Chapter 3.Simplex Method hand out last.pdf

Operations Research course material on simplex method 2020
COMPILED BY TSEGAY BERHE [ MSc IN PRODUCTION ENGINEERING & MANAGEMENT ] 1
Contents
3. Linear Programming: The Simplex Method............................................................................ 2
3.1. Solving an LP Problem by the Simplex Algorithm........................................................... 5
3.2. Simplex Method with More Than Two Variables............................................................. 8
3.3. Minimization Problems Case ......................................................................................... 10
3.3.1 Two Phase Method: Linear Programming................................................................. 11
3.3.2. Big M Method: Linear Programming......................................................................... 15
3.4. Special Case in Simplex Method.................................................................................... 18

3. Linear Programming: The Simplex Method
Realistic linear programming problems often have several decision variables and many constraints. Such
problems cannot be solved graphically; instead an algorithm such as the simplex procedures is used.
Simplex method is thus the most effective analytical method of solving linear programming problems.
Simplex algorithm – a general procedure for solving linear programming (LP) problems developed by
George Dantzig in 1947– requires the LP model to be in standard form.
The simplex method is an ITERATIVE or “step by step” method or repetitive algebraic approach that moves
automatically from one basic feasible solution to another basic feasible solution improving the situation
each time until the optimal solution is reached at.
Advantages and Characteristics of the simplex method
i. More realistic approach as it is not limited to problems with two decision variables
ii. Systematically examines basic feasible solutions for an optimal solution.
iii. Based on the solutions of linear equations (equalities) using slack variables to achieve
equality.
Properties of the LP model in standard form:
i. All the constraints are equations with non-negative right-hand-side (RHS) values.
ii. All the variables are non-negative.
iii. The objective of the problem is to either maximize or minimize the objective function value.
Objective Function
The maximization of a function is equivalent to the minimization of the negative of the same function.
Equivalence means that for the same set of constraints, the optimum values of the variables are the same
in both cases, although the values of the objective functions will appear with opposite signs but the same
magnitude.

𝑂𝑝𝑡𝑖𝑚𝑖𝑧𝑒(𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛), 𝑍 = ∑ 𝐶𝑖 ∗ 𝑋𝑖
𝑛
𝑖
Xi = decision variables
Ci = objective function coefficients
Constraints
𝑠𝑢𝑏𝑗𝑒𝑐𝑡𝑒𝑑 𝑡𝑜 𝑐𝑜𝑛𝑠𝑡𝑟𝑖𝑎𝑛𝑡𝑠
{
𝑎11𝑋1 + 𝑎12𝑋2 + . . . + 𝑎1𝑛𝑋𝑛 (≤, =, ≥)𝑏1
𝑎11𝑋1 + 𝑎12𝑋2 + . . . + 𝑎1𝑛𝑋𝑛 (≤, =, ≥)𝑏2
.
.
.
.
.
𝑎𝑚1𝑋1 + 𝑎𝑚2𝑋2 + . . . + 𝑎𝑚𝑛𝑋𝑛 (≤, =, ≥)𝑏𝑚
bi = constraint levels [RHS]
aij = constraint coefficients
1. A constraint of the type ≤ can be converted into an equation by adding a slack variable to the left side
of the constraint.
Ex.𝟒𝒙 + 𝟓𝒚 ≤ 𝟖 is converted into 𝟒𝒙 + 𝟓𝒚 + 𝒔𝟏 = 𝟖 in the standard form, with 𝑠1 ≥ 0.
If the constraint represents the limit on the availability of a resource, 𝑠1 will represent the slack or
unused amount of that resource.
2. A constraint of the type ≥ can be converted into an equation by adding a surplus variable to the left
side of the constraint.
Ex. 𝟐𝒙 + 𝟒𝒚 − 𝟔𝒛 ≥ 𝟗is converted into 𝟐𝒙 + 𝟒𝒚 − 𝟔𝒛 − 𝒔𝟐 = 𝟗 in the standard form, with 𝑠2 ≥
0.If the constraint represents the minimum production requirement, 𝑠2 will represent the items
produced beyond the minimum required.
3. The RHS can be made non-negative by multiplying both sides by -1, and the direction of an inequality
is reversed when both sides are multiplied by a negative number.
Ex.−𝟑𝒙 + 𝟕𝒚 ≤ −𝟏𝟓 is equivalent to 𝟑𝒙 − 𝟕𝒚 ≥ 𝟏𝟓 and is then converted into 𝟑𝒙 − 𝟕𝒚 − 𝒔𝟑 =
𝟏𝟓 in the standard form with 𝑠3 ≥ 0.

The Simplex Method;
The simplex algorithm considers the corner point feasible (CPF) solutions as the candidates for the optimal
mix. It starts at the origin, referred to as the starting solution, then moves to an adjacent corner point,
the choice dependent on which point will improve the value of the objective function more. The iterative
process is repeated by moving to another adjacent corner point, until the corner point where the value of
the objective function is optimized, is finally reached. The simplex algorithm is governed by two general
rules:
a. The next corner point must be adjacent to the current one. Thus, moving from one corner point to
the next involves moving along the edges of the solution area, and never across it.
b. The solution can never go back to a previously considered extreme point.
I. Maximization
Recall the example of the furniture manufacturer that produces wooden tables and chairs, formulated below:
Maximize: 𝑍 = 6𝑋 + 8𝑌
Subject to: 30𝑋 + 20𝑌 ≤ 300
5𝑋 + 10𝑌 ≤ 110
𝑋, 𝑌 ≥ 0
The standard form of the LP model is given by:
Maximize: 𝑍 = 6𝑋 + 8𝑌 + 0𝑆1 + 0𝑆2
Subject to: 30𝑋 + 20𝑌 + 𝑆1 = 300
5𝑋 + 10𝑌 + 𝑆2 = 110
𝑋, 𝑌, 𝑆1, 𝑆2 ≥ 0

