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11.5 Specified Study Period
What happens if the defender’s
remaining life is shorter than the study
period?
Focus on upgrading the defender and obtain
cost estimates in order to extend the
defender out to the study period.
These costs become part of retaining the
defender out to the prescribed study period.
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