How to Interpret the Matrix

8/11/2019 How to Interpret the Matrix

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How to interpret the matrix: In the following illustration, a detailed explanation of the different terms in the

simplex table, their meaning and what they represent is given. How we move

along from one point to another (as explained in the previous paragraph) to

reach the optimum solution will be explained clearly.

Illustration: D company produces two products X and Y. The requirements and constrains

with regard to its production is given below.

Product X Product Y

Material – 8 units

Labour – 6 hrs

Machine hours - 4

Contribution - 14

Material – 4 units

Labour – 8 hrs

Machine hours - 6

Contribution - 16During the forthcoming accounting period the availability of labour hrs will be

restricted to 2880 hrs. Material availability is limited to 3440 units. Themachine has the capacity to run for 2760 hrs. The marketing manager expects

the maximum sales potential for X is 420 units with respect to Product Y there

is no sales Limitations.

Formulate an LPP and find the optimum number of units to be produced.

Solution: The LPP is as follows:

Maximize C = 14x+16y subject to

8x + 4y ≤ 3440 ( Material constraint)

6x + 8y ≤ 2880 ( Labour constraint)

4x + 6y ≤ 2760 ( Machine hour constraint)

x ≤ 420 (Sales constraint)

When we convert these into equalities the equations become:

8x + 4y + S1 = 34406x + 8y + S2 = 2880

4x + 6y + S3 = 2760

x + S4 = 420



Slack variables: S1, S2, S3 and S4 are called slack variables. Slack variables represent the unused

amount of resources. Here S1represents unused amount of raw-materials, S2 -

unused labour hours, S3 – unused machine hours and S4 represents unutilized

sales potential.

Suppose we consider a situation when we are producing only 1 unit of x & y

each, we would be consuming 12 (i.e. 8+4) units of material. So, out of the 3440

units of material available, after the consumption of 12 units 3428 units will

remain unutilized. This is the material slack. This can be found out by

substituting x=1 & y=1 in the material constraint equation and solving for S1.Material constraint - 8x + 4y + S1 = 3440

= 8(1) + 4(1) + S1 = 3440 (substituting x=1, y=1)

= 12 + S1 = 3440

= S1 = 3428

S1 is material slack, so it gives the unutilized amount of material for the givensituation (1,1). Similarly substituting the values for a given situation in the other

equations will tell us how much of those resources remain unutilized.

Contents of the table: C j:

The first row is the C j row. It tells us how much profit we can get by increasingthe value of the variables by 1 unit. [Refer Table 1].We have 14 written on top

of x. This means 1unit of X will fetch us 14 Rs profit. Similarly, 1unit of Y will

fetch us 16 Rs profit. However the slack variables can never fetch any profit,

hence their values are 0 each.

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Iteration Table 1:

Each table explains the situation under a given circumstance. As already

explained, we always start with the origin in the simplex procedure. So the firsttable examines the situation at the origin (i.e. x=0 & y =0). In other words itexamines the situation when we are not producing anything.

(In the second column, we can find only the values S1, S2, S3 and S4. X & Y are

missing. This means that x & y are zero.

Iteration Table 1:

Solution:

The values in the ‘Solution’ column (3440, 2880 …) are the values of the

variables in the second column (S1, S2, S3 and S4). So we have to read the table

as S1=3440, S2 = 2880, etc. This is nothing but the slack when we are producing

nothing. (Remember the given situation for the first table is (0,0). (If we are not

producing anything, the entire 3440 units of materials remain unutilized. That’swhy S1= 3440)

Columns x, y, S1, S2, S3 and S4:

The values under these columns tell us the impact of the respective variables on

the values in the solution column (keeping other variables constant). In other

words it tells how many units of each of the resources will be consumed by each

of the products.

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Under Column x – Row S1, we have the number 8. This means – if we increase

the value of x by 1, the value of S1 will come down by 8. (Each unit of X

consumes 8 units of raw materials. So if we produce one unit of X, our material

slack reduces by 8).

In the same row under Column y, we have the number 4. This means each unit

of y will decrease the value of S1 by 4.

Similarly, Column y – Row S3 says 6. This means each unit of y will reduce the

value of S3 (machine hours slack) by 6 units from the current value of 2760.

Column S3 – Row S4 says 0. This is because S3 does not have any impact on thevalue of S4.

