Time Value of Money for the FE Civil Exam

NCEES FE Civil spec: Engineering Economics, A. Time value of money (e.g., equivalence, present worth, equivalent annual worth, future worth, rate of return). FE Reference Handbook: Engineering Economics chapter, factor table and interest tables.

A dollar today is worth more than a dollar next year because it can earn interest. Time value of money is the toolkit for moving cash flows to a common point in time so alternatives can be compared fairly. It is the largest part of the engineering economics section on the FE Civil exam and, once you know the interest factors, some of the fastest points available.

What the NCEES spec covers

Section A lists equivalence, present worth, equivalent annual worth, future worth and rate of return. In practice this means: converting between single sums (P, F), uniform series (A) and gradients (G); comparing alternatives by PW or annual worth; and finding or interpreting a rate of return.

Key concepts

Equivalence. Two cash flow patterns are equivalent at a given interest rate if they have the same present worth. Equivalence depends on the interest rate.
End-of-period convention. Uniform series payments A occur at the end of periods 1 through n, and the present worth of a series lands one period before the first payment.
Present worth (PW) brings everything to time zero. Annual worth (AW) spreads everything into an equal yearly amount, which is handy when alternatives have different lives.
Rate of return (ROR or IRR) is the interest rate that makes PW equal zero. Accept a project if its ROR is at least the minimum attractive rate of return (MARR).
Effective rate. If interest compounds m times per year at nominal rate r, the effective annual rate is $i_{e} = (1 + r / m)^{m} - 1$ .

Core formulas

(F / P, i, n) = (1 + i)^{n} (P / F, i, n) = (1 + i)^{- n}

(P / A, i, n) = \frac{( 1 + i ) ^{n} - 1}{i ( 1 + i ) ^{n}} (A / P, i, n) = \frac{i ( 1 + i ) ^{n}}{( 1 + i ) ^{n} - 1}

(F / A, i, n) = \frac{( 1 + i ) ^{n} - 1}{i} (A / F, i, n) = \frac{i}{( 1 + i ) ^{n} - 1}

(P / G, i, n) = \frac{( 1 + i ) ^{n} - 1}{i ^{2} ( 1 + i ) ^{n}} - \frac{n}{i ( 1 + i ) ^{n}}

Read the notation as "find P given A", and so on. The factors chain like fractions, for example $(P / A) (A / F) = (P / F)$ .

Where it is in the FE Reference Handbook

The Engineering Economics chapter has a table of factor names, symbols and formulas, followed by interest tables for a range of rates. If your rate and n are in the tables, reading the factor is faster than computing it. Otherwise use the formula; the two agree to the precision shown.

Worked example

A contractor can buy a machine for $50,000 that saves $9,000 per year for 8 years and has a $5,000 salvage value. At 8% interest, find the present worth and equivalent annual worth. Is it worth buying?

Step 1: factors at i = 8%, n = 8.

(P / A, 8%, 8) = \frac{1.0 8 ^{8} - 1}{0.08 ( 1.08 ) ^{8}} = 5.7466 (P / F, 8%, 8) = 1.0 8^{- 8} = 0.5403

Step 2: present worth.

P W = - 50, 000 + 9, 000 (5.7466) + 5, 000 (0.5403) = - 50, 000 + 51, 720 + 2, 701 = $4, 421

Step 3: annual worth. $(A / P, 8%, 8) = 0.17401$ , so

A W = 4, 421 (0.17401) = $769 per year

Both measures are positive, so the machine earns more than 8%. Solving for the rate that makes PW = 0 gives a rate of return of about 10.2%, which confirms the decision.

Common mistakes

Putting a uniform series PW at the time of the first payment instead of one period earlier.
Using a nominal rate with monthly periods without dividing by 12 (and multiplying n by 12).
Forgetting the salvage value, or treating it as a cost.
Comparing alternatives with different lives by PW without a common study period. Use AW instead.
Mixing up (A/P) and (P/A). If you are spreading a present cost into payments, you need A/P.

See the time value of money example for another solved problem.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

What is the approximate future value of $10,000 invested for 3 years at 5% interest compounded annually? See the worked solution or practice the Engineering Economics deck
What is the approximate present value of $5,000 received in 4 years at an annual discount rate of 6%? See the worked solution or practice the Engineering Economics deck
What is the approximate annual capital recovery for an initial cost of $20,000 over 5 years at 8% interest? Assume zero salvage value and equal end-of-year payments. See the worked solution or practice the Engineering Economics deck
A pump upgrade saves $12,000 at the end of each year for 10 years. At an interest rate of 6% per year, the present worth of the savings is most nearly: See the worked solution or practice the Engineering Economics deck
A loader costs $200,000, lasts 20 years and has a $20,000 salvage value. At 7% interest, its equivalent annual cost of ownership is most nearly: See the worked solution or practice the Engineering Economics deck
Maintenance on a bridge costs $2,000 in year 1 and increases by $500 each year through year 10. At 8%, the present worth of the maintenance costs is most nearly: See the worked solution or practice the Engineering Economics deck

Related Calculators and Examples

Frequently Asked Questions

Do I need to memorize interest factor formulas for the FE exam?

No. The FE Reference Handbook lists every factor formula and includes interest tables. You do need to know which factor converts what, and the end-of-period convention, so you can apply them quickly.

When should I use annual worth instead of present worth?

Annual worth is convenient when alternatives have different lives, because the annual worth over one life cycle equals the annual worth over repeated cycles. Present worth requires a common study period.

What is rate of return?

It is the interest rate at which the present worth of all cash flows equals zero. A project is acceptable if its rate of return is at least the minimum attractive rate of return.

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