Final(Case9)

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

This preview shows pages 1–3. Sign up to view the full content.

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)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online