Third time i have asked this question Hopefully i can get he

Third time i have asked this question... Hopefully i can get help on this try.

Your company needs a machine for the next seven? years, and you have two choices? (assume an annual interest rate of 7?%).Machine A costs ?$120000 and has an annual operating cost of ?$45000. Machine A has a useful life of seven years and a salvage value of ?$17000. Machine B costs ?$180000 and has an annual operating cost of ?$29000. Machine B has a useful life of five years and no salvage value.? However, the life of Machine B can be extended by two years with a certain amount of investment. If Machine? B\'s life is? extended, it will still cost ?$29000 annually to operate and still have no salvage value.

What would you pay at the end of year 5 to extend the life of Machine B by two? years?

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Solution

R = 7%

Useful life of machine A = 7 years

Useful life of machine B = 5 year (can be extended for 2 more years)

Let, Amount paid at the end of year 5 for machine B = X

If life is extended for 2 more years of machine B, then present worth of machine A should be equal to the present worth of the machine B.

120000 + 45000*(1-1/1.07^7)/.07 - 17000/1.07^7 = 180000 + 29000*(1-1/1.07^7)/.07 + X/1.07^5

351931.3 = 336289.4 + X*.713

X = (351931.3-336289.4)/.713

X = $21938.15

So, the amount of $21938.15 will be paid at the end of year 5 in machine B.