Engineering Economy Practice Problems with Solutions (25 Worked Examples)

This page collects 25 original engineering economy practice problems with complete worked solutions. Each solution is typed step by step, so you can follow the factor look-ups, the arithmetic and the decision rule. The problems cover the same core topics found in an introductory engineering economy course: time value of money, annuities and gradients, present, future and annual worth, cash flow diagrams, and discounted payback. If you need the theory first, start with Engineering Economics 101: cash flows, interest and rate of return.

How to use these engineering economy practice problems

• Try each problem on paper first, then compare your steps and answer with the worked solution.
• All rates are annual and cash flows occur at the end of each year unless stated otherwise.
• Factor values are rounded to four or five decimals; your answer may differ by a few cents or dollars depending on rounding.
• Notation: (F/P, i, n) means “find F given P at interest rate i over n periods,” and so on.

Engineering economy formula cheat sheet

To find	Given	Factor	Formula
F	P	(F/P, i, n)	(1 + i)n
P	F	(P/F, i, n)	(1 + i)−n
A	P	(A/P, i, n)	i(1 + i)n / [(1 + i)n − 1]
P	A	(P/A, i, n)	[(1 + i)n − 1] / [i(1 + i)n]
F	A	(F/A, i, n)	[(1 + i)n − 1] / i
A	F	(A/F, i, n)	i / [(1 + i)n − 1]
A	G (arithmetic gradient)	(A/G, i, n)	1/i − n / [(1 + i)n − 1]
P	G	(P/G, i, n)	(A/G)(P/A)
Effective rate	Nominal r, m periods	—	(1 + r/m)m − 1

Interest Basics & Rate of Return Practice Problems

These problems cover simple and compound interest, single-payment present and future worth, nominal versus effective rates, solving for time, and computing rate of return.

Problem 1: Simple vs. Compound Interest

Problem. A small shop borrows $8,000 for 5 years at 6% per year. Compare the total owed at the end of year 5 if the loan uses (a) simple interest and (b) interest compounded annually.

Solution.

1. Simple interest: F = P(1 + in) = $8,000(1 + 0.06 × 5) = $10,400.00.
2. Compound interest: F = P(1 + i)ⁿ = $8,000(1.06)⁵ = $10,705.80.
3. Difference = $305.80, the interest earned on previously accrued interest.

Answer. Simple: $10,400.00; compound: $10,705.80; compounding costs $305.80 more.

Problem 2: Future Value of a Single Deposit

Problem. A deposit of $12,500 earns 4.5% per year, compounded annually. What is the account worth after 8 years, and how much of that is interest?

Solution.

1. F = P(F/P, i, n) = $12,500(1.045)⁸.
2. (F/P, 4.5%, 8) = 1.4221.
3. F = $12,500 × 1.4221 = $17,776.26.
4. Interest earned = $17,776.26 − $12,500 = $5,276.26.

Answer. F = $17,776.26, including $5,276.26 of interest.

Problem 3: Present Worth of a Future Need

Problem. A plant must replace a press in 6 years at an expected cost of $40,000. If money earns 7% per year, how much must be set aside today?

Solution.

1. P = F(P/F, i, n) = $40,000(1.07)⁻⁶.
2. (P/F, 7%, 6) = 0.6663.
3. P = $40,000 × 0.6663 = $26,653.69.

Answer. Deposit $26,653.69 today.

Problem 4: Solving for the Number of Years

Problem. How many years will it take $5,000 to grow to $9,000 at 5% per year compounded annually? Compare the exact answer with the Rule of 72 estimate for doubling.

Solution.

1. F/P = 1.80 = (1.05)ⁿ.
2. n = ln(1.80) / ln(1.05) = 0.5878 / 0.0488 = 12.05 years.
3. Doubling time at 5%: exact n = ln 2 / ln 1.05 = 14.21 years; Rule of 72 gives 72 / 5 = 14.4 years.

Answer. About 12.0 years to reach $9,000. The Rule of 72 is within 0.2 year of the exact doubling time.

