Your company is deciding whether to invest in a new machine. The new machine wil
ID: 2782470 • Letter: Y
Question
Your company is deciding whether to invest in a new machine. The new machine will increase cash flow by $328366 per year. You believe the technology used in the machine has a 10-year life; in other words, no matter when you purchase the machine, it will be obsolete 10 years from today. The machine is currently priced at $1,760,000. The cost of the machine will decline by $110,000 per year until it reaches $1,320,000, where it will remain. The required return is 14%.
What is the NPV if the company decides to wait 2 years to purchases the machine? (Round answer to 2 decimal places. Do not round intermediate calculations)
Explanation / Answer
Year
Cash flow(C)
PV Factor F =1/(1+i)^n
PV = C x F
0
*$ (1,540,000)
1/(1+0.14)^0
1
$ (1,540,000.00)
1
$ 328,366
1/(1+0.14)^1
0.877192982
$ 288,040.35
2
$ 328,366
1/(1+0.14)^2
0.769467528
$ 252,666.97
3
$ 328,366
1/(1+0.14)^3
0.674971516
$ 221,637.70
4
$ 328,366
1/(1+0.14)^4
0.592080277
$ 194,419.03
5
$ 328,366
1/(1+0.14)^5
0.519368664
$ 170,543.01
6
$ 328,366
1/(1+0.14)^6
0.455586548
$ 149,599.13
7
$ 328,366
1/(1+0.14)^7
0.399637323
$ 131,227.31
8
$ 328,366
1/(1+0.14)^8
0.350559055
$ 115,111.67
NPV
$ (16,754.82)
*Cost of machine after two years = $ 1,760,000 – (2 x 110,000)
= $ 1,760,000 – 220,000 = $ 1,540,000
Year
Cash flow(C)
PV Factor F =1/(1+i)^n
PV = C x F
0
*$ (1,540,000)
1/(1+0.14)^0
1
$ (1,540,000.00)
1
$ 328,366
1/(1+0.14)^1
0.877192982
$ 288,040.35
2
$ 328,366
1/(1+0.14)^2
0.769467528
$ 252,666.97
3
$ 328,366
1/(1+0.14)^3
0.674971516
$ 221,637.70
4
$ 328,366
1/(1+0.14)^4
0.592080277
$ 194,419.03
5
$ 328,366
1/(1+0.14)^5
0.519368664
$ 170,543.01
6
$ 328,366
1/(1+0.14)^6
0.455586548
$ 149,599.13
7
$ 328,366
1/(1+0.14)^7
0.399637323
$ 131,227.31
8
$ 328,366
1/(1+0.14)^8
0.350559055
$ 115,111.67
NPV
$ (16,754.82)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.