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Reciprocal Allocation Method
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Transcript of Reciprocal Allocation Method
Introduction:
As part of management accounting, the use of departmental cost allocation method for assigning indirect costs, despite its old origins,
it is still widely used today.
Introduction:
One of the problems in using departmental cost allocation method, is the existence of reciprocal charges between service departments.
Through a practical example we will see how to solve this problem.
Departments
Labour Expenses Supplies Depreciation
A 14.000,00 4.800,00 7.600,00
B 10.625,00 3.125,00 5.375,00
X 5.125,00 5.875,00 4.000,00
Y 4.225,00 1.825,00 4.495,00
Example:
A company has four departments, two production departments (A and B) and two service departments (X and Y). The overhead analysis sheet provides the following costs:
Source Destination
A B X Y
Department X 45% 20% -- 35%
Department Y 42% 50% 8% --
Example:
Apart from the cost information, we have also the charge percentage of service departments:
x = 15.000+ 0,08yy = 10.545+ 0,35x
⎧⎨⎩
Proposed Solution:
x: Total Costs of Service Department X y: Total Costs of Service Department Y
x = 15.000+ 0,08yy = 10.545+ 0,35x
⎧⎨⎩
⇔−y = 10.545+ 0,35× 15.000+ 0,8y( )
⎧⎨⎪
⎩⎪
Proposed Solution:
x: Total Costs of Service Department X y: Total Costs of Service Department Y
x = 15.000+ 0,08yy = 10.545+ 0,35x
⎧⎨⎩
10.545 5.250 0,028y y−⎧
⇔ ⎨ = + +⎩
⇔
−y = 10.545+ 0,35× 15.000+ 0,8y( )
⎧⎨⎪
⎩⎪
Proposed Solution:
x: Total Costs of Service Department X y: Total Costs of Service Department Y
x = 15.000+ 0,08yy = 10.545+ 0,35x
⎧⎨⎩
10.545 5.250 0,028y y−⎧
⇔ ⎨ = + +⎩ ⇔
−0,972y = 15.795⎧⎨⎩
⇔
−y = 10.545+ 0,35× 15.000+ 0,8y( )
⎧⎨⎪
⎩⎪
Proposed Solution:
x: Total Costs of Service Department X y: Total Costs of Service Department Y
x = 15.000+ 0,08yy = 10.545+ 0,35x
⎧⎨⎩
10.545 5.250 0,028y y−⎧
⇔ ⎨ = + +⎩ ⇔
−0,972y = 15.795⎧⎨⎩
⇔
−y = 10.545+ 0,35× 15.000+ 0,8y( )
⎧⎨⎪
⎩⎪
⇔
−y = 16.250
⎧⎨⎩
Proposed Solution:
x: Total Costs of Service Department X y: Total Costs of Service Department Y
x = 15.000+ 0,08yy = 10.545+ 0,35x
⎧⎨⎩
10.545 5.250 0,028y y−⎧
⇔ ⎨ = + +⎩ ⇔
−0,972y = 15.795⎧⎨⎩
⇔
x = 15.000+ 0,08×16.250y = 16.250
⎧⎨⎩
⇔
−y = 10.545+ 0,35× 15.000+ 0,8y( )
⎧⎨⎪
⎩⎪
⇔
−y = 16.250
⎧⎨⎩
Proposed Solution:
x: Total Costs of Service Department X y: Total Costs of Service Department Y
x = 15.000+ 0,08yy = 10.545+ 0,35x
⎧⎨⎩
10.545 5.250 0,028y y−⎧
⇔ ⎨ = + +⎩ ⇔
−0,972y = 15.795⎧⎨⎩
⇔
x = 15.000+ 0,08×16.250y = 16.250
⎧⎨⎩
⇔
−y = 10.545+ 0,35× 15.000+ 0,8y( )
⎧⎨⎪
⎩⎪
⇔
x = 16.300y = 16.250
⎧⎨⎩
⇔−y = 16.250
⎧⎨⎩
Proposed Solution:
x: Total Costs of Service Department X y: Total Costs of Service Department Y
Description
Production Departments Service Departments
Total A B X Y
1st-Step Cost Allocation 26.400,00 19.125,00 15.000,00 10.545,00 71.070,00
Proposed Solution:
Description
Production Departments Service Departments
Total A B X Y
1st-Step Cost Allocation 26.400,00 19.125,00 15.000,00 10.545,00 71.070,00
Department X (1) 7.335,00 (3) 3.260,00 (16.300,00) (5) 5.705,00
Department Y (2) 6.825,00 (4) 8.125,00 (6) 1.300,00 (16.250,00)
( )( )( )
1 0,45 16.300
2 0,42 16.250
3 0,20 16.300
×
×
×
4( )0,50×16.250
5( )0,35×16.300
6( )0,08×16.250
Proposed Solution:
Description
Production Departments Service Departments
Total A B X Y
1st-Step Cost Allocation 26.400,00 19.125,00 15.000,00 10.545,00 71.070,00
Department X (1) 7.335,00 (3) 3.260,00 (16.300,00) (5) 5.705,00
Department Y (2) 6.825,00 (4) 8.125,00 (6) 1.300,00 (16.250,00)
2nd-Step Cost Allocation 40.560,00 30.510,00 0 0 71.070,00
( )( )( )
1 0,45 16.300
2 0,42 16.250
3 0,20 16.300
×
×
×
4( )0,50×16.250
5( )0,35×16.300
6( )0,08×16.250
Proposed Solution: