Final(Case9)

Final(Case9) - Annual Power Cost = $367,219.20 AW(6%,5) =...

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Joseph Murphy Case 9 #1 Install cost = $3500(87.8/0.134) Install cost = $2,293,283.58 # of poles = 1310/2 = 655 Annual Power Cost = 655*(2 bulbs/pole)*(0.4kw/bulb)*(12 hr/day)*(365 days/yr)*(0.08/kw-hr) Annual Power Cost = $183,609.60 AW(6%,5) = $43,588.92 Total Annual Cost = $2,336,872.50 B/C = $1,111,500/$2,336,872.50 B/C = 0.48 B/C = 0.48 #2 Unlighted Night to Day = 5/3 Lighted Night ot Day = 7/4 Unlighted Night to Day = 1.667 Lighted Night ot Day = 1.75 #3 Install cost = $2500(87.8/0.067) Install cost = $3,276,119.40 Annual Power Cost = 1310*(2 bulbs/pole)*(0.4kw/bulb)*(12 hr/day)*(365 days/yr)*(0.08/kw-hr)
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Unformatted text preview: Annual Power Cost = $367,219.20 AW(6%,5) = $87,177.84 Total Annual Cost = $3,363,297.24 B/C = $1,111,500/$3,363,297.24 B/C = 0.33 B/C = 0.33 #4 Unlighted = 199/379 Lighted = 839/2069 Unlighted = 0.525 Lighted = 0.406 Accidents Prevented = 199-(199*0.406) Accidents Prevented = 118 Accidents Prevented = 118 #5 B/C would have to equal 1 so, B = $1,456,030 @ $4500 per accident x = B/$4500 x= 324 839+324 = 1163 1163/0.525 = 2215 Accident Ratio = 839/2215 0.38 There would have to be 2215 accidents and the ratio would have to be 0.38....
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Final(Case9) - Annual Power Cost = $367,219.20 AW(6%,5) =...

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