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Reviewing Complex Flow Reviewing Complex Flow PerpetuityPerpetuity

Reviewing Complex Flow Reviewing Complex Flow PerpetuityPerpetuity

If there is a mix of recurring and non-If there is a mix of recurring and non-recurring or one-time cash flows that recurring or one-time cash flows that must be capitalized for perpetuity:must be capitalized for perpetuity:

1.) finding the 1.) finding the NPWNPW of all the of all the one-one-time, non-recurringtime, non-recurring cash flows ( cash flows (= = CCCCPart 1Part 1 ))

2.) finding the 2.) finding the Annual EquivalentAnnual Equivalent of of one cycleone cycle of all the of all the recurringrecurring cash flows, and then computing P cash flows, and then computing P ((= CC= CCPart 2Part 2) from the perpetuity ) from the perpetuity relationship relationship A = P(i)A = P(i)

3.) summing (1.) and (2.) to find the 3.) summing (1.) and (2.) to find the total capitalized cost: total capitalized cost:

CCCCTotalTotal = CC = CCPart 1Part 1 + CC + CCPart 2Part 2

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Perpetuity ExamplePerpetuity ExamplePerpetuity ExamplePerpetuity Example

A new football stadium will replace the A new football stadium will replace the current municipal stadium at an initial current municipal stadium at an initial cost of $250 million. cost of $250 million.

One year later, the old stadium will be One year later, the old stadium will be demolished at a net cost of $1.5 million. demolished at a net cost of $1.5 million.

Annual maintenance is expected to be Annual maintenance is expected to be $900 000, and every 15 years the $900 000, and every 15 years the skyboxes will be remodeled at a cost of skyboxes will be remodeled at a cost of $500 000. $500 000.

Find the capitalized cost if the discount Find the capitalized cost if the discount rate is 10%, compounded annually.rate is 10%, compounded annually.

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Perpetuity ExamplePerpetuity ExamplePerpetuity ExamplePerpetuity ExampleDIAGRAM:

0 1 2 3 n= yrs15 30

$500 K $500 K$1.5 M

$250 M

$900 K

CC1 = $250 000 000 + $1 500 000 (P|F,10%,1) = $ 251 363 650

CC2 = $900 000 + $500 000 (A|F,10%,15) = $ 9 157 500

0.10

CCT = CC1 + CC2 = $251 363 650 + $9 157 500 = $ 260 521 150

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Reviewing…NPWReviewing…NPWReviewing…NPWReviewing…NPWTwo approaches to handle

differing project lives:Common Multiple Period:

Projects are assumed to be repeated until a common multiple point in time is established.

Study Period: Select a study period for both projects and estimate cash flows to conform to the study period.

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Equivalent Equivalent Annual Worth AnalysisAnnual Worth Analysis

Equivalent Equivalent Annual Worth AnalysisAnnual Worth AnalysisEquivalent Annual Worth (EAW) can be used to compare projects.

Equivalent Annual Cost (EAC) can be used instead of EAW if revenues are not included.

EAC = – EAW

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EAW and Types of Projects:•Revenue projects are expected to make money at a rate at least as high as the MARR, so select largest EAW that is 0.•Service projects are “have to do” situations, so select largest EAW (lowest EAC).

Equivalent Annual Equivalent Annual Worth AnalysisWorth Analysis

Equivalent Annual Equivalent Annual Worth AnalysisWorth Analysis

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Capital Cost RecoveryCapital Cost RecoveryCapital Cost RecoveryCapital Cost Recovery

For a capital purchase (P) with a salvage value (S), the EAC can be calculated two ways:

1.P(A/P, i, n) – S (A/F, i, n)

2.(P – S) (A/P, i, n) + S( i )

Annual equivalentOpportunity

for loss of value cost

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Capital Cost RecoveryCapital Cost RecoveryCapital Cost RecoveryCapital Cost Recovery

Land is considered to have infinite life; it can be sold for its purchase price (neglecting inflation).

P = S and EAC = S*i

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Example 1Example 1Example 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 $45,000 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 Equivalent Annual Cost (EAC) for the proposed expansion using a MARR of 18% compounded annually.

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1st Costs: 200 000 Land

600 000 Building

70 000 Equipment

Annuals: 20 000 Maintenance

45 000 Energy

Increases: 3 000/yr Energy

Replacements: 70 000 Equipment (yr 10)

70 000 Equipment (yr 20)

Salvage: 100 000 Building (yr 30)

200 000 Land

Lifetime: 30 yrs

MARR: 18%, cpd annually

FIND: EAC (= – EAW)

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EAC = A0 + AN + AI + AR + AS

= (200 000 + 600 000 + 70 000) (A|P,18%,30)

+ (20 000 + 45 000)

+ 3 000 (A|G,18%,30)

+ [70 000 (P|F,18%,10) + 70 000 (P|F,18%,20)] (A|P,18%,30)

– (100 000 + 200 000) (A|F,18%,30)

= 157 731 + 65 000 + 16 034 + 2 888 – 390 = $241 263 / yr

EAC = ?

0 1 2 3 n=30 yrsDIAGRAM:

G=$3 K

$100 K10 20

$70 K

$70 K

$70 K

$600 K

$200 K

$20 K

$45 K

$200 K

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Example 2Example 2Example 2Example 2A 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 6% 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.)

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EAWF = (A|P,6%,50)[ – 400 000 – 20 000(P|F,6%,10) – 20 000(P|F,6%,20)

– 20 000(P|F,6%,30) – 20 000(P|F,6%,40)]

= – $26 807 / yr

Full Tunnel0 n=50 yrs

DIAGRAM:20 30

$400 K$20 K

10 40

$20 K$20 K$20 K

Half Tunnel

0 n=50 yrsDIAGRAM:

20 30

$200 K

10 40

$18 K$18 K$18 K$18 K

$18 K$18 K$300 K

EAWH = (A|P,6%,50)[ –200 000 –18 000(P|F,6%,10) – (300 000 +18 000)(P|F,6%,20)

– (18 000 +18 000)(P|F,6%,30) – (18 000 +18 000)(P|F,6%,40)]

= – $20 223 / yr CHOOSE THE HALF TUNNEL OPTION

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Example 3Example 3Example 3Example 3Your in-laws 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 just completed an engineering economics course as a part of your engineering degree, you knew immediately that your in-laws’ reasoning was incorrect.

Identify and describe the engineering economic principle involved.

Include in your explanation what it actually cost your in-laws 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)

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The economic principle is OPPORTUNITY COST:

EAC = (P – S)(A|P,i,N) + S(i)

= ($100 000 – $100 000)(A|P,6%,18) + $100 000(6%)

= $0 + $6 000

= $6 000 / year to live in the house for each of the past 18 years.

Effectively, you could tell your in-laws that if they had truly lived in the house for free, they could have spent $6 000 per year on better raising your spouse …

… but then, again, maybe you shouldn’t.

What is the Engr Econ What is the Engr Econ principle of Example 3?principle of Example 3?What is the Engr Econ What is the Engr Econ

principle of Example 3?principle of Example 3?