LINEAR PROGRAMMING – THE SIMPLEX METHOD
(1)
Problems involving both slack and surplus variables
A linear programming model has to be extended to comply with the requirements of the
simplex procedure, that is,
1. All equations must be equalities.
2. All variables must be present in all equations.
(I)
(II)
In the case of ‘ ≤ ’ constraints, slack variables are added to the actual
variables to make the equation an equality.
In the case of ‘ ≥ ’ constraints, surplus variables are used to make the
equation an equality.
A surplus variable is the difference between the total value of the true (decision) variables
and the number (usually, total resource available) on the right-hand side of the equation.
Thus, a surplus variable will always have a negative value.
Consider the following linear programming problem:
Minimise cost = 6 X 1 + 8 X 2
Subject to
X 1 + X 2 = 5000
≤ 1500
X1
X 2 ≥ 750
X1, X 2 ≥ 0
There are three constraints, consisting of one equality, one ‘smaller than or equal to’ and
one ‘greater than or equal to’.
The model should now be modified as follows:
X1 + X 2
+ S1
X1
X2
= 5000
= 1500
− S 2 = 750
X 1 , X 2 , S1 , S 2 ≥ 0
where S1 is a slack variable and S 2 is a surplus variable.
2
The presence of a surplus variable causes a problem when drawing the first simplex
tableau because of its negative value. In any maximisation problem, this tableau must satisfy the
following requirements:
1.
2.
3.
All the slack variables (and thus surplus variables as well) must form part of the
initial solution mix (basis).
The table must contain as many rows as there are constraints.
The elements in the columns of variables appearing in the basis must form a unit
vector.
If S 2 is included in the basis, the elements of the S 2 will be 0, 0 and –1 and thus not a
unit vector. Furthermore, the values of the true variables being 0 in the initial solution, this would
imply that S 2 would assume a negative value in the initial solution (it can be easily checked by
means of the third equality that, if X 2 = 0, S 2 = –750). This is definitely contrary to the nonnegativity restriction – all variables must have a positive value.
(2)
Artificial variables
This problem is solved by adding an artificial variable (denoted by Ai ) to the equation,
that is, a variable that has a positive value. Artificial variables are also used in equations which
are already equalities in order to comply with the requirements (1) and (2) above. Just remember
that an artificial variable has no significance pertaining to the solution of the problem – it is used
merely to find a solution mix in the first simplex tableau.
Although artificial variables will always form part of the initial solution mix, the
objective is to remove them as soon as possible by means of the simplex procedure. As long as an
artificial variable still appears in the solution mix, the final solution has not yet been found.
As any other variable, artificial variables are included in the objective function but the
value attached to them creates another problem. In minimisation problems, since the objective is
to find the lowest cost possible, a very large value is allocated to artificial variables. This value is
denoted by M where M is a very large positive value.
The above linear programming model is thus rewritten as
Minimise cost = 6 X 1 + 8 X 2 + 0S1 + 0 S 2 + MA1 + MA2
Subject to
1X 1 + 1X 2 + 0 S1 + 0 S 2 + 1A1 + 0 A2 = 5000
1X 1 + 0 X 2 + 1S1 + 0 S 2 + 0 A1 + 0 A2 = 1500
0 X 1 + 1 X 2 + 0 S1 − 1S 2 + 0 A1 + 1A2 = 750
X 1 , X 2 , S1 , S 2 , A1 , A2 ≥ 0
before using the simplex procedure.
3
Example 1 (slack variables only) [Maximisation]
Linpro Limited, a manufacturing enterprise, produces two products, namely product A
and product B. Both products are manufactured from the same raw material and processed by the
same machine. The available machine capacity is 600 hours and 10 000 kilogram of raw materials
are available. The production of one unit of product A utilises 30 minutes of machine time while
one unit of product B takes one hour to produce. To produce one unit of product A, 12.5 kg of
raw material is required whereas 10 kg is required for one unit of product B. The marginal
income of product A, of which a maximum of 700 units can be sold, amounts to R4 per unit.