3.1. Solving an LP Problem by the Simplex Algorithm
Step 0. Using the standard form, determine a starting basic feasible solution by setting n – m appropriate
variables at zero level [let the value of X and Y equals to zero and the value of the slack or surplus
variables equals the RHS].
Step 1. Select an entering variable from among the current non basic variables which, when increased
above zero, can improve the value of the objective function. If none exists, stop; the current basic solution
is optimal. Otherwise, go to step 2[ test optimality].
A variable that has the highest negative value in the Zj-Cj row or highest positive value in the Cj - Zj
(Maximization case) is the entering variable. A variable that has the most positive value in the Zj-Cj
row or the highest negative value Cj - Zj (Minimization case).
I. If all Zj - Cj ≥ 0 or Cj - Zj ≤ 0, then the basic feasible solution is optimal (Maximization case)
II. If all Zj - Cj ≤ 0, or Cj-Zj ≥0 then the basic feasible solution is optimal (Minimization case)
Step 2. Select a leaving variable by applying the minimum ratio test from among the current basic
variables that must be set to zero when the entering variable becomes basic. The row with the worst
negative/largest positive and minimum replacement ratio (or both maximization & minimization cases
respectively).
Minimum Ratio Test
i. Pick out each coefficient in the pivot column that is strictly positive (≥0).
ii. Divide each of these coefficients into the right-side entry for the same row.
iii. Identify the row that has the smallest of these ratios.
iv. The basic variable for that row is the leaving basic variable, so replace that variable by the entering
basic variable in the basic variable column of the next simplex tableau. Put a box around this row
and call it the pivot row. Also call the number that is in both boxes the pivot number.
Step 3. Determine the new basic solution by making the entering variable basic and the leaving variable
non basic. Go to step 1. Or Solve for the new BF solution by using elementary row operations (multiply or
divide a row by a nonzero constant; add or subtract a multiple of one row to another row) to construct a
new simplex tableau in proper form from Gaussian elimination below the current one, and then return to
the optimality test. The specific elementary row operations that need to be performed are listed below.
i. Divide the pivot row by the pivot number. Use this new pivot row in steps 2 and 3.
ii. For each other row (including row 0) that has a negative coefficient in the pivot column, add to
this row the product of the absolute value of this coefficient and the new pivot row.
iii. For each other row that has a positive coefficient in the pivot column, subtract from this row the
product of this coefficient and the new pivot row.
(𝐸𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑 𝑟𝑜𝑤 ÷ 𝑝𝑖𝑣𝑜𝑡) = (𝐸𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑟𝑒𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑟𝑜𝑤)
(
𝐸𝑙𝑒𝑚𝑒𝑛𝑡𝑠
𝑖𝑛 𝑛𝑒𝑤 𝑟𝑜𝑤
) = (
𝐸𝑙𝑒𝑚𝑒𝑛𝑡𝑠
𝑖𝑛 𝑜𝑙𝑑 𝑟𝑜𝑤
) − (
𝐼𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓
𝑜𝑙𝑑 𝑟𝑜𝑤
.
) (
𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔
𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛
𝑟𝑒𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑟𝑜𝑤
.
)

Illustration:
A manufacturing firm produces two products, A and B. Each of these products must be processed through
two different machines. One machine has 12 hours and the second machine has 8 hours of available
capacity. Each unit of product A requires two hours of time on both machines. Each unit of product B
requires three hours of time on the first machine and one hour on the second machine. The incremental
profit is $300 per unit of product A and $350 per unit of product B, and the firm can sell as many units of
each product as it can manufacture.
The objective of the firm is to maximize profits. The problem is to determine how many units of
product A and product B should be produced within the limits of available machine capacities.
Formulation
Let X1 – number of units of product A to be produced
X2 – number of units of product B to be produced
Maximize: Z = 300X1 + 350X2 (objective function/profit function –computes the total profit with
any given value of the basic variables X1 and X2)
Subject to: 2X1 + 3X2 ≤ 12 (constraint on the limited availability of Machine I)
2X1 + X2 ≤ 8 (constraint on the limited availability of Machine II)
X1 , X2 ≥ 0 (non-negativity constraints)
The standard form of the model is:
Maximize: Z = 300X1 + 350X2 + 0S1 + 0S2 (Since any unused machine time does not contribute to
profit, the coefficients of the slack variables in the objective function are zeros.)
Subject to: 2X1 + 3X2 + S1 = 12
2X1 + X2 + S2 = 8
X1 , X2, S1, S2 ≥ 0
Note: Only the functional constraints need to be transformed into equations. The non-negativity
constraints are left as inequalities, since they are not affected by the simplex solution method.
Setting up the initial tableau
In the simplex method, an initial solution needs to be established. The simplex starting solution for the firm
is to manufacture none of either product, that is, X1 = 0 and X2 = 0. This solution is technically feasible but
not financially attractive. It is represented in the graphical solution by the origin with coordinates (0,0).
Thus,
X1 = 0
X2 = 0
S1 = 12 – 2(0) – 3(0) = 12 hours unused for Machine I
S2 = 8 – 2(0) – 0 = 8 hours unused for Machine II

