# Equivalent annual worth for economics engineering

1) Consider the following estimates, and use an interest rate of 10% per year. The equivalent annual worth of alternative A is closest to:

A) $-25,130

B) $-37,100

C) $-41,500

D) $-42,900

2) Consider the following estimates, and use an interest rate of 10% per year. The equivalent annual worth of alternative B is closest to:

A) $-25,130

B) $-28,190

C) $-37,080

D) $-39,100

3) With an interest rate of 10% per year and given the following estimates, the annual worth of alternative F is closest to:

A) $32,600

B) $36,100

C) $39,020

D) $43,500

4) Given the following estimates, and with an interest rate of 10% per year, the annual worth of alternative G is closest to:

A) $10,000

B) $30,000

C) $36,000

D) $40,000

5) The first cost of a fairly large flood control dam is expected to be $5 million. The maintenance cost will be $60,000 per year, and a $100,000 outlay will be required every 5 years. If the dam is expected to last forever, its equivalent annual worth at an interest rate of 10% per year is closest to:

A) $-576,380

B) $-591,580

C) $-630,150

D) $-691,460

6) An investment of $50,000 resulted in uniform income of $10,000 per year for 10 years and a single amount of $5000 in year 5. The rate of return on the investment was closest to:

A) 10.6% per year

B) 14.2% per year

C) 16.4 % per year

D) 18.6 % per year

7) Five years ago, an alumnus of a small university donated $50,000 to establish a permanent endowment for scholarships. The first scholarships were awarded 5 years after the money was donated. If the amount awarded each year (i.e., the interest) is $5000, the rate of return earned on the fund is closest to:

A) 7.5% per year

B) 10% per year

C) 11% per year

D) 14% per year

8) When positive net cash flows are generated before the end of a project, and when these cash flows are reinvested at an interest rate that is less than the internal rate of return,

A) The resulting rate of return is equal to the internal rate of return.

B) The resulting rate of return is less than the internal rate of return.

C) The resulting rate of return is equal to the reinvestment rate of return.

D) The resulting rate of return is greater than the internal rate of return.

9) A $10,000 municipal bond due in 10 years pays interest of $400 per year. If an investor purchases the bond now for $9000 and holds it to maturity, the rate of return received by the investor will be closest to:

A) 3.5% per year

B) 4.2% per year

C) 5.3% per year

D) 6.9% per year

10) The difference between revenue and service alternatives is that service alternatives assume revenues are the same for all alternatives.

A) True

B) False

11) The rate of return for alternative X is 18% and for alternative Y is 16%, with Y requiring a larger initial investment. If a company has a minimum attractive rate of return of 16%,

A) The company should select alternative X.

B) The company should select alternative Y.

C) The company should conduct an incremental analysis between X and Y in order to select the correct alternative.

D) The company should select the do-nothing alternative.

12) When conducting a ROR analysis of mutually exclusive projects that have revenue estimates,

A) All the projects must be compared against the do-nothing alternative.

B) More than one project may be selected.

C) The project with the highest ROR should be selected.

D) An incremental investment analysis maybe necessary to identify the best one.

13) Consider the estimates below. If the alternatives are mutually exclusive and the MARR is 15% per year, the one(s) that should be selected is (are):

A) A

B) D

C) E

D) None of them

14) When a B/C analysis is conducted,

A) The benefits and costs must be expressed in terms of their present worth

B) The benefits and costs must be expressed in terms of their annual worth

C) The benefits and costs must be expressed in terms of their future worth

D) The benefits and costs can be expressed in terms of PW, AW, or FW.

15) An alternative has the following cash flows: benefits = $50,000 per year; disbenefits = $27,000 per year; costs = $25,000 per year. The B/C ratio is closest to:

A) 0.92

B) 0.96

C) 1.04

D) 2.00

16) In evaluating three mutually exclusive alternatives by the B/C method, the alternatives were ranked in terms of increasing total equivalent cost (A,B, and C, respectively), and the following results were obtained for the B/C ratios: 1.1, 0.9, and 1.3. On the basis of these results, you should:

A) Select A

B) Select C

C) Select A and C

D) Compare A and C incrementally.

17) The economic service life of an asset is:

A) The longest time that asset will still perform the function that it was originally purchased for.

B) The length of time that will yield the lowest annual worth of costs.

