Question

In: Accounting

Barstow Manufacturing Company has two service departments — product design and engineering support, and two production...

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.

Solutions

Expert Solution

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

  


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