Product B has a marginal income of R6 per unit. The company wants to determine what
combination of product A and product B must be produced in order to obtain maximum profit.
Formulate a linear programming model for the above problem and hence find the optimal
solution by using the simplex method.
Solution 1
We first identify our decision variables, objective function and thus write down the
constraints. Let the decision variables
X 1 = “number of units of product A to be manufactured” and
X 2 = “number of units of product B to be manufactured”
[Decision variables are normally those which appear in the objective function. It is always better
to use similar symbols instead of the traditional x and y to avoid confusion. We can, for example,
choose to denote our decision variables by the first letters of the names of the products. However,
for the simplex method, it is advisable to use indexed variables.]
The objective function may be formulated as follows (it can be obtained from the table
below):
Maximise Profit = 4 X 1 + 6 X 2
A table may also reveal to be helpful in understanding the problem better and to write down the
constraints easily as well. Consider the following:
Machine capacity (hrs)
Raw material (kg)
Demand (at most)
Marginal income
Product A
0.5
12.5
700
R4
Product B
1
10
R6
The constraints follow:
0.5 X 1 +
X 2 ≤ 600
12.5 X 1 + 10 X 2 ≤ 10000
X1
≤ 700
X1, X 2
≥0
Total available resource
600
10000
4
Now, we must ‘prepare’ this linear system of inequalities before using the simplex method. We
convert all inequalities to equalities by introducing slack variables. The new model is
Maximise Profit = 4 X 1 + 6 X 2 + 0 S1 + 0 S 2 + 0 S 3
subject to
0.5 X 1 + X 2 + S1
= 600
12.5 X 1 + 10 X 2
+ S2
+ S 3 = 700
X1
X1, X 2
= 10000
≥0
where S1 , S 2 and S 3 are slack variables.
Tableau of the initial solution
Solution mix
S1
S2
S3
Quantity
600
10000
700
X1
0.5
12.5
1
X2
1
10
0
S1
1
0
0
S2
0
1
0
S3
0
0
1
The first simplex tableau
Cj
0
4
6
0
0
0
Solution mix
S1
Quantity
600
X1
X2
S1
S2
S3
0.5
1
1
0
0
S2
S3
Zj
10000
12.5
10
0
1
0
700
1
0
0
0
1
0
0
0
0
0
0
4
6
0
0
0
0
0
Cj − Zj
[Explanation for notation
C j (row) = coefficients of variables in the objective function
C j (column) = coefficients of solution mix variables in the objective function
Z j = sum of the products of the C j column elements and corresponding variable column
elements. For example, the bold zero in the table is obtained as follows:
(0 × 1) + (0 × 10) + (0 × 0)
Note that the shaded cell gives the optimal profit.
]
5
The simplex procedure can be summarised as follows:
1.
Find the pivot column – the column that has the greatest positive entry in the
C j − Z j row. If there are no positive entries in a tableau, it means that the optimal
solution has already been reached.
Cj
0
0
0
0
0
Quantity
600
X1
X2
S1
S2
S3
0.5
1
1
0
0
10000
12.5
10
0
1
0
S3
Zj
700
1
0
0
0
1
0
0
0
0
0
0
4
6*
0
0
0
Cj − Zj
Find the pivot row. Divide each solution mix entry in the Quantity column by its
corresponding entry in the pivot column and record the result in the Ratio column.
The smallest non-negative ratio determines the pivot row.
Cj
0
0
0
4
6
0
0
0
Solution mix
S1
S2
Quantity
600
X1
X2
S1
S2
S3
Ratio
0.5
1
0
0
10000
12.5
1
10
0
1
0
600*
1000
S3
Zj
700
1
0
0
0
1
∞
0
0
0
0
0
0
4
6
0
0
0
Cj − Zj
3.
