Lecture 13

ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Chapter 8 Equivalent Annual Worth Analysis † Equivalent annual

worth criterion † Applying annual worth analysis † Mutually exclusive projects † Design economics

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Computing Equivalent Annual Worth $120 $80 0

1 2

$50

$189.43

$70 3

4

5

6

i = 12%

$100

A = $46.07 0

1

2

3

4

5

6

0

PW(12%) = $189.43

EAW(12%) = $189.43(A/P, 12%, 6) = $46.07 2

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Annual Worth Analysis † Principle: Measure investment worth on annual basis † Benefit:: 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 $700 $500

$1,000

$800

$800

$700 $400 $400 $500

$400 $400

$1,000 Repeating cycle 4

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

• First Cycle: PW(10%) = -$1,000 + $500 (P/F, 10%, 1) + . . . + $400 (P/F, 10%, 5) = $1,155.68 EAW(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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† 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.

Equivalent Annual Costs

Equivalent Annual Cost Capital costs + Operating costs

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Capital Cost Recovery S

† Definition: The cost of owning an

†

equipment is associated with two transactions—(1) its initial cost (I) and (2) its salvage value (S).

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Capital cost recovery : Taking into these sums, we calculate the capital recovery cost as:

I

N

0 1

2

CR(i) = I( A / P, i, N) − S(A / F, i, N)

3

N

CR(i)

= (I − S)(A / P, i, N) + iS

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Capital Cost Recovery i (1 + i ) N − 1 = F ( A / F ,i, N )

A= F

i (1 + i ) N (1 + i ) N − 1 = P ( A / P ,i, N )

A = P

S 0 N I 0 1

2

3

N

(A/F,i,N) = (A/P,i,N) - i

CR(i) = I( A / P, i, N) − S(A / F, i, N) = (I − S)(A / P, i, N) + iS

CR(i) 8

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Example - Capital Cost Calculation $50,000

† Given:

I = $200,000 N = 5 years S = $50,000 i = 20% † Find: CR(20%)

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$200,000 CR (i ) = ( I - S ) ( A / P, i, N ) + iS CR ( 20%) = ($200,000 - $50,000 ) ( A / P, 20%, 5) + (0.20)$50,000 = $60,157 9

Equivalent Annual 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.

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Equivalent Annual Worth 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). 11

Capital 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 for loss of value

Opportunity cost 12

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Capital 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 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 $45,000 the 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. 14

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Example 1 GIVEN: 1ST COST = $600,000 FOR BUILDING ANNUAL COST: A) ANNUAL MAINTENANCE: $20,000 B) ENERGY COST: $45,000 DURING FIRST YEAR INCREASING $3,000/YR SUBSEQUENTLY(LINEAR GRADIENT) STARTING AT YR 2 C) EQUIPMENT COST: $70,000 EVERY 10 YEARS ANNUAL REVENUE: $8,000 SALVAGE VALUE: $100,000 FOR BUILDING, $0 FOR EQUIPMENT PROPERTY VALUE: $200,000 LIFE TIME: 30 YEARS FOR PROJECT, 10 YEARS FOR EQUIPMENT MARR = 15%/YR, CPD ANNUALLY FIND: EAC (EQUIVALENT ANNUAL COST)

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EAC =

BUILDING COST ($600,000 - $100,000)(A|P,15%,30) + $100,000(0.15)

Example 1

+ 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) 16

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

EAC =

BUILDING COST ($500,000)(0.1523) + $100,000(0.15) + LAND COST

Example 1

$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 17

Example 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 hence for $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.) 18

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Example 2 GIVEN:

FULL CAPACITY

HALF CAPACITY

LIFETIME

50 YRS / 10 YR REPAIR

50 YRS / 20 YR EXPAND / 10 YR REPAIR

1ST COST

$400,000

$200,000 / $300,000

10 YR REPAIR

$20,000

$18,000 / $18,000

SALVAGE

-

-

INTEREST

5%/YR CPD. ANNUALLY

FIND: EACFULL AND EACHALF

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Example 2

EACFULL =

[$400,000

EACFULL =

+ $20,000(P|F,5%,10)

+ $20,000(0.6139)

+ $20,000(P|F,5%,20)

+ $20,000(0.3769)

+ $20,000(P|F,5%,30)

+ $20,000(0.2314)

+ $20,000(P|F,5%,40)]

+ $20,000(0.1420)]

x (A|P,5%,50)

x 0.0548 =

EACFULL = $23,415 / YR

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[$400,000

$427,284 x 0.0548 20

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Example 2

EACHALF =

[$200,000

EACHALF =

[$200,000

+ $18,000(P|F,5%,10)

+ $18,000(0.6139)

+ $318,000(P|F,5%,20)

+ $318,000(0.3769)

+ $36,000(P|F,5%,30)

+ $36,000(0.2314)

+ $36,000(P|F,5%,40)]

+ $36,000(0.1420)]

x (A|P,5%,50)

x 0.0548 =

EACHALF = $18,870 / YR

$344,347 x 0.0548

EACFULL = $23,415 / YR

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Example 3 Your 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. 22

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Example 3 GIVEN: LIFETIME = 18 YRS INTEREST = 6%/YR, CPD ANNUALLY FIRST COST = $100, 000 SALVAGE VALUE = $100,000 FIND EAC N = 18 YRS

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)

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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%) = $4,620.76 † Conclusion: Need an additional annual revenue in the amount of $220.76.

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ENGR 390 Lecture 13: Equivalent Annual Worth

Winter 2007

Summary † Equivalent 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. 25

† The equation for CR(i) is

CR(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. 26

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