(In the first table we are assuming x & y as zero. Hence we need to see only the

columns under x & y. The remaining columns will be zeros and ones. So we can

ignore them.)

Column CB

It indicates how much profit each variable in the second column earns. S1, S2,

S3 and S4 do not fetch any profit. Hence they are all zeros.

Row Z j:

This indicates how much profit is lost if we increase the values of x, y, S1, S2,

S3 and S4. As of now we do not lose any profit with the increase of thesevariables. Hence they are all zeros.

Row C j- Z j:

This gives the effective increase in profits when we increase the values of thevariables by 1.

C j tells us how much additional profit we can get. Z j tell us how much we lose.

So C j – Z j gives the net increase/decrease.

Each unit of x gives 14 profit (C j ) and what we lose is 0(Z j). Therefore the neteffect is an increase in the profit by 14. But if we produce y we can get an

additional profit of 16. So obviously we would choose to produce Y instead of

X.

(Graphically this can be interpreted as – we choose to move along the Y axis

instead of the X axis, as the rate of change is higher while moving towards A)

This is the reason why we choose the highest positive value among the C j-

Z j values and mark them.



Ratio:

Now that we choose to produce Y, the question is - how many units of y can we

produce?

Each of the constraints is a limiting factor that restricts the production of theunits. So we check, with the remaining units of each of the resources, how manyunits of y we can produce. We have 3440 units of materials. Each unit of y takes

4 units. So we can produce 3440/4=860units.

Labour – we have 2880 hrs left. Each unit of y takes 8. So we can produce

2880/8=360units.

So we divide the solution column (which tells how much units are left) with the

y column (which tells how many units are consumed) to know the ratio (i.e.

how many units can be produced).

The material constraint allows us to produce 860units. But the labour

constraints restricts us to just 360units. So the defining constraint would be the

one with the lowest ratio. That is the reason why we choose the minimum

positive ratio.

The sales constraint does not restrict the production of Y. Hence the ratio

is ∞. So ratio indicates the maximum units that can be produced keeping in

mind the particular constraint. The least ratio indicates the maximum qty that

can be produced.

(Graphical interpretation – the ratio indicates the maximum distance we can go

along the chosen line yet stay within the feasible region. It indicates that we

cannot go beyond the point A)

For all further calculations, the labour constraint is what matters the most, as it

has been completely used up. The other resources however are still available in

plenty. Hence we circle that row and take that as the point of reference for the

next table.

Iteration Table 2:

Now we have decided to produce 360units of Y. The second table examines the

situation when x=0 & y=360. Since all available units of labour is used up, the

labour slack (S2) is zero. So in this table, in the second column, Y replaces S2.



Iteration Table 2:

Solution:

When 360 units of Y are produced, it consumes 360*4=1440 units of material.

So out of 3440 only 2000 (i.e. 3440-1440) will be left. So the material slack for

the given situation is 2000. Hence the table reads S1 = 2000. Similarly S3=600& S4=420.

Columns x, y, S1, S2, S3 and S4:

The intersection in the first table is at 8. We divide the whole row by 8 and this

forms the second row in the second table. This row is called the main row.

If we have to increase the value of any other variable, say you want to produce

some units of x, you can do it only if you forego the production of some units of

Y. And how many units of Y need to be foregone depends on the units of labourconsumed by X & Y, as labour is completely exhausted.

1 unit of X requires 6 units of labour. One units of Y requires 8 units. So Xconsumes 3/4th the labour that Y consumes. So, to produce 1units of X we need

to forego the production of 3/4th unit of Y. We can say that 1 unit of X is

equivalent of ¾ unit of Y. This is what is given in the main row.

The main row gives the equivalent value of all other variables in terms of 1 unit

of Y.

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Main row: If we increase the value of X by 1, Y will come down by ¾. If we

increase the value of S2 by 1, the value of Y will come down by 1/8.

Row S1: If we increase the value of X by 1, S1 will come down by 5#

. If weincrease the value of S2 by 1, the value of S1 will go up by 1/2

##. (Positive

values indicate consumption/decrease in value. Negative values indicate

increase in value)

Note:#- To produce one unit of X, we will have to forego the production of 3/4

th unit

of Y. When we forego this, we will get back 3 units (3/4th of 4) of material. But

to produce one unit of x we consume 8 units. Thus the net effect is an additional

consumption of 8-3=5 units in the materials.

