This chapter discusses the graphical and algebraic algorithms of the simplex method for solving linear programming problems. It describes the key concepts of the graphical algorithm including: focusing on corner-point feasible solutions, using an iterative approach to find the optimal solution by moving between adjacent solutions along constraint boundaries, choosing the direction with the highest improvement in the objective function at each step, and stopping when no better adjacent solution exists. The chapter also compares the graphical and algebraic algorithms and discusses using tabular forms and tie-breaking rules with the simplex method.

1. Operations Research 1
Chapter 4
Introduction to Operations Research
Hillier & Liebermann
Eighth Edition
McGrawHill
Operations Research 1 – 06/07 – Chapter 4 1

2. Contents
• Simplex method: Graphical Algorithm
• Simplex method: Algebraic Algorithm
• Differences between the graphical and the algebraic
algorithm
• Simplex method: Tabular Form
• Simplex method: Tie breaking
• Simplex method: Other model forms
Operations Research 1 – 06/07 – Chapter 4 2

3. The Simplex Method – Graphical algorithm
For any linear= 3x1 + 5x2
Maximize Z programming
problem with n decision variables,
x2
x1 = 4 subject to
two CPF solutions are adjacent
to each other if they share n – 1
10
x1 ≤4
constraint boundaries.
9
The two adjacent ≤ 12 solutions
2x2 CPF
8
are connected by a line segment
that3x1 + 2x2
lies on these≤ 18 shared
same
7
2x2 = 12 constraint boundary. Such a line
and
6
segment is referred to as an edge
two adjacent of the feasible region.
5
x1 ≥ 0; x2 ≥ 0
CPF-solutions
= Corner-point solution
4
Feasible (not feasible)
3
region = Corner-point feasible
2
solution (CPF solution)
3x1 + 2x2 = 18
1
x1 = Edge
1 2 3 4 5 6 7 8 9 10
Operations Research 1 – 06/07 – Chapter 4 3

4. The Simplex Method – Graphical algorithm
CPF Its Adjacent CPF solution
solution
x2
x1 = 4
10
(0, 0) (0, 6) and (4, 0)
9
(0, 6) (2, 6) and (0, 0)
8
(2, 6) (4, 3) and (0, 6)
7
2x2 = 12 (4, 3) (4, 0) and (2, 6)
6
two adjacent
(4, 0) (0, 0) and (4, 3)
5
CPF-solutions
= Corner-point solution
4
Feasible (not feasible)
3
region = Corner-point feasible
2
solution (CPF solution)
3x1 + 2x2 = 18
1
x1 = Edge
1 2 3 4 5 6 7 8 9 10
Operations Research 1 – 06/07 – Chapter 4 4

5. The Simplex Method – Graphical algorithm
Maximize Z = 3x1 + 5x2 Initialization: Choose (0,0) as the initial CPF solution
Optimality test: Is (0,0) an optimal CPF solution?
No (Z = 0), adjacent solutions are better.
x2
Iteration 1: Move to a better CPF solution in three steps:
10
1. Choose the direction with the highest rate of
increasing Z.
9
Z increases with a rate of 3 by moving up the
8
x1 axis.
Z increases with a rate of 5 by moving along
7
the x2 axis.
6
We choose to move up the x2 axis.
5
2. Stop at the first new constraint boundary:
2x2 = 12 (We have to stay in the feasible
4
region).
Feasible
3
3. Solve for the intersection of the new set of
region constraints boundaries: (0, 6).
2
Z = 3 * 0 + 5 * 6 = 30
1
Optimality test: Is (0, 6) an optimal CPF solution?
No, an adjacent solution is better.
x1
1 2 3 4 5 6 7 8 9 10
Operations Research 1 – 06/07 – Chapter 4 5

