ACC423: Management Accounting II
Module4 (Week10)
Covering:
1. Regression Analysis
2. EOQ Analysis
3. Linear Programming
By:
E. P. Enyi, Ph.D, MBA, ACA, FAAFM, RFS, MFP, FIIA
Head, Dept of Accounting, Covenant University,
Ota, Nigeria
Regression Analysis
• Regression analysis is also called CURVE FITTING
• It is a statistical technique used for short or medium term activity
forecasts like next season’s projected sales, activity costs etc.
• It is also used in management accounting to establish the
relationship between two or more variables and to separate
cumulated costs into fixed and variable elements.
• There are primarily two main methods of finding the line of best fit;
which is the linear relationship between the variables under
consideration.
• It is a SIMPLE regression analysis when it involves only two
variables but it becomes a MULTIPLE regression analysis when
three or more variables are involved.
• To find the line of best fit, it is necessary to calculate a line which
minimizes the total of the squared deviations of the actual
observations from the calculated. This is known as the LEAST
SQUARES.
• The LEAST SQUARES method of linear regression is given as
y = a + bx
Where, a and b are constants and where a represents the fixed
element and b represents the slope of the curve; y in this case is the
dependent variable e.g. total sales while x is the independent
variable e.g. selling price per unit.
The following graph illustrates this:
Before you can find the value of y and x, you must first find the value of
a and b by solving the following two equations simultaneously:
∑y = an + b∑x
∑xy = a∑x + b∑x2
(1)
(2)
where n = number of observations or pairs of past data being used for
the analysis.
Example:
TMC Ltd has collected data for the past seven years on the sales of its
prime product named HODAC which is presented as follows:
No of Colors
1
2
3
4
5
6
7
Sales(‘000)
14
17
15
23
18
22
27
The management wish to know whether there is any relationship
between the number of colors included in the product and the total
sales volume and what would be the sales volume if 9 colors were
used.
Solution:
As a matter of fact, you can solve this problem using the visual fit
method but the solution will be less accurate than the more scientific
method of least squares. The outline of the least square solution is
as follows:
Colors(x)
1
2
3
4
5
6
7
Totals
28
Sales(y)
14
17
15
23
18
22
27
136
xy
14
34
45
92
90
132
189
596
x2
1
4
9
16
25
36
49
140
Here,
n = 7; ∑x = 28;
∑y = 136;
∑xy = 596;
∑x2 = 140
Such that:
136 = 7a + 28b
…….(1)
596 = 28a + 140b
…….(2)
Solving the equations simultaneously, we have:
a = 12
b = 1.86
Therefore, the regression line or linear relationship between the
number of colors in HODAC and the sales volume achieved is given
by the equation:
y = 12 + 1.86x
i.e. sales volume (‘000) = 12 + 1.86(No of colors used)
If 9 colors are used, then the expected sales volume will be:
y = 12 + (1.86 * 9) = 28.74 or 28,740 units of HODAC
For Class Work:
The following data were extracted from the production logbook of
Azundu Chemicals Ltd:
Batch
Units ((x) ‘000) Total Costs ((y) N’000)
a.
9
116
b.
16
211
c.
14
152
d.
38
410
e.
21
256
f.
25
298
g.
20
220
h.
15
180
i.
15
185
j.
11
129
Required: Separate the total costs into its fixed and variable elements.
If the company is to manufacture 30 units, how much will it costs?
How many units will a cost of N500,000 produce?
Economic Order Quantity (EOQ)
EOQ is a stock control tool. It is used in conjunction with other stock control
tools such as Re-Order Level, Minimum Stock Level, Maximum Stock
Level etc.
EOQ is defined as the optimal number of items to be placed in a single stock
replenishment order in order to achieve the least total over-all stock
holding costs per annum.
EOQ can be determined in two ways:By constructing a graph
By the use of a formula.
ASSUMPTION UNDERLYING THE USE OF EOQ
1.
That there is a known constant stock holding cost;
2.
That there is a known constant stock ordering cost;
3.
That the rates of demand are known;
4.
That the price per unit is known and constant;
5.
That replenishment is made instantaneously, i.e. batch is delivered whole
GRAPHICAL APPROACH
To calculate EOQ graphically, the following ingredients are necessary:
- Total costs per annum (i.e. ordering cost plus carrying costs)
- Number of orders required per annum (i.e. Annual DD / Order Qty)
- Average stock (i.e. Order Qty / 2)
Example:
A company uses 50000 bottles per annum which costs N10 each to
purchase. The ordering and handling costs are N150 per order and
carrying costs are 15% of the purchase cost per annum.
Required: Determine the Economic Order Quantity that will minimize
the total stock holding cost for the coming year; assuming all other
parameters remain as projected.
Solution:
Graphical solution is implemented by trying a combination of assumed
order quantities, e.g. 1000, 2000 etc. It is the quantity of each
assumed order batch size that will determine the number of orders
per annum.
