Foundations of Engineering Economy

Chapter 1

Foundations of Engineering Economy

Solutions to Problems

1.1 Time value of money means that there is a certain worth in having money and the

worth changes as a function of time.

1.2 Morale, goodwill, friendship, convenience, aesthetics, etc.

1.3 (a) Evaluation criterion is the measure of value that is used to identify “best”.

(b) The primary evaluation criterion used in economic analysis is cost.

1.4 Nearest, tastiest, quickest, classiest, most scenic, etc

1.5 If the alternative that is actually the best one is not even recognized as an alternative, it obviously will not be able to be selected using any economic analysis tools.

1.6 In simple interest, the interest rate applies only to the principal, while compound

interest generates interest on the principal and all accumulated interest.

1.7 Minimum attractive rate of return is the lowest rate of return (interest rate) that

a company or individual considers to be high enough to induce them to invest their money.

1.8 Equity financing involves the use of the corporation’s or individual’s own funds for making investments, while debt financing involves the use of borrowed funds. An example of equity financing is the use of a corporation’s cash or an individual’s savings for making an investment. An example of debt financing is a loan (secured or unsecured) or a mortgage.

1.9 Rate of return = (45/966)(100)

= 4.65%

1.10 Rate of increase = [(29 – 22)/22](100)

= 31.8%

1.11 Interest rate = (275,000/2,000,000)(100)

= 13.75%

1.12 Rate of return = (2.3/6)(100)

= 38.3%

1.13 Profit = 8(0.28)

= $2,240,000

1.14 P + P(0.10) = 1,600,000

1.1P = 1,600,000

P = $1,454,545

1.15 Earnings = 50,000,000(0.35)

= $17,500,000

1.16 (a) Equivalent future amount = 10,000 + 10,000(0.08)

= 10,000(1 + 0.08)

= $10,800

(b) Equivalent past amount: P + 0.08P = 10,000

1.08P = 10,000

P = $9259.26

1.17 Equivalent cost now: P + 0.1P = 16,000

1.1P = 16,000

P = $14,545.45

1.18 40,000 + 40,000(i) = 50,000

i = 25%

1.19 80,000 + 80,000(i) = 100,000

i = 25%

1.20 F = 240,000 + 240,000(0.10)(3)

= $312,000

1.21 Compound amount in 5 years = 1,000,000(1 + 0.07)5

= $1,402,552

Simple amount in 5 years = 1,000,000 + 1,000,000(0.075)(5)

= $1,375,000

Compound interest is better by $27,552

1.22 Simple: 1,000,000 = 500,000 + 500,000(i)(5)

i = 20% per year simple

Compound: 1,000,000 = 500,000(1 + i)5

(1 + i)5 = 2.0000

(1 + i) = (2.0000)0.2

i = 14.87%

1.23 Simple: 2P = P + P(0.05)(n)

P = P(0.05)(n)

n = 20 years

Compound: 2P = P(1 + 0.05)n

(1 + 0.05)n = 2.0000

n = 14.2 years

1.24 (a) Simple: 1,300,000 = P + P(0.15)(10)

2.5P = 1,300,000

P = $520,000

(b) Compound: 1,300,000 = P(1 + 0.15)10

4.0456P = 1,300,000

P = $321,340

1.25 Plan 1: Interest paid each year = 400,000(0.10)

= $40,000

Total paid = 40,000(3) + 400,000

= $520,000

Plan 2: Total due after 3 years = 400,000(1 + 0.10)3

= $532,400

Difference paid = 532,400 – 520,000

= $12,400

1.26 (a) Simple interest total amount = 1,750,000(0.075)(5)

= $656,250

Compound interest total = total amount due after 4 years – amount borrowed

= 1,750,000(1 + 0.08)4 – 1,750,000

= 2,380856 – 1,750,000

= $630,856

(b) The company should borrow 1 year from now for a savings of $656,250 –

$630,856 = $25,394

1.27 The symbols are F = ?; P = $50,000; i = 15%; n = 3

1.28 (a) FV(i%,n,A,P) finds the future value, F

(b) IRR(first_cell:last_cell) finds the compound interest rate, i

(c) PMT(i%,n,P,F) finds the equal periodic payment, A

(d) PV(i%,n,A,F) finds the present value, P

1.29 (a) F = ?; i = 7%; n = 10; A = $2000; P = $9000

(b) A = ?; i = 11%; n = 20; P = $14,000; F = 0

(c) P = ?; i = 8%; n = 15; A = $1000; F = $800

1.30 (a) PV = P (b) PMT = A (c) NPER = n (d) IRR = i (e) FV = F

1.31

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