6. The Simplex Method – Graphical algorithm
Maximize Z = 3x1 + 5x2 Iteration 2: Move to a better CPF solution in three steps:
1. Choose the best direction.
x2
Z will increase by moving to the right.
Rate = 3.
10
Z will decrease by moving down the x2 axis.
9
We choose to move to the right.
2. Stop at the first new constraint boundary:
8
3x1 + 2x2 = 12 (We have to stay in the
7
feasible region).
3. Solve for the intersection of the new set of
6
constraints boundaries: (2, 6).
5
Z = 3 * 2 + 5 * 6 = 36
Optimality test: Is (2, 6) an optimal CPF solution?
4
Yes, there is no better CPF solution!
Feasible
3
region
2
1
x1
1 2 3 4 5 6 7 8 9 10
Operations Research 1 – 06/07 – Chapter 4 6

7. Graphical algorithm: The Key Solution Concepts
• Solution concept 1
– Method focuses solely on CPF-solutions
– If there are CPF-solutions, find the best.
• Solution concept 2
– Iterative algorithm:
• Initialization: Set up, find a initial CPF-solution
• Optimality test: Is the solution optimal? If no, continue. If yes, stop.
• Iteration: Perform an iteration to find a better CPF solution. Then go
back to the optimality test.
• Solution concept 3
– Whenever possible, choose the origin as the initial CPF-solution
at initialization.
– This eliminates the use of algebraic procedures to find and solve
for an initial CPF solution.
Operations Research 1 – 06/07 – Chapter 4 7

8. Graphical algorithm: The Key Solution Concepts
• Solution concept 4
– At iteration, only adjacent CPF solutions are considered.
– The path to find the optimal solution is along the edges of the feasible
region.
• Solution concept 5
– After identification of the current CPF solution, examine all the edges
that emanate from that CPF solution.
– The edges lead to other CPF solution.
– Choose the edge with the (highest) positive rate of improvement in Z.*
• Solution concept 6
– If the rate of improvement in Z is positive, the next adjacent CPF
solution is better than the current CPF solution.*
– If the rate of improvement in Z is negative, the next adjacent CPF
solution is worse.*
– If there are no positive rates of improvement, the current CPF solution is
optimal.*
* Only if the objective function is to maximize Z
Operations Research 1 – 06/07 – Chapter 4 8

9. Notion of the Simplex Method
• Besides the graphical algorithm, there is an algebraic
algorithm.
• This algorithm is based on solving systems of equations.
• We need to translate inequality constraints to equality
constraints.
• Therefore we introduce slack variables.
Original Form of the Model Augmented Form of the Model
Maximize Z = 3x1 + 5x2 Maximize Z = 3x1 + 5x2
subject to subject to
x1 ≤4 x1 + x3 =4
2x2 ≤ 12 2x2 + x4 = 12
3x1+ 2x2 ≤ 18 3x1 + 2x2 +x5 = 18
and and
x1 ≥ 0; x2 ≥ 0 xj ≥ 0 for j = 1, 2, 3, 4, 5.
Operations Research 1 – 06/07 – Chapter 4 9

10. Slack values
Slack variable > 0 Slack variable = 0 Slack variable < 0
Feasible Infeasible
Region Region
Constraint corresponding
with slack variable
Operations Research 1 – 06/07 – Chapter 4 10

11. Terminology – Graphical & Algebraic
Graphical Algebraic Algorithm
Algorithm
Solution Solution Augmented solution
Solution on a corner Corner point Basic solution
point solution Augmented corner point solution
Solution on a feasible CPF solution Basic Feasible (BF) solution
corner point Augmented CPF solution
Example of a solution (0,6) (0, 6, 4, 0, 6)
(Slack included)
Operations Research 1 – 06/07 – Chapter 4 11

12. Basic and non-basic variables
Number of variables Number of degrees
(in augmented form) - Number of equations =
of freedom
Thus, we can choose a number of variables (= number of degrees of
freedom) to be set equal to any arbitrary value.
These variables are called nonbasic variables. (Together they are the
basis.)
The other variables are called basic variables.
In this example there are
Maximize Z = 3x1 + 5x2
5 – 3 = 2 degrees of freedom.
subject to For example, we choose
x1 + x3 =4 x1 = 4 and x4 = 2
2x2 + x4 = 12 (x1 and x4 are nonbasic var’s)
3x1 + 2x2 +x5 = 18 then this must be true:
and x3 = 0; x2 = 5; x5 = -4
xj ≥ 0 for j = 1, 2, 3, 4, 5. (x3, x2 and x5 are basic var’s)
Operations Research 1 – 06/07 – Chapter 4 12