##Similarly if we were to increase the value of S2 by 1, we need to decrease Y by

1/8th. When we do this we get 4*1/8=1/2 units of material back. But increasing

S2 does not consume any material. So the net effect is an increase in the

materials by ½ units which is indicated by -1/2 in the table.

Row S3: The values in the row S3 are the net effect on the solution values, after

considering the decreasing in the value of Y, which is now required to increase

any other variable.

Column CB

It indicates how much profit each variable in the second column earns. S1,

S3 and S4 do not fetch any profit. Hence they are all zeros. But Y fetches a profit

of Rs. 16.

Row Z j:

This indicates how much profit is lost if we increase the values of x, y, S1, S2,

S3 and S4.

If we increase X, we will be foregoing 3/4th unit of Y. Along with this we will

also be foregoing the profit from 3/4th unit of Y. That is ¾*16=12. So 12 is the

opportunity cost of producing one unit of X. (Opportunity cost - The

profit/opportunity foregone while making another choice)

Increasing one unit of Y is possible only if we decrease another unit of Y. So

we will lose 16. (Column X & S2 only needs to be considered. Others can beignored). Increasing one unit of S2 will lead to loss of profit to the extent of

1/8*16=2.



Row C j- Z j:

Each unit of x gives 14 profit (C j ) and what we lose(opportunity cost) is 12(Z j).

Therefore the net effect is an increase in the profit by 2. Each unit of S2 gives0profit (C j ) and what we lose is 2(Z j). Therefore the net effect is a decrease inthe profit by 2. So we choose the value with the highest positive integer - X.

(On producing Y you get 16, at the same time on foregoing you lose 16. So the

net effect is 0.)

Ratio: Now again the question is - how many units of X can we produce?

We check how many units of X we can produce before one of the resources run

out. We have 2000 units of materials. Each unit of X will consume an additional

5 units. So we can produce 2000/5=400units.

The sales constraint allows us to produce 420 units. But the labour material

constraint restricts us to just 400 units. So now the defining constraint would be

the material constraint (negative values to be ignore)

(Graphical interpretation – the ratio indicates that we wil have to stop at the

point B)

For the next table the material constraint is what matters, as it will becompletely used up. The other resources, (excluding labour) are now available.

Hence we circle that row and take that as the point of reference.

Final Matrix (Table 3): Now we are producing 400 units of X and 60 units of Y (to produce X we had

to decrease the production of Y). This table examines the situation when x=400

& y=60. Since all available units of material are used up, the material slack (S1)

is zero. So, in this table in the first column, X replaces S1. We do not find S1 &

S2 in the second row, indicating that they have been completely utilized. In every table two variables will take be assigned zero as the value. These arecalled non-basic variables as they do not appear in the basic solution. So the

variables that appear in the basic solution (second column) are the ones that

are not completely utilized.



Iteration Table 3:

Solution:When 400 units of X and 60units of Y are produced 800 units of machine hours

and 20 units of sales potential still remains unutilized. 6560 is the profit that can

be made in the current decision.

Columns x, y, S1, S2, S3 and S4:The interpretation is the same as the other table. Here we need to consider only

S1 and S2, the rest will be just zeros and ones. Column S1 indicates, to increase

the value of S1 by 1 unit (i.e. release 1 unit form production), we will have to

forego the production of X by 1/5 unit. And this will result in increase the

production of Y by 1/7th

unit and also S4 to increase by 1/5th

unit.

Similarly we can infer that we were to release one unit of labour (yet keep the

materials at zero), we wil have to decrease the production of Y by 1/5th unit and

increase the production of X by 1/10th unit. This will result in the increase of

S4 by 4/5

th

unit.

Row Z j:If we increase S1 or S2 we will lose profits. It is anyway quite obvious that we

will lose profits if we try to increase slack.

Row C j- Z j:

This row has only negative values and zeros. The negative values indicate that

increasing those variables will decrease the profit. Nothing can be done to

increase profits further, thus indicating that we have reached the optimumsolution.

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( Moving to either point A or C from the current point B will only decrease the

value of Z. so we decide to stay there)

So the optimum solution is to produce 400 units of x and 60 units of Y.

How to Interpret the Matrix

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