This solution contains only the slack variables S1 and S2. Substituting the values of the variables (X1 and
X2) and the slack variables in the objective function gives the following profit:
Profit (Z) = 300X1 + 350X2 + 0S1 + 0S2
= 300(0) + 350(0) + 0(12) + 0(8)
= 0
This first feasible solution is shown in the initial simplex tableau as
Coeff.objective
Coeff
B.v
Basic
variable[B.v] X1 X2 X3 S1 S2 basic solution Minimum ratio
0 S1
Coefficient of the variables in the constraints Right hand side
0 S2
0 S3
Zj
Cj-Zj
𝑍𝑗 = ∑ 𝐶𝑜𝑒𝑓𝑓. 𝐵. 𝑉𝑗 ∗ 𝑋𝑗
𝑗
1 ,
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑎𝑡𝑖𝑜 =
Corresponding 𝑏𝑎𝑠𝑖𝑐𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑣𝑎𝑙𝑢𝑒[𝑅𝐻𝑆]
Corresponding 𝑘𝑒𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑣𝑎𝑙𝑢𝑒[𝑃𝑖𝑣𝑜𝑡 𝑐𝑜𝑙𝑢𝑚𝑛]
Initial tableau
Cj 300 200 0 0
C.B.V B.V X1 X2 S1 S2 Quantity
Min
ratio
0 S1 2 3 1 0 12 6
0 S2 2 1 0 1 8 4
Zj 0 0 0 0
Cj-Zj 300 200 0 0
From the initial table the value of Cj-Zj ≥0, which shows the initial table is not optimal. Then the pivot
column [entering variable] is the one with the maximum Cj-Zj value, which is X1. Though, in order to
determine the Pivot row [ leaving variable] divide the RHS(quantity) by the corresponding column values
of X1. Then from this operation S2 because the minimum ratio along S2 is the minimum, Min {6,4} =4 .

Second tableau
Cj 300 200 0 0
Determining new value of each row
C.B.V B.V X1 X2 S1 S2 Quantity
Min
ratio
0 S1 0 2 1 -1 4 2 R1=R1
old
-2*Rnew
300 X1 1 0.5 0 0.5 4 8 Rnew
=R2
old
/key element
Zj 300 150 0 150
Min ratio=Quantity/X2 values along
column
Cj-Zj 0 50 0
-
150
From the Second tableau the value of Cj-Zj ≥0, which shows the initial table is not optimal. Then the pivot
column [entering variable] is the one with the maximum Cj-Zj value, which is X2. Though, in order to
determine the Pivot row [ leaving variable] divide the RHS(quantity) by the corresponding column values
of X2. Then from this operation S1 is the leaving variable due to minimum ratio value, Min {2,8} =2.
Third tableau
Cj 300 200 0 0
Quantity
C.B.V B.V X1 X2 S1 S2
Min
ratio
200 X2 0 1 0.5 -0.5 2
300 X1 1 0 -0.25 0.75 3
Zj 300 200 25 125 1300
Cj-Zj 0 0 -25 -125
From the third tableau the all the values of Cj-Zj ≤0, which shows the third tableau is the optimal table
with X1=3, X2=2, and Z =1300
3.2. Simplex Method with More Than Two Variables
➢ In previous section we discussed the simplex method of linear programming problem with two
decision variables. The simplex method computational procedure can be readily extended to
linear programming problems with more than two variables.
Example
Maximize
22X1 + 6X2 + 2X3
Subject to:
10X1 + 2X2 + X3 ≤ 100
7X1 + 3X2 + 2X3 ≤ 72

2X1 + 4X2 + X3 ≤ 80
X1, X2, X3 ≥ 0
➢ In order to solve this Lpp model through Simplex model, we should introduce slack
variables S1, S2 and S3 to change the inequalities signs in to equal sign equation. Thus,
the problem can be stated as
Maximize
22X1 + 6X2 + 2X3
Subject to:
10X1 + 2X2 + X3 + S1= 100
7X1 + 3X2 + 2X3+ S2= 72
2X1 + 4X2 + X3+ S3 = 80
X1, X2, X3, S1, S2, S3 ≥ 0
➢ Then from this standard for of the Lpp, the initial basic variables are S1, S2, S3 and the initial
basic solution of S1, S2, S3 are 100,72,80 respectively [ letting X1,X2 and X3 equals to zero]
INTIAL TABLE
Cj 22 6 2 0 0 0
CBV
Basic
Variable
X1 X2 X3 S1 S2 S3
Basic
solution
Min
Ratio
0 S1 10 2 1 1 0 0 100 10
0 S2 7 3 2 0 1 0 72 10.29
0 S3 2 4 1 0 0 1 80 40
Zj 0 0 0 0 0 0
Cj-Zj 22 6 2 0 0 0
1. Cj-Zj = 22 is the Largest positive value. Hence X1 should be taken as a basic
variable in the next iteration. [Entering variable or pivot column]
2. Calculate the minimum of the ratios
𝑀𝑖𝑛 ⌈
100
10
,
72
7
,
80
2
⌉ = 𝑀𝑖𝑛⌈10,10.29,40⌉ = 10
The variable S1 corresponding to which minimum occurs is made a non-basic variable [Leaving
row].
3. Then determine the new value of the entering variable and the other basic variable
values. [R represents row number]
𝑁𝑒𝑤 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒[𝑅𝑛𝑒𝑤
] =
𝐿𝑒𝑎𝑣𝑖𝑛𝑔 𝑟𝑜𝑤
𝑘𝑒𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
𝑵𝒆𝒘 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒃𝒂𝒔𝒊𝒄 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔[𝑅2
𝒏𝒆𝒘 = 𝑅2
𝒐𝒍𝒅 − 𝒄𝒓𝒐𝒔𝒔𝒑𝒐𝒎𝒅𝒊𝒏𝒈 𝒌𝒆𝒚 𝒗𝒂𝒍𝒖𝒆 ∗ 𝑅𝑛𝑒𝑤
𝑵𝒆𝒘 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒃𝒂𝒔𝒊𝒄 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔[𝑅3
𝒐𝒍𝒅 − 𝒄𝒓𝒐𝒔𝒔𝒑𝒐𝒎𝒅𝒊𝒏𝒈 𝒌𝒆𝒚 𝒗𝒂𝒍𝒖𝒆 ∗ 𝑅𝑛𝑒𝑤
𝑅𝑛𝑒𝑤
=
𝑅1
10
𝑅2
𝒐𝒍𝒅 − 𝟕 ∗ 𝑅𝑛𝑒𝑤
𝑅3
𝒐𝒍𝒅 − 𝟐 ∗ 𝑅𝑛𝑒𝑤

