Payback Period

n_{pb} = \frac{P}{A} (simple, uniform A); discounted: P = t = 1 \sum n A_{t} (1 + i)^{- t}

Simple payback is the time to recover the initial investment ignoring interest (P/A for uniform benefits). Discounted payback includes i and is longer. Payback ignores cash flows after recovery, so NPW/AW remain the proper selection methods; payback is a liquidity screen.

$P$: Initial investment
$A$: Uniform annual net benefit
$n_{p b}$: Payback period
$i$: Interest rate (discounted payback)

Step-by-step solved example

Machine costs $24,000 and saves $6,000 per year. Simple payback? If i = 10%, is discounted payback less than 5 years?

Simple payback: recover P from uniform A.

1. Simple
n = 24000/6000 = 4 years.
$n = 4 yr$
2. Discounted check
(P/A,10%,5) = 3.7908; 6000×3.7908 = $22,745 < 24,000. Not recovered in 5 years at 10%.

Answer: Simple n = 4 yr; discounted n > 5 yr at 10%

10 practice questions

0/10 correct

1.P = $10,000, A = $2,500/yr. Simple payback =

2.Simple payback ignores

3.Discounted payback compared with simple is

4.A project with huge benefits in year 20 and tiny early A has

5.If A = 0, simple payback is

6.Payback is best used as

7.Uneven cash flows: payback is the first n with

8.P = $8,000, savings $2,000, $3,000, then $4,000. Simple payback is

9.A shorter maximum-payback policy is

10.If discounted payback exists at MARR, then NPW is