Pricing
49/3335–8FE Reference Handbook 10.4 · Engineering Economics · P/G
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

Arithmetic Gradient Present Worth (P/G)

Present worth of a cash flow that starts at 0 at EOY 1 and increases by G each period. Superpose A + gradient. The G series is 0, G, 2G, …, (n−1)G.

Constant period-to-period change
Present worth of the gradient
Rate per period
Periods

Step-by-step solved example

Maintenance is $0 in year 1, then rises by $200/yr for n = 5 at i = 8%. Find P of the gradient.

P12345Gn
Arithmetic gradient: 0, G, 2G, … superimposed on a uniform A if present.

  1. 1. Cash flow

    Years 1–5: 0, 200, 400, 600, 800. That is G = 200.

  2. 2. Factor

    (P/G, 8%, 5) = 7.372.

  3. 3. Present

    P = 200 × 7.372 = $1,474.

Answer: P = $1,474

10 practice questions

0/10 correct

1.The FE gradient series has year-1 cash equal to

2.A cost of 100, 200, 300 over 3 years is

3.(P/G, i, 1) equals

4.Convert gradient to annual with

5.Geometric gradient uses a constant

6.If G is negative, P of the gradient is

7.n in P/G is the number of

8.Do not apply P/G to a series that is

9.Present worth of maintenance 0, 50, 100 at 10% over 3 years uses G =

10.P/G has units of