Chapter 6endustri.eskisehir.edu.tr/onur_kaya/İKT356/icerik/Ch6...Unequal Project Lives 1. Required...
Transcript of Chapter 6endustri.eskisehir.edu.tr/onur_kaya/İKT356/icerik/Ch6...Unequal Project Lives 1. Required...
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Chapter 6Annual Equivalence Analysis
Annual equivalent criterionApplying annual worth analysisMutually exclusive projects
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Annual Worth Analysis
Principle: Measure investment worth on annualbasis
Benefit: • Annual reports, yearly budgets• Seek consistency of report format• Determine unit cost (or unit profit)• Facilitate unequal project lifecomparison
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Annual Equivalent Worth
AE(i)=PW(i)*(A|P,i,N)
If AE(i) > 0 , acceptAE(i) < 0 , rejectAE(i) = 0 , remain indifferent
Since (A|P,i,N) > 0 for –1<i<∞AE(i) > 0 PW(i) > 0
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Computing Equivalent Annual Worth
$100$50
$80$120
$700
2 3 4 5 61
A = $46.07
2 3 4 5 61
AE(12%) = $189.43(A/P, 12%, 6)= $46.07
$189.43
00
PW(12%) = $189.43
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Annual Equivalent Worth - Repeating Cash Flow Cycles
$500$700 $800
$400 $400 $500$700
$800
$400 $400
$1,000 $1,000
Repeating cycle
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• First Cycle:
PW(10%) = -$1,000 + $500 (P/F, 10%, 1)+ . . . + $400 (P/F, 10%, 5)
= $1,155.68AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87
• Both Cycles:
PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)= $1,873.27
AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87
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Annual Equivalent Cost
When only costs are involved, the AE method is called the annual equivalent cost.Revenues must cover two kinds of costs: Operating costs and capital costs.
Capital costs
Operating costs
+
Ann
ual E
quiv
alen
t Cos
ts
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Capital and Operating CostsCapital costs are incurred by purchasing assets tobe used in production and service. Normally, they arenonrecurring (one-time) costs.Operating costs are incurred by the operation of physical plant or equipment needed to provideservice (e.g. labor and raw materials). Normally, theyrecur for as long as an asset is owned.Operating costs are on annual basis anyway. Annualequivalent of a capital cost is calledcapital recovery cost, CR(i).
Remember: (A|P, i, N) is called the capital recovery factor.
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Capital (Ownership) CostsDef: The cost of owning an
equipment is associated with two transactions—(1) its initial cost (I) and (2) its salvage value(S).Capital costs: Taking into these sums, we calculate the capital costs as:
since (A|F,i,N)=(A|P,i,N)-i
0 1 2 3 N
0N
I
S
CR(i)CR i I A P i N S A F i N
I S A P i N iS( ) ( / , , ) ( / , , )
( )( / , , )= −= − +
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Example - Capital Cost Calculation
Given:I = $200,000N = 5 yearsS = $50,000i = 20%
Find: CR(20%) $200,000
$50,000
50
CR i I S A P i N iSCR A P
( ) = ( - ) ( / , , ) + ( = ($200, - $50, ) ( / , )
+ (0.20)$5 , = $60,
20%) 000 000 20%, 50 000
157Annual cost of owning the asset at 20%
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Justifying an investment based on AE Method
Given: I = $20,000, S = $4,000, N = 5 years, i = 10%Find: see if an annual revenue of $4,400 is enough to cover the capital costs.Solution:CR(10%) = $4,620.76Conclusion: Need an additional annual revenue in the amount of $220.76.
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Example - Capital Cost Calculation for Mini Cooper
Given:I = $19,800N = 3 yearsS = $12,078i = 6%
Find: CR(6%) $19,800
$12,078
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( ) = ( - ) ( / , , ) + (6%) = ($19,800 - $12,078) ( / , 6%, 3)
+ (0.06)$12,078 = $3,613.55
CR i I S A P i N iSCR A P
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Applying Annual Worth Analysis
•Unit Cost (Profit) Calculation
• Unequal Service Life Comparison
• Minimum Cost Analysis
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Equivalent Worth per Unit of Time
0
1 2 3
$24,400$55,760
$27,340
$75,000 Operating Hours per Year
2,000 hrs. 2,000 hrs. 2,000 hrs.
