1. Prof. Dr. Olga Popova, OVGU Prof. Dr. Jörg Jablinski, OWL
Lecture
„Industrial Costing“
SS 2016/2017
Prof. Dr. Olga Popova, OVGU
Prof. Dr. Jörg Jablinski, OWL
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2. „Cost center accounting“ (Part 2)
Chapter 5
„Cost center accounting“ (Part 2)

3. Structure of the cost allocation sheet
with the direct method

4. Example: Internal cost allocation with direct method
Sum
Service cost center
Main cost center R E M P1 P2
Admin.
Distrib.
Reparation (hours)
1.600 h –
50 h
200 h
850 h
450 h
30h
20 h
Electricity(kWh)
kWh
35.000
20.000
25.000
Primary overhead costs of the service cost center reparation: ,- €
Primary overhead costs of the service cost center electricity: ,- €

5. Determination of the cost rates for internal cost allocation
The direct method doesn’t take the provided services from the service cost center reparation (50 hours) for the service cost center electricity and the provided services from electricity ( kWh) to reparation into account. Due to that fact the provided services have to be subtracted from the primary overhead cost of each service cost center.
Cost rate reparation : c R = ,− € h −50 h =41,73 €⁄h
Cost rate electricity: c R = ,− € kWh − kWh =0,095 €⁄kWh
In the internal cost allocation the primary overhead cost of the two service cost center reparation (R) and electricity (S), which are the part of the provided services for the main cost center, can be shifted to the main cost center with the two cos rates. After the internal cost allocation the service cost center can be described as empty and don’t contain any overhead cost.

6. Distribution of the service cost center
with the direct method
Service cost center
Main cost center R E M P1 P2
Admin.
Distrib.
Primary overhead costs
,- €
64.680,- €
53.937,- €
,- €
,- €
,- €
,- €
,- €
Assemble
,- €
8.346,- €
35.470,- €
18.778,- €
1.252,- €
834,- €
Reparation -
(200 h x
41,73 €/h)
(850 h x
(450 h x
(30 h x
(20 h x
,- €
1.909,- €
33.412,- €
14.320,- €
2.387,- €
electricity
( kWh X
( kWh
( kWh
( kWh
0,095 €/kWh)
∑ Overhead costs
0,- €
,- €
,- €
,- €
,- €
,- €

7. Structure of cost allocation sheet with the step-ladder method

8. Example: Internal cost allocation with the step-ladder method
Distribution of the service cost center costs with the step-ladder method
Sum
Service cost center
Main cost center R E M P1 P2
Admin.
Distrib.
Reparation (hours)
1.600 h –
50 h
200 h
850 h
450 h
30h
20 h
Electricity (kWh)
kWh
35.000
20.000
25.000
Primary overhead costs of the service cost center reparation: ,- €
Primary overhead costs of the service cost center electricity: ,- €
Premise: in the first instance the service cost center reparation is charged. No return debit is charged.

9. Determination of the cost rates for internal cost allocation
Cost rate reparation : c R = ,− € h =40,43 €⁄h
The numerator of the service cost center electricity changes. The secondary costs of the service cost center reparation with the amount of 2.021,- (50 h x 40, 43 €/h) are charged in the service cost center electricity. The kWh which are provided by the service cost center electricity for the service cost center reparation are not taken into account for the denominator, due to the fact that upstream cost center are not considered. By allocating the kWh with the main cost center the cost rate increases.
Cost rate elecricity: c S = ,− € ,−€ kWh − kWh =0,099 €⁄kWh

10. Distribution of the service cost center costs
with the step-ladder method
Service cost center
Main cost center R E M P1 P2
Admin.
Distrib.
Primary overhead costs
,- €
64.680,- €
53.937,- €
,- €
,- €
,- €
,- €
,- €
Assemble
,- €
2.021,- €
8.085,- €
34.361,- €
18.191,- €
1.21,-3€
809,- €
reparation
(50 h x
(200 h x
(850 h x
(450 h x
(30 h x
(20 h x
40,43 €/h)
Intermediate cut
,- €+
2.021,- €=
55.958,- €
,- €
1.981,- €
34.664,- €
14.856,- €
2.476,- €
1.981,-€
electricty
( kWh
( kWh
( kWh
( kWh x
0,099 €/kWh)
∑ Overhead costs
0,- €
,- €
,- €
,- €
,- €
,- €

