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57/3335–8FE Reference Handbook 10.4 · Engineering Economics
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

Equivalent Uniform Annual Worth

Convert every cash flow to a uniform annual series over the analysis life. Repeatability assumption: alternatives are compared over a common LCM or assumed to repeat at the same AW.

Annual worth
Present worth
Interest rate per period

Step-by-step solved example

P=−$80,000, A=+$20,000, n=6, i=8%, salvage 0. Is AW positive?

PAWn
Convert the whole cash-flow diagram to a uniform A.

  1. 1. Capital recovery

    (A/P,8%,6)=0.2163 → 80000×0.2163=$17,304.

  2. 2. AW

    AW=−17304+20000=+$2,696 > 0, accept.

Answer: AW ≈ +$2,700 (accept)

10 practice questions

0/10 correct

1.AW and NPW always

2.(A/P,i,n) equals

3.Different lives are handled by

4.A never-ending annual A has PW

5.If AW=0 then IRR

6.Salvage S enters AW as

7.Do-nothing AW is

8.i=10%, n=∞, P=1000: AW =

9.Inflation is handled by

10.Mutually exclusive choice: pick