• PW (15%) = $3553• AE (15%) = $3,553 (A/P, 15%, 3)
= $1,556
• Savings per Machine Hour= $1,556/2,000= $0.78/hr.
Note: 3553/6000=0.59/hour: instant savings in presentworth for each hourly use; does not consider the time over which the savings occur
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Equivalent Worth per Unit of Time(cont’d)
0
1 2 3
$24,400$55,760
$27,340
$75,000 Operating Hours per Year
1,500 hrs. 2,500 hrs. 2,000 hrs.
•Let C denote the equivalent annual savings per machine hour• $1,556=[(C)(1500)(P|F,15%,1)
+(C)(2500)(P|F,15%,2)+(C)(2000) (P|F,15%,3)] (A|P,15%,3)
C=$0.79/hr
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Breakeven Analysis
Problem:
At i = 6%, what should be the reimbursement rate per mile so that Sam can break even? $11,72539,000Total
$4,680$3,624$3,421
14,50013,00011,500
123
Total costs(Ownership
& Operating)
Miles Driven
Year(n)
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1711,500 miles13,000 miles14,500 milesExpected miles driven
$3,421$3,624$4,680Total of all costs
$971$1,003$988Total operating cost
110125135Parking and tolls
10010080Oil
522650688Gasoline and taxes
121315Accessories
273035Replacement tires
2008535Nonscheduled repairs
$2,450$2,621$3,693Total ownership cost
505778Registration and taxes
635635635Insurance
220153100Scheduled maintenance
$1,545$1,776$2,879Depreciation
Third YearSecond YearFirst Year
Breakeven Analysis
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• Equivalent annual cost of owning and operating the car[$4,680 (P/F, 6%, 1) + $3,624 (P/F, 6%, 2) + $3,421 (P/F, 6%, 3)] (A/P, 6%,3) = $3,933 per year
• Equivalent annual reimbursement Let X = reimbursement rate per mile [14,500X(P/F, 6%, 1) + 13,000X(P/F, 6%, 2) + 11,500 X (P/F, 6%, 3)] (A/P, 6%,3) = 13.058X
• Break-even value 13.058X = 3,933
X = 30.12 cents per mile
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Annual equivalent reimbursement as a function of cost per mile
4000
3000
2000
1000
0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
Annual equivalentcost of owning andoperating ($3,933)
Loss
Annual reimbursementamount
Minimumreimbursementrequirement ($0.3012)
Gain
Reimbursement rate ($) per mile (X)
Ann
ual e
quiv
alen
t ($)
13,058X
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Mutually Exclusive Alternativeswith Equal Project Lives
Standard PremiumMotor Efficient Motor
25 HP 25 HP$13,000 $15,60020 Years 20 Years$0 $018.65 kW per motor 18.65 kW per motor
89.5% 93%$0.07/kWh $0.07/kWh3,120 hrs/yr. 3,120 hrs/yr.
SizeCostLifeSalvageOutputEfficiencyEnergy CostOperating Hours
(a) At i= 13%, determine the operating cost per kWh for each motor.(b) At what operating hours are they equivalent?
