A machine costs $40,000 to purchase and $10,000 per year to operate. The machine has no...
A machine costs $40,000 to purchase and $10,000 per year to operate. The machine has no salvage life and a 10 year life. If i=10% per year compounded annually what is the present worth of the machine?
Find the present worth of the machine of Problem 7.8.
Ans. PW = -$I01 443.93 (the negative value indicates a cash outflow or cost)
Please show all work and formulas used thanks
Homework Answers
Initial cost = 40000
Maintenance cost = 10000
n = 10
i = 10% = 0.1
The present worth
= -40000-10000*((1+i)^n-1)/(i(1+i)^n)
= -40000-10000*(1+0.1)^10-1)/0.1(1+0.1)^10)
= -40000-10000*(1.59374/)0.1(2.59374))
= -40000-10000*(1.59374/0.259374)
= -40000-10000*(6.14456)
= -40000-61445.6
= -101445.6
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