" FOR NJOSH ONLY"

Engr. 610 Engineering Economic Analysis

Name: Abdulaziz Alharbi Summer 2016, Ins: Mutlu Ozer

Quiz-2; CFDs are required-no hand drawing; Use any computer drawing Software!

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Problem 1.

Because unintended lane changes by distracted drivers are responsible for 43% of all highway fatalities, Ford Motor Co. and Volvo launched a program to develop technologies to prevent accidents by sleepy drivers. A device costing $260 tracks lane markings and sounds an alert during lane changes. If these devices are included in 100,000 new cars per year beginning 3 years from now, what would be the present worth of their cost over a 10-year period at an interest rate of 8% per year?

Answer:

CFD:

8%

0 3 10

$100000*(260)

P=?

P = 100,000(260)(P/A,8%,8)(P/F,8%,2)

= $128,090,564.7

Problem 2.

Toyco Watercraft has a contract with a parts supplier that involves purchases amounting to $150,000 per year, with the first purchase to be made now, followed by similar purchases over the next 5 years. Determine the present worth of the contract at an interest rate of 12% per year.

Answer:

12%

$150000

P=?

P = 150,000 + 150,000(P/A,12%,5)

= $690,720

Problem 3.

Calculate the present worth in year 0 of the following series of disbursements. Assume that i = 9% per year.

Year	Disbursement, $	Year	Disbursement, $
0	0	6	5000
1	3500	7	5000
2	3500	8	5000
3	3500	9	5000
4	5000	10	5000
5	5000

Answer: CFD:

9%

   

0 1 3 4

10

P=?

$3500

$5000

P = 3500(P/A,9%,3) + 5000(P/A,9%,7)(P/F,9%,3)

= $28,291.963

Problem 4.

BKM Systems sales revenues are shown below. Calculate the equivalent annual worth (years 1 through 7), using an interest rate of 14% per year.

Year	Disbursement, $	Year	Disbursement, $
0		4	5000
1	4000	5	5000
2	4000	6	5000
3	4000	7	5000

Answer:

CFD:

14%

0 1 3 4 7

$4000

$5000

A = 4000 + 1000(F/A,14%,4)(A/F,14%,7)

= $4,458.6

Problem5.

Calculate the annual worth in years 1 through 10 of the following series of incomes and expenses, if the interest rate is 15% per year.

Year	Income, $/Year	Expense, $/Year
0	10,000	2000
1–6	800	200
7–10	900	300

Answer:

CFD:

10%

$10K $800 $900

$2K 1 6 7 10

$300

$200

A = 8000(A/P,15%,10) + 600

= $2,194.4

Problem 6.

Lifetime savings accounts, known as LSAs, would allow people to invest after-tax money without being taxed on any of the gains. If an engineer invests $10,000 now and $10,000 each year for the next 20 years, how much will be in the account immediately after the last deposit if the account grows by 12% per year?

Answer:

1 0 12% 20

$10K

 

F = 10,000(F/A,12%,21)

= $816,987

Problem 7.

Calculate the future worth (in year 11) of the following income and expenses, if the interest rate is 10% per year.

Year	Income, $	Expense, $
0	12,000	3000
1–6	800	200
7–11	900	200

Answer: CFD:

12%

$12K $800 $900

1 6 7 11

$3K $200 $200

 

F = 9000(F/P,10%,11) + 600(F/A,10%,11) + 100(F/A,10%,5)

= $37,407.13

Problem 8.

What is the equivalent worth in year 5 of the following series of income and disbursements, if the interest rate is 11% per year?

Year	Income, $	Expense, $
0	0	9000
1–5	6000	6000
6–8	6000	3000
9–14	8000	5000

Answer: CFD:

11%

 

$6000

0 1 5 6 8 9 14

$8000

$9K

$6000

$3000

$5000

Worth in year 5 = -9000(F/P,11%,5) + 3000(P/A,11%,9)

= $1,445.1

Problem 9.

Use the cash flow diagram below to calculate the amount of money in year 5 that is equivalent to all the cash flows shown, if the interest rate is 16% per year.

Answer:

Amt, year 5 = 1000(F/A,16%,4)(F/P,16%,2) + 2000(P/A,16%,7)(P/F,16%,1)

= $13,780.5

0 $1000 3 7 13yr

$2000

Problem 10

Find the value of x below such that the positive cash flows will be exactly equivalent to the negative cash flows, if the interest rate is 14% per year.

Answer:

Move all cash flows to year 9.

0 = -800(F/A,14%,2)(F/P,14%,8) + 700(F/P,14%,7) + 700(F/P,14%,4)

      –950(F/A,14%,2)(F/P,14%,1) + x – 800(P/A,14%,3)

0 = -800(2.14)2.8526) + 700(2.5023) + 700(1.6890)

      –950(2.14)(1.14) + x – 800(2.3216)

x = $6124.64

Problem 11.

Calculate the annual worth (years 1 through 7) of the following series of disbursements. Assume that i = 10% per year.

Year	Disbursement, $	Year	Disbursement, $
0	5000	4	5000
1	3500	5	5000
2	3500	6	5000
3	3500	7	5000

Answer:

A = 5000(A/P,10%,7) + 3500 + 1500(F/A,10%,4)(A/F,10%,7)

= $5,260.78 CFD:

1 3 4 7yr

$5000 $3500

$5000

Problem 12.

Exxon-Mobil is planning to sell a number of producing oil wells. The wells are expected to produce 100,000 barrels of oil per year for 8 more years at a selling price of $28 per barrel for the next 2 years, increasing by $1 per barrel through year 8. How much should an independent refiner be willing to pay for the wells now, if the interest rate is 11% per year?

Answer:

P = [2,800,000(P/A,11%,7) + 100,000(P/G,11%,7) + 2,800,000](P/F,11%,1)

= $15,507,024.85

Problem 13

Calculate the present worth (year 0) of a lease that requires a payment of $20,000 now and amounts increasing by 5% per year through year 10. Use an interest rate of 5% per year.

Answer: CFD:

P-1= 20,000{[1 – (1 + 0.05)^11/(1 + 0.05)^11]/(0.05 – 0.05)}.

= $220000

Problem 14.

The annual worth in years 4 through 8 of an amount of money x that will be received 2 years from now is $4000. At an interest rate of 12% per year, the value of x is closest to

Answer:

x = 4000(P/A,12%,5)(P/F,12%,1)

= $12874.54

Problem 15.

A company that manufactures hydrogen sulfide monitors is planning to make deposits such that each one is 5% larger than the preceding one. How large must the first deposit be (at the end of year 1) if the deposits extend through year 10 and the fourth deposit is $1250? Use an interest rate of 11% per year.

Answer:

11%

0 1 4 10

A4 = 1250

Ai = Ai(1-g)^i-1

A=1250(1-0.05)

= $1458

Problem 16.

The maintenance cost for a certain machine is $1000 per year for the first 5 years and $2000 per year for the next 5 years. At an interest rate of 9% per year, the annual worth in years 1 through 10 of the maintenance cost is closest to

Answer:

9%

0 1 5 6 10

1000

2000

P = 1000(P/A,9%,5) + 2000(P/A,9%,5)(P/F,9%,5)

= $1393.98

Problem 17.

If a sum of $5000 is deposited now, $7000 two years from now, and $2000 per year in years 6 through 10, the amount in year 10 at an interest rate of 8% per year will be closest to

Answer: CFD:

F = 5000(F/P,8%,10) + 7000(F/P,8%,8) + 2000(F/A,8%,5)

= $35484

10%

0 2 5 6 10

2000

500 7000