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52/3335–8FE Reference Handbook 10.4 · Engineering Economics · Depreciation
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

Straight-Line Depreciation

Equal annual depreciation D over depreciable life n. C is first cost, S is salvage used for books (often tax salvage). Book value after t years is C − tD. MACRS is tabulated separately in the handbook.

Annual depreciation
Initial cost (basis)
Salvage used in the method
Depreciable life
Book value at end of year t

Step-by-step solved example

C = $50,000, S = $5,000, n = 9 years. Find D and BV after 3 years.

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Book value declining linearly from P to S over n years.

  1. 1. Annual

    D = (50000 − 5000)/9 = $5,000 / yr.

  2. 2. Book value

    BV_3 = 50000 − 3(5000) = $35,000.

Answer: D = $5,000/yr, BV₃ = $35,000

10 practice questions

0/10 correct

1.SL depreciation of C = $9,000, S = $1,000, n = 4 is

2.Book value at t = n equals

3.If S = 0, BV after t years is

4.Depreciation is a

5.MACRS compared with SL typically gives

6.C = $20,000, D = $4,000/yr, S = 0 → n =

7.BV after 2 years: C = 12,000, S = 2,000, n = 5 is

8.Market value is not required to equal

9.Total SL depreciation over n years equals

10.On the FE, units of D must match