Engineering economy calculator

All eight discrete-compounding factors for any i and n, plus present worth of a cash flow.

Free, right in your browser. Every result shows its work.

Interest rate per period, i

Number of periods, n

Cash flow (leave any at 0)

Single sum F at year n

Uniform series A, years 1–n

$/yr

Gradient G (starts year 2)

$/yr

First cost at year 0 (optional)

Result

Present worth of the cash flow: $12,078.15
Net present worth (after first cost): $78.15
Equivalent uniform annual amount: $1,800.00
Future worth at year n: $26,075.81

1.A × (P/A, 8%, 10) = 1,800 × 6.7101
2.P = $12,078.15
3.Net = P − first cost = $12,078.15 − $12,000.00 = $78.15

All eight factors at i = 8%, n = 10
Factor	Name	Formula	Value
(F/P)	Single payment compound amount	(1 + i)ⁿ	2.1589
(P/F)	Single payment present worth	1 / (1 + i)ⁿ	0.46319
(A/P)	Capital recovery	i(1 + i)ⁿ / [(1 + i)ⁿ − 1]	0.14903
(P/A)	Uniform series present worth	[(1 + i)ⁿ − 1] / [i(1 + i)ⁿ]	6.7101
(A/F)	Sinking fund	i / [(1 + i)ⁿ − 1]	0.069029
(F/A)	Uniform series compound amount	[(1 + i)ⁿ − 1] / i	14.487
(P/G)	Uniform gradient present worth	[(1 + i)ⁿ − 1 − ni] / [i²(1 + i)ⁿ]	25.977
(A/G)	Uniform gradient uniform series	1/i − n / [(1 + i)ⁿ − 1]	3.8713

How it's tested on the exam

Shows up on: PE Mechanical · PE Civil · FE

Reading the factor name backwards. (P/A, i, n) converts an annual amount A INTO a present worth P — you multiply A by it. Say it as "P given A".
A rate that isn't in the printed table. The Handbook tables carry selected rates only; at an in-between rate, compute the factor from its formula (or here) rather than interpolating between table columns.
Gradient timing. The P/G factor assumes the first gradient amount arrives at the end of year 2, not year 1. A series that grows from year 1 is a uniform series A plus a gradient G.
Mixing periods and rates. A monthly payment needs a monthly rate and n in months; don't pair an annual rate with monthly periods.

Worked example

An equipment upgrade costs $12,000 and saves $1,800 a year for 10 years. At 8% interest, is it worth doing?

1.(P/A, 8%, 10) = [(1.08)¹⁰ − 1] / [0.08 × (1.08)¹⁰] = 6.7101
2.Present worth of savings = $1,800 × 6.7101 = $12,078
3.Net present worth = $12,078 − $12,000 = +$78

Marginally yes: net present worth is about +$78.

Where it is in the Handbook

Economic Analysis — Economic Factor Tables — PE Mechanical Reference Handbook
Engineering Economics — Interest Rate Tables — PE Civil Reference Handbook

Section names as printed in the current NCEES PE Reference Handbooks (Mechanical 2.1, Civil 2.2); numbering can shift between versions. On exam day you'll use your own approved calculator — here's which models NCEES allows.