Question

A hospital in The Upper Cumberland area bought a diagnostic machine at a cost of

$40,000. Maintenance cost is expected to remain constant throughout the life of this machine at $2,000 per year. The salvage value is estimated to be "0" at the end of the useful life of 10 years. Determine the economic life of this machine. MARR = 10%
A. 5 years
B. 1 year
C. 10 years
D. 7 years

Answer 1

To determine the economic life of the machine, we can use the annual worth method. The annual worth (AW) of the machine over its useful life of 10 years can be calculated as follows:

AW = P(A/P, i, n) + A(F/P, i, n)

where P is the initial cost of the machine, A is the annual maintenance cost, i is the MARR, and n is the useful life of the machine.

Substituting the given values, we get:

AW = 40000(A/P, 10%, 10) + 2000(F/P, 10%, 10)

Using the A/P and F/P factors from the tables, we get:

AW = 40000(0.162) + 2000(6.145)

AW = 6480 + 12290

AW = 18770

Since the AW is positive, the investment is economically justified. To find the economic life of the machine, we need to find the value of n that makes the AW equal to zero. We can use trial and error to find the value of n that makes the AW closest to zero. Trying n = 5, we get:

AW = 40000(A/P, 10%, 5) + 2000(F/P, 10%, 5)

AW = 40000(0.2638) + 2000(3.791)

AW = 10552 + 7582

AW = 18134

Trying n = 7, we get:

AW = 40000(A/P, 10%, 7) + 2000(F/P, 10%, 7)

AW = 40000(0.1895) + 2000(4.868)

AW = 7580 + 9736

AW = 17316

Therefore, the economic life of the machine is approximately 7 years, which is closest to D. 7 years.