Chapter 15
Allocation of Support Department
Costs, Common Costs, and Revenues
1
Cost allocation: what it means
z
z
z
z
Some production centers or departments provide
output required by other production centers
(Service departments).
The costs of these departments are allocated to
the internal users according to use and add to
their costs
so the costs of service departments providing
these products and services go indirectly into the
cost of saleable output
Service department‘s costs are allocated using
the actual utilization volume times an allocation
rate per unit
2
Allocating Support Departments Costs
z
An operating department (a production department in
manufacturing companies) adds value to a product or service
z A support department (service department) provides the
services that assist other operating and support departments in
the organization.
3
from general ledger
Additional cost
not out of pocket
Directly attributable
Structure of Service Department Cost Attribution
(Traceable
costs)
Support
departments
Secondary
Activities
Attribution to cost pools of the
centers; each activity has exactly
one cost driver
Usage × cost driver rate
Reciprocal services
Operating
departments
Primary
Activities
Product 1

Product n
4
Single-Rate and Dual-Rate Methods
z
The single-rate cost allocation method pools together all costs
in a cost pool.
z The dual-rate cost allocation method classifies costs in each
cost pool into two cost pools
Œ
a variable-cost cost pool and
Πa fixed-cost cost pool
z
Organizations commit to infrastructure costs on the basis of a
long-run planning horizon.
z Budgeted rates let the user department know in advance the
cost rates they will be charged
Œ
During the budget period, the supplier department, not the user
departments, bears the risk of any unfavorable cost variances.
ΠWhen actual rates are used for cost allocation, managers do not
know the rates to be used until the end of the budget period
ΠThe use of budgeted usage to allocate these fixed costs is
consistent with the long-run horizon.
5
Allocating Support Departments Costs
z
Direct method:
Œ
z
Allocates support department costs to operating
departments only.
Step-down (sequential allocation) method:
Œ
Allocates support department costs to other support
departments and to operating departments
Πcharge rates are calculated for support departments
according to a rank order. Those departments rank highest
that get the least from other departments
Πat each step support is charged only from the departments
whose charge rate has already been calculated
z
Reciprocal allocation method:
Œ
Allocates costs by services provided among all support
departments
Πsimultaneous equations approach
6
Example
z
The Canton Division of Smith Corporation has two
operating departments
Œ
Assembly and Finishing
and two support departments
Œ
Maintenance
• allocated using square feet.
• Total square feet = 255,000
ΠHuman Resources
• allocated using number of employees
• Total number of employees = 95
Maintenance
Budgeted costs
before allocations: $300,000
Square feet:
5,000
# of employees:
8
Human
Resources
$2,160,000
30,000
15
Assembly
Finishing
$1,700,000
110,000
48
$900,000
110,000
27 7
Direct Method
Support Departments
Maintenance
$300,000
0%
Operating Departments
110/220
$1,700,000
Assembly
24/72
$900,000
Finishing
0%
Human
Resources
$2,160,000
Original costs:
Maintenance Allocated:
Human Resources Allocated:
Total
Assembly
$1,700,000
150,000
1,440,000
$3,290,000
Finishing
$ 900,000
150,000
720,000
$1,770,000
8
Step-Down Method
z
Which support department should be allocated first?
Œ
Maintenance provides 12% of its services to Human Resources.
ΠHuman Resources provides 10% of its services to Maintenance.
z
Maintenance to Human Resources:
12% × $300,000 = $36,000
Œ
Maintenance to Assembly:
ΠMaintenance to Finishing:
z
44% × $300,000 = $132,000
44% × $300,000 = $132,000
Human Resources costs to be allocated become
$2,160,000 + $36,000 = $2,196,000
Œ
Human Resources to Assembly: 48 ÷ 72 × $2,196,000 =
$1,464,000
Œ Human Resources to Finishing: 24 ÷ 72 × $2,196,000 =
$732,000
9
xij = aijxij
Overhead Allocation Sheet
(Step down method)
Traceable costs
Secondary activities
j does not use > j
j=1
j=2
j=3
π1 x12
π1 x13
π2 x23
usage of
i by j
Primary activities
Costs from
general ledger
+
additional
non-out-ofpocket cost
S1
S2
S3
Total cost driver
volume, center j
Cost driver rate
... π3 x3j ...
