The HVAC engineer for a company that constructed one of the world’s tallest buildings requested that $500,000 be spent on software and hardware to improve efficiency of the environmental control system. This is expected to save $10,000 per year for 10 years in energy costs and $700,000 at the end of the 10 years in equipment refurbishment costs. Find the rate of return

Now let use the estimation procedure to determine i* mean while All income will be regarded as a single F in year 10 so that the P/F factor can be used.

Therefore The P/F factor is selected because most of the cash flow ($700,000) which already fits this factor and errors.

P =$500,000, n =10,

F =10(10,000) + 700,000 = $800,000. .

Now we can state that 500,000 =

800,000(P/F,i,10)(P/F,i,10) = 0.625

Roughly estimated i* is between 4% and 5%.

Let use 5% as the first trial because this approximate rate for the P/F factor is lower than the true value when the time value of money is considered.

At i* =5%, the IRR equation is

0 = -500,000 + 10,000(P/A,5%,10) + 700,000(P/F,5%,10)0 < $6946

The result is positive, indicating that the return is more than 5%.

Let Try i*= 6%.

0 = -500,000 + 10,000(P/A,6%,10) + 700,000(P/F,6%,10)0 > $-35,519

Since the interest rate of 6% is too high, linearly interpolate between 5% and 6%

i* = 5.00 + 6946/(6946 + 35519) = 5.16%

Therefore the RATE OF RETURN is 5.16%