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Engineering Economics 390. Courtesy of Atre, Sundar V Associate Professor, Oregon State University.
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ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 1
1
Chapter 8Equivalent Annual Worth Analysis
Equivalent annual worth criterionApplying annual worth analysisMutually exclusive projectsDesign economics
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Computing Equivalent Annual Worth
$100$50
$80$120
$700
2 3 4 5 61
EAW(12%) = $189.43(A/P, 12%, 6)= $46.07
A = $46.07
2 3 4 5 610
$189.43
0
PW(12%) = $189.43
i = 12%
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 2
3
Annual Worth Analysis
Principle: Measure investment worth on annual basisBenefit:: By knowing annual equivalent worth, we can:
• Seek consistency of report format• Determine unit cost (or unit profit)• Facilitate unequal project life comparison
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Equivalent Annual Worth - Repeating Cash Flow Cycles
$500$700 $800
$400 $400 $500$700
$800
$400 $400
$1,000 $1,000
Repeating cycle
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 3
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• First Cycle:
PW(10%) = -$1,000 + $500 (P/F, 10%, 1)+ . . . + $400 (P/F, 10%, 5)
= $1,155.68EAW(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
EAW(10%) = $1,873.27 (A/P, 10%,10) = $304.87
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Equivalent Annual Cost
When only costs are involved, the EA method is called the equivalent annual cost.Revenues must cover two kinds of costs: Operating costs and capital costs.
Capital costs
Operating costs
+
Equi
vale
nt A
nnua
l Cos
ts
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 4
7
Capital Cost Recovery
Definition: The cost of owning an equipment is associated with two transactions—(1) its initial cost (I) and (2) its salvage value (S).
Capital cost recovery : Taking into these sums, we calculate the capital recovery cost as:
0 1 2 3 N
0N
I
S
CR(i)CR i I A P i N S A F i NI S A P i N iS
( ) ( / , , ) ( / , , )( )( / , , )
= −= − +
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Capital Cost Recovery
0 1 2 3 N
0N
I
S
CR(i)CR i I A P i N S A F i NI S A P i N iS
( ) ( / , , ) ( / , , )( )( / , , )
= −= − +
A F ii
F A F i N
N=+ −
=( )( / , , )
1 1
A P i ii
P A P i N
N
N=+
+ −=
( )( )( / , , )
11 1
(A/F,i,N) = (A/P,i,N) - i
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 5
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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
157
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Equivalent Equivalent Annual Worth AnalysisAnnual Worth Analysis
Equivalent Annual Worth (EAW) can be used to compare projects.
Equivalent Annual Cost (EAC) can be used instead of EAW if revenues are not included.
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 6
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Equivalent Annual Equivalent Annual Worth AnalysisWorth Analysis
Project Selection:
• If all expenses and revenues are included, select largest EAW that is ≥ 0.
• If some or none of the revenues are included, select largest EAW (lowest EAC).
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Capital Cost RecoveryCapital Cost Recovery
For a capital purchase (I) with a salvage value (S), the EAC can be calculated two ways:
1. I(A/P, i, N) – S (A/F, i, N)
2. (I – S) (A/P, i, N) + S*i
Annual equivalent Opportunityfor loss of value cost
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 7
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Capital Cost RecoveryCapital Cost Recovery
Land is considered to have infinite life; it can be sold for its purchase price (neglecting inflation).
I = S and EAC = S*i
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Example 1Example 1
The Ragweed Pollination Company needs a new building to expand seed production. The building will cost $600,000, last for 30 years, and is expected to sell for $100,000. It will be built on property that costs $200,000, which will be sold with the building. Energy costs are projected to be $45,000the first year increasing by $3,000 each year after that. Needed equipment will cost $70,000 and last 10 years with no salvage value. It will be replaced with identical equipment 10 and 20 years from now. Annual maintenance is projected to be $20,000.
Determine the Annual Equivalent Cost for the proposed expansion using a MARR of 15% compounded annually.