Then by substitute the values at table 1;the second table looks as shown below.
table 2
Cj 22 6 2 0 0 0
CBV
Basic
Variable
X1 X2 X3 S1 S2 S3
Basic
solution
Min
Ratio
22 X1 1 0.2 0.1 0.1 0 0 10 50
0 S2 0 1.6 1.3 -0.7 1 0 2 1.25
0 S3 0 3.6 0.8 -0.2 0 1 60 16.67
Zj 22 4.4 2.2 2.2 0 0
Cj-Zj 0 1.6 -0.2 -2.2 0 0
2.1. Then table two shows that the value of the net value (Cj-Zj ≥0 which is 1.6) is the
Largest positive value, this shows that table 2 is not optimal table. Hence X2
should be taken as a basic variable in the next iteration. [Entering variable or
pivot column]
2.2. Referring from the minimum ratio [replacement ratio] values S2 is the leaving
variable which will be a non-basic variable for the next iteration. By performing the same
analysis as above table 3 will be as follow;
Table 3
Cj 22 6 2 0 0 0
CBV
Basic
Variable
X1 X2 X3 S1 S2 S3
Basic
solution
Min
Ratio
22 X1 1 0 -0.06 0.19 -0.13 0 9.75
6 X2 0 1 0.81 -0.44 0.63 0 1.25
0 S3 0 0 -2.13 1.38 -2.25 1 55.5
Zj 22 6 3.5 1.5 1 0 222
Cj-Zj 0 0 -1.5 -1.5 -1 0
Then table 3 shows that the value of the net value (Cj-Zj ≤0 which is optimal table), with
X1=9.75, X2=1.25, S3=55.5 and Z =222
3.3. Minimization Problems Case
Before discussing solution, let us have an idea of what is known as Big M Method, which is
generally used to solve minimization problems. While solving the linear programming problems
by graphical method, we have seen an iso-profit line is drawn and at the origin and then it is moved

away from the origin to find the optima point. Similarly, an iso -cost line is drawn away from the
origin in minimization problem and moved towards the origin to find the optimal point.
3.3.1 Two Phase Method: Linear Programming
• LPP, in which constraints may also have > and = signs, we introduce a new type of
variable, called the artificial variable. These variables are fictitious and cannot have any
physical meaning. Two Phase Simplex Method is used to solve a problem in which some
artificial variables are involved. The solution is obtained in two phases.
• In Two Phase Method, the whole procedure of solving a linear programming problem
(LPP) involving artificial variables is divided into two phases.
Phase I : Simplex method is applied to a specially constructed Auxiliary LPP leading to a
final simplex table containing a BFS to the original problem.
• Step 1: Assign a cost –1 to each artificial variable and a cost 0 to all other variables in the
objective function. [ In phase I, we form a new objective function by assigning zero to
every original variable (including slack and surplus variables) and -1 to each of the artificial
variables]
• Step 2: Construct the auxiliary LPP in which the new objective function Z*
is to be
maximized subject to the given set of constraints.
• Then we try to eliminate the artificial variables from the basis.
• Step 3: Solve the auxiliary problem by simplex method until either of the following three
possibilities arise:
➢ Max Z*
< 0 and at least one artificial variable appears in the optimum basis at a
positive level. In this case given problem does not have any feasible solution.
➢ Max Z*
= 0 and at least one artificial variable appears in the optimum basis at a zero
level. In this case proceed to Phase II.
➢ Max Z*
= 0 and no artificial variable appear in the optimum basis. In this case also
proceed to Phase II.
• The solution at the end of phase I serves as a basic feasible solution for phase II. In phase
II, the original objective function is introduced and the usual simplex algorithm is used to
find an optimal solution. The following are examples of Two Phase Method.
Example 1

