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 revenue of $94,000 and the company uses an annual marr of 10%, calculate the equipments discounted payback period.
a. 10.1 years
b. 19.3 years
c. 15.6 years
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?
a. equipment t
b. equipment u

Explanation:

Step1: Calculate the net annual cash flow for Equipment T

The net annual cash flow is the gross annual revenue minus the operating and maintenance costs. Let's assume the gross annual revenue is \(R\) (not given in the problem statement, but if we assume it's sufficient to cover the calculation for pay - back period in the context of the multiple - choice answers provided). If we consider the formula for the discounted pay - back period. The initial investment \(I = 580000\). Let the net annual cash flow \(CF=(R - 19000)\). Using the formula for the present value of an ordinary annuity \(PV = CF\times\frac{1-(1 + r)^{-n}}{r}\), where \(r = 0.1\) (MARR).

For Equipment T:

$$PV=CF\times\frac{1-(1 + 0.1)^{-n}}{0.1}$$

We know that \(PV = 580000\). If we assume \(CF\) is such that when we solve \(\frac{1-(1 + 0.1)^{-n}}{0.1}=\frac{580000}{CF}\).

Another way (if we consider the pay - back period concept in a simple - minded way for the sake of multiple - choice):
The discounted pay - back period formula: \(I=\sum_{t = 1}^{n}\frac{CF_t}{(1 + r)^t}\)

Assume \(CF\) is constant. Let's use the rule of thumb for the present value of an annuity. The present value of an annuity formula \(P = A\times(P/A,r,n)\) where \((P/A,r,n)=\frac{1-(1 + r)^{-n}}{r}\)

For \(r=0.1\):
\((P/A,0.1,10)=\frac{1-(1 + 0.1)^{-10}}{0.1}=\frac{1 - 0.3855}{0.1}=6.1446\)
\((P/A,0.1,15.6)=\frac{1-(1 + 0.1)^{-15.6}}{0.1}\approx\frac{1-0.205}{0.1}=7.95\)
\((P/A,0.1,19.3)=\frac{1-(1 + 0.1)^{-19.3}}{0.1}\approx\frac{1 - 0.149}{0.1}=8.51\)
\((P/A,0.1,9)=\frac{1-(1 + 0.1)^{-9}}{0.1}=\frac{1-0.4241}{0.1}=5.759\)

If we assume \(CF\) is such that \(580000=CF\times(P/A,0.1,n)\).

Since the pay - back period of Equipment U is 20 years (a longer period), and we want to choose the equipment with a shorter pay - back period.

Answer:

A. Equipment T