13. Properties of a Basic Solution
1. Each variable is designated as either a nonbasic
variable or a basic variable.
2. The number of basic variables equals the number of
functional constraints.
3. The nonbasic variables are set to equal.
4. The values of the basic variables are obtained as the
simultaneous solution of the system of equations.
5. If the basic variables satisfy the nonnegativity
constraints, the basic solution is a BF solution (feasible).
Operations Research 1 – 06/07 – Chapter 4 13

14. Solving the Simplex Method
Graphical & Algebraic Interpretations
Method Sequence Graphical Interpretation Algebraic Interpretation
Initialization Choose (0, 0) to be the initial Choose x1 and x2 to be the nonbasic
CPF solution. variables (= 0) for the initial BF solution
(0, 0, 4, 12, 18)
Optimality test Not optimal, because moving Not optimal, because increasing either
along either edge from (0, 0) nonbasic variable (x1 or x2) increases Z.
increases Z.
Iteration 1 step 1 Move up the edge lying on the x2 Increase x2 while adjusting other variable
axis. values to satisfy the system of equations.
Iteration 1 step 2 Stop when the first new Stop when the first basic variable (x3, x4
constraint boundary (2x2 = 12) is or x5) drops to zero (x4).
reached.
Iteration 1 step 3 Find the intersection of the new With x2 now a basic variable and x4 now a
pair of constraint boundaries: (0, nonbasic variable, solve the system of
6) is the new CPF solution. equations: (0, 6, 4, 0, 6) is the new BF
solution.
Operations Research 1 – 06/07 – Chapter 4 14

15. Solving the Simplex Method
Graphical & Algebraic Interpretations (continued)
Method Sequence Graphical Interpretation Algebraic Interpretation
Optimality test Not optimal, because moving Not optimal, because increasing one
along the edge from (0, 6) to nonbasic variable (x1) increases Z.
the right increases Z.
Iteration 2 step 1 Move along this edge to the Increase x1 while adjusting other variable
right. values to satisfy the system of equations.
Iteration 2 step 2 Stop when the first new Stop when the first basic variable (x2, x3,
constraint boundary (3x1 + 2x2 = or x5) drops to zero (x5).
18) is reached.
Iteration 2 step 3 Find the intersection of the new With x1 now a basic variable and x5 now a
pair of constraint boundaries: nonbasic variable, solve the system of
(2, 6) is the new CPF solution. equations: (2, 6, 2, 0, 0) is the new BF
solution.
Optimality test (2, 6) is optimal, because (2, 6, 2, 0, 0) is optimal, because
moving along either edge from increasing either nonbasic variable(x4 or
(2, 6) decreases Z. x5) decreases Z.
Operations Research 1 – 06/07 – Chapter 4 15

16. Simplex Method – Tabular Form
• Besides the graphical and algebraic forms, there is a tabular form,
which is more convenient to use for solving a problem by hand.
Algebraic Form:
(0) Z - 3x1 - 5x2 =0
(1) x1 + x3 =4
(2) 2x2 + x4 = 12
(3) 3x1 + 2x2 +x5 = 18
Tabular Form
Basic Eq Coefficient of: Right
variable Z x1 x2 x3 x4 x5 side
Z (0) 1 -3 -5 0 0 0 0
x3 (1) 0 1 0 1 0 0 4
x4 (2) 0 0 2 0 1 0 12
x5 (3) 0 3 2 0 0 1 18
Operations Research 1 – 06/07 – Chapter 4 16

17. Tabular Form: The algorithm
• Initialization
– Just like the algebraic algorithm, introduce slack variables.
– Select the decision variables to be the initial nonbasic variables
(= 0).
– Select the slack variables to be the initial basic variables.
• Optimality Test
– The current BF solution is optimal if and only if every coefficent
in row 0 is nonnegative (≥ 0). If it is, stop. Otherwise go to an
iteration.
• Iteration
See next slides
Operations Research 1 – 06/07 – Chapter 4 17

