Alpha Electronics can purchase a needed service for $130 per unit. The same service can be provided by equipment that costs $100,000 and that will have a salvage value of 0 at the end of 10 years. Annual operating costs for the equipment will be $7,000 per year plus $25 per unit produced. MARR is 12%/year.

a) Whats the annual worth if the expected production is 90units/year? 510units/year?

b)Determine the breakeven value for annual production that will return MARR on the investment in the new equipment.

Respuesta :

Answer:

a) Annual worth for 90 units/year = -7,550

Annual worth for 510 units/year = 36,550

b) The breakeven value for annual production that will return MARR on the investment in the new equipment is Q=235 units/year.

Explanation:

a) We can calculate the annual worth for any expected production substracting from the "purchased service" cost, the "equipment" costs. In the equipment cost, we considered a ten-year amortization of the equipment, that is 100,000/10=$10,000/year.

[tex]AW=C_1-C_2=(130Q)-(10,000+7,000+25Q)=105Q-17,000[/tex]

For Q=90, the annual worth is:

[tex]AW(90)=105*90-17,000=9,450-17,000=-7,550[/tex]

For Q=510, the annual worth is:

[tex]AW(510)=105*510-17,000=53,550-17,000=36,550[/tex]

b) We have to compare the two options (purchased service vs. equipment) in the same time span, so the two are evaluated over a 10 year period.

The purchased service option implies paying $130 per unit, so the cash flow each year is related linearly to the volume of production Q (units/year).

As the cash flow is constant for a certain level of production, we can use the annuity factor to calculate the present value PV.

The present value of this option is:

[tex]PV_1=\sum_{k=1}^{10}\dfrac{130Q}{(1+0.12)^k}=130Q*(\dfrac{1-(1+0.12)^{-10}}{0.12})\\\\PV_1=130Q*5.65=734.5Q[/tex]

The equipment option is more complex. We will consider the purchased in year 0 and the fixed and variable cost from year 1 to 10.

The present value is then:

[tex]PV_2=100,000+\sum_{k=1}^{10}\dfrac{7,000+25Q}{(1+0.12)^k}\\\\\\PV_2=100,000+(7,000+25Q)*(\dfrac{1-(1+0.12)^{-10}}{0.12})\\\\\\PV_2=100,000+(7,000+25Q)*5.65\\\\\\PV_2=100,000+7,000*5.65+5.65*25Q\\\\\\PV_2=100,000+39,550+141.25Q\\\\\\PV_2=139,550+141.25Q[/tex]

The breakeven value for annual production is the quantity for which both present values are equivalent:

[tex]PV_1=PV_2\\\\\\734.5Q=139,550+141.25Q\\\\(734.5-141.25)Q=139,550\\\\593.25Q=139,550\\\\Q=139,550/593.25=235.23\approx235[/tex]