Punto de equilibrio

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• Publicado : 10 de enero de 2011

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ANÁLISIS DEL PUNTO DE EQUILIBRIO

13.3 A metallurgical engineer has estimated that the capital investment for recovering valuable metals (nickel, silver, gold, etc.) from a copper refinery 'S wastewater stream will be \$15 million. The equipment will have a useful life of 10 years with no salvage value. The amount of metals currently discharged is 12,000 pounds per month. The monthly operatingcost is represented by \$(4,100,000) E1.8, where £ is the efficiency of metal recovery in decimal form. Determine the minimum removal efficiency required for the company to break even, if the average selling price of the metals is \$250 per pound. Use an interest rate of 1 % per month.

Set revenue at efficiency E equal to the total cost

12,000(E)(250) = 15,000,000(A/P,1%, 20) + (4,100,000)E1.83,000,000(E) = 15,000,000(0.01435) + (4,100,000)E1.8
3,000,000(E) – 4,100,000E1.8 = 215,250

Solve for E by trial and error:
at E = 0.55: 252,227 > 215,250
at E = 0.57: 219,409 > 215,250
at E = 0.58: 202,007 < 215,250

E = 0.572 or 57.2% minimum removal efficiency

13.11 A civil engineer has been promoted to Manager of engineered public systems.One of the products is an emergency intercept pump for potable water. If the tested water quality or volume varies by a preset percentage, the pump automatically switches to preselected options of treatments or water sources. The manufacturing process for the pump had the following fixed and variable costs over a 1-year period.

(a) Determine the minimum revenue per unit to break even at thecurrent production volume of 5000 units per year.
(b) If selling internationally and to large corporations is pursued, an increased production of 3000 additional units will be necessary. Determine the revenue per unit required if a profit goal of \$500,000 is set for the entire product line. Assume the cost estimates above remain the same.

FC = \$305,000 v = \$5500/unit

(a) Profit = (r – v)Q –FC
0 = (r – 5500)5000 – 305,000
(r – 5500) = 305,000 / 5000
r = 61 + 5500 = \$5561 per unit

(b) Profit = (r – v)Q – FC
500,000 = (r – 5500)8000 – 305,000
(r – 5500) = (500,000 + 305,000) / 8000
r = \$5601 per unit
BREAKEVEN ANALYSIS BETWEEN ALTERNATIVES

13.12 A Yellow Pages directory company must decide whether it should compose the adsfor its clients inhouse or pay a production company to compose them. To develop the ads inhouse, the company will have to purchase computers, printers, and other peripherals at a cost of \$12,000.
The equipment will have a useful life of 3 years, after which it will be sold for \$2000. The employee who creates the ads will be paid \$45,000 per year. In addition, each ad will have an average cost of\$8 to prepare for delivery to the printer. A total of 4000 ads are anticipated for the next few years. Alternatively, the company can outsource ad development at a fee of \$20 per ad regardless of the quantity. The current interest rate is 8% per year. What is the breakeven amount, and which alternative is economically better?

Let x = ads per year

-12,000(A/P,8%,3) – 45,000 + 2000(A/F,8%,3)–8x = -20x
-12,000(0.38803) – 45,000 + 2000(0.30803) = -12x
-49,040 = -12x

x = 4087 ads per year

At 4000 ads per year, select the outsource option at \$20 per ad for a total cost of \$80,000 versus the inhouse option cost of \$49,040 +8(4000) = \$81,040.

13.13 An engineering firm can lease a measurement system for \$1000 per month or purchase one for \$15,000. The leased systemwill have no monthly maintenance cost, but the purchased one will cost \$80 per month. At an interest rate of 0.5% per month, how many months must the system be required to break even?

Let n = number of months
-15,000(A/P, 0.5%, n) – 80 = -1000
-15,000(A/P, 0.5%, n) = -920
(A/P, 0.5%, n) = 0.0613

n is approximately 17 months

13.14 Two pumps can be used for...