Engineering economics and capital budgeting courses use a compact notation for time-value-of-money conversions: six factors that turn present amounts, future amounts and uniform series into one another, plus two more for cash flows that grow by a constant amount. This generator produces that table — one interest rate, every factor side by side, for as many periods as you need — in the same layout found in the appendices of engineering-economy textbooks and the FE exam reference handbook.
How to generate the table
- Enter the interest rate per period.
- Set the first and last period for the rows.
- Choose the decimal places.
- Tick gradient factors to add the P/G and A/G columns.
The result above the table lists every factor for the last period, labeled by name.
The six factors
| Factor | Name | Formula |
|---|---|---|
| F/P | Single payment compound amount | (1 + i)n |
| P/F | Single payment present worth | 1 ÷ (1 + i)n |
| F/A | Uniform series compound amount | [(1 + i)n − 1] ÷ i |
| A/F | Sinking fund | i ÷ [(1 + i)n − 1] |
| P/A | Uniform series present worth | [(1 + i)n − 1] ÷ [i(1 + i)n] |
| A/P | Capital recovery | i(1 + i)n ÷ [(1 + i)n − 1] |
The gradient factors are P/G = [(1 + i)n − in − 1] ÷ [i2(1 + i)n] and A/G = 1/i − n ÷ [(1 + i)n − 1].
Worked example: evaluating a machine
A machine costs $30,000, saves $6,000 a year for 10 years and has a $5,000 salvage value. The minimum attractive rate of return is 8%. From the row for n = 10 at 8%: P/A = 6.7101, P/F = 0.4632, A/P = 0.1490.
- Present worth of savings: $6,000 × 6.7101 = $40,260.49
- Present worth of salvage: $5,000 × 0.4632 = $2,315.97
- Net present worth: −$30,000 + $40,260.49 + $2,315.97 = $12,576.46
- Equivalent annual worth: $12,576.46 × 0.1490 = $1,874.26 a year
Both measures are positive, so the machine earns more than 8%.
Worked example: a rising cost
Maintenance costs $1,000 in year 1 and rises by $200 a year for 10 years. Treat it as a $1,000 uniform series plus a $200 gradient:
- Present worth: $1,000 × 6.7101 + $200 × 25.9768 = $11,905.45
- Equivalent uniform annual cost: $1,000 + $200 × 3.8713 = $1,774.26
Using the relationships to check your work
- Reciprocals: F/P × P/F = 1, F/A × A/F = 1 and P/A × A/P = 1 in every row.
- Capital recovery = sinking fund + rate: A/P − A/F = i.
- Chaining: F/A = P/A × F/P, so converting a series to the present and then to the future gives the same result as converting it straight to the future.
For grids across many rates, use the future value table, present value table or annuity payment table. For a full discounted cash flow analysis of uneven amounts, the NPV calculator does the arithmetic directly.
Factors are computed exactly and rounded to the decimals you choose. For study and analysis; not financial advice.
Frequently asked questions
What do the factor symbols mean?
They read as 'find / given'. F/P finds a future value given a present amount, P/F the reverse; F/A and P/A convert a uniform series A into a future or present value; A/F and A/P convert a single future or present amount into a uniform series. P/G and A/G handle a series that grows by a constant amount each period.
How are the six factors related?
They come in reciprocal pairs: P/F = 1 ÷ (F/P), A/F = 1 ÷ (F/A) and A/P = 1 ÷ (P/A). Also, A/P = A/F + i, so at 8% for 10 periods 0.1490 = 0.0690 + 0.08.
What is an arithmetic gradient?
A cash flow that increases by the same dollar amount G each period, starting from zero at the end of period 1 — for example, maintenance costs that rise $200 a year. P/G gives its present value per $1 of G, and A/G its equivalent uniform series.
Why does the table use one rate instead of a grid of rates?
Engineering-economics references print one page per interest rate with all factors side by side, because a single problem typically uses several factors at the same rate. Generate another table for a different rate.
Are these factors for end-of-period cash flows?
Yes. The standard factors assume uniform series payments at the end of each period and discrete compounding once per period. Adjust beginning-of-period series by multiplying by (1 + i).