A Table of Assumed Order Levels:
(a)
Order
Qty
(b)
(c)
(d)
(e)
(f)
Average Annual
Average
Stock
Total
No of
Ordering
Stock
Holding
Cost
Orders Costs
Cost p.a.
---------- ---------- ---------------------------------------Assumed
Dd/(a) (b)*150
(a)/2
(d)*1.50
(c)+(e)
---------- ---------- -----------------------------------------1000
50
7,500
500
750
8250
2000
25
3,750
1,000
1,500
5250
3000
16.67
2,500
1,500
2,250
4750
4000
12.5
1,875
2,000
3,000
4875
5000
10
1,500
2,500
3,750
5250
6000
8.5
1,250
3,000
4,500
5,750
------------------------------------------------------------------------------------------------------From the above table the graph can be prepared:
10
00
20
00
30
00
40
00
50
00
60
00
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
Orde Qty
No of Orders
Annual Order
Cost
Average Stock
Stk Holdg
Cost pa
TotalCosts
From the above graph, the EOQ is approximately 3,160 units as indicated by
the downward arrow.
EOQ BY FORMULA
The scientific formula for EOQ determination is given as:
EOQ = √(2*Co*D)/Cc
Where,
Co = Cost of order per batch
Cc = Carrying cost per item per annum
D = Annual Demand
Solution:
EOQ = √(2*150*50000)/1.5
= √10000000
= 3,162 units
There are other measurements of EOQ which are of interest to the
management accountant. These include:
1.
EOQ With Gradual Replenishment – where there is internal arrangement
for stock usage replenishment; and
2.
EOQ Where STOCK-OUTS Occur – where stocks can be completely
allowed to be used up before replenishment (this will involve additional
stock-out costs)
EOQ with Gradual Replenishment = √(2*Co*D)/(Cc(1-(D/R)))
Where,
1-(D/R) = rate of stock usage
R = rate of production per annum
Example:
Assume the company decides to make its own bottles with a machine
that can produce up to 250,000 units per annum, with other data
remaining the same, the EOQ with gradual replenishment will be:
EOQ = √(2*150*50000)/(1.5(1-(50000/250000)))
= √12500000
= 3,536 units per order
EOQ where STOCK-OUTS Occur = [√(2*Co*D)/Cc] * [√(Cc+Cs)/Cs]
Where, Cs = Cost of stock-out per item per annum.
CLASS WORK: Assuming that a courier service charges N0.75 to
make a quick delivery per item in addition to an admin cost of N0.25
per item for every stock-out orders, what is the new EOQ?
Linear Programming Techniques
LP is a mathematical technique concerned with the scientific allocation
of scarce resources to attain optimum value in decision making.
For LP to be effectively deployed:
a.
The problem must be capable of numerical translation;
b.
All factors involved in the problem must have linear relationship;
c.
The problem must provide alternative causes of action;
d.
There must be one or more restrictions on the factors involved.
USING THE LP SOLUTION TECHNIQUE
There are two stages involved in using the LP technique. The first is to
formulate the model and the second is to solve the problem using
the model formulated. It is important that the model formulation is
accurately done otherwise it will be a case of garbage in and
garbage out.
To formulate the model, you first determine the objective function, then
you determine the main body of the LP problem.
Methods:
There are two main methods of solving a LP problem:
1.
By graph – this is only possible when the resources are limited to two.
The graphical method cannot solve complex situations. Even so, the
solution provided will not be as accurate as the more scientific simplex
method.
2.
By Simplex Algorithm – this is a more scientific method and can handle
complex situations such as those involving more than two resource
combinations.
Maximization – is used when it is necessary to work out the solution (mix) that
gives the maximum benefit e.g. optimal profit.
Minimization – is used when it is necessary to work out the solution (mix) that
produces the least cost.
Example:
A firm makes two products fork and knife. Fork has a contribution of N3 per
unit and knife N4 per unit. The firm wishes to establish weekly production
plan which maximizes its total contribution. The production data are as
follows:
PER UNIT
Machining (Hrs) Labour (Hrs) Material (kgs)
4
4
1
2
6
1
Fork
Knife
Total Available
Per Week
100
180
40
Because of a trade agreement, sales of Fork are limited to a weekly
maximum of 20 units and to honor an agreement with an old
customer, at least 10 units of Knife must be sold per week.
Step 1: Formulate the LP model; this is given thus:
Maximize 3f + 4k
SUBJECT TO:
A
4f + 2k ≤ 100 (Machining Hr Constraint)
B
4f + 6k ≤ 180 (Labour Hr Constraint)
C
f + k ≤ 40 (Materials Constraint)
D
f
≤ 20 (Fork sales Constraint)
E
k ≥ 10 (Knife sales Constraint)
f, k ≥ 0
where, f = number of units of fork
k = number of units of knife
As this is a problem with only two unknowns (f and k) it can be solved
graphically.