Minimize z = -3X1 + X2 - 2X3
subject to
X1 + 3X2 + X3 ≤ 5
2X1 – X2 + X3 ≥ 2
4X1 + 3X2 - 2X3 = 5
X1, X2, X3 ≥ 0
Solution.
• If the objective function is in minimization form, then convert it into maximization form.
Changing the sense of the optimization
Any linear minimization problem can be viewed as an equivalent linear maximization
problem, and vice versa. Specifically:
Minimize ∑ 𝐶𝑗𝑋𝑗
𝑛
𝑗=1 = Maximize∑ −𝐶𝑗𝑋𝑗
𝑛
𝑗=1
If Z is the optimal value of the left-hand expression, then -Z is the optimal value of the
right-hand expression.
Maximize z = 3X1 – X2 + 2X3
subject to
X1 + 3X2 + X3 ≤ 5
2X1 – X2 + X3 ≥ 2
4X1 + 3X2 - 2X3 = 5
X1, X2, X3 ≥ 0
Converting inequalities to equalities
X
1
+ 3X
2
+ X
3
+ S1 = 5
2X
1
– X
2
+ X
3
– S2 = 2
4X
1
+ 3X
2
- 2X
3
= 5
X1, X2, X3, S1, S2&S3 ≥ 0
Where:
S1 is a slack variable &S2 is a surplus variable. The surplus variable S2 represents the
extra units.
Now, if we let x1, x2 and x3 equal to zero in the initial solution, we will have S1 = 5 and
S2 = -2, which is not possible because a surplus variable cannot be negative. Therefore, we
need artificial variables.
X1 + 3X2 + X4 +S1 = 5
2X1 – X2 + X3 – S2 + A1 = 2
4X1 + 3X2 - 2X3 + A2 = 5

X1, X2, X3, X4, X5, A1, A2 ≥ 0, Where A1 and A2 are artificial variables.
Phase 1 of Two Phase Method
• In this phase, we remove the artificial variables and find an initial feasible solution of the
original problem. Now the objective function can be expressed as
Maximize 0X1 + 0X2 + 0X3 + 0S1+ 0S2 + (–A1) + (–A2)
subject to
X1 + 3X2 + X3 + S1 = 5
2X1 – X2 + X3 – S2 + A1 = 2
4X1 + 3X2 - 2X3 + A2 = 5
• X1, X2, X3, X4, X5, A1, A2 ≥ 0
• Initial basic feasible solution
The initial basic feasible solution is obtained by setting X1 = X2 = X3 = S2 = 0, Then we shall
have A1 = 2, A2 = 5, S1 = 5
Table 1 REMOVING ARTIFICAL
VARIABLES
Cj 0 0 0 0 0 -1 -1
CBV
Basic
Variable
X1 X2 X3 S1 S2 A1 A2
Basic
solution
Min
Ratio
0 S1 1 3 1 1 0 0 0 5 5
-1 A1 2 -1 1 0 -1 1 0 2 1
-1 A2 4 3 -2 0 0 0 1 5 1.25
Zj -6 -2 1 0 1 -1 -1
Cj-Zj 6 2 -1 0 -1 0 0
From this table X1 is the entering variable and A1 is the leaving variable from the basic variable.
Then remove A1 from the whole table [along row wise and column wise and perform the next
iteration with the same analysis as the usual simple x method.
Table 2 REMOVING the second ARTIFICAL
VARIABLES
Cj 0 0 0 0 0 -1
CBV
Basic
Variable
X1 X2 X3 S1 S2 A2
Basic
solution
Min
Ratio
0 S1 0 3.5 0.5 1 0.5 0 4 1.14
0 X1 1 -0.5 0.5 0 -0.5 0 1 -2.00

-1 A2 0 5 -4 0 2 1 1 0.20
Zj 0 -5 4 0 -2 -1
Cj-Zj 0 5 -4 0 2 0
From table 2; X2 is the entering variable and A2 is the leaving variable from the basic variable.
Then remove A2 from the whole table [along row wise and column wise] which will be as follow
and perform the next iteration with the same analysis as the usual simple x method.
Table 3 BY REMOVING ARTIFICAL VARIABLES[Ready for Phase II]
Cj 3 -1 2 0 0
CBV
Basic
Variable
X1 X2 X3 S1 S2
Basic
solution
Min
Ratio
0 S1 0 0 3.3 1 -0.9 3.3 -3.67
3 X1 1 0 0.1 0 -0.3 1.1 -3.67
-1 X2 0 1 -0.8 0 0.4 0.2 0.50
Zj 3 -1 1.1 0 -1.3
Cj-Zj 0 0 0.9 0 1.3
Table 3 shows that all the artificial variables are removed from the table and this table helps to
perform Phase II: Phase two helps us to obtain the optimal values of the linear programming
problem model.
Then optimal values are determined through the usual simplex method. Then X2 is the leaving
variable and S2 is the entering variable. Then the value of the New basic variables are determined
using the usual simplex method. Which will be;
Table 4 ; Phase II
Cj 3 -1 2 0 0
CBV
Basic
Variable
X1 X2 X3 S1 S2
Basic
solution
Min
Ratio
0 S1 0 2.25 1.5 1 0 3.75 2.50
3 X1 1 0.75 -0.5 0 0 1.25 -2.50
0 S2 0 2.5 -2 0 1 0.5 -0.25

