Time Value of Money

FE Exam Weight: Engineering Economics accounts for 6-9 questions (~7% of the exam). These questions are highly formulaic — master the factor tables and they become easy points.

Core Concept

A dollar today is worth more than a dollar in the future because it can earn interest. All engineering economics problems revolve around moving cash flows through time using an interest rate.

Key Variables

The Six Standard Factors

Single Payment Factors

Future Worth Factor (F/P, i%, n): $F = P(1 + i)^n$

Present Worth Factor (P/F, i%, n): $P = F \cdot \frac{1}{(1 + i)^n}$

Uniform Series Factors

Sinking Fund Factor (A/F, i%, n): $A = F \cdot \frac{i}{(1 + i)^n - 1}$

Future Worth of Annuity (F/A, i%, n): $F = A \cdot \frac{(1 + i)^n - 1}{i}$

Capital Recovery Factor (A/P, i%, n): $A = P \cdot \frac{i(1 + i)^n}{(1 + i)^n - 1}$

Present Worth of Annuity (P/A, i%, n): $P = A \cdot \frac{(1 + i)^n - 1}{i(1 + i)^n}$

Summary Table

Exam Tip: The FE Reference Handbook contains factor tables for common interest rates. Learn to read them quickly — the factor name tells you the column to look in. For example, (P/A, 6%, 10) means "look in the P/A column at i = 6%, n = 10."

Gradient Series

Arithmetic Gradient

Cash flows increase by a constant amount G each period:

• Period 1: 0, Period 2: G, Period 3: 2G, ..., Period n: (n-1)G

$P_G = G \cdot \frac{(1+i)^n - in - 1}{i^2(1+i)^n}$

Geometric Gradient

Cash flows increase by a constant percentage g each period: $P = A_1 \cdot \frac{1 - (1+g)^n(1+i)^{-n}}{i - g} \quad (i \neq g)$

Example Problems

Example 1: You invest $10,000 at 8% annual interest. What is it worth in 5 years? $F = 10{,}000 \times (1.08)^5 = 10{,}000 \times 1.4693 = \$14{,}693$

Example 2: You need $50,000 in 10 years. How much must you deposit annually at 6% interest? $A = 50{,}000 \times (A/F, 6\%, 10) = 50{,}000 \times 0.07587 = \$3{,}794/\text{year}$

Example 3: A machine costs $20,000 and saves $4,000 per year for 8 years. At 10% interest, what is the present worth? $PW = -20{,}000 + 4{,}000 \times (P/A, 10\%, 8) = -20{,}000 + 4{,}000 \times 5.3349 = \$1{,}340$

Since PW > 0, the investment is economically justified.