Present worth and the time value of money

Engineering · Engineering economics · about 10 min

A dollar today is worth more than a dollar in five years, because today's dollar can be invested and earn interest. Engineers use this idea to compare projects with costs and savings spread over many years, by converting everything to its value today: its present worth (PW).

The key idea

At interest rate $i$ per period, a single amount $F$ received $n$ periods from now is worth today

P = \frac{F}{(1+i)^n}

and an equal payment $A$ every period for $n$ periods is worth

P = A\,\frac{(1+i)^n - 1}{i\,(1+i)^n}

A horizontal time line with evenly spaced tick marks. A dark blue up arrow at the far right end for the future amount. A dark red up arrow at the far left end for its value today, shorter than the other arrow. A thin curved arrow above the line runs from the right end back to the left end. — Present worth brings a future amount back to today.

Compounding and discounting

Investing $P$ at rate $i$ grows to $F = P(1+i)^n$ after $n$ periods: that's compounding. Running it backwards, from a future amount to today's value, is discounting. The rate used for this in engineering decisions is often the company's MARR, the minimum attractive rate of return.

Using PW to make decisions

For a single project: if its net present worth at the MARR is $\geq 0$ , it earns at least the MARR, so it's economically justified.
For mutually exclusive options with equal lives: pick the one with the highest PW (or the least negative, when comparing costs only).
Options with different lives must be compared over the same period, for example the least common multiple of their lives.

Worked example: a single payment and an annual series

At $8\%$ per year: (a) what is $10,000 received in $5$ years worth today? (b) what is $2,000 received at the end of every year for $5$ years worth today?

Two horizontal time lines, one above the other, each with tick marks at years 0 to 5. Top line: one tall dark blue up arrow at year 5 only. Bottom line: five equal, shorter dark blue up arrows, one at each of years 1, 2, 3, 4 and 5, and nothing at year 0. — The two payment patterns in this example.

Show the solution

(a) Single amount:

P = \frac{10\,000}{(1.08)^5} = \frac{10\,000}{1.4693} = \$6{,}805.83

(b) Uniform series. The series factor is

\frac{(1.08)^5 - 1}{0.08\,(1.08)^5} = 3.9927

P = 2000 \times 3.9927 = \$7{,}985.42

Even though (b) pays out $10,000 in total, the same as (a), it's worth more today because the money arrives sooner on average.

Common mistakes

Adding future amounts without discounting them

Money at different times can't be added directly. Bring every amount to the same point in time first.

Using the interest rate as a percentage in the formula

Use $i = 0.08$ , not $8$ .

Placing annual payments at the start of each year

The standard series formula assumes payments at the end of each period. Payments at the start need adjusting.

Check your understanding

1. As the interest rate rises, the present worth of a fixed future amount...

2. A project's net present worth at the MARR is $+\$ 1{,}200$. What does that tell you?

3. What is \ $1,000 received in one year worth today at$ 10\%$?

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