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Simplifying the ABC System: Equally Accurate Reduced ABC Systems

Selected activities and other information are provided for Patterson Company for its most recent year of operations.

						Expected Consumption Ratios
Activity		Driver		Quantity		Wafer A		Wafer B
7.   Inserting dies		Number of dies		2,000,000		0.7		0.3
8.  Purchasing materials		Number of purchase    orders		1,900		0.2		0.8
1.  Developing test programs		Engineering hours		11,000		0.25		0.75
3.  Testing products		Test hours		18,000		0.6		0.4
ABC assignments						$150,000		$150,000
Total overhead cost								$300,000

Required:

1. Form reduced system cost pools for activities 7 and 8. Do not round interim calculations. Round your final answers to the nearest dollar.

Inserting dies cost pool	$
Purchasing cost pool	$

2. Assign the costs of the reduced system cost pools to Wafer A and Wafer B. Do not round interim calculations. Round your final answers to the nearest dollar.

Wafer A	$
Wafer B	$

3. What if the two activities were 1 and 3? Repeat Requirements 1 and 2.

Form reduced system cost pools for activities 1 and 3.

Do not round interim calculations. Round your final answers to the nearest dollar.

Developing test programs cost pool	$
Testing products cost pool	$

Assign the costs of the reduced system cost pools to Wafer A and Wafer B.

Wafer A	$
Wafer B	$

Answer 1

SOLUTION

1. Inserting dies = 0.7 - 0.3 = 0.4

Purchasing materials = 0.2 - 0.8 = 0.6

Overhead costs should be assigned in 0.6:0.4

Inserting dies = $300,000 * 0.6 / 1 = $180,000

Purchasing materials = $300,000 * 0.4 / 1 = $120,000

2. Cost assigned to Wafer A

Inserting dies ($180,000*0.7)	126,000
Purchasing materials ($120,000*0.2)	24,000
Total Cost assigned to Wafer A	150,000

Cost assigned to Wafer B

Inserting dies ($180,000*0.3)	54,000
Purchasing materials ($120,000*0.8)	96,000
Total Cost assigned to Wafer B	150,000

3. Inserting dies = 0.25 - 0.75 = 0.50

Purchasing materials = 0.6 - 0.4 = 0.20

Overhead costs should be assigned in 0.2/0.7 : 0.5/0.7

Inserting dies = $300,000 * 0.2 / 0.7 = $85,714

Purchasing materials = $300,000 * 0.5 / 0.7 = $214,286

Cost assigned to Wafer A

Inserting dies ($85,714*0.25)	21,428
Purchasing materials ($214,286*06)	128,572
Total Cost assigned to Wafer A	150,000

Cost assigned to Wafer B

Inserting dies ($85,714*0.75)	64,286
Purchasing materials ($214,286*0.4)	85,714
Total Cost assigned to Wafer B	150,000