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 8
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Example 1Example 1GIVEN: 1ST COST = $600,000 FOR BUILDING ANNUAL COST: A) ANNUAL MAINTENANCE: $20,000
B) ENERGY COST: $45,000 DURING FIRST YEARINCREASING $3,000/YR SUBSEQUENTLY(LINEARGRADIENT) STARTING AT YR 2C) EQUIPMENT COST: $70,000 EVERY 10 YEARS
ANNUAL REVENUE: $8,000SALVAGE VALUE: $100,000 FOR BUILDING,
$0 FOR EQUIPMENTPROPERTY VALUE: $200,000LIFE TIME: 30 YEARS FOR PROJECT,
10 YEARS FOR EQUIPMENT MARR = 15%/YR, CPD ANNUALLYFIND: EAC (EQUIVALENT ANNUAL COST)
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Example 1Example 1
EAC = BUILDING COST
($600,000 - $100,000)(A|P,15%,30) + $100,000(0.15)
+ LAND COST
($200,000 - $200,000)(A|P,15%,30) + $200,000(0.15)
+ MAINTENANCE COST
$20,000
+ ENERGY COST
$45,000 + $3,000(A|G,15%,30)
+EQUIPMENT COST
[($70,000 – 0) + ($70,000 -0)(P|F,15%,10) + ($70,000-0)(P|F,15%,20)](A|P,15%,30)
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 9
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Example 1Example 1
EAC = BUILDING COST
($500,000)(0.1523) + $100,000(0.15)
+ LAND COST
$0 + $200,000(0.15)
+ MAINTENANCE COST
$20,000
+ ENERGY COST
$45,000 + $3,000(6.2066)
+ EQUIPMENT COST
[($70,000 + $70,000(0.2472) + $70,000(0.0611)](0.1523)
EAC = $91,150 + $30,000 + $20,000 + $45,000 + $18,620 + $13,948= $218,718
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Example 2Example 2
A 1000-foot tunnel must be constructed as part of a new aqueduct system for a major city. Two alternatives are being considered.One is to build a full-capacity tunnel now for $400,000. The other alternative is to build a half-capacity tunnel now for $200,000, and then to build a second parallel half-capacity tunnel 20 years hencefor $300,000. The cost of repair of the tunnel lining at the end of every 10 years is estimated to be $20,000 for the full capacity tunnel and $18,000 for each half-capacity tunnel.
Determine whether the full capacity tunnel or the half-capacity tunnel should be constructed now. Solve the problem by annual worth analysis, using an interest rate of 5% per year compounded annually, and a 50-year analysis period. (Note: There will be no tunnel lining repair at the end of the 50 years.)
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
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Example 2Example 2
5%/YR CPD. ANNUALLYINTEREST--SALVAGE
$18,000 / $18,000$20,00010 YR REPAIR$200,000 / $300,000$400,0001ST COST
50 YRS / 20 YR EXPAND / 10 YR REPAIR
50 YRS / 10 YR REPAIR
LIFETIMEHALF CAPACITYFULL CAPACITYGIVEN:
FIND: EACFULL AND EACHALF
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Example 2Example 2
EACFULL = [$400,000
+ $20,000(P|F,5%,10)
+ $20,000(P|F,5%,20)
+ $20,000(P|F,5%,30)
+ $20,000(P|F,5%,40)]
x (A|P,5%,50)
EACFULL = [$400,000
+ $20,000(0.6139)
+ $20,000(0.3769)
+ $20,000(0.2314)
+ $20,000(0.1420)]
x 0.0548
= $427,284 x 0.0548
EACFULL = $23,415 / YR
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
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Example 2Example 2
EACHALF = [$200,000
+ $18,000(P|F,5%,10)
+ $318,000(P|F,5%,20)
+ $36,000(P|F,5%,30)
+ $36,000(P|F,5%,40)]
x (A|P,5%,50)
EACHALF = [$200,000
+ $18,000(0.6139)
+ $318,000(0.3769)
+ $36,000(0.2314)
+ $36,000(0.1420)]
x 0.0548
= $344,347 x 0.0548
EACHALF = $18,870 / YR EACFULL = $23,415 / YR
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Example 3Example 3Your parents paid cash for a home they purchased 18 years ago for $100,000. They just sold it for $100,000. They were bragging that, neglecting such expenses as taxes, insurance, and utilities, it did not cost them anything to live in the house for the 18 years.
Having completed an engineering economics course as a part of your engineering degree, you know immediately that your parents’ reasoning was incorrect. Identify and describe the engineering economic principle involved.
Include in your explanation what it actually cost your parents each year to own the house provided that they value money at 6% per year compounded annually. Do not include changes in the purchasing power of money, and continue to neglect taxes, maintenance, insurance and utilities.
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
S.V. Atre 12
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Example 3Example 3GIVEN:
LIFETIME = 18 YRSINTEREST = 6%/YR, CPD ANNUALLYFIRST COST = $100, 000SALVAGE VALUE = $100,000
FIND EAC
EAC = (I-S )(A|P,i,N) + iS
= ($100,000 - $100,000)(A|P,6%,18) + $100,000(0.06)
= $6,000 (Method 1)EAC = I(A|P,i,N) - S(A|F,i,N)
= $100,000(A|P,6%,18) - $100,000(A|F,6%,18)
= $100,000(0.0924) - $100,000(0.0324)
= $100,000(0.0924 - 0.0324)
= $100,000(0.06)
= $6,000 (Method 2)
N = 18 YRS
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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.
ENGR 390 Lecture 13: Equivalent Annual Worth
Winter 2007
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25
SummaryEquivalent annual worth analysis, or EAW, is—along with present worth analysis—one of two main analysis techniques based on the concept of equivalence. The equation for EAW is
EAW(i) = PW(i)(A/P, i, N).AE analysis yields the same decision result as PW analysis.The capital recovery cost factor, or CR(i), is one of the most important applications of EAW analysis in that it allows managers to calculate an annual equivalent cost of capital for ease of itemization with annual operating costs.
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The equation for CR(i) isCR(i)= (I – S)(A/P, i, N) + iS,
where I = initial cost and S = salvage value.
EAW 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.5. Minimum cost analysis is easy to do when based on
annual equivalent worth.