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IrinaVladis [17]
4 years ago
13

Two currently owned machines are being considered for the production of a part. The capital investment associated with the machi

nes is about the same and can be ignored for purposes of this example. The important differences between the machines are their production capacities (production rate x available production hours) and their reject rates (percentage of parts produced that cannot be sold). Consider the following table: Machine A Machine B Production rate Hours available for production Percent parts rejected 100 parts/hour 7 hours/day 3% 130 parts/hour 6 hours/day 10% The material cost is $6.00 per part, and all defect-free parts produced can be sold for $12 each. (Rejected parts have negligible scrap value.) For either machine, the operator cost is $15.00 per hour and the variable overhead rate for traceable costs is $5.00 per hour. Assume that the daily demand for this part is large enough that all defect-free parts can be sold. Which machine should be selected?
Business
1 answer:
andrew11 [14]4 years ago
7 0

Answer:

Machine A's output in a day = 100 × 7 = 700

Rejected output of machine A =0.03 × 700 = 21

Thus defect free output of machine A = 700 – 21 = 679

Revenue from Using Machine A = 679 × 12 = 8148

Cost of Using Machine A = 6 × (Defect free output) + 15 × 7 + 5 × 7 = 6 × 679 + 105 + 35 = 4214

Thus, profit from using machine A = 8148 – 4214 = 3934

Machine B's output in a day =130 × 6 = 780

Rejected output of machine B = 0.1 × 780 = 78

Thus, defect free output of machine B = 780 – 78 = 702

Revenue from Using Machine B = 702 × 12 = 8424

Cost of Using Machine B = 6 × 702 + 15 × 6 + 5 × 6 = + 105 + 35 = 4212 + 120 = 4332

Thus profit from using machine B = 8424 – 4332 = 4029

Since the profit from using machine B is higher, Machine B should be selected.

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Amortization schedule is attached.

Explanation:

Key matrix

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5 0
3 years ago
Which of the following statements is false?
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6 0
3 years ago
Suppose that when the price of a certain commodity is p dollars per unit, then x hundred units will be purchased by consumers, w
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= x (P) - C(x)
= x(-0.05x+38) - (0.02x^{2} + 3x + 574.77)
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This profit equation is an equation of a parabola that opens downward (Since A=-0.07<0) and has its vertex at

x= -\frac{B}{2A}  = -\frac{35}{2 (-0.07)}  = 250

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P(250)=3800.23 hundred dollars

b. Profits are maximised at x=250 hundred units. The per unit price at this is,

p= -0.05x + 38&#10;= -0.05 (250) + 38&#10;= $25.5


7 0
3 years ago
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