Cambrige AS and A Level Accounting Notes (9706)/ ZIMSEC Advanced Accounting Level Notes: Reapportionment of service costs where there is interwork
 This is a solution to the question found here
 First we will use the repeated distribution method to redistribute service costs
 Lastly we will use the algebraic method to reapportion costs
 We have already looked at why we need to reapportion service costs among production departments here
Repeated Distribution Method
Overheads  Sewing  Finishing  Stores  Maintenance 
Already allocated  70 000  30 000  20 000  15 000 
Recharge stores  10 000  6 000  (20 000)  4 000 
  19 000  
Recharge maintenance  8 550  7 600  2 850  (19 000) 
2 850    
Recharge stores  1 425  855  (2 850)  570 
  570  
Recharge maintenance  257  228  85  (570) 
85    
Recharge stores  43  25  (85)  17 
  17  
Recharge stores  8  7  2  (17) 
2    
Recharge stores  1  1  (2)  
90 284  44 716     

 As shown the technique is simple you need to pick one service department and reapportion all its overheads to the other three departments in our case we first picked stores
 Then you pick the other department (in this case maintenance) and reapportion all its overheads to the other three departments
 The second step means there are now overheads costs in the stores department due to its share of the maintenance costs
 This cost has to be reapportioned again
 This process is repeated until there are immaterial costs in all service departments which can be reapportioned among the two production departments without the need to reapportion a share into the service departments
 In our case there remains $2 in the end which we split equally among the two departments (a result of rounding off $0.6
 We could have continued with the reapportioning but the benefits from continuing would not have been much
 You will have to use your discretion to determine when to stop reapportioning but its something you can do when amounts reach single digits
The Algebraic Method
 This involves the use of simultaneous equations to solve the problem of having to reapportion costs
 From our question we know that the total overheads in the stores department ought to be $20 000 plus 15% of the maintenance costs
 However we don’t know what the actual final for total overheads will be since they will also have to include a 20% share of stores cost which have already said ought to include a share of the maintenance costs
 From the above we have two unkowns i.e. the actual maintenance costs m and the actual stores costs s
 We can thus formulate two equations:
 s=20000+0.15m
 m=15000+02s
 Now there a number of ways to solve simultaneous equations but here we will use the substitution method
 Replace M in equation 1 to get:
 s=20000+2250+0.03s
 This becomes:
 0.97s=22250
 Which after we divide both sides by 0.97 gives:
 \text{s=\$22 938}
 Now that we know the value of s things get a little easier all you have to do is substitute the value of s into the second equation above:
 m=15000+0.2(22938)
 Solving this gives:
 \text{m=\$19 588}
 What all this means is that the total stores overheads including a share of maintenance overheads is $19 588
 It also means the total maintenance overheads including the share of stores overheads is $22 938
 Now we can reapportion these costs easily as shown below:
Overheads  Sewing  Finishing  Stores  Maintenance 
Already Allocated  70 000  30 000  20 000  15 000 
Recharge stores  11 469  6 881  (22938)  4 588 
Recharge maintenance  8 815  7 835  2 938  (19 588) 
Totals  90 284  44 716     
 Despite all appearances the algebraic is much faster and accurate that the repeated reapportionment method
 Wherever possible always use the algebraic method unless otherwise directed by the exam question
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