6
Solution mix
S1
S2
0
2.
4
Pivot
row
Select the pivot element in order to start applying basic row operations. The pivot
element is found at the intersection of the pivot row and the pivot column. It indicates
which variable will enter the solution mix and which one will leave. In the above
table, the pivot element is 1 (indicated in bold). We deduce that X 2 will enter the
solution mix and that S1 will leave. Why S1 ? Simply because, it is the solution mix
slack variable which is found in the first row.
6
The second simplex tableau
Now, we have to perform basic row operations in order to make the X 2 column entries
identical to those in the unit vector that was in the S1 column. In this context, we will consider
only two (out of the possible three) basic row operations:
1.
2.
Multiply a row by a non-zero constant.
Add a multiple of a row to another row.
Cj
0
4
6
0
0
0
Solution mix
S1
Quantity
600
X1
X2
S1
S2
S3
Ratio
0.5
1
1
0
0
S2
S3
Zj
10000
12.5
0
1
0
700
1
10
0
600*
1000
0
0
1
∞
0
0
0
0
0
0
4
6
0
0
0
0
0
Cj − Zj
Pivot
row
For example, if we have to make 10 (indicated in bold above) become 0, we have to take 10
subtract 10 times1 (the pivot element). In so doing, all the entries of the second row must also
subtract 10 times the entries in the first. Remember, we only consider the entries in the rows
involving the solution mix variables – in this case, S 2 and S 3 . The last entry in the pivot
column being a zero, we don’t need any further basic row operations – at least one good thing
here! The entries of the Z j and C j − Z j rows need to be recalculated.
The complete second simplex tableau looks like
Cj
6
0
0
4
6
0
0
0
Solution mix
X2
S2
Quantity
600
X1
X2
S1
S2
S3
0.5
1
1
0
0
4000
7.5
0
–10
1
0
S3
Zj
700
1
0
0
0
1
3600
3
6
6
0
0
1
0
–6
0
0
Cj − Zj
This procedure is repeated until there are no more positive entries in the C j − Z j row, in which
case we would have reached the optimal solution.
7
The third simplex tableau
Cj
6
0
0
4
6
0
0
0
X1
X2
S1
S2
S3
0.5
1
1
0
0
Ratio
1200
0
–10
1
0
533*
0
0
0
1
700
Solution mix
X2
S2
Quantity
600
S3
Zj
700
7.5
1
3600
3
6
6
0
0
1
0
–6
0
0
4000
Cj − Zj
Pivot
row
pivot column
The largest positive entry in the last row is 1 so that the pivot column is the X 1 column.
The ratios of Quantity: X 1 are recalculated and it is found that 533 is the smallest non-negative
value, meaning that the pivot row is the S 2 row. Thus, the pivot element is 7.5 (indicated in bold)
so that S 2 has to leave the solution mix for X 1 to enter. Logically, we don’t expect any other
tableaus to come up since both decision variables have entered the solution mix!
Using basic row operations, we first divide the second row entries by 7.5 to obtain a 1 in
its place before converting the 0.5 and 1 in the same column to zeros.
Cj
6
0
0
4
6
0
0
0
X1
X2
S1
S2
S3
0.5
1
1
0
0
0
–1.33
0.13
0
700
1
1
0
0
0
1
3600
3
6
6
0
0
1
0
–6
0
0
Solution mix
X2
Quantity
600
S2
S3
Zj
533
Cj − Zj
Basic row operations: ‘First row minus half second row’ and ‘third row minus second row’.
Cj
6
4
0
4
6
0
0
0
Solution mix
X2
X1
Quantity
333
X1
X2
S1
S2
S3
0
1
1.67
–0.067
0
533
1
0
–1.33
0.13
0
S3
Zj
167
0
0
1.33
–0.13
1
4130
4
6
4.67
0.13
0
0
0
–4.67
–0.13
0
Cj − Zj
8
As predicted, this is the final simplex tableau since there are no more positive entries in
the last row. We have therefore reached the optimal solution! Remember that the slack variable
S 3 is not a decision variable.
