Pricing
43/3335–8FE Reference Handbook 10.4 · Engineering Economics · P/F
1. Simple Interest2. Single-Payment Compound Amount (F/P)3. Single-Payment Present Worth (P/F)4. Uniform-Series Compound Amount (F/A)5. Sinking-Fund Factor (A/F)6. Capital-Recovery Factor (A/P)7. Uniform-Series Present Worth (P/A)8. Effective Interest Rate9. Arithmetic Gradient Present Worth (P/G)10. Net Present Worth11. Benefit–Cost Ratio12. Straight-Line Depreciation13. Geometric Gradient Present Worth14. Capitalized Cost (Perpetual PW)15. Payback Period16. Break-Even Quantity17. Equivalent Uniform Annual Worth18. Internal Rate of Return vs MARR19. MACRS Depreciation20. Declining-Balance Depreciation21. Inflation and Real Interest22. Expected Value of Alternatives

Handbook formula

Single-Payment Present Worth (P/F)

Discounts a future single cash flow F back to the present. (P/F, i, n) = (1+i)^{−n}.

Present worth
Future worth
Interest rate per period
Periods

Step-by-step solved example

How much must be invested now at 8% to have $5,000 in 4 years?

P12345An
A future F is brought back to t = 0 with (P/F).

  1. 1. Factor

    (P/F, 8%, 4) = (1.08)^{−4} = 0.7350.

  2. 2. Present

    P = 5000 × 0.7350 = $3,675.

Answer: P = $3,675

10 practice questions

0/10 correct

1.(P/F, 10%, 2) equals

2.P of $1,210 due in 2 years at 10% is

3.As n increases, P/F

4.P of $8,000 in 1 year at 6% is nearest

5.Discounting is the inverse of

6.(P/F, i, 0) equals

7.Higher i makes a future $1 today

8.F = $10,000, i = 8%, n = 0 → P =

9.The handbook symbol (P/F, i, n) is

10.P = F(1+i)^{−n} requires i as