Problem 5: Nominal vs. Effective Interest Rate

Problem. A credit line advertises 12% nominal annual interest. Find the effective annual rate if interest compounds (a) monthly and (b) quarterly. Which costs more?

Solution.

1. Effective rate: i_eff = (1 + r/m)^m − 1.
2. Monthly (m = 12): (1 + 0.12/12)¹² − 1 = 12.683%.
3. Quarterly (m = 4): (1 + 0.12/4)⁴ − 1 = 12.551%.

Answer. Monthly: 12.683%; quarterly: 12.551%. Monthly compounding costs more.

Problem 6: Rate of Return on a Single Investment

Problem. An investor puts $20,000 into a venture and receives a single payment of $29,500 at the end of year 6. Find the annual rate of return and decide whether it meets a MARR of 8%.

Solution.

1. Set P = F(P/F, i, n): $20,000 = $29,500(1 + i)⁻⁶.
2. (1 + i)⁶ = 1.4750, so i = 1.4750^1/6 − 1 = 6.69%.
3. Compare: 6.69% < 8% MARR.

Answer. Rate of return = 6.69%; does not meet the 8% MARR.

Problem 7: Equivalence of Cash Amounts

Problem. Is $10,000 received today equivalent to $13,000 received in 4 years if the interest rate is 7%? Which would you take?

Solution.

1. Convert the future amount to today: P = $13,000(P/F, 7%, 4) = $13,000 × 0.7629 = $9,917.64.
2. $9,917.64 < $10,000, so the two are not equivalent.

Answer. Take the $10,000 today; it is worth $82.36 more in present-value terms.

Annuities & Gradients Practice Problems

These problems use the uniform series factors (A/P, P/A, F/A, A/F) and the arithmetic and geometric gradient formulas.

Problem 8: Loan Payment (A given P)

Problem. A $25,000 equipment loan at 6.5% per year is repaid with 5 equal year-end payments. Find the annual payment and the total interest paid.

Solution.

1. A = P(A/P, i, n) = $25,000(A/P, 6.5%, 5).
2. (A/P, 6.5%, 5) = 0.24063.
3. A = $25,000 × 0.24063 = $6,015.86.
4. Total paid = 5 × $6,015.86 = $30,079.32; interest = $5,079.32.

Answer. Annual payment = $6,015.86; total interest = $5,079.32.

Problem 9: Savings Plan (F given A)

Problem. An engineer saves $3,000 at the end of each year for 10 years in an account paying 5% per year. What is the balance right after the last deposit?

Solution.

1. F = A(F/A, i, n) = $3,000(F/A, 5%, 10).
2. (F/A, 5%, 10) = 12.5779.
3. F = $3,000 × 12.5779 = $37,733.68.
4. Deposits total $30,000, so interest earned = $7,733.68.

Answer. Balance = $37,733.68.

Problem 10: Sinking Fund (A given F)

Problem. A company wants $50,000 available in 8 years to retire a bond. What equal year-end deposit must it make into a fund earning 4% per year?

Solution.

1. A = F(A/F, i, n) = $50,000(A/F, 4%, 8).
2. (A/F, 4%, 8) = 0.10853.
3. A = $50,000 × 0.10853 = $5,426.39.

Answer. Deposit $5,426.39 per year.

Problem 11: Present Worth of a Uniform Series (P given A)

Problem. A cost-saving upgrade will save $4,200 per year for 12 years. At 8% per year, what is the present worth of the savings, and what is the most the company should pay for the upgrade?

Solution.

1. P = A(P/A, i, n) = $4,200(P/A, 8%, 12).
2. (P/A, 8%, 12) = 7.5361.
3. P = $4,200 × 7.5361 = $31,651.53.

Answer. Present worth = $31,651.53; paying more than this loses money at an 8% MARR.

Problem 12: Arithmetic Gradient Maintenance Cost

Problem. Maintenance on a machine costs $1,000 at the end of year 1 and increases by $400 each year through year 6. At 9% per year, find (a) the present worth and (b) the equivalent uniform annual cost.

