oliviavaughan01
14.04.2020 •
Business
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
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Ответ:
5.16%
Explanation:
PW=0 equation.
0 = -500,000 + 10,000(P/A, i*,10) + 700,000(P/F, i*,10)
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%
Ответ:
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Explanation: