Engineering Economics

Compound-Interest Factors (Time Value of Money)

Evaluates any of the eight standard discrete compound-interest factors — (F/P), (P/F), (A/P), (P/A), (F/A), (A/F), (P/G), (A/G) — at an effective rate per period, with optional nominal-to-effective rate conversion and an optional amount to apply. Use for: 'find the future worth of $1,000 in 10 years at 5%', looking up or verifying interest-table factor values, capital recovery / sinking fund amounts, arithmetic-gradient conversions. Not for: comparing whole alternatives (worth-comparison) or rate-of-return problems (irr).

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

factorWhich factor: FP = (F/P) single-payment compound amount; PF = (P/F) single-payment present worth; AP = (A/P) capital recovery; PA = (P/A) uniform-series present worth; FA = (F/A) uniform-series compound amount; AF = (A/F) sinking fund; PG = (P/G) arithmetic-gradient present worth; AG = (A/G) gradient-to-uniform-seriesrate percent (optional)Effective interest rate per period, in PERCENT (5 means 5%). Provide either this OR nominalAnnualPercent + compoundingPerYear.nominal annual percent (optional)Nominal annual rate in percent (APR), converted to an effective annual rate using compoundingPerYear before the factor is evaluatedcompounding per year (optional)Compounding periods per year m (12 = monthly, 4 = quarterly); required with nominalAnnualPercentnNumber of periods n (years if the rate is effective annual). For P/G and A/G the gradient starts at period 2.amount (optional)Optional amount to multiply by the factor (currency-agnostic), e.g. a present value of 1000 with factor FP yields the future value

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

(F/P, 5%, 10) = 1.6289 — single payment compound amount

Given

factor: FP
rate percent: 5
n: 10

Assumptions

Discrete end-of-period compounding at a constant effective rate per period; uniform series start at the end of period 1 and arithmetic gradients at the end of period 2.
Factor values are reported to five significant figures, the precision of standard compound-interest tables.

Solution steps

Evaluate the single-payment compound amount factor
Substitute the effective rate per period (as a decimal) and the number of periods into the closed form.
(F/P) = (1 + i)^n
With i = 5 percent (decimal 0.05) and n = 10: (F/P, i, n) = 1.6289
= 1.6289

Results

Factor value (F/P, i, n)

1.6289

Where this answer was checked

source: Compound-interest factor closed form (NCEES FE Reference Handbook style) — (F/P, i, n) = (1 + i)^n
verified by: hand-recomputed
derivation: 1.05^10: 1.05^2 = 1.1025; ^4 = 1.1025^2 = 1.21550625; ^8 = 1.21550625^2 = 1.477455; ^10 = 1.477455·1.1025 = 1.628895. Table value 1.6289.

Example 2

(P/A, 8%, 20) = 9.8181 — uniform-series present worth

Given

factor: PA
rate percent: 8
n: 20

Assumptions

Discrete end-of-period compounding at a constant effective rate per period; uniform series start at the end of period 1 and arithmetic gradients at the end of period 2.
Factor values are reported to five significant figures, the precision of standard compound-interest tables.

Solution steps

Evaluate the uniform-series present worth factor
Substitute the effective rate per period (as a decimal) and the number of periods into the closed form.
(P/A) = ((1 + i)^n − 1) / (i·(1 + i)^n)
With i = 8 percent (decimal 0.08) and n = 20: (P/A, i, n) = 9.8181
= 9.8181

Results

Factor value (P/A, i, n)

9.8181

Where this answer was checked

source: Compound-interest factor closed form — (P/A, i, n) = (1 − (1 + i)^−n)/i
verified by: hand-recomputed
derivation: 1.08^20 = 4.660957; (P/F) = 1/4.660957 = 0.2145482; (P/A) = (1 − 0.2145482)/0.08 = 0.7854518/0.08 = 9.818147. Table value 9.8181.

Example 3

(A/G, 6%, 5) = 1.8836 — gradient to uniform series

Given

factor: AG
rate percent: 6
n: 5

Assumptions

Discrete end-of-period compounding at a constant effective rate per period; uniform series start at the end of period 1 and arithmetic gradients at the end of period 2.
Factor values are reported to five significant figures, the precision of standard compound-interest tables.

Solution steps

Evaluate the arithmetic-gradient to uniform series factor
Substitute the effective rate per period (as a decimal) and the number of periods into the closed form.
(A/G) = 1/i − n / ((1 + i)^n − 1)
With i = 6 percent (decimal 0.06) and n = 5: (A/G, i, n) = 1.8836
= 1.8836

Results

Factor value (A/G, i, n)

1.8836

Where this answer was checked

source: Compound-interest factor closed form — (A/G, i, n) = 1/i − n/((1 + i)^n − 1)
verified by: hand-recomputed
derivation: 1.06^5 = 1.3382256; (A/G) = 1/0.06 − 5/0.3382256 = 16.666667 − 14.783037 = 1.883630. Table value 1.8836.