# problemset7answer78 - Solve for i by trial and error or...

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7. In the context of this question, instead of using the equation PW(i)=0, we will use the equivalent one AW(i)=0 to make the calculation a bit simpler. The incremental rate is simply the rate of the cash flow we are given, the incremental cash flow. We have an annuity specified already with annual payment of 4000 for 8 years. Now let us take 4000 from the last amount of 12,000 to create an annuity of 9 years with 4000 per year. Thus the annual worth of this annuity is simply 4000. Given what we did above, in period 9 we are left with the amount of 12000-4000=8000. The AW of this amount spread over 9 years (starting with period 1) is 8000(A/F,i, 9) Finally, we want to express the initial flow, -22,000, as an annuity over the same 9 years. This gives us the annual worth of the -22,000 amount, -22,000(A/P,i, 8) Thus AW(i) = 0 becomes -22,000(A/P,i,9) + 4000 + (12,000 – 4000)(A/F,i,9) = 0
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Unformatted text preview: Solve for i by trial and error or Excel i = 14.3% (Excel) i > MARR; select alternative N 8. Here we have a geometric gradient series with base amount 8500 and gradient -500 to deal with. Why ? Because the first benefit at the end of year 1 is 9000 – 500(1) = 8500 After that the payments decrease by 500 in each year so the gradient is -500. Recall that we can decompose such as series into an annuity with payment 8500 and geometric gradient series with base amount zero and gradient -500. So we know immediately that 8500 is one number part of the AW of the project. Another part is the annual worth of the latter geometric series which is -500(A/G,i, 10). Finally, the annual worth of remaining amount of -40,000 is -40,000(A/P,i,10). Therefore we have: 0 = -40,000(A/P,i,10) + 8500 – 500(A/G,i,10) Solve for i by trial and error i = 10.5% is < MARR = 17% Select Z1...
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## This note was uploaded on 09/24/2011 for the course CPS 125 taught by Professor Panzer during the Winter '11 term at Ryerson.

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