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FinalPrepSolutions

# FinalPrepSolutions - FINALPREPSOLUTIONS 36 1 Scattergraph...

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FINALPREPSOLUTIONS 3–6 1. Yes, there appears to be a linear relationship. 2. Low: 700, \$2,628 High: 3,100, \$6,564 V = ( Y 2 Y 1 )/( X 2 X 1 ) = (\$6,564 – \$2,628)/(3,100 – 700) = \$3,936/2,400 = \$1.64 per visit F = \$6,564 – \$1.64(3,100) = \$1,480 OR F = \$2,628 – \$1.64(700) = \$1,480 Y = \$1,480 + \$1.64 X 3. Y = \$1,480 + \$1.64(1,900) = \$1,480 + \$3,116 = \$4,596 1 Scattergraph for Tanning Services 0 1,00 0 2,00 0 3,000 4,00 0 5,00 0 6,000 \$7,000 0 1,00 0 2,00 0 3,00 0 4,00 0 Number of tanning visits Cos t

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11–3 1. Supplier cost: First, calculate the activity rates for assigning costs to suppliers: Inspecting components: \$1,200,000/1,000 = \$1,200 per sampling hour Expediting work: \$960,000/100 = \$9,600 per order Reworking products: \$6,844,500/1,500 = \$4,563 per rework hour Warranty work: \$21,600,000/4,000 = \$5,400 per warranty hour Next, calculate the cost per component by supplier: Supplier cost: Grayson Lambert Purchase cost: \$144 × 200,000 ...................... \$28,800,000 \$129 × 800,000 ...................... \$103,200,000 Inspecting components: \$1,200 × 20 ............................ 24,000 \$1,200 × 980 .......................... 1,176,000 Expediting work: \$9,600 × 10 ............................ 96,000 \$9,600 × 90 ............................ 864,000 Reworking products: \$4,563 × 90 ............................ 410,670 \$4,563 × 1,410 ....................... 6,433,830 Warranty work: \$5,400 × 200 .......................... 1,080,000 \$5,400 × 3,800 ....................... 20,520,000 Total supplier cost .................... \$30,410,670 \$132,193,830 Units supplied ........................... ÷ 200,000 ÷ 800,000 Unit cost ................................ \$ 152.05 * \$ 165.24 * *Rounded to the nearest cent. The difference favors Grayson; furthermore, when the price concession is considered (\$135 – \$144), the cost of Grayson is \$143.05, which is much less than the Lambert component. Zavner should give serious consideration to accepting the contractual offer made by Grayson. The savings are in the mil- lions. 2
11–3 Concluded 2. To assign the lost sales cost, it would be helpful to know the number of de- fective units using the Grayson component versus those using the Lambert component. Warranty hours would act as a very good substitute driver. Us- ing this driver, the rate is \$4,500,000/4,000 = \$1,125 per warranty hour. The cost assigned to each component would be: Grayson Lambert Lost sales: \$1,125 × 200 ............ \$225,000 \$1,125 × 3,800 ......... \$4,275,000 This increases the cost of the Lambert component by \$4,275,000/800,000 = \$5.34 *. *Rounded. 11–4 1. Sales revenue = \$0.75 × 10,000,000 = \$7,500,000 for each customer type. ( Note : The total number of parts is the average order size times the number of sales orders.) Thus, the total customer-related activity costs are split equally: Cost allocation = 0.50 × \$5,900,000 = \$2,950,000 The profitability of each category is calculated as follows: Sales revenue ............................................................................. \$7,500,000 Less: Noncustomer-related cost (\$0.40 × 10,000,000) ........... 4,000,000 Less: Customer-related activity costs ..................................... 2,950,000 Customer profitability .......................................................... \$ 550,000 This profitability measure is suspect because the customer-related costs are assigned using revenues, a driver that is not causally related to the custom- er-related activity costs. This approach may actually have one set of custom- ers subsidizing the other.

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FinalPrepSolutions - FINALPREPSOLUTIONS 36 1 Scattergraph...

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