Question

Ratios Driver Wafer A Wafer B Activity 7. Inserting dies Quantity 1,800,000 Number of dies 0.7 0.3 Number of purchase 8. Purchasing materials 2,000 0.2 0.8 orders 0.25 0.75 Engineering hours Test hours 10,000 17,500 0.6 0.4 1. Developing test programs 3. Testing products ABC assignments Total overhead cost $145,000 $145,000 $290,000 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 II Wafer B

Answer 1

Solution

1) First calculate Global ratios for:

Wafer A = 0.50 ($145,000/$290,000)

Wafer B = 0.50 ($145,000/$290,000)

Now Equations are as follows:

Wafer A = 0.70 w ₁ + 0.20 w ₂ = 0.50

Wafer B = 0.30 w ₁ + 0.80 w ₂ = 0.50

Multiplying both sides of the first equation by 4, subtracting the second from the first, and solving, we obtain:

w ₁ = 0.60 and w ₂ = 0.40

Inserting dies cost pool: 0.6 × $290,000 = $174,000

Purchasing cost pool: 0.4 × $290,000 = $116,000

2. By using the consumption ratios, the same ABC cost assignment is realized with two drivers:

Wafer A: (0.70 × $ 174,000) + (0.20 × $ 116,000) = $ 147,000 + $ 28,000 = $ 145,000

Wafer B: (0.30 × $174,000 + (0.80 × $116,000) = $ 63,000 + $ 112,000 = $ 145,000

3. First calculate Global ratios for:

Wafer A = 0.50 ($145,000/$290,000)

Wafer B = 0.50 ($145,000/$290,000)

Equations:

Wafer A = 0.25 w ₁ + 0.60 w ₂ = 0.50 ..............(1)

Wafer B = 0.75 w ₁ + 0.40 w ₂ = 0.50 ..............(2)

Multiplying both sides of the first equation by 3, subtracting the second from the first, and solving, we obtain:

Wafer A = 0.25 x 3 w1 + 0.60 x 3 w2 = 0.50 x 3

Wafer A = 0.75 w1 + 1.80 w2 = 1.50

Less:Wafer B = 0.75 w1 + 0.40 w2 = 0.50

Net 0.00 w1 + 1.40 w2 = 1.00

By solving we get : w ₂ = 1/1.40 = 0.714

w1 = 1-0.714 = 0.286

Developing test programs cost pool = 0.286 × $290,000 = $82,857

Testing products cost pool = 0.714 × $290,000 = $207,143

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

Wafer A: (0.25 × $ 82,857) + (0.60 × $207,143) = $ 20,714+ $ 124,286= $ 145,000

Wafer B: (0.75 × $ 82,857) + (0.40 × $2,07,143) = $ 62,143 + $ 82,857 = $ 145,000