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Techniques & Controversial Issues

Chapter 6 Problem Solutions

James R. Martin, Ph.D., CMA

Professor Emeritus, University of South Florida

**SOLUTION 6-2**

1. Direct method:

P1 = 500,000 + (280/390)(S1) + (1,000/1,700)(S2)

= 500,000 + (280/390)(100,000) + (1,000/1,700)(200,000)

= 689,441.93

P2 = 300,000 + (110/390)(S1) + (700/1,700)(S2)

= 300,000 + (110/390)(100,000) + (700/1,700)(200,000)

= 410,558.07

Note: 689,441.93 + 410,558.07 = $1,100,000 total direct costs.

2. Step-down method:

S2 is closed first, therefore S2 = 200,000

S1 = 100,000 + (100/1,800)(S2)

= 100,000 + (100/1,800)(200,000) = 111,111.11

P1 = 500,000 + (280/390)(S1) + (1,000/1,800)(S2)

= 500,000 + (280/390)(111,111.11) + (1,000/1,800)(200,000)

= 690,883.19

P2 = 300,000 + (110/390)(S1) + (700/1,800)(S2)

= 300,000 + (110/390)(111,111.11) + (700/1,800)(200,000)

= 409,116.81

Note: 690,883.19 + 409,116.81 = $1,100,000

3. Reciprocal method:

S1 = 100,000 + (100/2,000)(S2)

S2 = 200,000 + (10/400)(S1) + (200/2,000)(S2)

P1 = 500,000 + (280/400)(S1) + (1,000/2,000)(S2)

P2 = 300,000 + (110/400)(S1) + (700/2,000)(S2)

Substitute S1 into the equation for S2 to solve for S2.

S2 = 200,000 + (10/400)(S1) + (200/2,000)(S2)

S2 = 200,000 + (10/400)[100,000 + (100/2,000)(S2)] + (200/2,000)(S2)

= 200,000 + 2,500 + (.00125)(S2) + (.1)(S2)

= 225,312.94

Then solve for S1, P1 and P2.

S1 = 100,000 + (100/2,000)(225,312.94)

= 111,265.65

P1 = 500,000 + (280/400)(111,265.65) + (1,000/2,000)(225,312.94)

= 690,542.43

P2 = 300,000 + (110/400)(111,265.65) + (700/2,000)(225,312.94)

= 409,457.58

Note: 690,542.43 + 409,457.58 = 1,100,000

**SOLUTION 6-3**

1. Direct method:

P1 = 6,000,000 + (1,000/1,500)(S1) + (800/1,200)(S2)

= 6,000,000 + (1,000/1,500)(500,000) + (800/1,200)(1,000,000)

= 7,000,000

P2 = 2,000,000 + (500/1,500)(S1) + (400/1,200)(S2)

= 2,000,000 + (500/1,500)(500,000) + (400/1,200)(1,000,000)

= 2,500,000

Note: 7,000,000 + 2,500,000 = $9,500,000 total direct costs.

2. Step-down method:

S2 is closed first, therefore S2 = 1,000,000

S1 = 500,000 + (1,000/2,200)(S2)

= 500,000 + (1,000/2,200)(1,000,000)

= 954,545.45

P1 = 6,000,000 + (1,000/1,500)(S1) + (800/2,200)(S2)

= 6,000,000 + (1,000/1,500)(954,545.45) + (800/2,200)(1,000,000)

= 7,000,000

P2 = 2,000,000 + (500/1,500)(S1) + (400/2,200)(S2)

= 2,000,000 + (500/1,500)(954,545.45) + (400/2,200)(1,000,000)

= 2,500,000

Note: 7,000,000 + 2,500,000 = 9,500,000

3. Reciprocal method:

S1 = 500,000 + (80/1,600)(S1) + (1,000/2,400)(S2)

S2 = 1,000,000 + (20/1,600)(S1) + (200/2,400)(S2)

P1 = 6,000,000 + (1,000/1,600)(S1) + (800/2,400)(S2)

P2 = 2,000,000 + (500/1,600)(S1) + (400/2,400)(S2)

Computer recommended for solution.

S1 = 1,010,830

S2 = 1,104,693

P1 = 7,000,000

P2 = 2,500,000

**SOLUTION 6-4**

1. b

2. c

3. d. B&O is closed first because it provides 9.8% (880/9,000) of itsâ€™ service to P&M while P&M provides only 1% (10/1,000) of it's service to B&O.

4. a since B&O receives no allocation from P&M,

5. c

6. d

7. b

8. a

9. b

10. c

11. c because X1 and X2 consume the same proportion of machine hours (20% and 80% respectively) in each department.

12. d

**SOLUTION 6-5**

1. a

2. c

3. a

4. c

5. d

6. d

7. d

8. b

**SOLUTION 6-6**

1. d

2. b

3. a

4. c

5. e Use materials as an allocation basis in P1 and machine hours in P2.