PDF Chapter 6 Equivalent Annual Worth - OUP

Chapter 6

Equivalent Annual Worth

6-1 Deere Construction just purchased a new track hoe attachment costing $12,500. The CFO, John, expects the implement will be used for five years when it is estimated to have a salvage value of $4,000. Maintenance costs are estimated to be $0 the first year and will increase by $100 each year thereafter. If a 12% interest rate is used, what is the equivalent uniform annual cost of the implement?

a. $2,925 b. $2,975 c. $3,015 d. $3,115

Solution

EUAC = 12,500(A/P, 12%, 5) - 4,000(A/F, 12%, 5) + 100(A/G, 12%, 5) = $3,015.40

The answer is c.

6-2 The survey firm of Myers, Anderson, and Pope (MAP) LLP is considering the purchase of a piece of new GPS equipment. Data concerning the alternative under consideration are presented below.

First Cost Annual Income Annual Costs Recalibration at end of Year 4 Salvage Value

$28,000 7,000 2,500 4,000 2,800

If the equipment has a life of eight years and MAP's minimum attractive rate of return (MARR) is 5%, what is the annual worth of the equipment?

Solution

EUAC = 28,000(A/P, 5%, 8) - 4,500 - 4,000(P/F, 5%, 4)(A/P, 5%, 8) - 2,800(A/F, 5%, 8) = -$47.63

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Chapter 6 Annual Cash Flow Analysis

Ronald McDonald decides to install a fuel storage system for his farm that will save him an

estimated 6.5 cents/gallon on his fuel cost. He uses an estimated 20,000 gallons/year on his farm.

Initial cost of the system is $10,000 and the annual maintenance the first year is $25 and increases

by $25 each year thereafter. After a period of 10 years the estimated salvage is $3,000. If money

is worth 12%, is it a wise investment?

Solution

EUAC = 10,000(A/P, 12%, 10) + 25 + 25(A/G, 12%, 10) = $1,884.63

EUAB = 20,000(.065) + 3,000(A/F, 12%, 10) = $1,471.00

EUAW = -$413.63 not a wise investment

6-4 The incomes for a business for five years are as follows: $8,250, $12,600, $9,750, $11,400, and $14,500. If the value of money is 12%, what is the equivalent uniform annual benefit for the fiveyear period?

Solution

PW = 8,250(P/F, 12%, 1) + 12,600(P/F, 12%, 2) + 9,750(P/F, 12%, 3) + 11,400(P/F, 12%, 4) + 14,500(P/F, 12%, 5)

= $39,823

EUAB = 39,823(A/P, 12%, 5) = $11,047

6-5 At an interest rate of 10% per year, the perpetual equivalent annual cost of $70,000 now, $100,000 at the end of year six, and $10,000 per year from the end of year ten through infinity is closest to:

a. $16,510 b. $24,200 c. $31,500 d. $37,630

Solution

P = 70,000 + 100,000(P/F, 10%, 6) + 10,000(P/A, 10%, )(P/F, 10%, 10) = $165,110

A = 165,110(A/P, 10%, ) = $16,511

The answer is a.

6-6 The state engineer estimates that the cost of a canal is $680 million. The legislative analyst estimates the equivalent annual cost of the investment for the canal to be $20.4 million. If the analyst expects the canal to last indefinitely, what interest rate is he using to compute the

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equivalent annual cost (EAC)? If the canal lasts only 50 years, what interest rate will the analyst

be assuming if he believes the EAC to be the same $20.4 million?

Solution

a) A = P(A/P, i%, n) For n = , (A/P, i%, ) = i A = P(i)

i = A/P = = .03 or

i = 3%

b)

A = P (A/P, i%, 50)

(A/P, i%, 50) = = .03

Searching interest tables at n = 50

i = 1.75%

6-7 What uniform annual payment for 12 years is equivalent to receiving all of the following:

$ 3,000 at the end of each year for 12 years 20,000 today

4,000 at the end of 6 years 800 at the end of each year forever

10,000 at the end of 15 years

Use an 8% interest rate.

Solution

A1 = $3,000 A2 = 20,000(A/P, 8%, 12) = $2,654 A3 = 4,000(P/F, 8%, 6)(A/P, 8%, 12) = $334.51 A4 = (800/.08)(A/P, 8%, 12) = $1,327 A5 = 10,000(P/F, 8%, 15)(A/P, 8%, 12) = $418.27

6-8 For the following cash flow diagram, which equation properly calculates the uniform equivalent?

a. A = 100(A/P, i, 3) + 100(A/F, i, 3) b. A = 100(A/P, i, 15) c. A = 100(A/P, i, 15) + 100(A/F, i, 3) d. A = 100(A/F, i, 3) + 100(A/F, i, 15)

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Chapter 6 Annual Cash Flow Analysis

0

3

6

9

12

15

100

100

100

100

100

100

Solution

The correct equation is (c).

6-9 A project has a first cost of $75,000, operating and maintenance costs of $10,000 during each year of its 8 year life, and a $15,000 salvage value. What is its equivalent uniform annual cost (EUAC) if the interest rate is 12%?

Solution

EUAC = 75,000(A/P, 12%, 8) + 10,000 - 15,000(A/F, 12%, 8) = $23,878.00

6-10 A recent engineering graduate makes a donation of $20,000 now and will pay $375.00 per month for 10 years to endow a scholarship. If interest is 9%, what annual amount can be awarded? Assume the first scholarship will be bestowed at the end of the first year after full funding.

Solution

P = 20,000 + 375.00(P/A, ?%, 120) = 49,603.25

A = Pi

= 49,603.25(.09) = $4464.29 scholarship

6-11 A rich folk singer has donated $500,000 to endow a university professorial chair in Bohemian Studies. If the money is invested at 8.5%, how much can be withdrawn each year, ad infinitum, to pay the Professor of B.S.?

Solution

A = 500,000(A/P, 8.5%, ) = 500,000(.085) = $42,500

6-12 A foundation supports an annual seminar on campus by using the earnings of a $50,000 gift. It is felt that 10% interest will be realized for 10 years, but that plans should be made to anticipate an interest rate of 6% after that time. What uniform annual payment may be established from the beginning, to fund the seminar at the same level into infinity?

Solution P

Chapter 6 Annual Cash Flow Analysis

95

A = ?

n = 10 i = 10%

n = i = 6%

P'

Assume first seminar occurs at time of deposit.

P' = A/i = A/.06 P = A + A(P/A, 10%, 10) + P'(P/F, 10%, 10) 50,000 = A + 6.145A + (A/.06) x .3855 13.57A = 50,000 A = $3,684.60

6-13 Given:

A = ?

P = $12,000,000

n = i = 3%

Find: A

Solution

A = Pi = 12,000,000(0.03) = $360,000

6-14 A project requires an initial investment of $10,000 and returns benefits of $6,000 at the end of every 5th year thereafter. If the minimum attractive rate of return (MARR) is 10%, the equivalent uniform annual worth is closet to

a. -$17.20 b. -$1,600 c. -$5,000 d. -$8,410

Solution

Year

0

5

Cash Flow ($) -10,000 6,000

10 6,000

15 6,000

20 6,000

25.... ... 6,000 6,000

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