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

Answer:

Option C (6.47%) is the right answer.

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%) = $-1,000,000 + 150,000(\frac{P}{A} , 5 \ percent, 9)$

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

                 = $66,200$

On taking, i = 10%

⇒ PW(10%) = $-1,000,000 + 150,000(\frac{P}{A} , 10 \ percent, 9)$

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

                  = $-136,150$

By interpolation, we get

⇒ $i=5 \ percent + [\frac{66,220-0}{(66,200-(-136,150)} ]\times (10 \ percent - 5 \ percent)$

⇒   $=0.05 + (\frac{66,220}{202,350} )\times 0.05$

⇒   $=0.0663$

i.e.,

⇒   $= 6.63 \ percen t$