Engineering Economics

Define the time value of money concept.

A dollar today is worth more than a dollar tomorrow because of its earning potential through interest or investment. It forms the basis for all economic equivalence equations.

Find the future value of a single payment of $5,000 in 4 years at 8% interest.

Use compound amount formula [HB].
F = P(1+i)^n
F = 5000(1.08)^4 = 6,803.

Find the present worth of $10,000 received 5 years from now at 6% interest.

Use single payment present worth [HB].
P = F(1+i)^{-n}
P = 10000(1.06)^{-5} = 7,472.

An investment of $3,000 per year for 8 years earns 10%. Find the future worth.

Use uniform series compound amount factor [HB].
F = A\frac{(1+i)^n - 1}{i}
F = 3000\frac{(1.10)^8 - 1}{0.10} = 25,752.

Find the annual payment required to repay a $15,000 loan over 6 years at 8%.

Use capital recovery formula [HB].
A = P\frac{i(1+i)^n}{(1+i)^n - 1}
A = 15000\frac{0.08(1.08)^6}{(1.08)^6 - 1} = 3,248.

A project requires an initial investment of $10,000 and returns $2,500 per year for 5 years at 8%. Find the Present Worth.

Use uniform series present worth factor [HB].
P = A\frac{(1+i)^n - 1}{i(1+i)^n}
P = 2500\frac{(1.08)^5 - 1}{0.08(1.08)^5} = 9,985.
Compare to cost → PW ≈ -$15 → Not justified.

A $4,000 investment returns $1,200 per year for 4 years. Find the rate of return (IRR).

Trial and error:
At i=10%, PW = +$29. At i=11%, PW = -$36 ⇒ IRR ≈ 10.5%.

A $20,000 loan is repaid with $5,100 annual payments for 5 years. Find the implied interest rate.

Use capital recovery relation A/P and trial. At 9%, A = $5,092 → rate ≈ 9%.

Compute the future worth of $2,000 invested annually for 10 years at 6%.

F = 2000\frac{(1.06)^{10} - 1}{0.06} = 26,558. [HB]

You need $20,000 in 5 years. How much should you deposit annually at 6%?

Rearrange uniform series compound amount.
A = F\frac{i}{(1+i)^n - 1}
A = 20000\frac{0.06}{(1.06)^5 - 1} = 3,530. [HB]

A cash flow increases by $500 each year for 6 years (starting at year 1). Find PW at 8%.

Use gradient present worth factor [HB].
P = G\left[\frac{(1+i)^n - i n - 1}{i^2(1+i)^n}\right]
P = 500\left[\frac{(1.08)^6 - 1 - 0.08(6)(1.08)^6}{0.08^2(1.08)^6}\right] = 2,230.

A project costs $12,000 and yields $3,500 annually for 5 years. Find PW at 10%.

P = 3500\frac{(1.10)^5 - 1}{0.10(1.10)^5} - 12000 = 1,225.
Positive PW → Accept.

A project costs $5,000 and yields $1,200 per year for 6 years. Find FW at 8%.

F = 1200\frac{(1.08)^6 - 1}{0.08} - 5000(1.08)^6 = 4,121.
Accept since FW > 0.

Find the payback period for a $10,000 investment generating $2,500 annually.

Payback = \frac{10000}{2500} = 4 \text{ years.}

A machine costs $15,000 and has a salvage value of $3,000 after 6 years. Find annual depreciation (SL).

D = \frac{P - S}{N} = \frac{15000 - 3000}{6} = 2000.

Compute book value after 4 years using straight-line depreciation from above.

BV = P - nD = 15000 - 4(2000) = 7000.

A $40,000 asset has a 5-year life, 10% rate, and MACRS depreciation. What is the first-year depreciation?

Multiply basis by MACRS rate (20%).
D_1 = 40000(0.20) = 8000.

Find present worth of a future maintenance cost of $1,000 per year escalating 3% annually for 5 years at 10%.

Use geometric gradient approximation.
P = A_1\frac{1 - \left(\frac{1+g}{1+i}\right)^n}{i - g}
P = 1000\frac{1 - \left(\frac{1.03}{1.10}\right)^5}{0.10 - 0.03} = 3,922. [HB]

Find the effective annual rate for a nominal 12% compounded monthly.

i_{eff} = (1 + \frac{0.12}{12})^{12} - 1 = 0.1268 = 12.68\%. [HB]

Find equivalent annual worth (AW) of a project costing $10,000 with PW = $1,200 at i = 8%, n = 5.

