Question

A simple, direct space heating system is currently being used in a professional medical office complex. An upgraded “variable air-volume system” retrofit can be purchased and installed for $200,000 (investment cost). Its power savings in the future will be 500,000 kilo-Watt hours per year over its estimated life of 8 years. The cost of electricity is $0.10 per kilo-Watt hour. If the firm’s cost of capital is 12% per year and the residual value of the system in 8 years is $20,000, should the new system be purchased?

*recommended, PW = 56,458

2. Confirm your recommendation by applying the IRR method.

Answer 1

Initial Cost = $200,000

Power savings 500,000 kilo-Watt hour per year

Life = 8 years

Cost of electricity = $0.10 per kilo-watt hour

Annual power savings = 500,000 * $0.10 = $50,000

Salvage Value = $20,000

Cost of Capital (MARR) = 12%

PW = -$200,000 + $50,000 (P/A, 12%, 8) + $20,000 (P/F, 12%, 8)

PW = -$200,000 + $50,000 (4.9676) + $20,000 (0.40388) = $56,457.6 or $56,458

As the NPW is greater than ZERO the new system can be purchased.

Now, calculating the IRR using the trial and error method.

Let the rate of interest is 19%. Calculate the PW of the system at 19%.

PW = -$200,000 + $50,000 (P/A, 19%, 8) + $20,000 (P/F, 19%, 8)

PW = -$200,000 + $50,000 (3.95436) + $20,000 (0.24867) = $2,691

The PW is positive. Increase the rate of interest to get negative PW. Increase the rate of interest to 20%. Calculate PW at 20%.

PW = -$200,000 + $50,000 (P/A, 20%, 8) + $20,000 (P/F, 20%, 8)

PW = -$200,000 + $50,000 (3.83716) + $20,000 (0.23257) = -$3,490

Using the interpolation and calculating IRR

IRR = 19% + [2,691 – 0 ÷ 2,691 – (-3,490)]

IRR = 19.44%

IRR is greater than MARR (given cost of capital). Hence, the system can be purchased.