A recently installed machine earns the company revenue at a continuous rate of 60000t + 45,000 dollars per yr during the first six months of operation and at a continous rate of 75000 dollars per yr after the first six months. The cost of the machine is $150,000, interest rate is 7% compounded continuously and t is the time in years since it was installed.
a.) Find present value of revenue earned by machine during first yr of operation
b.) Find how long it will take the machine to pay for itself, how long will it take for the present value of the revenue to equal the cost of the machine?
Here is one way.
a.) Find present value of revenue earned by machine during first yr of operation.
The first 6 months, or from t=0 to t=0.5:
R(t) = 60,000t +45,000
R(0.5) = 60,000(0.5) +45,000 = 75,000 -----**
The next 6 months, or from t=0.5 to t=1. Or, let t' = (t+0.5)
R(t') = 75,000t'
So, for the first 6 months of t',
R(0.5) = 75,000(0.5) = 37,500
Therefore, for the first year of the machine,
R = 75,000 +37,500 = 112,500 dollars ---------------answer.
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b.) Find how long it will take the machine to pay for itself, how long will it take for the present value of the revenue to equal the cost of the machine?
With interest at 7% per annum, compounded continuously, the cost ot the machine in T years is:
C(T) = (150,000)[e^(rT)]
C(T) = (150,000)[e^(0.07T)] ---------------(1)
The Revenue in T years, where T is some year after t=1,
R(T) = 112,500 +(75,000)(T-1) ---------------(2)
(1) = (2),
(150,000)[e^(0.07T) = 112,500 +(75,000)(T-1)
Divide both sides by 75,000,
2[e^(0.07T)] = 1.5 +T -1
2[e^(0.07T)] = T +0.5
2[e^(0.07T)] -T -0.5 = 0 -----(3)
Hard to solve for T.
Use iteration, Newton's Method.
You should get T is about 1.7606 after the 3rd iteration if your T seed is 1.7
Therefore, in a little more than one and 3/4 years, the Revenue would equal the cost of the machine. --------answer.