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Solution:
(a) Operating cost per kWh per unit
Determine total input power
Conventional motor:
input power = 18.65 kW/ 0.895 = 20.838kW
PE motor:
input power = 18.65 kW/ 0.930 = 20.054kW
Input power = output power% efficiency
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Determine total kWh per year with 3120 hours of operation
Conventional motor:
3120 hrs/yr (20.838 kW) = 65,018 kWh/yr
PE motor:
3120 hrs/yr (20.054 kW) = 62,568 kWh/yr
Determine annual energy costs at $0.07/kwh: Conventional motor:
$0.07/kwh × 65,018 kwh/yr = $4,551/yrPE motor:
$0.07/kwh × 62,568 kwh/yr = $4,380/yr
Solution:
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Capital cost:Conventional motor:
$13,000(A/P, 13%, 20) = $1,851PE motor:
$15,600(A/P, 13%, 20) = $2,221
Total annual equivalent cost:Conventional motor:
AE(13%) = $4,551 + $1,851 = $6,402Cost per kwh = $6,402/58,188 kwh = $0.1100/kwh
(where 58188=18.65*3120 is the total output power per year)PE motor:
AE(13%) = $4,380 + $2,221 = $6,601Cost per kwh = $6,601/58,188 kwh = $0.1134/kwh
Solution:
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Savings from switching from conventional to PE motors?
Incremental capital cost required=2221-1851=370Incremental energy savings=4551-4380=171Hence, at 3120 annual operating hours, a loss of 199 for each motor.
Solution:
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(b) Break-evenoperatinghours = 6,742
Solution:
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Mutually Exclusive Alternatives with Unequal Project Lives
1. Required service period = Indefinite
2. Each alternative replaced by identical assetwith same costs and performance
Then,
We may solve for AE of each project based on its initial life span rather than on LCM of project lives.
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Model A:0 1 2 3
$12,500$5,000 $5,000
$3,000
Model B:0 1 2 3 4
$15,000$4,000 $4,000 $4,000
$2,500
Mutually Exclusive Alternatives with Unequal Project Lives
Required servicePeriod = Indefinite
Analysis period =LCM (3,4) = 12 years
(Least common multiple)
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Model A:
$12,500$5,000 $5,000
$3,000
0 1 2 3
• First Cycle:PW(15%) = -$12,500 - $5,000 (P/A, 15%, 2)
- $3,000 (P/F, 15%, 3)= -$22,601
AE(15%) = -$22,601(A/P, 15%, 3) = -$9,899• With 4 replacement cycles:
PW(15%) = -$22,601 [1 + (P/F, 15%, 3) + (P/F, 15%, 6) + (P/F, 15%, 9)]
= -$53,657AE(15%) = -$53,657(A/P, 15%, 12) = -$9,899
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Model B:
$15,000$4,000 $4,000
$2,500
0 1 2 3 4
$4,000
• First Cycle:PW(15%) = - $15,000 - $4,000 (P/A, 15%, 3)
- $2,500 (P/F, 15%, 4)= -$25,562
AE(15%) = -$25,562(A/P, 15%, 4) = -$8,954• With 3 replacement cycles:
PW(15%) = -$25,562 [1 + (P/F, 15%, 4) + (P/F, 15%, 8)]= -$48,534
AE(15%) = -$48,534(A/P, 15%, 12) = -$8,954
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IF:1. The service of the selected alternative required on a continuous basis2. Each alternative replaced by identical asset with same costs andperformance
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SummaryAnnual equivalent worth analysis, or AE, is—along with present worth analysis—one of two main analysis techniques based on the concept of equivalence. The equation for AE is
AE(i) = PW(i)(A/P, i, N).AE analysis yields the same decision result as PW
analysis.
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The capital recovery cost factor, or CR(i), is one of the most important applications of AE analysis in that it allows managers to calculate an annual equivalent cost of capital for ease of itemization with annual operating costs.
The equation for CR(i) isCR(i)= (I – S)(A/P, i, N) + iS,
where I = initial cost and S = salvage value.
Summary
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AE analysis is recommended over NPW analysis in many key real-world situations for the following reasons:1. In many financial reports, an annual equivalent value
is preferred to a present worth value.2. Calculation of unit costs is often required to determine
reasonable pricing for sale items.3. Calculation of cost per unit of use is required to
reimburse employees for business use of personal cars.
