Question

Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typical 60-watt incandescent light bulb costs $.54 and lasts for 1,000 hours. A 15-watt LED, which provides the same light, costs $3.85 and lasts for 12,000 hours. A kilowatt hour of electricity costs $.130. A kilowatt-hour is 1,000 watts for 1 hour. However, electricity costs actually vary quite a bit depending on location and user type. An industrial user in West Virginia might pay $.04 per kilowatt-hour whereas a residential user in Hawaii might pay $.25.

You require a return of 10 percent and use a light fixture 500 hours per year. What is the break-even cost per kilowatt-hour? (Do not round intermediate calculations and round your answer to 6 decimal places, e.g., 32.161616.)

Answer 1

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 of the light bulb
The number of watts used by the bulb per hour, WPH = W / 1,000
And the kilowatt hours used per year, KPY = WPH × H
The electricity cost per year, ECY = KPY × C
The NPV of the decision to buy the light bulb, NPV = -P - ECY(PVIFAR%,t)
And the EAC is:
EAC = NPV / (PVIFAR%,t)
Substituting, we get:
EAC = [-P - (W / 1,000 × H × C)PVIFAR%,t] / PVIFAR%,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:
[-$0.54- (60 / 1,000 × 500 × C)PVIFA10%,2] / PVIFA10%,2
= [-$3.85 - (15 / 1,000 × 500 × C)PVIFA10%,24] / PVIFA10%,24
[-$0.54 - $52.066C] / 1.7355 = [-$3.85 - $67.386C] / 8.9847
-$4.85 - $467.80C = -$6.68 - $116.95C
$350.85C = $1.83
C = $0.005216
So, unless the cost per kilowatt hour is extremely low, it makes sense to use the CFL.