S1
S2
S3
Sj
x1
x2
x3
x3
π1 =S1 /x1
π2
π3
π3
Cost driver rates
One line for each
kind of input used.
Entries in each
line sum up to the
respective amount
in the cost
recording column
Sum of row i = Si
Sum of column j:
total cost of center j
Start
10
Reciprocal
M
Maintenance
–
Human Resources 10%
HR
12%
–
Total Cost(j) = traceable cost (j) +
A
44%
60%
F
44%
30%
Σall i αij × Total cost(i)
where αij denotes j‘s share of total i‘s service
M = 300,000 + 0.10 HR
HR = 2,160,000 + 0.12 M
M – 0.10 HR = 300,000
– 0.12 M +
HR = 2,160,000
10
M – HR = 3,000,000
– 0.12 M + HR = 2,160,000
9.88 M = 5,160,000
M = 5,160.000 / 9.88 = 522,267
HR = 2,160,000 + 0.12× 522,267
= 2,222,672
11
Reciprocal
Before
allocation:
Allocation:
Allocation:
Total
M
HR
$300,000
(522,267)
222,267
$2,160,000
62,672
($2,222,672)
A
$1,700,000
229,797
1,333,603
$3,263,400
F
$ 900,000
229,797
666,802
$1,796,599
Total cost Assembly Department: $3,263,400
Total cost Finishing Department: $1,796,599
12
Budgeting requirements (cont‘d)
z
Then the total volumes of cost drivers are
determined by the system of equations
xi = yi + ∑ aij ⋅ x j
j
(One equation for each
secondary activity i)
The same system can be written as a matrix
equation:
(I – A)x = y
13
Cost driver rates πi
z
Activity account balance
Activity j
Traceable costs
KPj
Service delivered
Cost of secondary
activities
πj ·xj
∑π
i
i
Î
K Pj + ∑ π i ⋅ xij = π j ⋅ x j
i≠ j
⋅ xij
Ð
K Pj + ∑ π i ⋅ aij ⋅ x j = π j ⋅ x j
i≠ j
π j − ∑ π i ⋅ aij = k
i≠ j
(I – AT)π = k
P
j
Í
K
P
j
xj
Ð
+ ∑ π i ⋅ aij = π j
i≠ j
=: kPj
14
Example: „Fall River Company“*)
z
Service centers: Power Department, Water Department; Production
centers: Divisions 1 und 2.
z Data:
Units of service provided to:
Units of service
provided from:
Power
Water
Power
20
Water
traceable costs
Div.1
Div.2
Total
70
80
70
240
30
10
70
50
160
$ 4.9
$ 1.25
Activity account balances:
240 π1 = 20 π1 + 30 π2 + 4.9
160 π2 = 70 π1 + 10 π2 + 1.25
⇒
220 π1 – 30 π2 = 4.9
– 70 π1 + 150 π2 = 1.25
*) Kaplan/Atkinson, Advanced Management Accounting, 3rd ed. p.74-76 and 80-81.
Numbers modified.
15
Calculation
direct solution
220 π1 – 30 π2 = 4.9
– 70 π1 + 150 π2 = 1.25
| ×5
1100 π1 – 150 π2 = 24.5
+
– 70 π1 + 150 π2 = 1.25
1030 π1
= 25.75
π1 = 0.025
π2 =
1.75+1.25
150
= 0.02
Power and water cost:
Div. 1: $3.4 mill., Div. 2: $2.75 mill.
