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 0 1,00 2,00 3,000 4,00 5,00 6,000 $7,000 0 1,00 2,00 3,00 4,00 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
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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
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This note was uploaded on 02/08/2010 for the course ACTG 3000 taught by Professor C during the Spring '10 term at Oregon State.

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FinalPrepSolutions - FINALPREPSOLUTIONS 36 1. Scattergraph...

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