11. System of emitting and receiving cost center
x 1.2 c1
C 2
Cost center 1
Cost center 2
x 2.1 c2
x 3.1 c3
x 3.2 c3
x 1.3 c1
x 2.3 c2
Cost center 3
C 3
Cj: Primary cost of cost center j
cj: Cost per unit of the cost center j
xj: Quantity ouput of cost ceter j
xi,j: Quantity oupt of cost center i at the cost center j
i= 1…n: Index of the emitting cost center
j= 1…n: Index of the receiving cost center

12. Example: Internal cost allocation with the equation method
System of equations
Input in EUR =
Output in EUR
Primary costs of the considered cost center+ secondary cost of the other cost center
Total capacity of the considered cost center C1 +
x 1,1 c1
x 2,1 c2
...
x n,1 cn
x1 c1 C2
x 1,2 c1
x 2,2 c2
x n,2 cn
x2 c2 C3
x 1,3 c1
x 2,3 c2
x n,3 cn
x3 c3 . Cn
x 1,n c1
x 2,n c2
x n,n cn
xn cn
Example: Internal cost allocation with the equation method
Sum
Service cost center
Main cost center R E M P1 P2
Admiin.
Distrib.
Reparation (hours)
1.600 h –
50 h
200 h
850 h
450 h
30 h
20 h
Electricity (kWh)
kWh
35.000
20.000
25.000
Primary overhead costs of the service cost center reparation: ,- €
Primary overhead costs of the service cost center electricity: ,- €

13. System of the emitting and receiving service cost center
64.680,- €
Service cost center for reparation
50 h
Service cost center for electricity
53.937,- €
kWh
1.550 h
kWh
to all main service cost centers

14. 1. Step: Setting up the equation
The service cost center reparation delivered and received something from the service cost center electricity. Therefore it is necessary to set up an input- and an output- equation for both. The input of the service cost center reparation is defined by the primary cost
(64.680,- €) and the secondary cost of the service cost center electricity KWh were delivered to the reparation. The output of the reparation amounts hours.
Service cost center reparation: ,- € · cE = · cR
The primary cost of the service cost center electricity amounts ,- € an in addition plus 50 hours of reparation, which were not converted to Euro. The output is evaluated with KWh.
Service cost center electricity : ,- € + 50 · cR = · cE

15. Step: Transforming the equation 1:
cR = 40,425 € + 21,875 cE
Step: Inserting into equation 2:
53.937,- € + [50 · (40,425 € + 21,875 cE)] = cE
53.937,- € ,25 € ,75 cE = kE
55.958,25 € = ,25 cE
cE = 0, €/kWh
Step: Inserting into equation 1:
64.680,- € kWh · 0, €/kWh = cR
1.600 cR = ,- € cR = 42,469 €/h
Solving the system of equations leads to the following results:
Cost rate reparation: cR = 42,469 €/h
Cost rate electricity: cE = 0, €/kWh

16. Solution for the example:
Internal cost allocation with the equation method
Service cost center
Main cost center R E M P1 P2
Admin.
Distrib.
Primary overhead costs
,- €
64.680,- €
53.937,- €
,- €
,- €
,- €
,- €
,- €
Intermediate cut
64.680,- €+
3.270,- €=
67.950,- €
Assemble
,- €
2.123,- €
8.494,- €
36.099,- €
19.111,- €
1.274,- €
849,- €
Reparation
(50 h x
42,469 €/h)
(200 h x
(850 h x
(450 h x
(30 h x
(20 h x
Íntermediate cut
53.937,- €+
2.123,- €=
56.060,- €
3.270,- €
,- €
1.869,- €
32.702,- €
14.015,- €
2.335,- €
1.869,-€
electricity
(35.000
kWh X
(20.000
( kWh
0,0934 €/kWh)
∑ Overhead costs
0,- €
,- €
,- €
,- €
,- €
,- €

17. Direct material costs (DMC)
The production costs of the production and the production costs of the sales can be determined by the following way:
Direct material costs (DMC)
+ Material overheads (MO)
+ Direct production costs (DPC)
+ Production overheads (PO)
+ Special production costs (SPCProd)
= Production costs of the production (PCProd)
+ Decreasing inventory
Increasing inventory
Capitalized work
= Production costs of sales (PCSales)

18. Internal cost allocation