⎛ x11
⎜
x
A=⎜ 1
⎜ x21
⎜ x
⎝ 1
matrix calculus
x12 ⎞ ⎛ 20
70 ⎞
⎟ ⎜
x2 ⎟ ⎜ 240 160 ⎟⎟
=
x22 ⎟ ⎜ 30 10 ⎟
⎜
⎟
x2 ⎟⎠ ⎝ 240 160 ⎠
⎛ 220 − 30 ⎞
⎜
⎟
T
240
240
⎟
(I − A ) = ⎜
−
70
150
⎜
⎟
⎜
⎟
⎝ 160 160 ⎠
⎛ 120 16 ⎞
⎜
⎟
T −1
103
103
⎟
(I − A ) = ⎜
56
352
⎜
⎟
⎜
⎟
⎝ 103 309 ⎠
⎛ 120 16 ⎞⎛ 4.9 ⎞
⎟ ⎛ 0.025 ⎞
⎜
⎟⎜
T −1 P
240
103
103
⎟ = ⎜⎜
⎟⎜
⎟⎟
(I − A ) k = ⎜
1
.
25
56
352
⎟ ⎝ 0.02 ⎠
⎜
⎟⎜
⎟
⎜
⎟⎜
⎝ 103 309 ⎠⎝ 160 ⎠
16
Why the matrix calculus is useful
z
z
The numerical data required are provided in the
accounting data base and can automatically
downloaded into the matrix A and a vector of
traceable unit costs kP.
Spreadsheet software usually offers the function of
matrix inversion
¾
z
For larger problems an LP algorithm may be used
so the cost driver rates can be determined
automatically.
17
Interpretation of R:=(I − A)-1
z
z
Consider the equation for required total output of service i as a
function of external requirements y:
xi (y) = Σj rij yj
Differentiate this function w.r.t. yj . You get:
∂ xi (y)
= rij
∂ yj
z
z
This means: rij represents the additional total output
of service i required per additional unit of external
output requirement of service j.
Therefore the matrix R is sometimes called the total
requirements matrix)
18
Interpretation of R:=(I − A)-1
z
In particular:
Œ
If you purchase one unit of the service i externally (reduce
external demand by one unit)
Œ
then you need rii units less to be procured internally.
z
Or, in other words: if you reduce internal procurement
of the service by one unit, you need to buy only 1/ rii
units from external sources.
z
Since the function xi (y) is linear, this is globally true.
19
Interpretation of R:=(I − A)-1
z
z
This means:
If you close down service center i then you
Œ can save the total reciprocal cost πi xi for this center but
Πneed xi/rii units of the service externally
You will break even if the external procurement price
pi satisfies:
pi xi / rii= pi xi
i.e. you may pay at most an external price of
pi = πi rii
20
Interpretation of RT := (I − AT)-1
z
So we get the reciprocal cost per unit as a function
of the traceable cost:
cj (kP) = Σi kiP rij
z
Similarly to the above:
∂ cj (kP)
∂ kiP
= rij
21
Reciprocal method: Extension
z
Dual rate system for assigning committed costs:
Peak load pricing
¾
z
Assigning
• committed cost according to capacity reservations by
users
• flexible cost according to actual usage
if a service is outsourced:
¾
the reciprocal method shows the effect of cost drivers
on required total volume (capacity) for all service
departments
22
Allocating Common Costs
z
1.
Two methods for allocating common cost
Stand-alone cost allocation method
Œ
2.
Incremental cost allocation method
Œ
Œ
3.
actual cost is allocated in the ratio of stand-alone costs
a sequence of cost objects is defined
each object bears the incremental cost according to the
sequence
Shapley Value
Œ
Œ
the average of incremental costs over all possible
sequences is charged to the object
the Shapley Value can be justified based on a set of
plausible axioms
23
An Example
z
Three divisions (1,2,3) in an organization need typing
services
¾
Let K(i,...,j) denote the cost for the typing service when
divisions (i,...,j) pool their typing services.