Step 2: Draw the axis of the graph ensuring that you calibrate as appropriate:
f
0
k
Step 3: Remove the inequalities from the equations and solve for each
constraint as follows:
Constraint Equation
A
4f + 2k = 100
B
4f + 6k = 180
C
f + k = 40
D
f = 20
E
k = 10
In A, where f=0, then k=100/2=50;
where k=0, f=100/4=25
In B, where f=0, then k=180/6=30;
where k=0, f=180/4=45
In C, where f=0, then k=40;
where k=0, f=40
In D, f=20
In E, k=10
These are then plotted in the given graph below to produce the
following image:
Fork
60
50
40
Fork
Knife
30
20
10
0
0
20
40
60
80 Knife
LP Model Graph Showing the FEASIBLE REGION
THE FEASIBLE REGION
The feasible region is the portion of the graph which falls within the
boundary of possible solutions. It is the only area of the graph which
satisfied all the conditions imposed by the various constraints. It is
usually shaded to distinguish it from other non-feasible area. It is
from this region that a set of possible solutions are outlined. In this
graph, there are 4 possible outcomes from which we choose the one
with the optimal contribution as follows:
Point
Fork(N3)
Knife(N4)
Total
I
20 * 3
10 * 4
100
II
15 * 3
20 * 4
125 (Choice)
III
0*3
30 * 4
120
IV
0*3
10 * 4
40
From the above analysis, we can see that the combination which
produced the highest benefit is point II, i.e. the one obtained at the
rightmost upward projection of the feasible area on the graph. This
is translated to mean that a production combination of 15 forks and
20 knives per week should be pursued by the firm as this is the mix
that will give maximum contribution.
MINIMIZATION
Provided that there are two unknowns (i.e. two products), we can also
use the graphical method to solve minimization problem, but, with
the following differences:
a.
In minimization problem the limitations are of the greater than or
equal to type (≥) rather than the less than or equal to type (≤);
b.
The feasible region is found above all or most of the limitations
rather than below them;
c.
The normal objective is to minimize cost so that the objective
function line represent cost rather than profit or contribution and
because the objective is to minimize cost, the optimum point will
be found from the cost line furthest to the left which still touches
the feasible region (the opposite of the maximization method).
Example:
A firm is to market a new fertilizer which is to be a mixture of two
ingredients A and B. The properties of the two ingredients are
given in the table below:
INGREDIENT ANALYSIS
NITRPHOS COST
OGEN LIME
PHATE
/KG
30%
40%
10%
N12
10%
45%
5%
N8
BONE
MEAL
Ingredient A
20%
Ingredient B
40%
It has been decided that:
a.
The fertilizer will be sold in bags containing a minimum of 100kg
b.
It must contain at least 15% nitrogen, 8% phosphate and 25% bone
meal.
The firm wishes to meet the above requirements at the minimum cost possible
SOLUTION
Set the objective function as follows:
MINIMIZE 12a + 8b
Subject to:
A
a+
b ≥ 100
B
0.3a + 0.1b ≥ 15
C
0.1a + 0.05b ≥ 8
D
0.2a + 0.4b ≥ 25
a ≥ 0 and b ≥ 0
(Weight Constraint)
(Nitrogen Constraint)
(Phosphate Constraint)
(Bone Meal Constraint)
Where,
a = kgs of ingredient A
b = kgs of ingredient B
Remove the inequalities and solve thus:
In A, if b=0 then a=100;
if a=0 then b=100
In B, if b=0 then a=15/0.3=50;
if a=0 then b=15/0.1=150
In C, if b=0 then a=8/0.1=80;
if a=0 then b=8/0.05=160
In D, if b=0 then a=25/0.2=125;
if a=0 then b=25/0.4=62.5
The above computations are then graphed as follows:
Ingredient
A
180
160
140
Feasible Region
120
100
80
60
40
20
0
Ing A
Ing B
0
50
100
150
200
Ingredient B
Reading from the graph:
Point
Ing A (N12)
Ing B (N8)
Total (N)
I
125 * 12
0*8
1500
II
75 * 12
30 * 8
1140
III
60 * 12
40 * 8
1040 (Lowest = Optimal)
IV
0 * 12
160 * 8
1280
The above implies that the best combination of ingredients that would
minimize total cost (i.e. result in the least cost) is 60 kg of A and 40 kg of B.
SIMPLEX ALGORITHM METHOD
This is a step-by-step arithmetic technique whereby one moves
progressively from a zero position to the maximum possible contribution. It
utilizes the solution method of matrix algebra. The steps involved in the
simplex algorithm method are numerous and the underlying logic is
complex and cannot be adequately covered within the time scheduled for
this lecture. The student is advised to study this topic independently.