QUESTION IMAGE
Question
a mining company is performing an economic analysis between two equipment. the initial price for equipment t is $580,000 and has a service life of 23 years with operating and maintenance costs of $19,000 per year. if the equipment can generate gross annual revenues of $94,000 and the company uses an annual marr of 10%, calculate the equipments discounted payback period.
o a. 10.1 years
o b. 19.3 years
o c. 15.6 years
o d. 9.0 years
if the alternative equipment u has a payback period of 20 years, which of the two equipment should be preferred based on the payback period?
o a. equipment t
o b. equipment u
Step1: Calculate net annual cash - flow for Equipment T
Net annual cash - flow \(CF_T=94000 - 19000=\$75000\)
Step2: Set up the discounted pay - back period formula for Equipment T
The initial investment \(P = 580000\). The formula for the present value of an ordinary annuity is \(P = CF\times\frac{1-(1 + r)^{-n}}{r}\), where \(r = 0.1\) (MARR) and \(CF = 75000\). So, \(580000=75000\times\frac{1-(1 + 0.1)^{-n}}{0.1}\)
Step3: Simplify the equation
First, \(\frac{580000\times0.1}{75000}=1-(1.1)^{-n}\). Then, \(\frac{58000}{75000}=1-(1.1)^{-n}\), and \(0.7733 = 1-(1.1)^{-n}\). So, \((1.1)^{-n}=1 - 0.7733=0.2267\)
Step4: Solve for \(n\)
Take the natural logarithm of both sides: \(-n\ln(1.1)=\ln(0.2267)\). Then, \(n=-\frac{\ln(0.2267)}{\ln(1.1)}\)
Using \(\ln(0.2267)\approx - 1.489\) and \(\ln(1.1)\approx0.0953\), \(n=\frac{1.489}{0.0953}\approx15.6\) years
Since the discounted pay - back period of Equipment T is approximately \(15.6\) years and the pay - back period of Equipment U is \(20\) years. A shorter pay - back period is better.
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A. Equipment T