1. Chapter 3 The Simplex Method and Sensitivity Analysis
Doua Nassar

2. Simplex Method Also called simplex technique or simplex algorithm
It is universal that is any linear model for which the solution exists can be solved by simplex method
The simplex method attempts to move from one corner point of the solution space to a better corner point until the optimum is found
Doua Nassar

3. 3.1 LP model in equation form (standard form)
All variables are non-negative
The right-hand side of each constraint is
non-negative
All constraints are expressed as equations (=)
Objective function may be of maximization or minimization type
Unrestricted variables or unconstrained variables
Doua Nassar

4. 3.1.1 Converting Inequalities into Equations
with Non-negative Right-hand side
Two types of variables:
i) Slack or unused variable (inequality <)
ii) Surplus or extra variable (inequality >)
Doua Nassar

5. i) Slack or unused variable (inequality <):
- The difference between the Right-Hand Side (R.H.S) and Left-Hand Side (L.H.S) of
the (inequality <) constraint thus yields the unused or slack amount of the resource.
R.H.S - L.H.S = slack or unused amount
- To convert the (inequality <) to an equation (=), a non-negative slack variable is
added to the left-hand side of the constraint.
Example of Reddy Mikks model:
Constraint of raw material M1,
(L.H.S) 6x1 + 4x2 < (R.H.S)
Let s1 be the slack or unused amount of raw material M1,
(L.H.S) 6x1 + 4x2 + s1 = (R.H.S)
s1 > 0 (non-negative slack variable)
or (6x1 + 4x2 ) = s1
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6. L.H.S - R.H.S = surplus or extra amount
ii) Surplus or extra variable (inequality >):
- The difference between the Left-Hand Side (L.H.S) and Right-Hand Side(R.H.S)
of the (inequality >) constraint thus yields the surplus or extra amount of the
resource.
L.H.S - R.H.S = surplus or extra amount
- To convert the (inequality >) to an equation (=),a non-negative surplus variable is
subtracted from the left-hand side of the constraint.
Example:
(L.H.S) 5x1 + 7x2 > (R.H.S)
Let s2 be the surplus or extra amount ,
(L.H.S) 5x1 + 7x2 - s2 = (R.H.S)
s2 > 0 (non-negative surplus variable)
or 5x1 + 7x = s2
Doua Nassar

7. Right-hand side of the constraint to be non-negative: Example:
First method:-
(L.H.S) - 3x1 + 2x2 > (R.H.S)
Convert to equation (=),
(L.H.S) - 3x1 + 2x2 - s1 = (R.H.S)
s1 > 0
Multiply both sides of the equation by -1 to get non-negative right-hand side ,
3x1 - 2x2 + s1 = 15
Second method:-
Multiply both sides of the constraint by -1 to get non-negative right-hand side ,
(L.H.S) 3x1 - 2x2 < (R.H.S)
3x1 - 2x2 + s1 = 15
Both equation and equation are same. 1 2 1 2
Doua Nassar

8. 3.1.2 Dealing with unrestricted or unconstrained variables
- Unrestricted or unconstrained variables can be either positive , negative or zero.
If a variable is unrestricted or unconstrained, it is expressed as the difference between two non-negative variables.
- If x2 is unrestricted variable then it can be expressed as the difference between y1 non-negative variable and y2 non-negative variable,
x2 = y1 – y2 where y1 > 0 and y2 > 0
Value of x2 unrestricted variable is either positive ,
zero or negative depending upon whether y1 is larger , equal to or
smaller than y2 .
Doua Nassar

9. Example:
Express the following LP problem in the standard form:
Maximize z = 4x1 + 6x2 + 2x3
subject to
2x1 + 5x2 < 6
x1 + 9x2 - 7x < 8
4x1 - 10x2 + 2x3 > 3
x1 > 0 , x2 > 0
Solution:
Here x1 and x2 are restricted to be non-negative variables while x3 is unrestricted variable.
Let x3 = y1 – y2 where y1 > 0 and y2 > 0 , then above LP problem can be written as,
Maximize z = 4x1 + 6x2 + 2(y1 – y2)
2x1 + 5x2 < 6
x1 + 9x2 - 7(y1 – y2) < 8
4x1 - 10x2 + 2(y1 – y2) > 3
x1 > 0 , x2 > 0, y1 > 0, y2 > 0
Doua Nassar

10. subject to Maximize z = 4x1 + 6x2 + 2y1 – 2y2 2x1 + 5x2 < 6or
Maximize z = 4x1 + 6x2 + 2y1 – 2y2
subject to
2x1 + 5x2 < 6
x1 + 9x2 - 7y1 + 7 y < 8
4x1 - 10x2 + 2y1 –2 y2 > 3
x1 > 0 , x2 > 0, y1 > 0, y2 > 0
Including slack and surplus variables, the standard form is ,
2x1 + 5x2 + s1 = 6
x1 + 9x2 - 7y1 + 7 y2 + s = 8
4x1 - 10x2 + 2y1 –2 y2 - s3 = 3
x1 > 0 , x2 > 0, y1 > 0, y2 > 0 , s1 > 0 , s2 > 0 , s3 > 0
Doua Nassar

11. 3.2 Transition from Graphical to Algebraic Solution
- LP Solution by Graphical method is obtained
by the corner points of the feasible space
satisfying all the constraints.
- For transition from graphical to algebraic
method , the solution space is represented by
m simultaneous linear equations and
n non- negative variables.
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12. In algebraic representation:
The number of equations m is always less than or equal to the number of variables n .
If m=n , and the equations are consistent, the system has only one solution.
e.g. The equation x = 2 has m=n=1 , the solution is
unique.
If m < n (which represents the majority of LPs), then the system of equations , again if consistent, will yield an infinite number of solutions .
e.g. The equation x + y =1 has m=1 and n=2, it yields an
infinite number of solutions (any point on the line
x + y = 1 is a solution).
If the number of equations m is larger than the number of variables n, then at least m – n equations must be redundant.
Doua Nassar

13. Below gives a parallel steps for the algebraic
and graphical method of solutions:
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14. Algebraic Determination of Corner Points
- The LP solution space is represented algebraically by finding the
candidates (corner points) for the optimum.
- The corner points are determined by the simultaneous linear
equations as follows:
In a set of m x n (m < n) where m = number of equations and n =number of
non-negative variables) ,
* set (n – m) variables equal to zero
* solve the m equations for the remaining from n variables
* if solution is unique, or there may be more
solutions , it is called basic solution , but they
must correspond to the corner points (feasible
or infeasible) of the solution space.
* the maximum number of Corner Points or basic solutions (feasible or
infeasible) can be determined by the formula,
Doua Nassar

15. To understand the above points, let us solve. Example 3.2-1
Consider the following LP with two variables:
Maximize z = 2x1 + 3x2
subject to
2x1 + x2 < 4
X1 + 2x2 < 5
X1 , x2 > 0
Note:
Example is explained by the file names Example part 1 , Example part 2 and Example part 3 .
Doua Nassar

16. Some important definitions:- Solution:
A set of variables xj (j = 1,2,3,4,……n) is called a
solution to the LP problem if it satisfies the constraints.
Feasible Solution:
feasible solution to the LP problem if it satisfies the constraints
as well as non-negativity restrictions xj > 0 (j = 1,2,3,4,……n) , si > 0 (i = 1,2,3,4,……m) .
Non-Basic Variables:
Set (n – m) variables equal to zero are called non-basic variables.
Basic Variables:
The values of remaining from n variables (some of them may be zero) are called basic variables.
Doua Nassar

17. A solution obtained by setting (n – m) variables equal to zero
Basic Solution:
A solution obtained by setting (n – m) variables equal to zero
and solving of remaining from n variables (some of them may
be zero) is called a basic solution . (Together non-basic variables and basic variables give a basic solution).
Basic Feasible Solution:
A basic solution which is feasible, is said to be a basic feasible
solution.
Non-Degenerate Basic Feasible Solution:
It is a basic feasible solution in which all the basic variables are
positive or non-zero ( value of basic variable > 0) .
Degenerate Basic Feasible Solution:
It is a basic feasible solution in which one or more of the basic variables are equal to zero ( value of basic variable = 0) .
Doua Nassar

18. Convex Set in Linear Programming (LP):
Optimal Solution:
A basic feasible solution is said to be optimal if it optimizes the objective function.
Convex Set in Linear Programming (LP):
The feasible solution space is said to form a convex set, if for any two feasible points in the set ,the line segment joining any two feasible points lies entirely in the set.
Extreme Point (Corner Point):
Every basic feasible solution of a LP is an extreme point of the convex set of feasible solutions and every extreme point is a basic feasible solution of the set of the constraints.
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19. Convex set
Non-convex set
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20. Algebraically method finds all the basic
3.3 The Simplex Method
Algebraically method finds all the basic
solutions or corner points of the LP problem.
The Simplex method only finds few of the
basic solutions or corner points of the LP
problem.
Doua Nassar

21. 3.3.1 Iterative Nature of the Simplex Method For maximize z:
Normally , the simplex method starts at the origin (0,0)
where x1 = 0 , x2 = 0 . Both x1 and x2 are non-basic
variables.
At this starting point the value of the objective
function z = 0 .
In an objective function z it is to be decided which of
the two non-basic variables above their current zero
values can improve (increase) the value of z.
In simplex method one non-basic variable is increased
at a time.
The non-basic variable which has the largest rate of
improvement in z is selected.
Doua Nassar

22. Both x1 and x2 are non-basic variables.
For minimize z:
Both x1 and x2 are non-basic variables.
In an objective function z it is to be decided which of the two non-basic variables can improve (decrease) the value of z.
In simplex method one non-basic variable is decreased at a time.
The non-basic variable which has the smallest rate of improvement in z is selected.
- Either x1 or x2 will be selected in the objective function z which has the smallest rate of improvement .
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23. - The objective function z shows that an increase
Example 3.2-1:
Maximize z = 2x1 + 3x2
- The objective function z shows that an increase
in either x1 or x2 or both above their current
zero values will improve(increase) the value of z.
The rate of improvement in z is 2 for x1 and 3 for x2 .
We therefore select to increase x2 , the variable with the largest rate of improvement.
Doua Nassar

24. Doua Nassar

25. Figure 3.4 Iterative process of the simplex method of Example 3.2-1
Doua Nassar

26. The path of the simplex method is thus defined as A B C .
The figure 3.4 shows that the value of x2 will be increased from point A origin until corner point B is reached.
At point B simplex method will then increase the value of x1 to reach the improved corner point C, which is the optimum.
The path of the simplex method is thus defined
as A B C .
- Each corner point along the path is associated
with an iteration.
Simplex method moves alongside the edges of the solution space which means that the method cannot cut across the solution space, for example going from A to C directly cannot be done.
Doua Nassar

27. The following table summarizes these representations:
We need to make transition from graphical solution to the algebraic solution by showing how the points A , B and C are represented by their basic and non-basic variables .
The following table summarizes these representations:
Corner point Basic variables Non-basic (zero) variables
A s1 , s x1, x2
B s1 , x x1, s2
C x1, x s1 , s2
Doua Nassar

28. In simplex method there are entering and leaving variables.
We can notice that there is a change pattern in the basic and non-basic variables as the solution moves along the path A B C .
In simplex method there are entering and leaving variables.
Entering variable means that variable which is going from non-basic variable to basic variable.
Leaving variable means that variable which is going from basic variable to non-basic variable.
Doua Nassar

29. From A to B x2 is the entering variable because it enters the basic solution .
From A to B s2 is the leaving variable because it leaves the basic solution .
From B to C x1 is the entering variable because it enters the basic solution .
From B to C s1 is the leaving variable because it leaves the basic solution .
Corner point Basic variables Non-basic (zero) variables
A s1 , s x1, x2
B s1 , x x1, s2
C x1, x s1 , s2
Note:
Path of three examples has been explained by the file name Three
examples of path .
Doua Nassar

30. 3.3.2 Computational Details of the Simplex Algorithm
- The computational details of a simplex iteration explains rules
for determining the entering and leaving variables as well as
for stopping the computations when the optimum solution has
been reached.
Example 3.3-1
Note:
Example has been explained by the file names Example
3.3-1 part 1 , Example part 2 , Example part 3 ,
Example part 4 and Example part 5 .
Doua Nassar

31. Find all the basic solutions to the following
Example 3.3-2
Find all the basic solutions to the following
problem. Also find which of the basic solutions
are (i) basic feasible solution (ii) non-degenerate
basic feasible solution (iii) optimal basic feasible
solution.
Maximize z = x1 + 3x2 + 3x3
subject to
x1 + 2x2 + 3x3 = 4
2x1 + 3x2 + 5x3 = 7
x1 , x2 , x3 > 0 1 2
Doua Nassar

32. - Here m = 2 ( number of equations) , n = 3
Solution:
- Here m = 2 ( number of equations) , n = 3
(number of non-negative variables) .
- Total number of corner points or basic solutions are,
C = 3!___ = !___ = X 2 X 1_ = 3
2!(3-2)! ! 1! X 1 X 1
- Number of non-basic variables are,
n – m = 3 – 2 = 1
- When x1 = 0 (non-basic variable), then x2 and x3 are basic variables.
Put x1 = 0 in equations and to
get values of x2 and x3 ,
x1 + 2x2 + 3x3 = 4
2x1 + 3x2 + 5x3 = 7 1 2 1 2
Doua Nassar

33. Now equations become, 2x2 + 3x3 = 4 3x2 + 5x3 = 7
Multiply eq by 3 and eq by 2 ,
6x2 + 9x3 = 12
6x2 + 10x3 = 14
Subtract eq from eq ,
+ 6x2 + 9x3 = +12
___________________
x3 = 2 1’ 2’ 1’ 2’
1’’
2’’
1’’
2’’
2’’
1’’
Doua Nassar

34. Put x3 = 2 in eq to get value of x2, 2x2 + 3x3 = 4 2x2 + 3(2) = 4
Therefore x1 = 0 , x2 = -1 and x3 =2 1’ 1’
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35. Put x2 = 0 in equations and to get values of x1 and x3 ,
- When x2 = 0 (non-basic variable), then x1 and x3 are basic variables.
Put x2 = 0 in equations and to
get values of x1 and x3 ,
x1 + 2x2 + 3x3 = 4
2x1 + 3x2 + 5x3 = 7
Now equations become,
x1 + 3x3 = 4
2x1 + 5x3 = 7 1 2 1 2 1’ 2’
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36. Put x3 = 1 in eq to get value of x1, x1 + 3x3 = 4 x1 + 3(1) = 4
Multiply eq by 2 ,
2x1 + 6x3 = 8
Subtract eq from eq ,
2x1 + 6x3 = 8
+ 2x1 + 5x3 = + 7
___________________
x3 = 1
Put x3 = 1 in eq to get value of x1,
x1 + 3x3 = 4
x1 + 3(1) = 4
x = 4
x1 = 4 – 3
x1 = 1
Therefore x1 = 1 , x2 = 0 and x3 = 1 1’
1’’ 2’
1’’
1’’ 2’ 1’ 1’
Doua Nassar

37. Put x3 = 0 in equations and to get values of x1 and x2 ,
- When x3 = 0 (non-basic variable), then x1 and x2 are basic variables.
Put x3 = 0 in equations and to
get values of x1 and x2 ,
x1 + 2x2 + 3x3 = 4
2x1 + 3x2 + 5x3 = 7
Now equations become,
x1 + 2x2 = 4
2x1 + 3x2 = 7
Multiply eq by 2 ,
2x1 + 4x2 = 8 1 2 1 2 1’ 2’ 1’
1’’
Doua Nassar

38. Put x2 = 1 in eq to get value of x1, x1 + 2x2 = 4 x1 + 2(1) = 4
Subtract eq from eq ,
2x1 + 4x2 = 8
+ 2x1 + 3x2 = + 7
__________________
x2 = 1
Put x2 = 1 in eq to get value of x1,
x1 + 2x2 = 4
x1 + 2(1) = 4
x = 4
x = 4 – 2
x = 2
Therefore x1 = 2 , x2 = 1 and x3 = 0 2’
1’’
1’’ 2’ 1’ 1’
Doua Nassar

39. Optimal basic feasible solution is number 3 which gives
Serial Non-basic Basic Basic Basic Non-degenerate Objective
number variables variables solution feasible basic feasible value
solution solution z_____
x (x2, x3) (-1,2) No No
x (x1, x3) (1 ,1) Yes Yes
x (x1, x2) (2, 1) Yes Yes
Optimal basic feasible solution is number 3 which gives
maximize z = 5 where x1 = 2 , x2 = 1 and x3 = 0 .
Doua Nassar

40. Example: Find the solution of the LP problem algebraically.
Max Z=2x1+3x2
ST.
2x1 + x2 < 4
x x2 < 5
x1 , x2 > 0
Doua Nassar

41. Doua Nassar

42. Solve the following LP problem by simplex method. Max z = 5x1 - 4x2
Example:
Solve the following LP problem by simplex
method.
Max z = 5x x2
subject to
6x x2 < 24
x x2 < 6
-x1 + x2 < 1
x2 < 2
x1 , x2 > 0
Doua Nassar

43. Solution:
1- Convert constraints to equation or standard form.
Max z = 5x x2
subject to
6x x2 + s1 = 24
x x2 + s2 = 6
-x1 + x2 + s3 = 1
x2 + s4 = 2
x1 , x2 , s1 , s2 , s3 , s4 > 0
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44. 2- Put the coefficient of equation in a table
Sol.
RHS S4 S3 S2 S1 X2 X1
Basic 24 1 4 6 2 -1 -4 -5 z
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45. 3- Determine Entering variable
3- Determine Entering variable. Entering variable - max negative number in the objective
Entering variable
Sol.
RHS S4 S3 S2 S1 X2 X1
Basic 24 1 4 6 2 -1 -4 -5 z
Doua Nassar

46. 4- Find the Leaving variable Calculate Sol col. / Pivot col
4- Find the Leaving variable Calculate Sol col. / Pivot col. (entering variable) Take the smallest positive value
Entering variable
Sol/
Pivot col.
Sol.
RHS S4 S3 S2 S1 X2 X1
Basic 4 24 1 6 2 -1
No Need -4 -5 z
Leaving variable
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47. 5- Find the Pivot value , Pivot Element Pivot Value= Entering Variable Col Leaving Variable Row Pivot Element = Leaving variable Row / Pivot value Pivot Value = 6 Pivot Element = (6, 4, 1, 0, 0, 0, 24) / =( 1, 4/6, 1/6, 0, 0, 0, 4)
Doua Nassar

48. 6- Gauss-Jordan row operations:
1. Replace the leaving variable in the Basic column with the
entering variable.
2. All other rows , including z:
New row = Current row – {(Pivot column co-efficient) X (New pivot row)}
- s2 -row:
(1, 2, 0, 1, 0, 0, 6) – (1)(1 , 4/6, 1/6, 0,0,0,4)
= (0, 4/3, -1/6, 1, 0, 0, 2)
Make it for all rows THE RESULT…..
Doua Nassar

49. Sol.
RHS S4 S3 S2 S1 X2 X1
Basic 4
1/6
2/3 1 2
-1/6
4/3 5
5/3 20
5/6
-2/3 z
Now take a look to Z row if you get a positive and zero values stop work
IF not Repeat all steps from step 3 until you reach positive or zeros values
Doua Nassar

50. Sol/ Pivot col. Sol. RHS S4 S3 S2 S1 X2 X1 Basic
Entering variable
Sol/
Pivot col.
Sol.
RHS S4 S3 S2 S1 X2 X1
Basic 6 4
1/6
2/3 1
1.5 2
-1/6
4/3 3 5
5/3
No need 20
5/6
-2/3 z
Leaving variable
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51. Now let see Z Row we find all values is positive and zeros
Sol.
RHS S4 S3 S2 S1 X2 X1
Basic 3
-1/2
1/4 1
3/2
3/4
-1/8
5/2
-5/4
3/8
-3/4
1/8 21
1/2 z
Now let see Z Row we find all values is positive and zeros
SO Stop work
We reach the optimal solution
X1= 3
X2= 3/2
Z= 21
Doua Nassar

52. What about Minimization
What about Minimization?? Simplest way to solve SEE the attached pdf file for minimization objective
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53. 3.3.3 Summary of the Simplex Method
Optimality condition for maximization objective:
- The entering variable in a maximization
objective problem is the non-basic variable
having the most negative co-efficient in the
z-row.
The optimum is reached at the iteration where
all the z-row co-efficients of the non-basic
variables are non-negative (all co-efficients of the
non-basic variables in the z-row > 0) .
Doua Nassar

54. Optimality condition for minimization objective:
- The entering variable in a minimization
objective problem is the non-basic variable
having the most positive co-efficient in the
z-row.
The optimum is reached at the iteration where
all the z-row co-efficients of the non-basic
variables are non-positive (all co-efficients of the
non-basic variables in the z-row < 0) .
Doua Nassar

55. Feasibility condition: For both the maximization and the minimization
problems the leaving variable is the basic
variable associated with the smallest
non-negative ratio.
Gauss-Jordan row operations:
1. Pivot row:
(a) Replace the leaving variable in the Basic column with the
entering variable.
(b) New pivot row = Current pivot row ÷ Pivot element
2. All other rows , including z:
New row = Current row – {(Pivot column co-efficient) X (New pivot row)}
Doua Nassar

56. The steps of the Simplex method are:
- Determine a starting basic feasible solution.
Step 2
Select an entering variable using the optimality condition . Stop
if there is no entering variable , then 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.
Doua Nassar