Interpretation of the final simplex tableau
1.
The final (optimal) solution is X 1 = 533 and X 2 = 333 so that the optimal profit is
R4130. In fact, this solution mix corresponds to a corner-point (vertex) on graph.
Note that each number in the columns of the variables occurring in the solution mix
represents a unit vector.
2.
The entry at the intersection of the S1 column and the C j − Z j row, –4.67,
represents the amount by which the total profit will decrease for every unit of S1
added to the solution mix, that is, R4.67. If the available machine hours are reduced
by one hour, X 2 will be reduced by 1.67 units in the solution mix whereas X 1 will
be increased by 1.33 units.
Decrease in profit = R6 × 1.67 = R10.00
Increase in profit = R4 × 1.33 = R 5.33
Net decrease in profit
= R 4.67
3.
The entry at the intersection of the S1 column and the Z j row, 4.67, represents the
amount by which the total profit will increase if one additional machine hour is made
available for the production of product A and production B (that is, 601 machine
hours). The shadow price with respect of one machine hour is thus R4.67.
4.
The entry at the intersection of the S 2 column and the Z j row, 0.13, is the shadow
price with respect to one kilogram of raw material. If the available raw material is
increased by one kilogram, the total profit will increase by 13 cents.
5.
The entry at the intersection of the S 3 column and the Z j row, 0, is the shadow
price with respect to the demand for product A. Since the current solution mix does
not fully satisfy the demand for product A, ( S 3 = 1.67) , an increase in the demand for
this product will have no influence on the total profit.
9
Example 2 (slack and artificial variables) [Minimisation]
A producer of fertiliser receives an order to supply exactly 5000 kg of fertiliser,
consisting of a special mixture of phosphate and ammonium nitrate. The mixture must not contain
more than 1500 kg of phosphate and not less than 750 kg of ammonium nitrate. The cost per
kilogram of phosphate and ammonium nitrate are R6 and R8 respectively. The producer would
like to determine the ratio in which raw materials must be combined so as to limit costs to a
minimum.
Formulate a linear programming model for the above problem and hence find the optimal
solution by using the simplex method.
Solution 2
We first identify our decision variables, objective function and thus write down the constraints.
Let the decision variables
X 1 = “number of kg of phosphate” and
X 2 = “number of kg of ammonium nitrate”
The objective function may be formulated as follows:
Minimise Cost = 6 X 1 + 8 X 2
The constraints are
X 1 + X 2 ≤ 5000
X1
≤ 1500
X 2 ≥ 700
X1, X 2 ≥ 0
[Note that this the very same example on artificial variables mentioned above.]
The above linear programming model is thus modified as
Minimise cost = 6 X 1 + 8 X 2 + 0S1 + 0 S 2 + MA1 + MA2
Subject to
1X 1 + 1X 2 + 0 S1 + 0 S 2 + 1A1 + 0 A2 = 5000
1X 1 + 0 X 2 + 1S1 + 0 S 2 + 0 A1 + 0 A2 = 1500
0 X 1 + 1 X 2 + 0 S1 − 1S 2 + 0 A1 + 1A2 = 750
X 1 , X 2 , S1 , S 2 , A1 , A2 ≥ 0
before using the simplex procedure.
10
Tableau of the initial solution
Solution mix
A1
S1
A2
Quantity
5000
1500
750
X1
1
1
0
X2
1
0
1
S1
0
1
0
S2
0
0
–1
A1
1
0
0
A2
0
0
1
The first simplex tableau
Cj
M
0
M
6
8
0
0
M
M
Solution mix
A1
Quantity
5000
X1
X2
S1
S2
A1
A2
1
1
0
0
1
0
S1
A2
Zj
1500
1
0
1
0
0
0
750
0
1
0
–1
0
1
5750M
M
2M
0
–M
M
M
6–M
8–2M
0
M
0
0
Cj − Zj
This time, to identify the pivot column, we have to find the largest negative entry in the C j − Z j
row (since this is a minimisation problem). This is the X 2 column because, remember, M is a
large positive number so that 8–2M is more negative than 6–M.