C) The length of time that will yield the lowest present worth of costs.

D) The time required for its market value to reach the originally estimated salvage value.

18) In a replacement study conducted last year, it was determined that the defender should be kept for 3 more years. Now, however, it is clear that some of the estimates that were made for this year and next year were in error. The proper course of action is to:

A) Replace the existing asset now.

B) Replace the existing asset 2 years from now, as was determined last year.

C) Conduct a new replacement study using the new estimates.

D) Conduct a new replacement study using last year's estimates.

19) When all future cash flows are expressed in constant-value dollars, the rate that should be used in the factor equations is the:

A) Market interest rate.

B) Inflation rate.

C) Inflated interest rate.

D) Real interest rate.

20) An assembly line robot arm with a first cost of $25,000 in 1985 had a cost of $29,860 in 1992. If the M&S equipment cost index was 789.6 in 1985 and the robot cost increased exactly in proportion to the index, the value of the index in 1992 was closest to:

A) Less than 800

B) 832.3

C) 914.6

D) More than 925

21) A 50-hp turbine pump was purchased for $2100. If the exponent in the cost capacity equation has a value of 0.76, a 200-hp turbine could be expected to cost about:

A) Less than $5000

B) $5980

C) $6020

D) More than $6100

22) A machine with a 10-year life is to be depreciated by the MACRS method. The machine has a first cost of $30,000 with a $5000 salvage value. Its annual operating cost is $7000 per year. The depreciation charge in year 3 is nearest to:

A) $3600

B) $4320

C) $5860

D) $7120

23) An asset with a first cost of $50,000 is to be depreciated by the straight-line method over a 5-year period. The asset will have annual operating costs of $20,000 and a salvage value of $10,000. According to the straight line method, the book value at the end of year 3 will be closest to:

A) $8000

B) $20,000

C) $24,000

D) $26,000

24) An asset with a first cost of $50,000 is depreciated by the straight line method over a 5-year life. Its annual operating cost is $20,000, and its salvage value is expected to $10,000. The book value at the end of year 5 will be nearest to:

A) $0

B) $8000

C) $10,000

D) $14,000

25) An asset had a first cost of $50,000 an estimated salvage value of $10,000 and was depreciated by the MACRS method. If its book value at the end of year 3 was $21,850 and its market value was $25,850, the amount of depreciation charged against the asset up to that time was closest to:

A) $18,850

B) $21,850

C) $25,850

D) $28,150

26) Calculate the annual worth (years 1 through 10) of the following series of disbursements. Assume that i = 12% per year.

A) $-6817

B) $-5817

C) $-4817

D) $-7817

#### Solution Preview

Hello!

Here are your answers.

Question 1 - Answer is B

In order to find the equivalent annual worth, we first find the present worth of machine A:

PV = -50000 - 20000/1.1 - 20000/1.1^2 - 10000/1.1^3 = $92,223

[notice that the last payment is -$10,000 rather than -$20,000, because of thesalvage value of $10,000]

Now, the equivalent annual worth is the annual payment for a number of periods equal to the duration of the machine (3 periods) such that their present worth is the same as the present worth of the machine.

As before, we proceed first by finding the present worth of $1 on the next 3 years:

PV = 1/1.1 + 1/1.1^2 + 1/1.1^3 = $2.48685

Therefore, the annual payment whose PV is $92,223 is 92223/2.48685 = $37,084

So that's the equivalent annual worth of this machine.

Question 2 - Answer is A

The idea is the same as in question 1. We first find the present worth of this machine:

PV = -80000 -10000/1.1 - 10000/1.1^2 - 10000/1.1^3 - 10000/1.1^4 - 10000/1.1^5 + 15000/1.1^6

The result is $109,440.75. We now find the present worth of a $1 cash flow each year for 6 years:

PV = 1/1.1 + 1/1.1^2 + 1/1.1^3 + 1/1.1^4 + 1/1.1^5 + 1/1.1^6

The result is $4.35526. Therefore, the equivalent annual worth of machine B is 109440.75/4.35526 = $25,128.40

Question 3 - Answer is C

The idea is once again the same as in the previous question. We first find the present worth of the machine; then we find the present worth of a $1 cash flow (for a number of years equal to the duration of the machine), and then we divide the former by ...

#### Solution Summary

The value of an asset at the end of its life is determined.