Pricing
46/3335–8FE Reference Handbook 10.4 · Engineering Economics · A/P
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

Capital-Recovery Factor (A/P)

Uniform annual amount A that recovers a present investment P over n periods at i, including return on capital. EUAC of a first cost.

Equivalent uniform annual cost
Present investment
MARR or interest rate
Life in periods

Step-by-step solved example

Machine costs $10,000. Recover capital in 5 years at 8%. Find A (no salvage).

P12345An
Recover a present P as a uniform A over n periods.

  1. 1. Factor

    (A/P, 8%, 5) = 0.08(1.08)^5 /[(1.08)^5 − 1] = 0.2505.

  2. 2. Annual

    A = 10000 × 0.2505 = $2,505.

Answer: A = $2,505 / yr

10 practice questions

0/10 correct

1.(A/P, 10%, 2) is nearest

2.A to recover $1,000 in 2 years at 10% is nearest

3.A/P equals A/F plus

4.EUAC of a first cost uses

5.If salvage S occurs at n, EUAC is

6.As n → ∞, A/P approaches

7.(A/P, i, 1) equals

8.Capital recovery includes

9.P = $8,000, i = 0%, n = 4 → A =

10.A/P and P/A are