Solution.

1. Base amount: $1,000(P/A, 9%, 6) = $1,000 × 4.4859 = $4,485.92.
2. Gradient: G(P/G, 9%, 6) = $400 × 10.0924 = $4,036.95.
3. PW = $4,485.92 + $4,036.95 = $8,522.87.
4. Annual equivalent: A = $1,000 + $400(A/G, 9%, 6) = $1,000 + $400 × 2.2498 = $1,899.92.

Answer. PW = $8,522.87; equivalent annual cost = $1,899.92.

Problem 13: Geometric Gradient Revenue

Problem. A product line earns $30,000 at the end of year 1, and revenue grows 5% per year through year 7. At 10% per year, find the present worth of the revenue stream.

Solution.

1. P = A₁ × [1 − ((1 + g)/(1 + i))ⁿ] / (i − g).
2. ((1.05)/(1.10))⁷ = 0.7221.
3. P = $30,000 × (1 − 0.7221) / (0.10 − 0.05) = $166,761.00.

Answer. Present worth = $166,761.00.

Problem 14: Deferred Annuity

Problem. A lease pays $6,000 at the end of each year from year 4 through year 10 (7 payments), with nothing in years 1 to 3. At 8% per year, find the present worth at time 0.

Solution.

1. Value the 7-payment series one year before its first payment (at t = 3): $6,000(P/A, 8%, 7) = $6,000 × 5.2064 = $31,238.22.
2. Discount back 3 years: × (P/F, 8%, 3) = × 0.7938.
3. PW = $31,238.22 × 0.7938 = $24,797.91.

Answer. Present worth = $24,797.91.

Present, Future & Annual Worth Practice Problems

These problems apply present worth (PW), future worth (FW), annual worth (AW) and capitalized cost to compare alternatives at a stated MARR.

Problem 15: Present Worth Comparison of Two Machines

Problem. Two machines have equal 5-year lives. Machine A: first cost $50,000, operating cost $9,000/yr, salvage $10,000. Machine B: first cost $72,000, operating cost $6,000/yr, salvage $16,000. At MARR = 10%, which is the better choice?

Solution.

1. PW_A = −50,000 − 9,000(P/A, 10%, 5) + 10,000(P/F, 10%, 5) = −50,000 − 34,117 + 6,209 = −$77,908.
2. PW_B = −72,000 − 6,000(P/A, 10%, 5) + 16,000(P/F, 10%, 5) = −72,000 − 22,745 + 9,935 = −$84,810.
3. The larger (less negative) PW is preferred.

Answer. Select Machine A: PW = −$77,908 versus −$84,810.

Problem 16: Future Worth of a Project

Problem. A project costs $30,000 now and returns $9,500 at the end of each year for 6 years. At MARR = 12%, find the future worth at year 6 and decide whether to accept.

Solution.

1. FW = −30,000(F/P, 12%, 6) + 9,500(F/A, 12%, 6).
2. (F/P, 12%, 6) = 1.9738; (F/A, 12%, 6) = 8.1152.
3. FW = −59,215 + 77,094 = $17,880.

Answer. FW = $17,880; accept (FW > 0).

Problem 17: Annual Worth with Salvage Value

Problem. A delivery van costs $45,000, has a 6-year life, a salvage value of $8,000 and annual operating costs of $3,500. At 10% per year, find the annual worth (equivalent annual cost).

Solution.

1. Capital recovery: 45,000(A/P, 10%, 6) = 45,000 × 0.22961 = $10,332.
2. Salvage credit: 8,000(A/F, 10%, 6) = 8,000 × 0.12961 = $1,037.
3. AW = −10,332 − 3,500 + 1,037 = −$12,795.

Answer. Annual worth = −$12,795 per year (an equivalent annual cost of $12,795).

Problem 18: Alternatives with Different Lives (AW Method)

Problem. Pump X: cost $18,000, 4-year life, $2,500/yr operating cost, $3,000 salvage. Pump Y: cost $26,000, 7-year life, $1,800/yr operating cost, $5,000 salvage. At 8%, which is cheaper per year?

