Question

PROBLEM

A machine that costs $ 40,000 has a useful life of 8 years and it is estimated that the residual value at the end of its useful life is $ 5,000. This machine, which will be used to make pieces of complex geometry and will have an annual operation and maintenance cost of $ 8,000. The operator of this machine receives $ 15.00 per hour and the machine consumes power at a rate of $ 1.15 per hour. It has also been estimated that each piece needs an average of 48 minutes of machine time to be manufactured.

Note: Although the operator must remain with the machine for as long as the machine is processing parts, he or she can perform other functions the rest of the time.

1. If the annual production has been forecast close to two thousand pieces (2,000), calculate the total unit cost (per piece) considering the value of the money over time if the company uses an annual MARR of 15%.

2. If the pieces can be sold for $ 30 each and we can presume that this price will remain fixed for the length of the study period (8 years),

a) Determine the number of pieces that must be produced and sold so that the company that invests in this alternative neither wins nor loses money.

b) under the presumption of an annual volume of 2,000 pieces, determine the number of years that must elapse before the company recovers its investment if we consider the value of money over time ("discounted payback").

Answer 1

		Annual Depreciation=(40000-5000)/8	$4,375
		Annual Operations and maintenance	$8,000
		Annual Operation time =(2000*48)/60 Hour	1,600
		Cost of Operator=$15.00*1600	$24,000
		Cost of power=$1.15*1600	$1,840
		Total Annual Cost	$39,815
		Present Value (PV) of Cost:
		(Cost)/((1+i)^N)
		i=Discount Rate=MARR=15%=0.15
		N=Year of Cost incurred
	N	Year	1	2	3	4	5	6	7	8
	A	Cost	$39,815	$39,815	$39,815	$39,815	$39,815	$39,815	$39,815	$39,815	SUM
	B=A/(1.15^N)		$34,622	$30,106	$26,179	$22,764	$19,795	$17,213	$14,968	$13,016	$178,663
		Total Costs	$178,663
		Total production quantity =2000*8	16,000
		Unit Cost=178663/16000	$11.17
	2	Number of pieces required to be sold
		Annual Fixed costs:
		Annual Depreciation=(40000-5000)/8	$4,375
		Annual Operations and maintenance	$8,000
		Total Fixed costs	$12,375
		Variable Costs for 2000 units:
		Cost of Operator=$15.00*1600	$24,000
		Cost of power=$1.15*1600	$1,840
		Variable Costs for 2000 units:	$25,840
		Variable Cost per unit=25840/2000	$12.92
		Contribution per unit =$30-$12.92=	$17.08
		Break even quantity =Fixed Costs/Unit contribution	724.53162	(12375/17.08)
	2a	Number of pieces to make neither profit nor loss	725	(Rounded to whole number)
	2b	Number of years to recover investment:
		Revenue per year=2000*30=	$60,000
		Cash Costs per year (Excluding depreciation)	$35,440	(39815-4375)
		Net Cash Flow per year=60000-35440=	$24,560
		Cash Flow in year 0=Investment	($40,000)
		Cash Flow in year 8=24560+5000(Residual Value)	$29,560
	N	Year	0	1	2	3	4	5	6	7	8
	A	Cash Flow	-$40,000	$24,560	$24,560	$24,560	$24,560	$24,560	$24,560	$24,560	$29,560
	B=A/(1.15^N)	Present Value Of Cash Flow	-$40,000	$21,357	$18,571	$16,149	$14,042	$12,211	$10,618	$9,233	$9,663
		Cumulative Present Value of Cash Flow	-$40,000	-$18,643	-$73	$16,076	$30,118	$42,329	$52,947	$62,180	$71,843
		Discounted Payback =2+(73/16149)=	2.0045	years