Cj
M
0
M
6
8
0
0
M
M
Solution mix
A1
Quantity
5000
X1
X2
S1
S2
A1
A2
1
1
0
0
1
0
S1
A2
Zj
1500
1
0
1
0
0
0
750
0
1
0
–1
0
1
5750M
M
2M
0
–M
M
M
6–M
8–2M*
0
M
0
0
Cj − Zj
Pivot column
11
To identify the pivot row, the ratios Quantity: X 2 have to be calculated and the Ratio
column created. Again, the smallest non-negative entry determines the pivot row.
Cj
M
0
M
6
8
0
0
M
M
Solution mix
A1
S1
Quantity
5000
X1
X2
S1
S2
A1
A2
1
1
0
0
1
0
1500
1
0
1
0
0
0
Ratio
5000
∞
A2
Zj
750
0
1
0
–1
0
1
750*
5750M
M
2M
0
–M
M
M
6–M
8–2M
0
M
0
0
Cj − Zj
Pivot
row
The pivot element is therefore 1 (indicated in bold above), that is, X 2 enters and A2 leaves the
solution mix. By means of basic row operations, we make the X 2 column become a unit vector
and calculate the values of Z j and C j − Z j accordingly to obtain the following tableau:
The second simplex tableau
Cj
M
0
6
8
0
0
M
M
Solution mix
A1
S1
Quantity
4250
X1
X2
S1
S2
A1
A2
1
0
0
1
1
–1
1500
1
0
1
0
0
0
X2
Zj
750
0
1
0
–1
0
1
4250M + 6000
M
8
0
M–8
M
–M+8
6–M
0
0
–M+8
0
2M–8
8
Cj − Zj
The pivot column, row (the ratios are recalculated) and element are again identified.
Cj
M
0
8
Solution
mix
A1
S1
X2
Zj
Cj − Zj
6
8
0
0
M
M
Quantity
X1
X2
S1
S2
A1
A2
Ratio
4250
1
0
0
1
1
–1
4250
1500
0
1
0
0
0
1500*
750
1
0
1
0
–1
0
1
∞
4250M + 6000
M
8
0
M–8
M
–M+8
6–M*
0
0
–M+8
0
2M–8
12
After applying the relevant basic row operations, we obtain
The third simplex tableau
Cj
M
6
6
8
0
0
M
M
Solution mix
A1
Quantity
2750
X1
X2
S1
S2
A1
A2
0
0
–1
1
1
–1
X1
X2
Zj
1500
1
0
1
0
0
0
750
0
1
0
–1
0
1
2750M + 15000
6
8
6–M
M–8
M
–M+8
0
0
M–6
–M+8
0
2M–8
8
Cj − Zj
The procedure is repeated since there is still an artificial variable in the solution mix (that is, a
negative entry in the last row, –M+8; the pivot column, row (the ratios are recalculated) and
element are again identified.
Cj
M
6
6
8
0
0
M
M
Solution mix
A1
X1
Quantity
2750
X1
X2
S1
S2
A1
A2
0
0
–1
1
–1
1500
1
0
1
1
0
0
0
X2
Zj
750
0
1
0
–1
0
1
2750M + 15000
6
8
6–M
M–8
M
–M+8
0
0
M–6
–M+8*
0
2M–8
8
Cj − Zj
The fourth (and final) simplex tableau
Cj
0
6
8
6
8
0
0
M
M
Solution mix
S2
Quantity
2750
X1
X2
S1
S2
A1
A2
0
0
–1
1
1
–1
X1
X2
Zj
1500
1
0
1
0
0
0
3500
0
1
–1
0
1
0
37000
6
8
–2
0
8
0
0
0
2
0
M–8
M
Cj − Zj
Ratio
2750*
∞
–750
13
Interpretation of the final tableau
In order to minimise cost, the mix must consist of 1500 kg phosphate ( X 1 ) and 3500 kg
of ammonium nitrate ( X 2 ). In this mix, the minimum requirement set for ammonium nitrate will
be exceeded by 2750 kg ( S 2 ). The minimum total cost amounts to R37000.
