Calculate the eight standard engineering economy interest factors, convert nominal to effective rates, and apply a factor to a given amount.
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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
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
(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)