Question

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

the previous question

A new production system for a factory is to be purchased and installed for $164,546. 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 $8,058 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. = -12,608.45

Answer 1

Cost of new production system = 164,546

MARR = 15%

N = 6 years

Year	Annual saving	Present value of annual saving	Rough work to calculate present value of saving
1	30,000.00	26,086.96	[30,000 / 1.15^1]
2	30,000.00	22,684.31	[30,000 / 1.15^2]
3	30,000.00	19,725.49	[30,000 / 1.15^3]
4	30,000.00	17,152.60	[30,000 / 1.15^4]
5	30,000.00	14,915.30	[30,000 / 1.15^5]
6	30,000.00	12,969.83	[30,000 / 1.15^6]
		113,534.48

Present value of salvage at the end of year 6 = [8,058 / 1.15^⁶] = 3,483.69

Net present value of new production system = -Cost of new production system + Present value of annual saving + Present value of salvage = - 164,546 + 113,534.48 + 3,484.69 = - 47,527.82

Annual worth is calculated as: Net present value of new production system / {[1 - (1 + MARR)^^-N] / MARR} = - 47,527.82 / {[1 - (1 + 0.15)^^-6] / 0.15} = - 12,558.61