Zj 3 2.25 -1.5 0 0
Cj-Zj 0 -3.25 3.5 0 0
From table 4 S1 is the leaving variable and X3 is the entering variable,and this table is not
optimale it requires further iteration.
Table 5; Phase II
Cj 3 -1 2 0 0
CBV
Basic
Variable
X1 X2 X3 S1 S2
Basic
solution
Min
Ratio
2 X3 0 1.5 1 0.667 0 2.5
3 X1 1 1.5 0 0.33333 0 2.5
0 S2 0 5.5 0 1.33333 1 5.5
Zj 3 7.5 2 2.33333 0 12.5
Cj-Zj 0 -8.5 0 -2.33 0
From table 5 the net values along Cj-Zj are all zeros or negative value this indicates
that table 5 is the optimal table with
X1=2.5,X2=0,X3=2.5 and Z=12.5
3.3.2.Big M Method: Linear Programming
If equation i does not have a slack (or a variable that can play the role of a slack), an artificial
variable, Ai, is added to form a starting solution similar to the all-slack basic solution. However,
because the artificial variables are not part of the original problem, a modeling “trick” is needed
to force them to zero value by the time the optimum iteration is reached (assuming the problem
has a feasible solution). The desired goal is achieved by assigning a penalty defined as:
𝐴𝑟𝑡𝑖𝑓𝑖𝑐𝑖𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 = {
−𝑀, 𝑖𝑛 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠
+𝑀, 𝑖𝑛 𝑚𝑎𝑥𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠
The big m method is a modified version of the simplex method in linear programming (LP) in
which we assign a very large value (M) to each of the artificial variables.
Key Steps of the big M method
Big M Method: Introducing slack, surplus, and artificial variables to form the modified problem

1. If any problem constraints have negative constants on the right side, multiply both sides by
-1 to obtain a constraint with a nonnegative constant. (remember to reverse the direction of
the inequality if the constraint is an inequality).
2. Introduce a slack variable for each ≤ constraint of the form
3. Introduce a surplus variable and an artificial variable in each ≥ constraint.
4. Introduce an artificial variable in each = constraint.
5. For each artificial variable A, add –MA to the objective function [Maximization problem]
or add +MA to the Minimization objective function. Use the same constant M for all
artificial variables.
Big M Method: solving the problem
1. Form the preliminary simplex tableau for the modified problem.
2. Use row operations to eliminate the M’s in the bottom row of the preliminary simplex
tableau in the columns corresponding to the artificial variables. The resulting tableau is the
initial simplex tableau.
3. Solve the modified problem by applying the simplex method to the initial simplex tableau
found in the second step.
4. Relate the optimal solution of the modified problem to the original problem.
a) if the modified problem has no optimal solution, the original problem has no
optimal solution.
b) if all artificial variables are 0 in the optimal solution to the modified problem,
delete the artificial variables to find an optimal solution to the original problem
c) if any artificial variables are nonzero in the optimal solution, the original problem
has no optimal solution.
Big M Method Example for illustration :
Maximize z = x1 + 5x2
subject to
3x1 + 4x2 ≤ 6
x1 + 3x2 ≥ 2
x1, x2 ≥ 0
Solution.
Converting inequalities to equalities
By introducing surplus variables, slack variables and artificial variables, the standard form
of LPP becomes.
Maximize Z= X1 + 5X2 + 0S1 + 0S2– MA1
subject to
3X1 + 4X2 + S1 = 6
X1 + 3X2 – S2 + A1 = 2
Xi ≥ 0, S1, S2& A1 ≥ 0
Where:
S1 is a slack variable
S2 is a surplus variable.
A1 is an artificial variable.
Initial basic feasible solution

X1 = X2 = S2 = 0
A1 = 2, S1 = 6
Initial table
Cj 1 5 0 0 -M
CBV BV X1 X2 S1 S2 A1
Basic
solution
Min
ratio
0 S1 3 4 1 0 0 6 1.50
-M A1 1 3 0 -1 1 2 0.67
Zj -M -3M 0 M -M
Cj -Zj 1+M 5+3M 0 0-M 0
Then from the initial table X2 is the entering variable to the basic solution whereas
A1 is the leaving variable from the basic solution. Then the new values of the basic
variables are performed as the usual simplex method.

Table 2
Cj 1 5 0 0 -M
Basic
solution
Min
ratio
0 S1 1.67 0 1 1.33 -1.33 3.33 2.50
5 X2 0.33 1 0 -0.33 0.33 0.67 -2.00
Zj 1.67 5 0 -1.67 1.67
Cj -Zj -0.67 0.00 0.00 1.67 (-M-1.67)
Then from table 2 S2 is the entering variable to the basic solution whereas S1 is the leaving
variable from the basic solution. Then the new values of the basic variables are performed as
the usual simplex method.
Table 3
Cj 1 5 0 0 (-M)
Basic
solution
Min
ratio
0 S2 1.25 0 0.75 1 -1.00 2.50
5 X2 0.75 1 0.25 0 0 1.50
Zj 3.75 5 1.25 0 0 7.50
Cj -Zj -2.75 0.00 -1.25 0.00 (-M+0)
From the result of table 3 the net values along Cj –Zj row are zeros and negative value
(Cj -Zj≤0), then this indicates that this table is the optimal table with;
X1=0,X2=1.5,S1=0,S2=2.5,Z=7.5
3.4. Special Case in Simplex Method
✓ Infeasible solution
✓ Multiple solution (infinitely many solution)
✓ Unbounded solution
✓ Degenerated solution
1. Infeasible solution; If artificial variable exists in the optimal table, it is infeasible.
A situation with no feasible solution may exist if the problem was formulated improperly.
Infeasibility comes about when there is no solution that satisfies all of the problems constraints.in
simplex method an infeasibility solution is indicated at the final tableau. In its all Cj-Zj row entries

