Time Value of Money
Key Takeaways
- A dollar today is worth more than a dollar later because it can earn interest; every cash flow must be moved through time at rate i before it can be compared.
- The two single-payment factors are F = P(1+i)^n (F/P) and P = F/(1+i)^n (P/F); the exponent n must match the compounding period of i.
- The four uniform-series factors (P/A, A/P, F/A, A/F) all build on (1+i)^n − 1 and convert between a level annuity A and a lump sum P or F.
- The FE Reference Handbook prints these factors in interest tables; (P/A, 6%, 10) means read the P/A column at i = 6%, n = 10.
- Effective annual rate i_eff = (1 + r/m)^m − 1 converts a nominal rate r compounded m times per year; never plug a nominal rate straight into an annual factor.
- Cash flows occur at end-of-period by convention, and timing errors (an extra or missing period) are the most common engineering-economics mistake on the FE.
FE Exam Weight: Engineering Economics accounts for 6–9 questions (~7% of the FE Other Disciplines exam, 110 questions total in a 6-hour appointment). The topic is intensely formulaic — once you can read the factor tables in the NCEES FE Reference Handbook and track timing, these become some of the fastest points on the test.
Core Concept: Equivalence
Money has a time value. A dollar today can be invested to earn interest, so it is worth more than a dollar received later. Every engineering-economics problem is really one question: move each cash flow to a common point in time at the interest rate i, then compare. Two cash-flow streams are equivalent if they have the same value at the same point in time and the same i.
The standard convention on the FE is end-of-period cash flow: a payment in year n lands at the end of year n, and time 0 is now (the present). Drawing a quick cash-flow diagram — an arrow up for receipts, down for disbursements — prevents the single most common error, an off-by-one in n.
The Five Core Variables
| Symbol | Name | Meaning |
|---|---|---|
| P | Present worth | Single lump sum at time 0 |
| F | Future worth | Single lump sum at time n |
| A | Annual amount | Equal end-of-period payment in a uniform series |
| i | Interest rate | Effective rate per compounding period |
| n | Number of periods | Count of compounding periods |
The golden rule: i and n must use the same period. If interest compounds monthly, i is the monthly rate and n is a number of months — not years.
The Six Standard Compound-Interest Factors
The handbook tabulates all six. Memorize the algebraic forms so you can verify a table value or handle an interest rate the table skips.
Single-Payment Factors
Compound amount (F/P, i%, n) — grow a present sum forward:
Present worth (P/F, i%, n) — discount a future sum back:
Uniform-Series (Annuity) Factors
Present worth of annuity (P/A, i%, n):
Capital recovery (A/P, i%, n) — the loan-payment factor:
Future worth of annuity (F/A, i%, n):
Sinking fund (A/F, i%, n):
How to Read the Factor Name
The notation (X/Y, i%, n) reads "find X, given Y." So (P/A, 6%, 10) gives the multiplier you apply to A to get P at i = 6% over 10 periods. In the handbook table for i = 6%, go to row n = 10 and read the P/A column. Each factor and its reciprocal are paired: (P/F) and (F/P) are reciprocals, as are (P/A) and (A/P), and (F/A) and (A/F).
| Find | Given | Factor | Notation |
|---|---|---|---|
| F | P | Single-payment compound amount | (F/P, i%, n) |
| P | F | Single-payment present worth | (P/F, i%, n) |
| P | A | Uniform-series present worth | (P/A, i%, n) |
| A | P | Capital recovery | (A/P, i%, n) |
| F | A | Uniform-series compound amount | (F/A, i%, n) |
| A | F | Sinking fund | (A/F, i%, n) |
Nominal vs. Effective Interest — A Classic Trap
A nominal annual rate r compounded m times per year is not the rate you use in an annual factor. Convert first:
Example: 12% nominal compounded monthly gives a monthly rate of 1% and an effective annual rate i_eff = (1.01)^12 − 1 = 0.1268, or 12.68%. If a problem says "12% per year compounded monthly" and asks for an annual result, either work in months (i = 1%, n in months) or use i_eff = 12.68% with n in years — never plug 12% into an annual factor.
The Arithmetic Gradient
When a series grows by a constant amount G each period (0 in year 1, G in year 2, 2G in year 3, …, (n−1)G in year n), convert it to a present worth with the gradient factor (P/G, i%, n): A growing series is split into a base annuity A handled by (P/A) plus a gradient piece handled by (P/G). The gradient itself starts in period 2, not period 1 — another timing trap.
Worked Present-Worth Examples
Example 1 — Future worth (F/P). Invest $10,000 at 8% per year for 5 years. F = 10,000 × (F/P, 8%, 5) = 10,000 × (1.08)^5 = 10,000 × 1.4693 = $14,693.
Example 2 — Present worth of an annuity (P/A). A machine saves $4,000/year for 8 years; i = 10%. The present worth of those savings is P = 4,000 × (P/A, 10%, 8) = 4,000 × 5.3349 = $21,340. If the machine costs $20,000 today, the net present worth is +$1,340, so the purchase is justified.
Example 3 — Sinking fund (A/F). You need $50,000 in 10 years at 6%. A = 50,000 × (A/F, 6%, 10) = 50,000 × 0.07587 = $3,794/year.
Example 4 — Capital recovery (A/P). Repay a $100,000 loan over 15 years at 7%. A = 100,000 × (A/P, 7%, 15) = 100,000 × 0.10979 = $10,979/year. Always confirm the factor against (1+i)^n: (1.07)^15 = 2.7590, so A/P = 0.07(2.7590)/(2.7590 − 1) = 0.10979. ✔
A nominal rate of 8% per year compounded quarterly has what effective annual interest rate?
Which factor converts a single present amount P into an equivalent uniform annual series A (the loan-payment factor)?
An investment of $20,000 returns $5,000 per year for 6 years at i = 8%. The net present worth is approximately:
You deposit $2,000 at the end of each year into an account earning 5% annually. How much is in the account immediately after the 4th deposit?