Question

A new production system for a factory is to be purchased and installed for $132,625. This system will save approximately 300,000 kWh of electric power each year for a 6-year period. Assume the cost of electricity is $0.10 per kWh, and factory MARR is 15% per year, and the salvage value of the system will be $8,351 at year 6. Using the AW method to analyzes if this investment is economically justified

A- calculate the AW of the above investment and insert the result below.

B- Based on the AW value you got in the previous question, is this investment economically justified or not? type you explanation below

Answer 1

Solution:

Given,

Initial investment (P) = $132,625

Approximate electricity power to be saved every year =  300,000 kWh

Cost of electricity = $0.10 per kWh

Therefore, annual saving (C) = 300,000 x 0.10

= $ 30,000

MARR (i)= 15% per year

Life of the system (n) = 6 years

Salvage value of the system at the 6th year (S)= $8,351

The cashflow diagram is as follows:

A) We know that the annual worth (AW) method is commonly used for comparing alternatives. AW means that all incomes and disbursements are converted into an equivalent uniform annual (end-of-period) amount, which is the same each period.

Therefore, AW = -P (A/P, i, n) + C + S (A/F, i, n)

= -132,625(A/P, 15%, 6) + 30,000 + 8,351 (A/F, 15%, 6)

= (-132,625 x 0.2642) + 30,000 + (8,351 x 0.1142)

= -35,039.525 + 30,000 + 953.6842

= -$4,085.8408

= -$4,085.84

B) Based on the AW value in the previous question, is this investment economically not justified because the anuual worth obtained is a negative value. So it means that the by installing the machine, the factory will incur annual loss $4,085.84.