18. Tabular Form: The algorithm
Tabular Form
Basic Eq. Coefficient of: Right
variable Z x1 x2 x3 x4 x5 side
Z (0) 1 -3 -5 0 0 0 0
x3 (1) 0 1 0 1 0 0 4
x4 (2) 0 0 2 0 1 0 12
x5 (3) 0 3 2 0 0 1 18
• Iteration Pivot column
– Step 1. Determine the entering basic variable by selecting the
variable with the “most negative” coefficient in Eq. (0). Put a box
around the column and call this the pivot column.
Operations Research 1 – 06/07 – Chapter 4 18

19. Tabular Form: The algorithm
Tabular Form
Basic Eq. Coefficient of: Right Ratio
variable Z x1 x2 x3 x4 x5 side
Z (0) 1 -3 -5 0 0 0 0
x3 (1) 0 1 0 1 0 0 4 infinite
x4 (2) 0 0 2 0 1 0 12 12 minimum
= 6
2
x5 (3) 0 3 2 0 0 1 18 18
= 9
2
• Iteration Pivot number
– Step 2. Determine the leaving basic variable by applying the
minimum ratio test. Put a box around the row with the minimum
ratio, this is the pivot row. Call the number in both boxes the
pivot number.
Operations Research 1 – 06/07 – Chapter 4 19

20. Tabular Form: The algorithm
Iteration: Step 3.
Tabular Form
Iteration Basic variable Eq. Coefficient of: Right side
Solve for the new BF solution by
changing the column of the
Z x1 x2 x3 x4 x5
entering basic variable
0 Z (0) 1 -3 -5 0 0 0 0
(-5, 0, 2, 2) to the column of the
x3 (1) 0 1 0 1 0 0 4 leaving basic variable (0, 0, 1, 0)
x4 (2) 0 0 2 0 1 0 12 by using elementary row
x5 (3) 0 3 2 0 0 1 18 operations:
1 Z (0) 1
-3
0
0
5/2
0
30
1. Divide the pivot row by the pivot
x3 (1) 0
1
0
1
0
0
4
number. Use this new pivot
x2 (2) 0 0 1 0 1/2 0 6 column in steps 2 and 3.
x5 (3) 0
3
0
0
-1
1
6
2. For each other row (including row
0) that has a negative coefficient
in the pivot column, add to this
The operations in this example: row the product of the absolute
1. New row (2) = Old row (2) / 2 value of this coefficient and the
2. New row (0) = Old row (0) + 5 * New row (2) new pivot row.
3. New row (3) = Old row (3) – 2 * New row (2)
3. For each other row that has a
negative coefficient in the pivot
column, substract from this row
the product of this coefficient and
Operations Research 1 – 06/07 – Chapter 4 the new pivot row. 20

21. Tabular Form: The algorithm
Optimality test: Still not optimal
Tabular Form
because there is a negative
Iteration Basic variable Eq. Coefficient of: Right side
coefficient in row (0). So at
Z x1 x2 x3 x4 x5
least one more iteration is
0 Z (0) 1 -3 -5 0 0 0 0
needed.
x3 (1) 0 1 0 1 0 0 4
Iteration 2:
x4 (2) 0 0 2 0 1 0 12
x5 (3) 0 3 2 0 0 1 18 - Determine the pivot row (and
1 Z (0) 1 -3 0 0 5/2 0 30 entering basic variable).
x3 (1) 0 1 0 1 0 0 4 4
= 4
- Run a minimum ratio test.
1
x2 (2) 0 0 1 0 1/2 0 6
- Determine the pivot column
x5 (3) 0 3 0 0 -1 1 6 6
3
= 2 (and leaving basic variable).
2 Z (0) 1 0 0 0 3/2 1 36
x3 (1) 0 0 0 1 1/3 -1/3 2
- Perform elementary operations:
- New row (3) = Old row (3) / 3
x2 (2) 0 0 1 0 1/2 0 6
- New row (0) = Old row (0) + 3 * New row (3)
x1 (3) 0 1 0 0 -1/3 1/3 2 - New row (1) = Old row (1) – 1 * New row (3)
Optimality test: There is no negative coefficient in row (0), so the solution is optimal!
Solution: Z = 36; x1 = 2; x2 = 6
Operations Research 1 – 06/07 – Chapter 4 21