will be the proper sign to imply optimality but an artificial variable (A) will still be in the solution
matrix.
Minimization
case
Cj 5 8 0 0 M
Basic
solution
Min
ratio
5 X1 1 1 -2 3 0 200
8 X2 0 1 1 2 0 100
M A2 0 0 0 -1 1 20
Zj 5 8 -2 31-M M
Cj -Zj 0 0 2 M-31 0
Even though all Cj –Zj are positive or 0[Cj –Zj ≥0][ i.e. the criterion for an optimal solution in a
minimizations case, no feasible solution is possible because an artificial variable (A2) remains in
the solution mix.
2. Multiple solution (infinitely many solution); If there is a zero under one or more non basic
variables in the last tableau (optimal solution tableau), then there is a multiple optimal
solution.
Maximization case
Cj 3 2 0 0
CBV BV X1 X2 S1 S2 Basic solution Min ratio
3 X2 1.5 1 1 0 6
0 S2 1 0 0.5 1 3
Zj 3 2 2 0
Cj -Zj 0 0 -2 0 Multiple solution
Max Z=3X1+2X2
X1=0, X2=6, S2=3 Z=12 Or
X1=3, X2=1.5, Z=12
N.B; the Cj-Zj value of the non-basic variable X1 is zero. Thus there is alternative optimal solution.

3. Unbounded solution; In simplex method, the condition of unbounds will be discovered prior
to reaching the final tableau. We will note that problem when trying to decide which variable
to remove from the solution mix is impossible. The procedure in unbounded solution is to divide
each basic solution value (quantity) column number by the corresponding pivot column
number.
The row with the smallest positive ratio is replaced. But, when determining the leaving variable of
any tableau, if there is no positive ratio (all the entries in the pivot column are negative and zeroes)
or undefined, then the solution is unbounded.
Maximization
case
Cj 6 9 0 0
CBV BV X1 X2 S1 S2
Basic
solution Min ratio
9 X2 -1 1 2 0 30 -30
0 S1 -2 0 -1 1 10 -5
Zj -9 9 18 0
Cj -Zj 15 0 -18 0 Unbounded
From this table we can determine which non basic variable will be entering into the basic
solution [X1] but we can’t determine which variable is the leaving variable [with negative
minimum or replacement ration]. then such problem is called problem with unbounded solution.
4. Degenerated solution [Tie for leaving basic variable] ; To select the outgoing variable, we
have to select the lowest ratio or limiting ratio in the replacement ratio column. However,
some times during the phase of solution the ration may be equal. This situation in linear
programming problem is known as degeneracy.
To solve degeneracy, the following method are used.
Method 1;
a) select any on row as you please. If you are lucky, you will get optimal solution otherwise,
the problem cycles or
b) identify the rows which are having same ratio say for example S1 and S2 rows having the
same ratio. In each case select the row, which contains the variable with the smaller
subscript. That the row contain S1 is selected as a leaving row. Additionally, suppose that
the row of variables X and Y are having the same ratio, then select the row containing X
as the key row.

Method 2;
a) divide the element of unit matrix by corresponding element of key column. Then verify the
ratios column wise in unit matrix starting from left to right. Once the ratios are unequal
the degeneracy is solved. Select the minimum ratio and the row containing that element is
the Key row [this is done to the rows that are in tie].
b) If the degeneracy is not solved by method 2 (a) then divide the elements of the main matrix
by the corresponding element in the key column and verify the ratios. Once the ratios are
un equal, select the lowest ratio [ this should be done only to rows that are in tie]
Example
Max Z=3X+9Y
S.T
X+4Y≤8
X+2Y≤4
X &Y≥0
Maximization
case
Cj 3 9 0 0
CBV BV X1 X2 S1 S2
Basic
solution Min ratio
0 S1 1 4 1 0 8 2
0 S2 1 2 0 1 4 2
Zj 0 0 0 0
Cj -Zj 3 9 0 0 DEGENERACY
From this table the entering variable is known which is X2. But the leaving variable is not known
because of tie in the leaving row. Then apply method 2 (a) by dividing the unit matrix by
corresponding key column value.
Maximization
case
Cj 3 9 0 0
CBV BV X1 X2 S1 S2
Basic
solution Min ratio
0 S1 1 4 1/4 0/4 8 2
0 S2 1 2 0/2 1/2 4 2
Zj 0 0 0 0
Cj -Zj 3 9 0 0 DEGENERACY
0 0

S1 S2
1/4 0/4
0/2 1/2
Then from this 0/2=0 is the smallest, then S2 is the leaving row from the basic solution.
5. Tie for entering variables [key column]
In order to break this, tie the selection for the key column can be made arbitrarily. However,
the number of solution can be minimized by adapting the following roles.
a) If there is a tie between two decision variables, then the selection can be made
arbitrarily.
b) If there is a tie between a decision variable and a slack [surplus variable], then select
the decision variable to enter into basics first.
c) If there is a tie between slack or surplus variable, then the selection can be made
arbitrarily.

