llw nehe IME 440/640-Homework 2 Due: Tuesday, February 6, in class st ar 1. The
ID: 2258338 • Letter: L
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llw nehe IME 440/640-Homework 2 Due: Tuesday, February 6, in class st ar 1. The pumping cost for delivering water from the Ohio River to Wheeling Steel for cooling hot-rolled steel was $4million for the first 5 years starting from year 2. An effective energy conservation program resulted in a reduced cost million in year 7, $1.74 million in year 8, and amounts decreasing by S30,000 each year through year 15. determine the equivalent values listed to $1.77 below at 16% per year. the equivalent annual series of the pumping costs over the last 10 years. a) b) the equivalent annual series of the pumping costs from year 3 to 10. c) Equivalent worth of the project at the end of year 6. d) the initial cash flow of the equivalent geometric series with increase rate of 15% starting from year 3 and ending in year 20. lar thermal eleetric)Explanation / Answer
1. a) The equivalent annual pumping cost over the last 10 years need to calculate the present value of the pumping costs from Year 6(end) to Year 15(end). The sum of these PVs is then equated to equivalent annual cash flow of $ K per year using the 10-year annuity formula. This value $ K is the equivalent annual series of pumping costs for the last 10 years (assuming that project ends in year 15). Also. the Present Value of each pumping cost from year 6 to year 15 is calculated at the end of year 5(or beginning of year 6)
Therefore, PV of Pumping Costs (at end of Year 5) = PV(5)
= 1400000 / (1.16)^(1) + 1770000 / (1.16)^(2) + 1740000 / (1.16)^(3) + .......................+ 1530000 / (1.16)^(10)
PV(5) = $ 7873783
This Present Value of pumping cost needs to be spread into an equivalent stream of equal annual cash flows starting from Year 6 and running upto Year 15.
Let this equal annual cash flow be $ K
Therefore, 7873783 = K x (1/0.16) x [1-{1/(1.16)^(10)}]
K = $ 1629094.231
b) Similar to part(a), the annual pumping costs (which is not uniform through time) needs to be collapsed down into an equivalent total present value. The resultant present value is then equated to an equal annual cash flow stream running from Year 3 (end) to Year 10(end).
The Present Value is calculated at the end of year 2 (or beginning of year3) .
PV(2) = 1400000 / (1.16)^(1) + 1400000 / (1.16)^(2) + 1400000 / (1.16)^(3) + 1400000 / (1.16)^(4) + 1770000 / (1.16)^(5) + 1740000 / (1.16)^(6) + 1710000 / (1.16)^(7) + 1680000 / (1.16)^(8)
PV(2) = $ 6591834
This Present Value of pumping cost needs to be spread into an equivalent stream of equal annual cash flows starting from Year 3 and running upto Year 10.
Let this equal annual cash flow be $ K
6591834 = K x (1/ 0.16) x [1-{1/(1.16)^(8)}]
K = $ 1517600.105
c) The Equivalent Worth of the project at the end of Year 6 is the difference between the Present Worth (at end of Year 6) of the cost benefits(accrued owing to the energy conservation program) and the Present Worth(at the end of Year 6) of the pumping costs.
Equivalent Worth =EW(6) = PWB(6) - PWC(6)
PWB(6) = 0 / (1.16) + 30000 / (1.16)^(2) + 30000 / (1.16)^(3) + .......+ 30000 / (1.16)^(9) = $ 112334.2
PWC(6) = 1770000 / (1.16) + 1740000 / (1.16)^(2) + ........+ 1530000 / (1.16)^(9) = $ 7733589
EW(6) = PWB(6) - PWC(6) = 112334.2 - 7733589 = $ - 7621255
d) The entire cash flow of pumping costs that begins with $ 1400000 in year 2 and ends with $ 1530000 in year 15, has to be replaced with an increasing equivalent geometrice series that has an increase rate of 15%, begins in Year 3 and ends in Year 20.
For doing the same, one needs to calculate the Present Value of the Pumping Costs at t=0. This present value is then equated to an increasing geometric series of 15% increase, beginning in Year 3 and ending in Year 20.
PV(0) = 1400000 / (1.16)^(2) + 1400000 / (1.16)^(3) +..........+ 1770000 / (1.16)^(7) + 1740000 / (1.16)^(8) +...............+ 1530000 / (1.16)^(15) = $ 7125925
As the geometric cash flow begins in year 3 (end) and not year 0(end) one needs to calculate the present value of the cash flow at end of year 2.
The same can be calculated as PV(2) = PV(0) x (1.16)^(2) = 7125925 x (1.16)^(2) = $ 1
Let the first(initial) cash flow of the geometric series be A(1)
Therefore, 9588645 = A(1) x [1 - {(1.15)^(18) / (1.16)^(18)}] / (0.16 - 0.15)
A(1) = $ 664455.5516
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