Question

The Oklahoma Department of Transportation (ODOT) is looking at the purchase of a new technology for asphalt paving of major highways. The new machine will cost $1,000,000 and is expected to provide a net revenue of $150,000 per year for 9 years. There is no salvage value for this machine. If MARR=10% per year, what is the exact IRR for this project?

A. 15.15%
B. 13.68%
C. 6.47%
D. 2.14%

Answer 1

Option C (6.47%) is the right answer.

Step-by-step explanation:

The given values are:

New machine's cost,

= $1,000,000

Net revenue,

= $150,000

Time,

= 9 years

MAAR,

= 10% per year

Now,

On taking, i = 5%

⇒ PW(5%) =
`-1,000,000 + 150,000((P)/(A) , 5 \ percent, 9)`

=
`-1,000,000 + 1,066,200`

=
`66,200`

On taking, i = 10%

⇒ PW(10%) =
`-1,000,000 + 150,000((P)/(A) , 10 \ percent, 9)`

=
`-1,000,000 + 863,85 0`

=
`-136,150`

By interpolation, we get

⇒
`i=5 \ percent + [(66,220-0)/((66,200-(-136,150)) ]* (10 \ percent - 5 \ percent)`

⇒
`=0.05 + ((66,220)/(202,350) )* 0.05`

⇒
`=0.0663`

i.e.,

⇒
`= 6.63 \ percen t`