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CHAPTER 10 B-189 25. A kilowatt hour is 1,000 watts for 1 hour. A 60-watt bulb burning for 500 hours per year uses 30,000 watt hours, or 30 kilowatt hours. Since the cost of a kilowatt hour is \$0.101, the cost per year is: Cost per year = 30(\$0.101) Cost per year = \$3.03 The 60-watt bulb will last for 1,000 hours, which is 2 years of use at 500 hours per year. So, the NPV of the 60-watt bulb is: NPV = –\$0.50 – \$3.03(PVIFA 10%,2 ) NPV = –\$5.76 And the EAC is: EAC = –\$5.83 / (PVIFA 10%,2 ) EAC = –\$3.32 Now we can find the EAC for the 15-watt CFL. A 15-watt bulb burning for 500 hours per year uses 7,500 watts, or 7.5 kilowatts. And, since the cost of a kilowatt hour is \$0.101, the cost per year is: Cost per year = 7.5(\$0.101) Cost per year = \$0.7575 The 15-watt CFL will last for 12,000 hours, which is 24 years of use at 500 hours per year. So, the NPV of the CFL is: NPV = –\$3.50 – \$0.7575(PVIFA 10%,24 ) NPV = –\$10.31 And the EAC is: EAC = –\$10.85 / (PVIFA 10%,24 ) EAC = –\$1.15 Thus, the CFL is much cheaper. But see our next two questions. 26. To solve the EAC algebraically for each bulb, we can set up the variables as follows: W = light bulb wattage C = cost per kilowatt hour H = hours burned per year P = price the light bulb The number of watts use by the bulb per hour is: WPH = W / 1,000 And the kilowatt hours used per year is: KPY = WPH × H

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B-190 SOLUTIONS The electricity cost per year is therefore: ECY = KPY × C The NPV of the decision to but the light bulb is: NPV = – P – ECY(PVIFA R%,t ) And the EAC is: EAC = NPV / (PVIFA R%,t ) Substituting, we get: EAC = [–P – (W / 1,000 × H × C)PVIFA R%,t ] / PFIVA R%,t We need to set the EAC of the two light bulbs equal to each other and solve for C, the cost per kilowatt hour. Doing so, we find: [–\$0.50 – (60 / 1,000 × 500 × C)PVIFA 10%,2 ] / PVIFA 10%,2 = [–\$3.50 – (15 / 1,000 × 500 × C)PVIFA 10%,24 ] / PVIFA 10%,24 C = \$0.004509 So, unless the cost per kilowatt hour is extremely low, it makes sense to use the CFL. But when should
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