Solution.

1. Compare on annual worth so each pump is evaluated over its own life (the repeatability assumption).
2. AW_X = −18,000(A/P, 8%, 4) − 2,500 + 3,000(A/F, 8%, 4) = −5,435 − 2,500 + 666 = −$7,269.
3. AW_Y = −26,000(A/P, 8%, 7) − 1,800 + 5,000(A/F, 8%, 7) = −4,994 − 1,800 + 560 = −$6,234.

Answer. Choose Pump Y: AW = −$6,234 versus −$7,269.

Problem 19: Capitalized Cost of a Long-Life Structure

Problem. A bridge deck costs $200,000 to build, needs $5,000 per year of maintenance forever, and requires a $30,000 resurfacing every 10 years. At 6% per year, find the capitalized cost.

Solution.

1. Perpetual maintenance: 5,000 / 0.06 = $83,333.
2. Resurfacing as an annual cost: 30,000(A/F, 6%, 10) = 30,000 × 0.07587 = $2,276.04; capitalized: ÷ 0.06 = $37,934.
3. CC = 200,000 + 83,333 + 37,934 = $321,267.

Answer. Capitalized cost = $321,267.

Cash Flow Diagram Practice Problems

These problems build the habit of drawing and reading cash flow diagrams: inflows as up arrows, outflows as down arrows, and one period between tick marks.

Problem 20: Build a Cash Flow Table from a Description

Problem. A firm buys a test stand for $12,000 at time 0. It saves $3,000 per year in years 1 to 3, and in year 4 it saves $3,000 and is sold for $2,500. Construct the cash flow table and the net cash flow for each year.

Year	Outflow	Inflow	Net cash flow
0	$12,000	$0	−$12,000
1	$0	$3,000	$3,000
2	$0	$3,000	$3,000
3	$0	$3,000	$3,000
4	$0	$5,500	$5,500

Solution.

1. Draw a time line from 0 to 4. Costs point down (negative); receipts point up (positive).
2. Year 4 has two cash flows at the same point: +$3,000 savings and +$2,500 sale, for a net of +$5,500.
3. Net cash flows (year 0 to 4): −$12,000, $3,000, $3,000, $3,000, $5,500.
4. Sum of net cash flows (undiscounted) = $2,500.

Answer. Net cash flows are −$12,000, $3,000, $3,000, $3,000, $5,500.

Problem 21: Evaluate a Cash Flow Diagram

Problem. Using the net cash flows from the test-stand problem above, find the present worth at 8% per year.

Solution.

1. PW = Σ CF_t(P/F, 8%, t).
2. Year 0: −$12,000 × 1.0000 = −$12,000.00.
3. Year 1: $3,000 × 0.9259 = $2,777.78.
4. Year 2: $3,000 × 0.8573 = $2,572.02.
5. Year 3: $3,000 × 0.7938 = $2,381.50.
6. Year 4: $5,500 × 0.7350 = $4,042.66.
7. PW = −$226.04.

Answer. PW = −$226.04; the stand is not justified at 8%.

Problem 22: Same Loan, Two Perspectives

Problem. A bank lends $10,000 at 7% per year, repaid in 3 equal year-end payments. Draw the cash flow from the borrower’s and the lender’s point of view and find the payment.

Year	Interest	Principal	Ending balance
1	$700.00	$3,110.52	$6,889.48
2	$482.26	$3,328.25	$3,561.23
3	$249.29	$3,561.23	$0.00

Solution.

1. A = $10,000(A/P, 7%, 3) = $10,000 × 0.38105 = $3,810.52.
2. Borrower: +$10,000 at t = 0, then −$3,810.52 at t = 1, 2, 3.
3. Lender: −$10,000 at t = 0, then +$3,810.52 at t = 1, 2, 3.
4. The diagrams are mirror images; the sign depends on whose perspective you take, and the rate of return is the same (7%) for both.