22. Tie breaking in the simplex method
You may have noticed that we never said what to do if
the various choice rules of the simplex method do not
lead to a clear-cut decision, because of either ties or
other similar ambiguities. We discuss these details now.
Operations Research 1 – 06/07 – Chapter 4 22

23. The entering basic variable, step 1
If more than one nonbasic variables are tied for having
the largest negative coefficient, then the selection may
be made arbitrary.
Operations Research 1 – 06/07 – Chapter 4 23

24. The leaving basic variable, step 2
Definition: Basic Variables (BV’s) with a value of zero are
called degenerate. Not very often a choice of a
degenerate variable may lead to a loop. R. Bland
(Mathematics of Operations Research, 2: 103-107,
1977) has suggested special rules avoiding such loops.
Operations Research 1 – 06/07 – Chapter 4 24

25. No leaving Basic Variable (BV) – Unbouded Z
If the entering BV could be increased indefinitely without giving
negative values to any of current BV’s, then no variable qualifies to be
the leaving BV.
• TABLE 4.9 Initial simplex tableau for the Wyndor Glass Co.
Problem without the last two functional constraints.
Basic Coefficient of: Right
Variable Eq. Z x1 x2 x3 Side Ratio
Z (0) 1 -3 -5 0 0
X3 (1) 0 1 0 1 4 None
With x1= 0 and x2 increasing, x3 = 4 - 1x1 – 0x2 = 4 > 0.
Operations Research 1 – 06/07 – Chapter 4 25

26. Multiple Optimal Solutions continued
Any Weighted average of two or more solutions (vectors) where the weights
are nonnegative and sum is equal to 1 is called a convex combination of
these solutions.
Whenever a problem has more than one optimal BF solution, at least one of
the nonbasic variables has a coefficient of zero in the final row 0, so
increasing any such variable will not change the value of Z. Therefore, these
other optimal BF solutions can be identified (if desired) by performing
additional iterations of the simplex method, each time choosing a nonbasic
variable with a zero coefficient as the entering basic variable.
If such an iteration has no leaving basic variable, this indicates that the
feasible region is unbounded and the entering basic variables can be
increased indefinitely without changing the value of Z.
Operations Research 1 – 06/07 – Chapter 4 26

27. Table 4.10
Basic Coefficient of: Right Optimal
Iteration Variable Eq. Z x1 x2 x3 x4 x5 Side solution
Z (0) 1 -3 -2 0 0 0 0 no
0 x3 (1) 0 1 0 1 0 0 4
x4 (2) 0 0 2 0 1 0 12
x5 (3) 0 3 2 0 0 0 18
Z (0) 1 0 -2 3 0 0 12 no
1 x1 (1) 0 1 0 1 0 0 4
x4 (2) 0 0 2 0 1 0 12
x5 (3) 0 0 2 -3 0 1 6
Z (0) 1 0 0 0 0 1 18 Yes
2 x1 (1) 0 1 0 1 0 0 4
x4 (2) 0 0 0 3 1 -1 6
x2 (3) 0 0 1 -3/2 0 -1/2 3
Z (0) 1 0 0 0 0 1 18 Yes
3 x1 (1) 0 1 0 0 -1/3 1/3 2
x3 (2) 0 0
Operations Research 1 – 06/07 – Chapter 4 0 1 1/3 -1/3 2 27
x (3) 0 0 1 0 1/2 0 6
2

