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Engineering Economics covers the time value of money, present and annual worth, rate of return, depreciation, and project comparison. Exams are formula-driven but detail-sensitive: compounding period mismatches and sign conventions cause most lost marks. Timed practice with worked solutions builds the cash-flow-diagram discipline that prevents them.

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Topics typically tested

  • Cash flow diagrams & compound interest factors
  • Present worth & annual worth analysis
  • Internal rate of return
  • Benefit-cost analysis
  • Depreciation methods
  • Breakeven & replacement analysis

Sample Engineering Economics exam questions

Three original questions in the style of a real Engineering Economics exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choicePresent worth & annual worth analysis

A company is considering a machine that costs $50,000 now. It is expected to save $12,000 per year in labour, received at the end of each year for 6 years, after which the machine can be sold for a salvage value of $8,000. The company's minimum attractive rate of return (MARR) is 10% per year, compounded annually. What is the present worth of the investment?

  1. A.

    $2,263

  2. B.

    $6,779

  3. C.

    $10,263

  4. D.

    $30,000

Worked solution
Answer:B

Draw the cash flow diagram: a $50,000 outflow at time 0, six equal $12,000 inflows at the ends of years 1 to 6, and a single $8,000 inflow at the end of year 6 (at the same time as the last saving). Discount every cash flow to time 0 at 10%.

Compound interest factors at i=10%i = 10\%, n=6n = 6:

(P/A,10%,6)=(1+i)n−1i(1+i)n=1.16−10.1(1.16)=0.77160.17716=4.3553(P/A, 10\%, 6) = \frac{(1+i)^n - 1}{i(1+i)^n} = \frac{1.1^6 - 1}{0.1(1.1^6)} = \frac{0.7716}{0.17716} = 4.3553 (P/F,10%,6)=11.16=11.7716=0.5645(P/F, 10\%, 6) = \frac{1}{1.1^6} = \frac{1}{1.7716} = 0.5645

Present worth:

PW=−50,000+12,000(4.3553)+8,000(0.5645)=−50,000+52,264+4,516=6,780PW = -50{,}000 + 12{,}000(4.3553) + 8{,}000(0.5645) = -50{,}000 + 52{,}264 + 4{,}516 = 6{,}780

(Carrying unrounded factors gives $6,779.) The present worth is positive, so the machine earns more than the 10% MARR and is acceptable. Equivalently, the annual worth is AW=PW(A/P,10%,6)=6,779(0.2296)≈1,557AW = PW(A/P, 10\%, 6) = 6{,}779(0.2296) \approx 1{,}557, about $1,557 per year.

The tempting answer $2,263 forgets the salvage value entirely, and $10,263 adds the $8,000 salvage at face value without discounting it six years; both are common slips. $30,000 ignores the time value of money altogether (72,000 + 8,000 − 50,000).

PW≈PW \approx $6,779 (positive, so the machine is acceptable)

Question 2Short answerCash flow diagrams & compound interest factors

A student deposits $200 at the end of every month for 4 years into a savings account that pays a nominal annual interest rate of 6% compounded monthly. No other deposits or withdrawals are made. How much is in the account immediately after the last deposit?

Worked solution
Answer:

$10,820 ($10,819.57 to the cent)

Match the compounding period to the cash flow period. A nominal 6% per year compounded monthly is an effective rate per month of

i=0.0612=0.005=0.5% per monthi = \frac{0.06}{12} = 0.005 = 0.5\%\ \text{per month}

and 4 years of monthly deposits is n=4(12)=48n = 4(12) = 48 periods. The cash flow diagram is a uniform series of 48 deposits of $200 at the ends of months 1 to 48. The future worth is wanted at month 48, the moment of the last deposit, so the standard F/AF/A factor applies with no extra compounding.

(F/A,0.5%,48)=(1+i)n−1i=1.00548−10.005=1.27049−10.005=54.098(F/A, 0.5\%, 48) = \frac{(1+i)^n - 1}{i} = \frac{1.005^{48} - 1}{0.005} = \frac{1.27049 - 1}{0.005} = 54.098 F=A (F/A,0.5%,48)=200(54.098)=10,819.6F = A\,(F/A, 0.5\%, 48) = 200(54.098) = 10{,}819.6

The student deposited 48 × $200 = $9,600 in total, so about $1,220 is interest. That is reasonable: the average deposit sits in the account for about two years at roughly 6% per year, and 9,600 × 0.06 × 2 ≈ 1,150 before compounding.

Using the annual rate of 6% with n=4n = 4 years would be wrong here, because both the deposits and the compounding are monthly.

F≈F \approx $10,820

Question 3Worked problemDepreciation methods

A piece of production equipment is purchased for $120,000. It has an estimated useful life of 5 years and an estimated salvage value of $20,000 at the end of that life. (a) Using straight-line depreciation, find the annual depreciation charge and the book value at the end of year 3. (b) Using double-declining-balance (DDB) depreciation, find the depreciation charge in year 3 and the book value at the end of year 3. (c) Still using DDB, find the depreciation charge in year 4, given that the book value is not allowed to fall below the salvage value.

Worked solution
Answer:

(a) $20,000 per year, BV₃ = $60,000; (b) D₃ = $17,280, BV₃ = $25,920; (c) D₄ = $5,920

(a) Straight line spreads the depreciable amount (cost minus salvage) evenly over the life:

Dt=B−SN=120,000−20,0005=20,000 per yearD_t = \frac{B - S}{N} = \frac{120{,}000 - 20{,}000}{5} = 20{,}000\ \text{per year}

Book value falls by that amount each year, so after three years BV3=120,000−3(20,000)=BV_3 = 120{,}000 - 3(20{,}000) = $60,000.

(b) DDB applies a fixed rate of 2/N=2/5=0.402/N = 2/5 = 0.40 to the book value at the start of each year (not to cost minus salvage), so the charges shrink geometrically:

D1=0.40(120,000)=48,000,BV1=72,000D2=0.40(72,000)=28,800,BV2=43,200D3=0.40(43,200)=17,280,BV3=25,920\begin{aligned} D_1 &= 0.40(120{,}000) = 48{,}000, & BV_1 &= 72{,}000\\ D_2 &= 0.40(72{,}000) = 28{,}800, & BV_2 &= 43{,}200\\ D_3 &= 0.40(43{,}200) = 17{,}280, & BV_3 &= 25{,}920 \end{aligned}

The same result follows directly from BV3=B(1−0.40)3=120,000(0.216)=BV_3 = B(1 - 0.40)^3 = 120{,}000(0.216) = $25,920. Since $25,920 is still above the $20,000 salvage, no adjustment is needed in year 3.

(c) The unadjusted year-4 charge would be 0.40(25,920)=0.40(25{,}920) = $10,368, which would drop the book value to $15,552, below salvage. Depreciation stops at salvage, so the year-4 charge is limited to the remaining depreciable amount:

D4=BV3−S=25,920−20,000=5,920D_4 = BV_3 - S = 25{,}920 - 20{,}000 = 5{,}920

After that BV4=BV_4 = $20,000 and the year-5 charge is zero. Over the five years DDB writes off the full $100,000, the same total as straight line, but front-loads it: 48% of it in year 1 versus 20% under straight line.

(a) $20,000 per year, BV3=BV_3 = $60,000; (b) D3=D_3 = $17,280, BV3=BV_3 = $25,920; (c) D4=D_4 = $5,920

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