4. Make-or-buy decisions usually require the development of unit costs for the various alternatives.
Summary
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Example 1:Susan wants to buy a car that she will keep forthe next four years. She can buy a Honda Civicat 15000 and then sell it for 8000 after fouryears. If she bought this car, what would be her annual ownership cost (capital recovery cost)? Assume that her interest rate is 6%.
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2.2500)8000)(06.0()4%,6,|()800015000(or
2.2500)4%,6,|(8000)4%,6,|(15000%)6(
=+−
=−=
PA
FAPACR
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Example 2:The following cash flows represent the potential annualsavings associated with two different types of productionprocesses, each of which requires an investment of 12000. Assuming an interest rate of 15%, complete thefollowing tasks: (a) Determine the equivalent annualsavings for each process. (b) Determine the hourlysavings for each process assuming 2000 hours of operation per year. (c) Determine which process shouldbe selected.
635045603635022804
635068402635091201
-12000-120000BAn
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( )
B.Select (c)
hour/07.12000
08.2147 :B Process
hour/95.02000
14.1893 :A Process
(b)
08.2147)4%,15,|()4%,15,|(635012000%)15(
14.1893)4%,15,|(15.1
228015.1
456015.1
684015.1
912012000%)15(
(a)
432
=
=
=+−=
=⎟⎠⎞
⎜⎝⎛ ++++−=
PAAPAE
PAAE
B
A
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Example 3:Norton Auto Parts, Inc. is considering two different forklifttrucks for use in its assembly plant.
Truck A costs 15000 and requires 3000 annually in operating expenses. It will have a 5000 salvagevalue at the end of its 3-year service life. Truck B costs 20000 and requires 2000 annually in operating expenses. It will have a 8000 salvagevalue at the end of its 4-year service life.
The firm’s MARR is 12%. Assuming that the trucks areneeded for 12 years and no significant changes areexpected in the future price and functional capacity of both trucks, select the most economical truck based on AE analysis.
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.economical more is BTruck
4.69102000)8000)(12.0()4%,12,|()800020000(%)12(
77633000)5000)(12.0()3%,12,|()500015000(%)12(
=++−=
=++−=
PAAE
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B
A
Capital Recovery Cost
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Example 4:Your company needs a small front loader for handling bulk materials at the plant. It can be leased from the dealer for three years for $4050 per year including all maintenance. It can also be purchased for $14,000. You expect the loader to last for six years and to have a salvage value of $3000. You predict that maintenance will cost $400 the first year and increase by $200 per year in each year after the first. Your MARR is 15%. Use annual equivalence analysis to determine whether to lease or buy the loader. What is the shortest project life for which the annual worths you have calculated are exactly correct?
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( )
6) of multipleinteger an (or years 6for correct Exactly
LEASE.
46.4175 )6%,15,|()6%,15,|(3000)6%,15,|(200)6%,15,|(40014000%)15(
=−++= PAFPGPAPAE
Present worth
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Example 5:Consider the following cash flows for two types of models. Bothmodels will have no salvage value upon their disposal (at the end of their respective service lives). The firm’s MARR is known to be 15%. (a) Notice that models have different service lives. However, model A will be available in the future with the same cash flows. Model B is available now only. If you select model B now, you will have toreplace it with model A at the end of year 2. If your firm uses thepresent worth as a decision criterion, which model should be selectedassuming that your firm will need either model for an indefiniteperiod? (b) Suppose that your firm will need either model for only twoyears. Determine the salvage value of model A at the end of year twothat makes both models indifferent (equally likely).
3500310000350021000035001-15000-60000
BAnProject cash flow
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4354.20721257)2%,15,|()2%,15,|(35006000
(b)
47.5653)2%,15,|(15.015.8721257%)15(
33.581415.015.872%)15(
(a)
18.773)2%,15,|(1257%)15(15.872)3%,15,|(2.1991%)15(
1257)2%,15,|(1000015000%)15(2.1991)3%,15,|(35006000%)15(
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