In: Economics
What is the payback period for a project with the following characteristics, given a minimum attractive rate of return (MARR) of 12%?
First Cost $20,000
Annual Benefits $8,000
Annual Maintenance $2,000 in year 1, then increases by $500 per year
Salvage value $2,000
Useful Life 10 years
First Cost = $20,000
Annual Benefits = $8,000
Annual Maintenance = $2,000 in year 1 and then it increases by $500 per year
Salvage value = $2,000
Life = 10 years
Calculate the Pay-back Period.
Step 1 – Calculate the Annual Net Cash Flow
Annual Net Cash Flow = Annual Benefits – Annual Maintenance Costs
Year |
Annual Benefits |
Annual Maintenance |
Net Benefits |
1 |
8000 |
2000 |
6000 |
2 |
8000 |
2500 |
5500 |
3 |
8000 |
3000 |
5000 |
4 |
8000 |
3500 |
4500 |
5 |
8000 |
4000 |
4000 |
6 |
8000 |
4500 |
3500 |
7 |
8000 |
5000 |
3000 |
8 |
8000 |
5500 |
2500 |
9 |
8000 |
6000 |
2000 |
10 |
8000 |
6500 |
1500 |
Step 2 – Calculate the discounted Pay-back Period at MARR of 12%
Year |
CF |
PV Factor |
DCF |
CCF |
0 |
$-20,000 |
1 |
$-20,000 |
$-20,000 |
1 |
$6,000 |
0.89 |
$5,357.14 |
$-14,642.86 |
2 |
$5,500 |
0.8 |
$4,384.57 |
$-10,258.29 |
3 |
$5,000 |
0.71 |
$3,558.9 |
$-6,699.39 |
4 |
$4,500 |
0.64 |
$2,859.83 |
$-3,839.56 |
5 |
$4,000 |
0.57 |
$2,269.71 |
$-1,569.85 |
6 |
$3,500 |
0.51 |
$1,773.21 |
$203.36 |
In between 5th and 6th year the initial cost is recovered. Using interpolation
Pay-Back Period = 5 + [$-1,569.85 – 0 ÷ $-1,569.85 – ($203.36)] * 1
Pay-Back Period = 5.89 years
The discounted pay-back period will be 5.89 years.
Note – The simple or conventional payback period will be
Year |
CF |
NCF |
0 |
$-20,000 |
$-20,000 |
1 |
$6,000 |
$-14,000 |
2 |
$5,500 |
$-8,500 |
3 |
$5,000 |
$-3,500 |
4 |
$4,500 |
$1,000 |
Pay-Back Period = 3 + (3,500 ÷ 4,500) = 3.78 years.