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 $.49 and lasts for 1,000 hours. A 15-watt LED, which provides the same light, costs $3.60 and lasts for 12,000 hours. A kilowatt hour of electricity costs $.125. 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 11 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.)

Light emitting diode (LED) light bulbs have become required in recent years, but do they make financial sense? Suppose a typical 60-watt incandescent light bulb costs $.51 and lasts for 1,000 hours. A 15-watt LED, which provides the same light, costs $3.70 and lasts for 12,000 hours. A kilowatt-hour is 1,000 watts for 1 hour. Suppose you have a residence with a lot of incandescent bulbs that are used on average 500 hours a year. The average bulb will be about halfway through its life, so it will have 500 hours remaining (and you can’t tell which bulbs are older or newer).

If you require a return of 10 percent, at what cost per kilowatt-hour does it make sense to replace your incandescent bulbs today? (A negative answer should be indicated by a minus sign. Do not round intermediate calculations and round your answer to 6 decimal places, e.g., 32.161616.)

Martin Enterprises needs someone to supply it with 121,000 cartons of machine screws per year to support its manufacturing needs over the next five years, and you’ve decided to bid on the contract. It will cost you $800,000 to install the equipment necessary to start production; you’ll depreciate this cost straight-line to zero over the project’s life. You estimate that, in five years, this equipment can be salvaged for $148,000. Your fixed production costs will be $445,000 per year, and your variable production costs should be $10.20 per carton. You also need an initial investment in net working capital of $71,000. If your tax rate is 22 percent and you require a return of 10 percent on your investment, what bid price should you submit? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)

A five-year project has an initial fixed asset investment of $300,000, an initial NWC investment of $28,000, and an annual OCF of −$27,000. The fixed asset is fully depreciated over the life of the project and has no salvage value. If the required return is 11 percent, what is this project’s equivalent annual cost, or EAC? (A negative answer should be indicated by a minus sign. Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)

Answer 1

	BREAK EVEN COST PER KW HOUR
	Assume Break even cost=X
	Consumption for incandescent bulb per year	30000 watt hour	(500*60)
	Consumption for incandescent bulb per year	30 unit
	Consumption for LED bulb per year	7500 watt hour	(500*15)
	Consumption for LED bulb per year	7.5 unit
	Annual Savings ofelectricity with LED bulb	22.5 units	(30-7.5)
	Life of LED buld =12000/500	24	years
	Present value of Savings of $1 per year for 24 years	$8.35	(Using PV function of excelwith Rate =11%, Nper=24, Pmt=-1)
	Present Value of totalsavings =$8.35*22.5*X	$187.83	X
	Number of incandescent Bulb required	12	(Every 2 years)
	Interest Rate in 2 years =(1.11^2)-1=	0.2321
	Number of 2 year period	12
	Expense per year	$0.49
	Present value of 12 numbers bulbs	$1.94	(Using PV function of excelwith Rate =0.2321, Nper=12, Pmt=-0.49)
	Cost of LED Bulb	$3.60
	IncrementalCost of LED bulb =3.60-1.94	$1.66
	For Break Even :187.83X=1.66
	BREAK EVEN COST PER KW HOUR=1.66/187.83=	0.008844732	Dollar
	COST OF REPLACEMENT PER BULB	$3.70
	Present value of Savings of $1 per year for 24 years	$8.98	(Using PV function of excelwith Rate =10%, Nper=24, Pmt=-1)
	Present Value of Savings=8.98*22.5*X
	Present Value of Saving in electricity	$202.16	X
	X=Cost of electricity per unit
	Cost of Loss of 500 hoursaverage life	0.255	($0.51/2)
	COST PER KWHour=X
	202.16X=3.7+0.255
	X=3.955/202.16=	$0.0195640	Dollar