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CHAPTER 10
B189
25.
A kilowatt hour is 1,000 watts for 1 hour. A 60watt 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 60watt bulb will last for 1,000 hours, which is 2 years of use at 500 hours per year. So, the NPV of
the 60watt 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 15watt CFL. A 15watt 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 15watt 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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View Full DocumentB190
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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 Spring '06
 Tapley
 Finance

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