Decision Inputs (Data) Cost Structure Alternative #1 Cost Structure Alternative
ID: 2378548 • Letter: D
Question
Decision Inputs (Data)
Cost Structure Alternative #1
Cost Structure Alternative #2
Delivery price (i.e., revenue) per package
$60
$60
Variable cost per package delivered
$48
$30
Contribution margin per unit
$12
$30
Fixed costs (per year)
$600,000
$3,000,000
(1) Assume that for the coming year total fixed costs are expected to increase by 10% for each of the two alternatives. What is the new break-even point, in terms of number of deliveries, for each decision alternative? By what percentage did the break-even point change for each case? How do these figures compare to the percentage increase in budgeted fixed costs?
(2) Assume an average income-tax rate of 40%. What volume (number of deliveries) would be needed to generate an after-tax profit, ?A, of 5% of sales for each alternative?
(3) Consider the original data in the problem. Construct a graph for each of the two alternatives depicting pre-tax profit, ?B, as function of volume (number of deliveries per year). Clearly label the profit equation for each alternative.
(4) Based on the graphs prepared in (9), which decision alternative do you think is the more profitable one for this business?
(5) Based on the original data and the graphs prepared above in (10), which decision alternative is more risky to the business? Explain. (Hint: Think about, and define in your answer, the notion of
Decision Inputs (Data)
Cost Structure Alternative #1
Cost Structure Alternative #2
Delivery price (i.e., revenue) per package
$60
$60
Variable cost per package delivered
$48
$30
Contribution margin per unit
$12
$30
Fixed costs (per year)
$600,000
$3,000,000
Explanation / Answer
let units = x alt 1 alt 2 SALES 60x 60x Variable cost 48x 30x Contribution margin 12x 30x Fixed costs (per year) 600000 3000000 net income 12x-600000 30x-3000000 tax @ 40% (12x-600000)0.4 (30x-3000000)0.4 after tax profit (12x-600000)0.6 (30x-3000000)0.6 after tax profit = 5% of sales (60x) 0.05 (60x) 0.05 equating these two equations (12x-600000)0.6 = (60x) 0.05 (30x-3000000)0.6 = (60x) 0.05 x = 85715 x = 120000
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