¾ Cost for stand-alone provision of typing
K(1) = 10;
K(2) = 20; K(3) = 30
ΠCosts with pooled typing services
K(1,2) = 25; K(1,3) = 35; K(2,3) = 32
K(1,2,3) = 35
z
Stand-alone method:
¾
C(1) = 35/6 = 5.83; C(2) = 35/3 = 11.66; C(3) = 35/2 = 17.5
24
Incremental Method and Shapley Value
z
z
Application to the example
Cost Function K(1) = 10; K(2) = 20; K(3) = 30
K(1,2) = 25; K(1,3) = 35; K(2,3) = 32
K(1,2,3) = 35
z
Cost increments occuring when divisions join the
pool in a certain order for each possible
sequence
Sequence Div. 1 Div. 2 Div. 3
= Table of allocations
by incremental method:
Shapley-Value:
Stand-alone method:
1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1
Σ/6
10
10
5
3
5
3
6
15
0
20
20
0
2
9½
10
25
10
12
30
30
19 ½
5.83; 11.66; 17.5
Revenues and Bundled Products
z
z
A bundled product is a package of two or more
products (or services) sold for a single price.
Bundled product sales are also referred to as “suite
sales.” The individual components of the bundle
also may be sold as separate items at their own
“stand-alone” prices.
Examples
Banks
Checking
ƒ Safety deposit boxes
ƒ Investment advisory
ƒ
Hotels
Lodging
ƒ Food and beverage
services
ƒ Recreation
ƒ
Tours
Transportation
ƒ Lodging
ƒ Guides
ƒ
26
Revenue Allocation Methods
z
z
English Languages Institute buys English language
software programs locally and then sells them in
Mexico and Central America
English sells the following programs: Grammar,
Translation, and Composition
¾
These programs are offered stand-alone or in a bundle
Stand-alone
Unit
Price
Cost
Grammar
$255
$180
Translation
$ 85
$ 45
Composition $185
$ 95
Bundle (Suites)
Grammar + Translation
Grammar + Composition
Grammar + Translation + Composition
Price
$290
$350
$410
27
Revenue Allocation Methods
z
1.
The two main revenue allocation methods
The stand-alone method with alternative weights
Selling prices
b) Unit costs
c) Physical units
d) Stand-alone product revenues
a)
2.
3.
The incremental method with alternative sequences
The Shapley Value
28
Stand-Alone Revenue Allocation Method
z
Consider the Grammar and Translation suite, which sells for
$290 per copy.
z 1a) Grammar:
$290× 255/(255 + 85) = $217.50
Translation: $290× 85/(255 + 85) = $72.50
1b) Grammar:
$290× 180/(180 + 45) = $232
Translation: $290× 45/(180 + 45) = $58
1c) Grammar:
$290/2 = $145
Translation: $290/2 = $145
1d) Assume that the stand-alone revenues in 2003
Grammar $734,400; Translation $81,600, Composition $133,200.
Grammar:
Translation:
$734,400 ÷ $816,000 = 0.90, $290 × 0.90 = $261
$81,600 ÷ $816,000 = 0.10, $290 × 0.10 = $29
29
Incremental Revenue Allocation Method
z
The first-ranked product is termed the primary
product in the bundle
¾
z
z
z
If the suite selling price exceeds the stand-alone price of the
primary product, the primary product is allocated 100% of
its stand-alone revenue.
The second-ranked product is termed the first
incremental product
The third-ranked product is the second incremental
product, and so on.
Assume that Grammar is designated as the primary
product:
¾
Grammar and Translation suite selling price = $290 per copy
¾ Allocated to Grammar: $255
¾ Remaining to be allocated: ($290 – $255) = $35 > Translation
30
Shapley Value, compared to other methods
z
z
z
z
z
z
Grammar, Translation, Composition:
Grammar, Composition Translation:
Translation, Grammar, Composition:
Translation, Composition, Grammar:
Composition, Grammar, Translation:
Composition, Translation, Grammar:
Shapley Value:
Stand alone:
G
255
255
205
140
165
140
193.33
199.14
T
45
60
85
85
60
85
70
66.38
C
110
95
120
185
185
185
146.67
144.48
31