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HOW TO USE THE RECIPROCAL ALLOCATION METHOD By Paulino Silva

Reciprocal Allocation Method

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HOW TO USE THE RECIPROCAL ALLOCATION METHOD

By Paulino Silva

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:

QUESTION: Present the two-stage cost allocation process.

Example:

1st STEP: Elimination of reciprocal charges.

Proposed Solution:

x: Total Costs of Service Department X y: Total Costs of Service Department Y

Proposed Solution:

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

2nd STEP: Two-stage cost allocation process.

Proposed Solution:

Description

Production Departments Service Departments

Total A B X Y

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

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:

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Acknowledgements: