Simplex Method Explained

_____________________Dr. Atif Shahzad
BE, MECHANICAL ENGINEERING
UNIVERSITY OF ENGINEERING & TECHNOLOGY, TAXILA, PAKISTAN, 2000
MCS, SOFTWARE ENGINEERING
SZABIST,, ISLAMABAD, PAKISTAN, 2003
MS, AUTOMATION & PRODUCTION SYSTEMS
ECOLE CENTRALE DE NANTES, NANTES, FRANCE, 2007
PhD, AUTOMATION & APPLIED INFORMATICS
UNIVERSITE DE NANTES, NANTES, FRANCE, 2011
EMAIL: atifshahzad@Gmail.com
TEL: +92-333-5219846, +92-51-5179755
LINKEDIN: pk.linkedin.com/in/dratifshahzad
1 Dr. Atif Shahzad 15-Oct-15

OPTIMIZATION
MODELS
COURSE OBJECTIVES
2Dr. Atif Shahzad
15-Oct-15

OPTIMIZATION
MODELS
TODAY’S LECTURE
Simplex method
3Dr. Atif Shahzad
15-Oct-15

LP Problem
1–5
15-Oct-15Dr. Atif Shahzad

Converting LP to standard form
1
All Inequalities
to be
EQUATIONS
2
.
All variables
To be
NON-NEGATIVE
1–6

LP Standard Form
Inequalities to Equations
 Slack Variable: Variable added to a  constraint to convert it to an
equation (=).
 A slack variable represents unused resources
 A slack variable contributes nothing to the objective function value.
 Surplus Variable: Variable subtracted from a  constraint to convert
it to an equation (=).
 A surplus variable represents an excess above a constraint
requirement level.
 Surplus variables contribute nothing to the calculated value of the
objective function.
1–7

LP Standard Form
Dealing with unrestricted variables
 Unrestricted Variable:A variable that is unbounded and can take
any value.
 BUT, standards form requires that the variables be nonnegative
 So, What to do??
 Write an unrestricted variable 𝑥 as:
 𝑥 = 𝑥+- 𝑥−, where 𝑥+, 𝑥−≥ 0
 If 𝑥+> 0 and 𝑥−= 0 then 𝑥+ represents a slack
 If 𝑥−
> 0 and 𝑥+
= 0 then 𝑥−
represents a surplus
1–8

Basic & Non-Basic Variables
 Consider a System A𝑥 = 𝑏 of 𝑚 linear equations in 𝑛
variables
 Assume 𝑛 ≥ 𝑚
 Set 𝑛 − 𝑚 variables equal to ZERO
(Non-Basic Variables)
 Solve for remaining 𝑚 variables
(Basic Variables)
1–9

Basic & Non-Basic Variables
Example
𝑥1 + 𝑥2 = 3
−𝑥2 + 𝑥3 = −1
 𝑛 = 3, 𝑚 = 2, 𝑛 − 𝑚 =1
 If NBV={𝑥3} then BV={𝑥1, 𝑥2}
 Setting 𝑥3 =0, we get {𝑥1, 𝑥2} ={2,1}
 {𝑥1, 𝑥2, 𝑥3}={2,1,0} is a basic solution.
 If NBV={𝑥2} then BV={𝑥1, 𝑥3} THEN ??
 If NBV={𝑥1} then BV={𝑥2, 𝑥3} THEN ??
1–10

Basic Feasible Solution
 Basic Solution(BS) : This solution is obtained by setting any
n variables (among m+n variables) equal to zero and
solving for remaining m variables, provided the determinant
of the coefficients of these variables is non-zero.
 Such m variables are called basic variables and remaining
n zero valued variables are called non basic variables.
 Basic Feasible Solution(BFS) : It is a basic solution which
also satisfies the non negativity restrictions, i.e., all variables
are nonnegative.
1–11

Basic Feasible Solution
 BFS are of two types:
Degenerate BFS: If one or more basic variables
are zero.
Non-Degenerate BFS: All basic variables are non-
zero.
 Optimal BFS: BFS which optimizes the objective function.
1–12

Optimality & Feasibility Conditions
Conditions
FEASIBILITY
For both the maximization and
the minimization problems, the
leaving variable is the basic
variable associated with the
smallest nonnegative ratio (with
strictly positive denominator).
OPTIMALITY
The entering variable in a
maximization (minimization)
problem is the nonbasic
variable having the most
negative (positive) coefficient
in the z-row.
Tiesarebrokenarbitrarily.
Tiesarebrokenarbitrarily.
The optimum is reached at the
iteration where all the z-row
coefficients of the non basic
variables are nonnegative
(non-positive ).
1–14

Simplex Method
Algorithm
 Step 1. Determine a starting basic feasible solution.
 Step 2. Select an entering variable using the optimality
condition. Stop if there is no entering variable; the last
solution is optimal. Else, go to step 3.
 Step 3. Select a leaving variable using the feasibility
condition.
 Step 4. Determine the new basic solution by using the
appropriate Gauss-Jordan computations. Go to step 2.
1–15

Example
max z = 13𝑥1 + 11𝑥2
4𝑥1 + 5𝑥2 ≤ 1500
5𝑥1 + 3𝑥2 ≤ 1575
𝑥1 + 2𝑥2 ≤ 420
𝑥1, 𝑥2 ≥ 0
16
Dr. Atif Shahzad 15-Oct-15

Step 0:
Standard Form
Convert all the inequality constraints into equalities by the use of slack
variables.
Let S1, S2 , S3 be three slack variables.
Model can rewritten as:
Z - 13x1 -11x2 = 0
Subject to constraints:
4x1+5x2 + S1 = 1500
5x1+3x2 + S2 = 1575
x1+2x2 +S3 = 420
x1, x2, S1, S2, S3 > 0
17

Step 1
Initial BFS
 Find the Initial BFS.
 Choose Basic variables
 S1, S2, S3
 One Feasible solution that satisfies all the
constraints is:
x1= 0, x2= 0, S1= 1500,
S2= 1575, S3= 420 and Z=0.
So, NBV={𝑥1, 𝑥2} & BV={𝑆1, 𝑆2, 𝑆3}
And BFS is {𝑥1, 𝑥2, 𝑆1, 𝑆2, 𝑆3}={0,0,1500,1575,420}
1–18

Step 1:
Tableaux
Row No. Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500
2 S2 0 5 3 0 1 0 1575
3 S3 0 1 2 0 0 1 420
Set up an initial table.
19

Step 2:
Entering Variable
Row
NO.
Basic
Variable
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500
2 S2 0 5 3 0 1 0 1575
3 S3 0 1 2 0 0 1 420
Choose the most negative number from row 0 (i.e Z row).
Therefore, x1 is an entering variable.
Iteration 1
20

Step 3:
Leaving variable
Row
NO.
Basic
Variable
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500 375
2 S2 0 5 3 0 1 0 1575 315
3 S3 0 1 2 0 0 1 420 420
Calculate Ratio = Sol col. / 𝑥1 col. (𝑥1 > 0)
Choose minimum Ratio. That variable(i.e., S2) is
a leaving variable.
Iteration 1
21

Step 4:
Find New BFS (Pivoting)
Row
NO.
Basic
Variable
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500 375
2 S2 0 5 3 0 1 0 1575 315
3 S3 0 1 2 0 0 1 420 420
• x1 becomes basic variable
• S2 becomes non basic variable.
Iteration 1
22

Step 4:
Find New BFS (New Pivot Row)
Row
NO.
Basic
Variable
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500 375
2 S2 0 5 3 0 1 0 1575 315
3 S3 0 1 2 0 0 1 420 420
0 5 3 0 1 0 15750 1 3/5 0 1/5 0 375
5
New Pivot row =
Current Pivot row
Pivot Element
=
Iteration 1
23

Step 4:
Find New BFS (New Other Rows)
Row
NO.
Basic
Variable
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500
2 S2 0 1 3/5 0 1/5 0 375
3 S3 0 1 2 0 0 1 420
1 0 -16/5 0 13/5 0 4095
-13 0 1 3/5 0 1/5 0 375
Iteration 1
New Other row Current row its Pivot col Coeff.x New pivot row
= 1 -13 -11 0 0 0 0
−
−
24

Step 4:
New Tableaux
Row
No.
Basic
Varia
ble
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240
2 x1 0 1 3/5 0 1/5 0 315
3 S3 0 0 7/5 0 -1/5 1 105
Iteration 1
• x1 enters and becomes basic variable.
• S2 leaves and becomes non basic variable.
• We have a new table.
25

Step 2:
Entering Variable
Row
NO.
Basic
Varia
ble
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240
2 x1 0 1 3/5 0 1/5 0 315
3 S3 0 0 7/5 0 -1/5 1 105
Iteration 2
26

Step 3:
Leaving variable
Row
NO.
Basic
Variab
le
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2
Calculate Ratio = Sol col. / 𝑥2 col. (𝑜𝑛𝑙𝑦 𝑤ℎ𝑒𝑟𝑒 𝑥2 > 0)
Choose minimum Ratio. That variable(i.e., S3) is
a leaving variable.
27

Step 3:
Leaving variable
Row
No.
Basic
Variab
le
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2
• x2 becomes basic variable
• S3 becomes non basic variable.
28

Row
No.
Basic
Variab
le
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2Step 4:
Find New BFS (New Pivot Row)
0 0 7/5 0 -1/5 1 750 0 1 0 -1/7 5/7 375/7
7/5
New Pivot row =
Current Pivot row
Pivot Element
=
29

Row
No.
Basic
Variab
le
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2Step 4:
Find New BFS (New Other Row)
-16/5 0 0 1 0 -1/7 5/7 375/7
New Other row Current row its Pivot col Coeff. New pivot row
= 1 0 -16/5 0 13/5 0 4095
−
−
x
30

Row
No.
Basic
Variab
le
Z x1 x2 S1 S2 S3
0 Z 1 0 0 0 15/7 16/7 4335
1 S1 0 0 0 1 -3/7 -13/7 45
2 x1 0 1 0 0 2/7 -3/7 270
3 S3 0 0 1 0 -1/7 5/7 75
Iteration 2Step 4:
New Tableaux
• x2 enters and becomes basic variable.
• S3 leaves and becomes non basic variable.
• We have a new table.
31

Step 4:
RESULT
Next Table is :
Row
NO.
Basic
Variab
le
Coefficients of: Sol.
Z x1 x2 S1 S2 S3
0 Z 1 0 0 0 15/7 16/7 4335
1 S1 0 0 0 1 -3/7 -13/7 45
2 x1 0 1 0 0 2/7 -3/7 270
3 x2 0 0 1 0 -1/7 5/7 75
Optimal Solution is : x1= 270, x2= 75, Z= 4335
32

Iterations Done
x1 x2 S1 S2 S3
Sol.
Z -13 -11 0 0 0
S1 4 5 1 0 0 1500
S2 5 3 0 1 0 1575
S3 1 2 0 0 1 420
Z 0 -3 1/5 0 2 3/5 0 4095
S1 0 2 3/5 1 - 4/5 0 240
x1 1 3/5 0 1/5 0 315
S3 0 1 2/5 0 - 1/5 1 105
Z 0 0 0 2 1/7 2 2/7 4335
S1 0 0 1 - 3/7 -1 6/7 45
x1 1 0 0 2/7 - 3/7 270
x2 0 1 0 - 1/7 5/7 75
Optimal Solution is :
x1= 270, x2= 75, Z= 4335
33

Example 2
Max. Z = 3x1+5x2+4x3
2x1+3x2 < 8
2x2+5x3 < 10
3x1+2x2+4x3 < 15
x1, x2, x3 > 0
34

Cont…
Let S1, S2, S3 be the three slack variables.
Modified form is:
Z - 3x1-5x2-4x3 =0
2x1+3x2 +S1= 8
2x2+5x3 +S2= 10
3x1+2x2+4x3+S3= 15
x1, x2, x3, S1, S2, S3 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8,
S2= 10, S3= 15 and Z=0.
35

Cont…
Basic
Variable
Z x1 x2 x3 S1 S2 S3
Z 1 -3 -5 -4 0 0 0 0
S1 0 2 3 0 1 0 0 8 8/3
S2 0 0 2 5 0 1 0 10 5
S3 0 3 2 4 0 0 1 15 15/2
Therefore, x2 is the entering variable and S1 is the
departing variable.
36

Cont…
Basic
Variable
Z x1 x2 x3 S1 S2 S3
Z 1 1/3 0 -4 5/3 0 0 40/3
x2 0 2/3 1 0 1/3 0 0 8/3 -
S2 0 -4/3 0 5 -2/3 1 0 14/3 14/15
S3 0 5/3 0 4 -2/3 0 1 29/3 29/12
departing variable.
37

Cont…
Basic
Variable
Z x1 x2 x3 S1 S2 S3
Z 1 -11/15 0 0 17/15 4/5 0 256/15
x2 0 2/3 1 0 1/3 0 0 8/3 4
x3 0 -4/15 0 1 -2/15 1/5 0 14/15 -
S3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41
departing variable.
38

Cont…
Basic
Variable
Z x1 x2 x3 S1 S2 S3
Z 1 0 0 0 45/41 24/41 11/41 765/41
x2 0 0 1 0 15/41 8/41 -10/41 50/41
x3 0 0 0 1 -6/41 5/41 4/41 62/41
x1 0 1 0 0 -2/41 -12/41 15/41 89/41
Optimal Solution is : x1= 89/41, x2= 50/41, x3=62/41, Z= 765/41
39

Example 3:
Minimization
Min.. Z = x1 - 3x2 + 2x3
3x1 - x2 + 3x3 < 7
-2x1 + 4x2 < 12
-4x1 + 3x2 + 8x3 < 10
x1, x2, x3 > 0
40

Cont…
Convert the problem into maximization problem
Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -
Z
3x1 - x2 + 3x3 < 7
-2x1 + 4x2 < 12
-4x1 + 3x2 + 8x3 < 10
x1, x2, x3 > 0
41

Cont…
Let S1, S2 and S3 be three slack variables.
Modified form is:
Z’ + x1 - 3x2 + 2x3 = 0
3x1 - x2 + 3x3 +S1 = 7
-2x1 + 4x2 + S2 = 12
-4x1 + 3x2 + 8x3 +S3 = 10
x1, x2, x3 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 =
10
42

Cont…
Basic
Variable
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 1 -3 2 0 0 0 0
S1 0 3 -1 3 1 0 0 7 -
S2 0 -2 4 0 0 1 0 12 3
S3 0 -4 3 8 0 0 1 10 10/3
departing variable.
43

Cont…
Basic
Variable
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 -1/2 0 2 0 3/4 0 9
S1 0 5/2 0 3 1 1/4 0 10 4
x2 0 -1/2 1 0 0 1/4 0 3 -
S3 0 -5/2 0 8 0 -3/4 1 1 -
departing variable.
44

Cont…
Basic
Variable
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 0 0 13/5 1/5 8/10 0 11
x1 0 1 0 6/5 2/5 1/10 0 4
x2 0 0 1 3/5 1/5 3/10 0 5
S3 0 0 0 11 1 -1/2 1 11
Optimal Solution is : x1= 4, x2= 5, x3= 0,
Z’ = 11 Z = -11
45

Example 4
Max.. Z = 3x1 + 4x2
x1 - x2 < 1
-x1 + x2 < 2
x1, x2 > 0
46

Cont…
Let S1 and S2 be two slack variables
.
Modified form is:
Z -3x1 - 4x2 = 0
x1 - x2 +S1 = 1
-x1 + x2 +S2 = 2
x1, x2, S1, S2 > 0
Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2
and Z=0.
47

Cont…
Basic
Variable
Z x1 x2 S1 S2
Z 1 -3 -4 0 0 0
S1 0 1 -1 1 0 1 -
S2 0 -1 1 0 1 2 2
departing variable.
48

Cont…
Basic
Variable
Z x1 x2 S1 S2
Z 1 -7 0 0 4 8
S1 0 0 0 1 1 3 -
x2 0 -1 1 0 1 2 -
x1 is the entering variable, but as in x1 column every no. is less than equal to
zero, ratio cannot be calculated. Therefore given problem is having a
unbounded solution.
49

Example :
Solving in a compact form
max z = 40𝑥1 + 88𝑥2
2𝑥1 + 8𝑥2 ≤ 60
5𝑥1 + 2𝑥2 ≤ 60
𝑥1, 𝑥2 ≥ 0
50

Step 0:
Standard Form
Convert all the inequality constraints into equalities by the use of slack
variables.
Let S1, S2 be three slack variables.
Model can rewritten as:
z - 40x1 -88x2 = 0
2x1+8x2 + S1 = 60
5x1+2x2 + S2 = 60
x1, x2, S1, S2> 0
51

1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
52

1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
53

1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
60
2
60
5
54

1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
60
2
60
5
55

1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
𝑅1 + 8𝑅3
𝑅2 − 0.4𝑅3
56

1
0
0
0
0
5
−72
7.2
2
0
1
0
8
−0.4
1
480
36
60
𝑅1 + 8𝑅3
𝑅2 − 0.4𝑅3
57

1
0
0
0
0
5
−72
7.2
2
0
1
0
8
−0.4
1
480
36
60
𝑅1 + 10𝑅2
𝑅3 − 2
7.2
𝑅2
58

1
0
0
0
0
5
0
7.2
0
10
1
− 1
3.6
8
−0.4
1
0.9
840
36
50
𝑅1 + 10𝑅2
𝑅3 − 2
7.2
𝑅2
59

1
0
0
0
0
5
0
7.2
0
10
1
− 1
3.6
8
−0.4
1
0.9
840
36
50
𝑅1 + 10𝑅2
𝑅3 − 2
7.2
𝑅2
𝒛 𝒙 𝟏 𝒙 𝟐
𝟖𝟒𝟎 𝟓𝟎
𝟓
𝟑𝟔
𝟕.𝟐
60

Dr. Atif Shahzad
THANK YOU FOR YOUR INTEREST
15-Oct-15

Simplex Method Explained

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