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47/3335–8FE Reference Handbook 10.4 · Engineering Economics · P/A
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

Uniform-Series Present Worth (P/A)

Present worth of n uniform end-of-period amounts A. Used for capitalized benefits or costs.

Present worth of the series
Uniform amount
Rate per period
Number of amounts

Step-by-step solved example

Annual benefit $1,200 for 6 years at 7%. Find P.

P123456An
Present worth of a uniform series A.

  1. 1. Factor

    (P/A, 7%, 6) = [(1.07)^6 − 1]/[0.07(1.07)^6] = 4.7665.

  2. 2. Present

    P = 1200 × 4.7665 = $5,720.

Answer: P = $5,720

10 practice questions

0/10 correct

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

2.P of A = $100 for 2 years at 10% is nearest

3.A perpetuity (n = ∞) has P =

4.P/A is the reciprocal of

5.As i increases, P/A

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

7.If i = 0, P/A equals

8.Annual O&M of $4,000 for 8 years at 6% is converted to P with

9.Capitalized cost of a perpetual A is

10.First payment of an ordinary P/A series occurs at