OpenSpecEngineering

Solves the five time-value-of-money quantities against each other. Give any four of present worth P, future worth F, a uniform amount A per period, the interest rate i and the number of periods n, and the page finds the fifth. Engineers use it to size a loan payment, find the rate a deal really earns, or see how long savings take to reach a target.

The governing equation

P·(1 + i)ⁿ + A·[(1 + i)ⁿ − 1]/i + F = 0

P is a single amount at time zero, F a single amount at the end of period n, and A an equal amount at the end of every period from 1 to n. i is the interest rate per period, as a decimal in the equation and in percent on the page. The equation says the cash flows, all moved to the end of period n, add to zero [1], [2]. At a zero rate, the A term becomes A·n.

P, F and A come out in closed form. The rate and the number of periods have no closed form, so the page finds them by bisection: i between −100 % and 1,000 %, and n up to 2,000 periods. It reports an error when no value in that range balances the flows.

Signs, periods and common mistakes

• Use a sign convention: money you receive is positive and money you pay is negative. A loan you take is a positive P, and its payments are negative. If every amount has the same sign, no rate balances them.
• The rate and the period must match. For monthly payments, enter the monthly rate (a 6 % nominal annual rate is 0.5 % a month) and n in months.
• The rate solved for is the rate per period, so a monthly rate is not an annual rate. The effective annual rate is (1 + i)^m − 1 for m periods a year. The page’s second rate row repeats the per-period rate and does not do this conversion.
• A is paid at the end of each period. For payments at the start of each period (an annuity due), multiply A’s present worth by (1 + i).
• Leave unused amounts at 0. Results show money to six significant figures, so a payment of 4,833.20 appears as -4,833.2.

Worked example

A plant finances a 250,000 machine with a five-year loan at 6 % a year, compounded monthly, paid off in equal monthly payments. Find the payment.

1. Choose Solve For Uniform amount (A).
2. Enter Present Worth (P) 250000, the loan received, and Future Worth (F) 0.
3. Enter Interest Rate per Period (%) 0.5 and Number of Periods (n) 60.
4. Press Solve. The Solution shows: Uniform amount A: -4,833.2

The payment is 4,833.20 a month, negative because the plant pays it. Check: (1.005)⁶⁰ = 1.34885, so A = 250,000 × 0.005 × 1.34885 / 0.34885 = 4,833.2. Over the loan the plant pays 60 × 4,833.20 = 289,992, so the interest is about 40,000.