In: Finance
A standard electric motor is rated at 10 horsepower (HP) and costs $800. Its fullload
efficiency is specified to be 87%. A newly designed, high efficiency motor of
the same size has an efficiency of 93%, but costs $1,200. It is estimated that the
motors will operate at a rated 10-HP output for 2,500 hours a year, and the cost of
energy will be $0.07 per kilowatt-hour. Each motor is expected to have 12-year
life. At the end of 12 years, the first motor will have a salvage value of $50, and
the second motor will have a salvage value of $100. Consider the MARR to be
8%. (Note: 1HP = 0.7457 KW)
1- What is the net present value (NPW) of the cash flows associated with the high
efficiency motor?
2- Which of the motors should be selected when comparing their net present worth
(NPW)?
Step 1: Identification of Alternatives
Alternative 1: Standard Motor
Alternative 2: High Efficiency Motor
Step 2: Table showing Data for both the alternatives
Particulars | Standard Motor | High Efficiency Motor |
Cost of Motor | $800.00 | $1,200.00 |
Efficiency | 87% | 93% |
Operating Hours | 2500 | 2500 |
Cost of Energy Per KWH | $0.07 | $0.07 |
Life (years) | 12 | 12 |
Salvage at the end of Yr 12 | $50.00 | $100.00 |
MARR | 8% | 8% |
1HP = 0.7457KW |
Step 3: Calculations of Cash Flows from Both the Alternatives
Particulars | Standard Motor | High Efficiency Motor |
Cash Outflows | ||
Cost of Motor | $(800.00) | $(1,200.00) |
Cash Inflows | ||
Efficiency | 87% | 93% |
Operating Hours | 2500 | 2500 |
Output of motor in HP (units) (note1) | 21750 | 23250 |
Output in KW (units) (A) (note 2) | 16218.975 | 17337.525 |
Cost of Energy Per KWH (B) | $0.07 | $0.07 |
Annual Revenue ((Year 1-12 years) (A)*(B) | $1,135.33 | $1,213.63 |
Salvage at the end of (Year-12) | $50.00 | $100.00 |
Notes:
1. Output of 10-HP Motor = Efficiency x Operating Hours x 10HP
2. Output in KW = Output in HP * 0.7457
Step 4: Calculation of Net Present Value associated with High Efficiency Motor
NPV= Present Value of Cash Inflows - Present Value of Cash Outflows
Particulars | High Efficiency Motor | Period | PVF @8% | Present Value |
Cash Outflows | ||||
Cost of Motor | $(1,200.00) | 0 | 1 | $(1,200.00) |
Cash Inflows | ||||
Annual Revenue | $1,213.63 | 1-12 | 7.536 | $9,145.89 |
Salvage at the end of year 12 | $100.00 | 12 | 0.397 | $39.70 |
NPV | $7,985.59 |
Step 5: Calculation of Net Present Value associated with Standard Motor
Particulars | Standard Motor | Period | PVF @8% | Present Value |
Cash Outflows | ||||
Cost of Motor | $(800.00) | 0 | 1 | $(800.00) |
Cash Inflows | ||||
Annual Revenue ((Year 1-12 years) | $1,135.33 | 1-12 | 7.536 | $8,555.83 |
Salvage at the end of (Year-12) | $50.00 | 12 | 0.397 | $19.85 |
NPV | $7,775.68 |
Part a)
Net present value (NPV) of the cash flows associated with the High efficiency motor = $7,985.59
Part b)
Net present value (NPV) of the cash flows associated with the Standard motor = $7,775.68
Net present value (NPV) of the cash flows associated with the High efficiency motor = $7,985.59
Decision: Since the NPV of High Efficiency motor is higher as compared to standard Motor, we should select Option 2 as it provides more Benefits