In: Accounting
Mushed’s String Company has two service departments, Maintenance and Personnel. Maintenance Department costs of $320,000 are allocated on the basis of budgeted maintenance-hours. Personnel Department costs of $80,000 are allocated based on the number of employees. The costs of operating departments A and B are $160,000 and $240,000, respectively. Data on budgeted maintenance-hours and number of employees are as follows: | |||||
Support Departments | Production Departments | ||||
Maintenance | Personnel | A | B | ||
Budgeted costs | $320,000 | $80,000 | $160,000 | $240,000 | |
Budgeted maintenance-hours | NA | 200 | 240 | 160 | 600 |
Number of employees | 10 | NA | 40 | 120 | 170 |
Required: | |||||
Allocate the service department costs to the production departments using | |||||
1. The Direct Allocation Method | |||||
2. The Step Down Method | |||||
3. The Reciprocal Method | |||||
Solution:
It is given that:
Budgeted costs | $ 160,000 | $ 240,000 | $ 320,000 | $ 80,000 | |
A | B | Maintenance | Personnel | Total | |
Maintenance | 240 | 160 | 0 | 200 | 600 |
Personnel | 40 | 120 | 10 | 0 | 170 |
Answer 1:
Direct Allocation Method:
A | B | Maintenance | Personnel | |
Operating costs | $ 160,000 | $ 240,000 | $ 320,000 | $ 80,000 |
Maintenance | $ 192,000 | $ 128,000 | $ (320,000) | |
Personnel | $ 20,000 | $ 60,000 | $(80,000) | |
Total | $ 372,000 | $ 428,000 | $ - | $ - |
Answer 2:
Step Down Method:
A | B | Maintenance | Personnel | |
Operating costs | $ 160,000 | $ 240,000 | $ 320,000 | $ 80,000 |
Maintenance | $ 128,000 | $ 85,333 | $ (320,000) | $ 106,667 |
Total | $ 288,000 | $ 325,333 | $ - | $ 186,667 |
Personnel | $ 46,667 | $ 140,000 | $ - | $(186,667) |
Total | $ 334,667 | $ 465,333 | $ - | $ - |
Answer 3:
Reciprocal Method:
Let the total cost of | Maintenance | be x. |
Let the total cost of | Personnel | be y. |
Now,
So, x = 320000 + 0.059y | ---------Equation 1 |
y = 80000 + 0.33x | ---------Equation 2 |
and,
Solving Equation 1 and 2: |
Substitute value of 'y' from equation 2 in Equation 1- |
So, x = 320000 + 0.059 (80000 + 0.33x) |
So, x = $ 331200 |
The total cost of Maintenance is $ 331200 |
and,
Putting the value of x in equation 2- |
So, y = 80000 + 0.33 * 331200 |
So, y = $ 190400 |
The total cost of Personnel is $ 190400 |
and,
A | B | Maintenance | Personnel | |
Total cost | $ 160,000.00 | $ 240,000.00 | $ 320,000.00 | $ 80,000.00 |
Maintenance | $ 132,480.00 | $ 88,320.00 | $(331,200.00) | $ 110,400.00 |
Total cost | $ 292,480.00 | $ 328,320.00 | $ (11,200.00) | $ 190,400.00 |
Personnel | $ 44,800.00 | $ 134,400.00 | $ 11,200.00 | $(190,400.00) |
Total cost | $337,280.00 | $462,720.00 | $ (0.00) | $ - |
In case of any doubt, please feel free to comment.