14
(3)
Extraordinary situations
The following circumstances may also arise when the simplex method is applied.
CASES
HOW TO RECOGNISE THEM IN A SIMPLEX TABLEAU
Multiple optimal solutions
If the index of a variable not contained in the solution mix is zero
in the final tableau, it means that there is more than one optimal
solution. Mathematically, the inclusion of this variable in the
solution mix will not change the value of the objective function.
Degeneracy
One or more constraints are redundant – as a result of this, two
equally small non-negative ratios may occur in the table when the
pivot row is to be identified. If the incorrect one is selected, the
only problem is that an extra iteration will be carried out.
Infeasibility
This case is easily recognised if there is still an artificial variable
in the solution mix whilst all the indices indicate that an optimal
solution has been found.
Unbounded solutions
When identifying the pivot row, all the ratios will be either
negative or ∞ . It means that one or more variables can increase
indefinitely without the constraints being exceeded in a
maximisation problem.
15
(4)
Key concepts
Artificial variable
A variable that is only used in equations and ‘ ≥ ’ constraints to find a
feasible initial solution.
Index row C j − Z j
This represents the net amount by which the objective function will
increase (if positive) or decrease (if negative) if one additional unit of a
variable is added to the solution mix.
Lower limit for
shadow prices
The level to which a limited resource can be reduced with a constant
reduction in the profit for each unit by which the resource is reduced.
Pivot column
The column with the largest positive (negative) entry in the C j − Z j row
in the case of a maximisation (minimisation) problem.
Pivot row
The row with the smallest non-negative ratio between the quantity column
and the number in the pivot column. The ratio represents the maximum
quantity of the pivot variable that can be included in the next solution
mix. The variable in the pivot row is replaced by the pivot variable in the
next simplex tableau.
Pivot variable
The variable that will enter the solution mix.
Principal row
The new row in the next simplex tableau that replaces the pivot row in the
previous tableau. The value of the principal row is determined by dividing
the pivot row by the pivot number.
Shadow prices
The values of the slack variables in the Z j row. A shadow price is the
value of one additional unit of resource that is already being fully utilised.
Slack variable
A variable added to a ‘ ≤ ’ constraint to change it to an equality. It
represents the unused portion of the resource.
Solution mix
A combination of variables in the objective function representing a
possible solution to the linear programming problem.
Surplus variable
A variable added to a ‘ ≥ ’ constraint to change it to an equality. It
represents the quantity by which the utilisation of a resource exceeds the
prescribed minimum.
Upper bound for
shadow prices
The level to which a limited resource can be reduced while the profit
decreases by a constant amount with the removal of each unit of the
resource.
Z j row
It represents the total value of the solution mix in the quantity column. It
further shows the amount by which the total value will be reduced if one
additional unit of a slack variable is added to the solution mix.
16
Exercises
Questions for discussion
1.
Briefly explain the purpose of the simplex method as well as the procedure that is
followed.
2.
Enumerate the points of similarity between the simplex method and the graphical
method. Under what circumstances will preference be given to each method?
3.
Define the following terms and indicate under what circumstances each is used:
(a) slack variable
(b) surplus variable
(c) artificial variable
4.
How will you go about selecting the following after the first simplex tableau has been
completed?
(a) the pivot column
(b) the pivot row
(c) the pivot element
5.
Explain fully why the row with the smallest ratio is identified as the pivot row.
6.
