AIN SHAMS UNIVERSITYFACULTY OF ENGINEERINGNew Program in Engineering (Freshmen) -Fall 2009Engi neeri ng Economics (H UM N240)

Ass.#5

Present Worth, Capitalized- Co st Evaluati onand Annual Worth Evaluation

1. An Investor is trying to decide whether he should invest the $30,000 he receivedfrom the sale of his boat in the stock market or in a small fast food restaurant withthree other partners. If he buys the stock, he will receive 3500 shareg which paydividends of $l per share each quarter. He expects the stock to appreciate to $40,000six years from now. If he invests in the restaurant, he will have to put up another$10,000 one year from now, but starting 2 years from now, his share of the profit willbe $9,000 per year for 5 years, after which he will receive $35,000 from the sale ofthe business. Using a PW analysis and an interest rate l2Yo per year compoundedquarterly, which investment should he make?

2.Two methods can be used for producing a certainmachine part. Method 1 costs$20,000 initially and will have $5,000 salvage value after 3 years. The operating costwith this method is $ 8500 per year. Method 2 has an initial cost of$ 15,000, but itwill last only 2 years. Its salvage value is $3,000. The operating cost for method 2 is $7,000 per year. If the minimum attractive rate of return is 12Yo per year, whichmethod should be used on the basis of present worth analysis?

3 Compare the two plans below using the PW method atl: l3Yo per yearPlan A Plan B

Machine I Machine ?First cost, $Annual operating cost, $Salvage value, $Life. years

10,000500

1,00040

30,000100

5,00040

5,000200-200

20,,

4. compare the alternatives shown below on the basis of present-worthcomparison. The interest rate is 16Yo per year:

Ahernative Rl Alternative R2First cost, $ 147,000 56,000Annual operating cost, $ 11,00cin year 1: increasing 30,000 in year 1:increasing

Salvage value, $

by $500 per year5,000

by $1000 per year2,000

a-tLife. vears 6

5. An alumnus of private university desire to establish a permanent scholarship in hisname. He'plans to donate $20,000 per year for 10 years starting iyear from now and

leave $100,000 when he dies. If the alumnus dies 15 years from now, how muchmoney will be in the account immediately after the $ 100,000 deposit, assuming theaccount earned interest atarate of9o/o per year.

6. Compare the alternatives shown below on the basis of their capitalizedcosts. Use i : I4o/o per year.

Alternative U Alternative W

First cost, $Annual operating cost, $Salvage value, $

8,500,0008,0005,000

50,000,0007,0002,000

Life- vears 5

7. Find the annual worth amount (per month) of a truck which had a first cost of $38,000, an operating cost of $ 2,000 per month, and a salvage value of $ 1 1,000 after4 years at an interest rate of 9Yo per year compounded monthly. Use salvage sinkingfund method.

8. Machines that have the following costs are under consideration for acontinuous production process. Using an interest rate l2Yo per year,

determine which alternative should be selected on the basis of an annualworth analysis:

Machine G Machine HFirst cost, $Annual operating cost, $Salvage value, $

62,00015,0008.000

77,00021,00010,000

6Lif

9. The Mining Company is considering purchasing a machine which costs

$30,000 and is expected to last 11 years, with a $3,000 salvage value. Theannual operating expenses are expected to be $8,000 for the first 3 years, but owing toincreased use, the operating costs will increase by $2000 per year for the next 8 years.

Alternatively, the company can purchase a highly automated machine at a cost of$58,000. This machine will last only 6 years because of its higher technology and

deiicate design, and its salvage value will be $15,000. Because it is so automated, itsoperating cost will be $4,000 per year. If the. company's minimum athactive rate ofreturn is 18% per year, which machine should be selected on the-basis of an annualworth analysis?

10. Determine the annual worth of the following cash flows at an interestrate of l2oh per year:

Year012-67-12 13 onCash Flow, $ -50,000 -6,000 -2,000 +3,000 +4,000

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