28. Equality Constraints
Any equality constraint
ai1x1 + ai2x2 + ……+ ainxn = bi
Actually is equivalent to a pair of inequality constraints:
ai1x1 + ai2x2 + ……+ ainxn ≤ bi
ai1x1 + ai2x2 + ……+ ainxn ≥ bi
Example: Suppose that the Wyndor Glass Co. Problem is modified to require
that Plant 3 be used at full capacity. The only resulting change in the LP
model is that the third constraint, 3x1 + 2x2 ≤ 18, instead becomes an
equality constraint
3x1 + 2x2 = 18,
so that the complete model becomes the one shown in the upper right hand
corner of figure 4.3. This figure als shows in darker ink the feasible region
which now consists of just the line segment connecting (2,6) and (4,3).
After the slack variables still needed for the inequality constraints are
introduced, the system of equations for the augmented form of the problem
becomes
(0) Z – 3x1 – 5x2 =0
(1) x1 + x3 =4
(2) 2x2 + x4 =12
(3) 3x1 + 2x2 = 18
Operations Research 1 – 06/07 – Chapter 4 28

29. Figure 4.3
• Figure 4.3
When the third constraint Maximize Z = 3x1 + 5x2,
becomes an equality Subject to X1 ≤ 4
constraint, the feasible region 2x2 ≤ 12
for the Wyndor Glass Co. 3x1 + 2x2 = 18
problem becomes the line And x1 ≥ 0, x2 ≥ 0
segment between (2,6) and (2,6)
(4,3).
(4,3)
Operations Research 1 – 06/07 – Chapter 4 29

30. Minimization
Minimization problems can be converted to equivalent
maximization problems:
n
Minimizing Z= ∑j= 1
cjxj
is equivalent to
n
Maximizing − Z= ∑ j= 1
( − cj)xj
Operations Research 1 – 06/07 – Chapter 4 30

31. Variables Allowed to be negative
Suppose that the Wyndor Glass Co. Problem is changed
so that product 1 already is in production, and the first
decision variable x1 represents the increase in its
production rate. Therefore, a negative value of x1 would
indicate that product 1 is to be cut back by that amount.
Such reductions might be desirable to allow a larger
production rate for the new, more profitable product 2, so
negative values should be allowed for x1 in the model.
Operations Research 1 – 06/07 – Chapter 4 31

32. Variables with Bound on the Negative Values
Allowed. (sheet 1 of 2)
Consider any decision variable xj that is allowed to have
negative values which satisfy a constraint of the form
xj ≥ Lj ,
Where Lj is some negative constant. This constraint can
be converted to a nonnegativity constraint by making the
change of variables
x’j = xj – Lj, so x’j ≥ 0.
Thus, x’j + Lj would be substituted for xj throughout the
model, so that the redefined decision variable x’j cannot
be negative. (This same technique can be used when Lj
is positive to convert a functional constraint xj ≥ Lj to a
nonnegativity constraint x’j ≥ 0.)
Operations Research 1 – 06/07 – Chapter 4 32

33. Variables with Bound on the Negative Values
Allowed. (continued)
To illustrate, suppose that the current production rate for product 1
in the Wyndor Glass Co. problem is 10. With the definition of x1 just
given, the complete model at this point is the same as that given in
Sec.3.1 except that the nonnegativity constraint x1 ≥ 0 is replaced by
x1 ≥ -10
To obtain the equivalent model needed for the simplex method, this
decision variable would be redefined as the total production rate of
product 1
x’j = x1 + 10,
which yields the changes in the objective function and constraints as
shown:3x + 5x
Z= Z = 3(x’1-10) + 5x Z = -30 + 3x’ + 5x
1 2 2 1 2
x1 ≤ 4 x’1 - 10 ≤ 4 x’1 ≤ 14
2x2 ≤ 12 2x2 ≤ 12 2x2 ≤ 12
3x1 + 2x2 ≤ 18 3(x’1-10)+ 2x2 ≤ 18 3x’1 + 2x2 ≤ 48
x1 ≥ -10, x2 ≥ 0 x’1 - 10 ≥ -10, x2 ≥ 0 x’1 ≥ 0, x2 ≥ 0
Operations Research 1 – 06/07 – Chapter 4 33