**Note 1** E’s costs are apportioned directly as no reciprocal service is involved.

Note 2 It doesn’t really matter which of the two remaining cost centres you start with.

Note 3 On the last reapportionment, D’s overheads are apportioned on the basis of 75/95 to A and 20/95 to B. The reciprocal service to C is ignored as, by now, it is not material.

Algebraic approach

Firstly, we can setup the overhead re-apportionment process as a set of equations.

Let:

A = the total overhead $ apportioned to department A

B = the total overhead $ apportioned to department B, etc

Then:

A = 80,000 + 0.40 C + 0.75 D + 0.30 E

B = 100,000 + 0.50 C + 0.20 D + 0.70 E

C = 10,000 + 0.05 D

D = 20,000 + 0.10 C

E = 4,000

If you remember your school maths, you will note that the equations for C and D are simultaneous – ie C is a function of D, and D is a function of C. These two equations must be solved first. Various approaches are possible to solve simultaneous equations but substitution is probably quickest.

Substituting the D equation into the C equation:

C = 10,000 + 0.05 (20,000 + 0.10 C)

Multiplying out the bracket:

C = 10,000 + 1000 + 0.005 C

Collecting terms:

0.995 C = 11,000

C = 11,055.3

Substituting into the D equation:

D = 20,000 + 0.10 × 11,055.3

D = 21,105.5

Finally, plugging these values into the equations for A and B, the total overhead apportioned to each of the production cost centres is:

A = 80,000 + 0.40 × 11,055.3 + 0.75 × 21,105.5 + 0.3 × 4,000

A = 101,451.2

B = 100,000 + 0.50 × 11,055.3 + 0.20 × 21,105.5 + 0.7 × 4,000

B = 112, 548.8

These results, as they should be, are quite close to the repeated distribution approach.

Test your understanding

The following question is representative of questions on this topic that you might experience in the Paper F2 exam.

A company has two production cost centres (V and W) and two service cost centres (X and Y). The following overheads have been apportioned and allocated to the four cost centres.