In: Accounting
Barstow Manufacturing Company has two service departments — product design and engineering support, and two production departments — assembly and finishing. The distribution of each service department's efforts to the other departments is shown below:
FROM | TO | ||||||||||
Design | Support | Assembly | Finishing | ||||||||
Design | 0 | % | 10 | % | 30 | % | 60 | % | |||
Support | 20 | % | 0 | % | 45 | % | 35 | % | |||
The direct operating costs of the departments (including both variable and fixed costs) were as follows:
Design | $ | 140,000 |
Engineering Support | $ | 160,000 |
Assembly | $ | 550,000 |
Finishing | $ | 840,000 |
The total cost accumulated in the finishing department using the reciprocal method is(calculate all ratios and percentages to 4 decimal places, for example 33.3333%, and round all dollar amounts to the nearest whole dollar):
Multiple Choice
$1,007,449.
$627,143.
$1,890,000.
$682,551.
$1,062,857.
Answer)
Calculation of total cost of finishing department using Reciprocal Method
Departmental overheads before allocation |
Producing Departments |
Service Departments |
Total |
||
Assembling |
Finishing |
Product Design |
Engineering Support |
||
$550,000 |
$840,000 |
$140,000 |
$160,000 |
$1,690,000 |
|
Allocation of overheads: |
|||||
Product Design (30%, 60%, 10%) |
52653 |
105306 |
-175510 |
17551 |
0 |
Engineering Support (45%, 35%, 20%) |
79898 |
62143 |
35510 |
-177551 |
0 |
Total departmental overhead |
$682,551 |
$1,007,449 |
$0 |
$0 |
$1,690,000 |
Therefore, total cost allocated to finishing department is $ 1,007,449.
Working Note:
Department |
Departmental overhead Before distribution |
Services Provided |
|
Product design |
Engineering Support |
||
Product Design |
$140,000 |
0% |
20% |
Engineering Support |
$160,000 |
10% |
0% |
Assembling |
$550,000 |
30% |
45% |
Finishing |
$840,000 |
60% |
35% |
Total Department overhead |
$1,690,000 |
100% |
100% |
Let the total cost of Product design department be “x” and that of Engineering support department be “y”.
Equations:
x = $ 140,000 + 0.20 y ….. Equation 1
y = $ 160,000 + 0.10x ….. Equation 2
Substituting the value of y in equation 1 we get
x = $ 140,000 + 0.20 ($ 160,000 + 0.10x)
x = $ 140,000 + $ 32,000 + 0.02x
x – 0.02x = $ 172,000
x = $ 175,510 (approximately)
Substituting the value of x in Equation 2 we get,
y = $ 160,000 + 0.10 X $175,510
y = $ 177,551