A = P\frac{i(1+i)^n}{(1+i)^n - 1}
A = 1200\frac{0.08(1.08)^5}{(1.08)^5 - 1} = 301.
Annual worth = +$301 ⇒ Accept. [HB]

A company invests $25,000 and saves $6,000 per year for 6 years. Find the rate of return.

Try i = 15% → PW = +558; i = 17% → PW = -143 ⇒ IRR ≈ 16.6%.

Compare two alternatives: A: $15,000 now; B: $8,000 now + $4,000 in 4 years at 10%. Which has lower PW?

PW_B = 8000 + 4000(1.10)^{-4} = 10,728 → Choose B.

Find the future worth of an increasing series of payments: $1,000, $1,200, $1,400, $1,600, $1,800 at 10%.

Add each individually:
FW = 1000(1.10)^4 + 1200(1.10)^3 + 1400(1.10)^2 + 1600(1.10)^1 + 1800 = 7,323.

Find equivalent uniform annual cost (EUAC) of an asset: cost = $12,000, life = 5 years, i = 8%, no salvage.

A = P\frac{i(1+i)^n}{(1+i)^n - 1}
A = 12000\frac{0.08(1.08)^5}{(1.08)^5 - 1} = 3,009.
EUAC = $3,009.

Compare two designs: A: cost $20k, life 6 yrs, EUAC = $4.2k; B: cost $28k, life 8 yrs, EUAC = $4.1k. Which is better?

Lower EUAC → B preferred.

A company borrows $50,000 at 9% to be repaid quarterly over 5 years. Find payment.

Use compound interest quarterly.
i = 0.09/4 = 0.0225, n = 20
A = 50000\frac{0.0225(1.0225)^{20}}{(1.0225)^{20} - 1} = 3,185.

A project requires $50,000 initial and yields $12,000 annually for 6 years. Find the PW at 10%.

P = 12000\frac{(1.10)^6 - 1}{0.10(1.10)^6} - 50000 = 1,876.
Positive PW → Accept.

Two options: A = $10k now, B = $5k now + $6k in 3 years. i = 8%. Which is better (PW)?

PW_B = 5000 + 6000(1.08)^{-3} = 9,771.
Choose B (lower cost).

Find the break-even point where two processes’ costs are equal. A = $1000 + 8x, B = $400 + 12x.

Set equal: 1000 + 8x = 400 + 12x → x = 150.
Break-even = 150 units.

Determine payback time for project with uneven cash flows: year 1 $3k, year 2 $4k, year 3 $5k, cost $10k.

Cumulative recovery after 3 years = $12k → Payback between 2 and 3 years.

A loan of $100,000 at 6% is repaid with equal payments over 10 years. Find total interest paid.

A = 100000\frac{0.06(1.06)^{10}}{(1.06)^{10}-1} = 13,587.
Total = $135,870 → Interest = $35,870.

If inflation is 3% and nominal rate is 10%, find the real interest rate.

i_{real}=\frac{1+i_{no\min al}}{1+f}-1=\frac{1.10}{1.03}-1=0.068=6.8 [HB]

A project has a PW of $5,000, what is its Benefit/Cost ratio if initial cost = $20,000?

B/C = \frac{PWB}{PWC} = \frac{25000}{20000} = 1.25
Since >1 → Accept.

If an investment has FW = $3,000, cost = $2,500, interest = 10%, n = 4, find its annual worth.

A = (FW - Cost)\frac{i}{(1+i)^n - 1}
A = 500\frac{0.10}{(1.10)^4 - 1} = 118.
Annual worth = +$118 ⇒ Accept.

Compute the equivalent present worth of a continuous inflow of $1000 per year for 5 years at 10%.

P = \int_0^5 1000e^{-0.10t}dt = 1000\frac{1 - e^{-0.5}}{0.10} = 3,935. [HB]

A project has a benefit-cost ratio (B/C) = 0.9 at i = 8%. What does this imply?

B/C < 1 → Project’s present benefits < costs → Reject.

A project has PW = $0 at 12%. What does that mean?

Internal rate of return (IRR) = 12%. It’s the break-even interest rate.

Find the sinking fund payment to replace a $50,000 asset in 10 years at 6%.

A = F\frac{i}{(1+i)^n - 1}
A = 50000\frac{0.06}{(1.06)^{10} - 1} = 3,682. [HB]

If an asset costs $30,000 and annual savings = $5,000 for 8 years, find the payback period and comment.

Payback = 30000/5000 = 6 \text{ years.}
Accept only if life ≥ 6 years.
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