# 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

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#### 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.

Engineering Economics

1) The time it would take for money to double at a simple interest rate of 10% per year is closest to:

A) 5 Years

B) 7 Years

C) 10 Years

D) 12 Years

2) At a compound interest rate of 10% per year, $10,000 one year ago is equivalent to how much 1 year from now?

A) $8264

B) $9091

C) $11,000

D) $12,000

3) In most engineering economy studies, the best alternative is the one which:

A) Will last the longest time

B) Is easiest to implement

C) Costs the least

D) Is most politically attractive

4) The present worth of an investment of $20,000 in year 10 at an interest rate of 12% per year is closest to:

A) $6440

B) $7560

C) $8190

D) $10,3000

5) The amount of money that could be spent now in lieu of spending $50,000 seven years from now at an interest rate of 18% per year is closest to:

A) $15,700

B) $17,800

C) $19,300

D) $25,100

6) Income from a precious metals mining operation has been decreasing uniformly for 5 years. If income in year 1 was $100,000 and it decreased by $10,000 per year through year 5, the present worth of the income at 10% per year is closest to:

A) $310,500

B) $352,200

C) $379,100

D) $447,700

7) Income from sales of a certain oil additive has been averaging $100,000 per year. At an interest rate of 18% per year, the present worth of the income for 5 years is closest to:

A) $296,100

B) $312,700

C) $328,400

D) $335,100

8) A manufacturing company wants to have $100,000 available in 5 years to replace a production line. The amount of money that would have to be deposited each year at an interest rate of 10 % per year would be closest to:

A) $12,380

B) $13,380

C) $16,380

D) $26,380

9) The present worth of $5000 in year 3, $10,000 in year 5, and $10,000 in year 8 at an interest rate of 12% per year is closest to:

A) $12,100

B) $13,300

C) $14,900

D) $16,200

10) If $10,000 is borrowed now at 10% per year interest, the balance after a $4000 payment is made 5 years from now will be closest to:

A) Less than $13,500

B) $14,300

C) $16,100

D) More than $17,000

11) The present sum needed to provide for an annual withdrawal of $1000 for 25 years beginning 5 years from now at an interest rate of 10% per year is closest to:

A) Less than $6500

B) $7800

C) $8500

D) More than $9000

12) If a company wants to have $100,000 in a contingency fund 10 years from now, the amount the company must deposit each year in years 1 through 5, at an interest rate of 10% per year, is closest to:

A) Less than $8000

B) $8420

C) $9340

D) More than $10,000

13) An interest rate of 1% per month is:

A) A nominal rate

B) A simple rate

C) A effective rate with an unknown compounding period

D) An effective rate with a known compounding period

14) An interest rate of effective 14% per year, compounded weekly, is:

A) An effective rate per year

B) An effective rate per month

C) A nominal rate per year

D) A nominal rate per month

15) An interest rate of 2% per month is the same as:

A) 24% per year

B) A nominal 24% per year, compounded monthly

C) An effective 24% per year, compounded monthly

D) Both (a) and (b)

16) An interest rate of 15% per year, compounded monthly, is nearest to:

A) 1% per month

B) 15.12% per year

C) 16.08% per year

D) 16.92% per year

17) An environmental testing company needs to purchase $40,000 worth of equipment 2 years from now. At an interest rate of 20% per year, compounded quarterly, the present worth of the equipment is closest to:

A) $27,070

B) $27,800

C) $26,450

D) $28,220

18) A steel fabrication company invested $800,000 in a new shearing unit. At an interest rate of 12% per year, compounded monthly, the monthly income required to recover the investment in 3 years is closest to:

A) $221,930

B) $31,240

C) $29,160

D) $26,570

19) Consider the following estimates. The cost of money is 10% per year. The present worth of machine X is closest to?

A) $-68,445

B) $-97,840

C) $-125,015

D) $-223,120

20) Given the following estimates with the cost of money being 10% per year, the capitalized cost of machine Y is closest to:

A) $-30,865

B) $-97,840

C) $-308,650

D) $-684,445

21) The present worth of an alternative that provides infinite service is called its:

A) Net present value

B) Discounted total cost

C) Capitalized cost

D) Perpetual annual cost

22) In comparing alternatives with different lives by the present worth method, it is necessary to:

A) Compare them over a period equal to the life of the longer-lived alternative.

B) Compare them over a time period of equal service.

C) Compare them over a period equal to the life of the shorter-lived alternative.

D) Find the present worth over one life cycle of each alternative.

23) The upgraded version of a machine has a first cost of $20,000, an annual operating cost of $6000, and a salvage value of $5000 after its 8-year life. At an interest rate of 10% per year, the capitalized cost is closest to:

A) $-9,312

B) $-10,006

C) $-93,120

D) $-100,060

24) Find the capitalized cost of a present cost of $30,000, monthly costs of $1000, and periodic costs every 5 years of $5000. Use an interest rate of 12% per year, compounded monthly.

A) $-80,000

B) $-136,100

C) $-195,200

D) $-3,600,000

25) 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