X Company is considering replacing one of its machines in order to save operatin
ID: 2511107 • Letter: X
Question
X Company is considering replacing one of its machines in order to save operating costs. Operating costs with the current machine are $68,000 per year; operating costs with the new machine are expected to be $38,450 per year. The new machine will cost $156,000 and will last for five years, at which time it can be sold for $3,000. The current machine will also last for five more years but will not be worth anything at that time. It cost $45,000 four years ago, but its current disposal value is only $6,000. Assuming a discount rate of 8%, what is the incremental net present value of replacing the current machine?
b. Assume the following two changes: 1) both machines will last for six more years, 2) the salvage value of the new machine after six years will be zero. If X Company replaces the current equipment, what is the approximate internal rate of return?
Explanation / Answer
New Machine Cost 156000 Curent cost of old machine 6000 Net outflow 150000 (156000-6000) Operating Cost with curent Machine 68000 Operating Cost with new machine 38450 Annual Net Saving 29550 (68000-38450) Expected Life 5 years Salavage Value new machine 3000 Cost of capital 8% A a b c=a*b Year Amount PV Factor@8% Present Value 0 -150000 1 (150,000) 1 to 5 29550 3.9927 =PV(8%,5,-1,0,0) Present value annuity @8% for 5years 117,985 5 3000 0.6806 =(1/1.08)^5 Present value @8% for 5th year 2,042 NPV (29,974) Since incremental NPV is negative machine should not be replaced. B a b c=a*b Year Amount PV Factor@8% Present Value 0 -150000 1 (150,000) 1 to 6 29550 4.6229 =PV(8%,6,-1,0,0) Present value annuity @8% for 6years 136,606 NPV (13,394) Since incremental NPV is negative machine should not be replaced. IRR Calculation Year Cashflow 0 -150000 1 29550 2 29550 3 29550 4 29550 5 29550 6 29550 IRR 5% =IRR(C325:C331)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.