Describe the lost important differences between maximisation and minimisation
problems in the application of the simplex method.
7.
Why should an artificial variable appear in the optimal solution mix and what should
the decision-maker do if this happens?
8.
Explain what additional information is made available by the final simplex tableau.
17
Linear programming problem
Toskry Limited manufactures three types of videocassette tapes:
The L-500 with a recording time of two hours
The L-750 with a recording time of three hours
The P-2000 with a recording time of eight hours
The production manager consults you to help him determine the number of units of each type
of cassette tape to be manufactured per week, given an available production capacity of 450 hours
per week. The cassette tapes are packed in containers of 100 cassettes each, which must be
considered as a unit for production purposes.
The production manager has the following information available:
Production cost per unit (rands)
Production time per unit (mins)
L-500
L-750
P-2000
30
15
45
30
70
60
The following information is provided by the marketing manager:
The demand for the L-500 is virtually indefinite and it is dispatched to the trade as
it becomes available, without taking permanent orders.
There is a permanent order of 200 units of the L-750 per week. However, if more is
produced, it could be sold without difficulty.
The P-2000 is specially manufactured for Video Scene Limited, which has placed
an order of 150 units per week.
The profit margin of the company is the same for all three types of tapes.
Required
(a)
(b)
(c)
Formulate a linear programming model.
Construct an extended model that can be solved by means of the simplex method.
Find the optimal solution.
Solution to the linear programming problem
Minimise Cost = 30 X 1 + 45 X 2 + 70 X 3
Subject to
0.25 X 1 + 0.5 X 2 + X 3 ≤ 450
X2
≥ 200
X 3 = 150
X1, X 2 , X 3 ≥ 0
18
Minimise Cost = 30 X 1 + 45 X 2 + 70 X 3 + 0S1 + 0S 2 + MA1 + MA2
Subject to
0.25 X 1 + 0.5 X 2 + X 3 + S1
= 450
− S 2 + A1
X2
= 200
+ A2 = 150
X3
X 1 , X 2 , X 3 , S1 , S 2 , A1 , A2 ≥ 0
Tableau of the initial solution
Solution mix
S1
Quantity
450
X1
X2
X3
S1
S2
A1
A2
0.25
0.5
1
1
0
0
0
A1
A2
200
0
1
0
0
–1
1
0
150
0
0
1
0
0
0
1
The first simplex tableau
Cj
0
30
Solution mix
S1
Quantity X 1
450
0.25
A1
A2
Zj
M
M
45
70
0
0
M
M
X2
X3
S1
S2
A1
A2
0.5
1
1
0
0
0
Ratio
900
0
0
–1
1
0
200*
1
0
0
0
1
∞
200
0
150
0
1
0
350M
0
M
M
0
–M
M
M
30
45–M*
70–M
0
M
0
0
30
45
70
0
0
M
M
Cj − Zj
The second simplex tableau
Cj
0
45
M
Solution mix
S1
X2
Quantity
350
X1
X2
X3
S1
S2
A1
A2
0.25
0
1
1
0.5
–0.5
0
200
0
1
0
0
–1
1
0
A2
Zj
150
0
0
1
0
0
0
1
9000+150M
0
45
M
0
–45
45
M
30
0
70–M*
0
45
M–45
0
Cj − Zj
Ratio
350
∞
150*
19
The third and final simplex tableau
Cj
0
30
45
70
0
0
M
M
Solution mix
S1
Quantity
200
X1
X2
X3
S1
S2
A1
A2
0.25
0
0
1
0.5
–0.5
–1
X2
X3
Zj
200
0
1
0
0
–1
1
0
150
0
0
1
0
0
0
1
19500
0
45
70
0
–45
45
70
30
0
0
0
45
M–45
M–70
45
70
Cj − Zj
Optimal solution:
X 2 = 200 , X 3 = 150
Minimum profit = R 19 500
Courtesy Mr. Rajesh Gunesh
(E-mail: [email protected])