Answer. Payment = $3,810.52. The loan balance reaches $0 after year 3.

Discounted Payback Period Practice Problems

Discounted payback is the time needed for the present worth of net cash flows to recover the initial investment. It accounts for the time value of money, unlike simple payback.

Problem 23: Discounted Payback with Uniform Savings

Problem. A system costs $60,000 and saves $18,000 at the end of each year. At 10%, find the discounted payback period and compare it with the simple payback.

Solution.

1. Simple payback = 60,000 / 18,000 = 3.33 years.
2. Set 60,000 = 18,000(P/A, 10%, n). Then (1.10)ⁿ = 1 / (1 − P·i/A) = 1 / (1 − 0.3333) = 1.5000.
3. n = ln(1.5000) / ln(1.10) = 4.25 years.
4. Check: PW at n = 4: 57,058 (short of 60,000); PW at n = 5: 68,234 (recovered).

Answer. Discounted payback ≈ 4.25 years, so it is recovered during year 5; the simple payback of 3.33 years is too optimistic.

Problem 24: Discounted Payback with Uneven Cash Flows

Problem. A project costs $50,000 and returns $14,000, $16,000, $18,000, $15,000, $12,000 in years 1 to 5. At MARR = 12%, find the discounted payback period using linear interpolation within the payback year.

Year	Net cash flow	P/F at 12%	Discounted flow	Cumulative
0	−$50,000	1.0000	−$50,000.00	−$50,000.00
1	$14,000	0.8929	$12,500.00	−$37,500.00
2	$16,000	0.7972	$12,755.10	−$24,744.90
3	$18,000	0.7118	$12,812.04	−$11,932.85
4	$15,000	0.6355	$9,532.77	−$2,400.08
5	$12,000	0.5674	$6,809.12	$4,409.04

Solution.

1. Discount each net cash flow and accumulate until the total turns positive (see table).
2. The cumulative total crosses zero during year 5.
3. Fraction of the year = unrecovered balance ÷ that year’s discounted flow, so the discounted payback ≈ 4.35 years.

Answer. Discounted payback ≈ 4.35 years (project is recovered within its 5-year life).

Problem 25: Why Payback Alone Can Mislead

Problem. Project A cash flows (years 0 to 5): −$40,000, $20,000, $20,000, $10,000, $0, $0. Project B: −$40,000, $10,000, $10,000, $10,000, $10,000, $40,000. At 10%, compute each discounted payback and present worth, then explain which project to choose.

Solution.

1. Project A discounted payback ≈ 2.70 years; Project B ≈ 4.33 years (B is recovered in its final years).
2. PW_A = $2,224; PW_B = $16,536.
3. Discounted payback ignores every cash flow after the payback point, so it favors quick recovery over total value.

Answer. A pays back faster (2.70 vs. 4.33 years), but B has the higher PW ($16,536 vs. $2,224). Use payback as a liquidity screen and PW or AW as the decision measure.

Engineering economy FAQ

What is engineering economy?

Engineering economy is the method engineers use to compare the costs and benefits of projects over time, using interest and equivalence so that cash flows in different years can be compared fairly.

What is the difference between present worth and annual worth?

Present worth converts every cash flow to a single value today. Annual worth converts the same cash flows to an equal yearly amount. Both rank alternatives the same way when they are evaluated correctly at the same MARR.

What is discounted payback period?

The discounted payback period is the time needed for the present worth of a project’s net cash flows to repay the initial investment. It accounts for interest, so it is always longer than simple payback.

What is MARR?

The minimum attractive rate of return (MARR) is the lowest rate of return a company will accept for a project. A project is acceptable when its rate of return meets or exceeds MARR, or equivalently when its PW or AW is positive at MARR.

How do I choose between alternatives with different lives?

Compare annual worth over each alternative’s own life, or compare present worth over a common time period such as the least common multiple of the lives, assuming each alternative repeats.

Keep learning

Review the theory in Engineering Economics 101, then return here to test yourself. More practice problem sets for other engineering topics are on the way.