3.5. Revised simplex method
Question: why revised simplex method has been developed?
Answer: few disadvantage of the usual simplex method leads to develop revised simplex method.
They are;
➢ It is very time consuming
➢ Many number of iterations has to follow
➢ It does not give inverse and simplex multipliers
Two forms of revised simplex method
1. Standard form I; in this form identity (basic) matrix is obtained after introducing slack
variables only.
2. Standard form II; in this form artificial variable are needed for an initial identity (basics)
matrix.
Basic steps in revised simplex method
Step 1. Express the problem in revised simplex form
a. All bi ≥ 0 [Right side of the constraints of the inequality]
b. The objective function should be maximization
c. Convert inequality in to equations [by introducing slack variables]
d. The objective function should be treated as first equation
Step 2. The equation should be represented in tableau matrix form
✓ Let ai represents the non-basic variable [additional table]
✓ β0 coefficient of the objective function value on each equation
✓ β1,2,3……=respective coefficient of each slack variables at each equation
Step 3. Compute Δj and check for optimality
a. To compute Δj and to check the optimality of the solution
𝛥𝑗 = 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑤 𝑜𝑓 𝛽−1
𝑜𝑟 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑤 𝑜𝑓 𝑛𝑜𝑛 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 ∗ 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑒
𝛥𝑗 = 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑤 𝑜𝑓 𝛽−1
𝑜𝑟 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑤 𝑜𝑓 𝑛𝑜𝑛 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 ∗ [𝑎1,𝑎2,…]
b. If Δj ≥0 then it indicates the optimal solution otherwise search for the entering
and leaving vector
c. To find entering and leaving vector
i. Δk = Min [Δ1, Δ2,.......]
ii. To find the leaving variable put the values into the table and compute min
[XB/ Xk]
d. Then compute the improved table as the usual simplex method [searching the new
value of the basic variables]
e. Repeat the above steps until Δj ≥0

Example;
Max Z=X1+2X2
s.t
X1+X2 ≤ 3
X1+2X2 ≤ 5
3X1+X2 ≤ 6
X1, X2 ≥ 0
Solution
Step 1. Express the problem in revised simplex form
a. All bi ≥ 0 and
b. The problem is a maximization problem.
c. Convert inequality in to equations [by introducing slack variables]
X1+X2+S1 = 3
X1+2X2 +S2= 5
3X1+X2 +S3 =6
X1, X2 and S1, S2, S3≥ 0
d. The objective function should be treated as first equation
Z-X1-2X2=0
Step 2. The equation should be represented in tableau matrix form
Note; S1,S2 and S3 are the initial basic variable with the value of 3,5,6 respectively.
And X1 and X2 are the non-basic variable represented by a1 and a2 respectively.
Note; additional variables (a1 and a2) helps in finding Δj and entering vector [ak ]
a. 𝛥𝑗 = 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑤 𝑜𝑓 𝛽−1
𝑜𝑟 𝑓𝑖𝑟𝑠𝑡 𝑟𝑜𝑤 𝑜𝑓 𝑛𝑜𝑛 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 ∗ 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑒
(𝛥1, 𝛥2) = [1 0 0 0] ∗ (
−1 −2
1 1
1 2
3 0
) = (−1, −2)
𝛥1 = −1 and 𝛥2 = −2 both Δ1&2≤ 0 which is not optimal then it needs further solution
improvement .
b. To find entering and leaving vector
Δk = Min [Δ1, Δ2] =Min [-1, -2] =-2, then the k value is 2, which indicates that X2 or a2 is
the entering variable
c. To find the leaving variable put the values into the table and compute min [XB/ X2]
XK Min[XB /XK]
K a1 a2
Z 1 0 0 0 0 -1 -2
S1 0 1 0 0 3 1 1
S2 0 0 1 0 5 1 2
S3 0 0 0 1 6 3 0
Additional table
β1
-1
β0
(1)
β1 β2 β3 XB
B.variable

From the min ratio the minimum value is related to S2 and this variable is the leaving
variable from the basic variable [ignore the negative ratio] and 2 is the key element value.
d. Compute the improved table as the usual simplex method [searching the new value of the
basic variables] by taking the intermediate coefficient matrix.
New value
Tableau
XK=X2 Min[XB/XK]
K=2 a1 a2
Z 1 0 0 0 0 -2 _ -1 -2
S1 0 1 0 0 3 1 3 1 1
S2 0 0 1 0 5 2 2.5 1 2
S3 0 0 0 1 6 1 6 3 0
Additional table
B1
-1
B.variable β0
(1)
β1
(1)
β2
(1)
β3
(1)
XB
0 0 0 0 -2
1 0 0 3 1
0 1 0 5 2
0 0 1 6 1
β1
(1)
β2
(1)
β3
(1)
XB X2
Intedrmidated coefficent matrix
oprations applied
0 1 0 5 -2 R1←R1+ R3 new
1 - 1/2 0 1/2 1 R1←R2- R3 new
0 1/2 0 2.5 2 R3←1/2R3 old
0 - 1/2 1 3.5 1 R4←R4- R3 new
β1
(1)
β2
(1)
β3
(1)
XB X2
XK=X Min[XB/XK]
K= a1 a4
Z 1 0 1 0 5 -1 0
S1 0 1 - 1/2 0 1/2 1 0
X2 0 0 1/2 0 2.5 1 1
S3 0 0 - 1/2 1 3.5 3 0
B1
-1
B.variable β0
(2)
β1
(2)
β2
(2)
β3
(2)
XB
Additional table
non basic variable

(𝛥1, 𝛥4) = [1 0 1 0] ∗ (
−1 0
1 0
1 1
3 0
) = (0,1)
Δj ≥ 0, which indicates that this is the optimal value, with Z=5, X2=2.5 and X1=0

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