The document discusses the simplex method for solving linear programming problems. It begins by introducing the simplex method and explaining that it is an iterative procedure that examines corner points to find an optimal solution. It then defines key terms like slack, surplus, and artificial variables. The majority of the document outlines the step-by-step procedures for using the simplex method to solve maximization and minimization problems. It also discusses how to handle special cases that may arise, such as degeneracy, unbounded problems, and infeasible problems.

1. LINEAR PROGRAMMING
SIMPLEX METHOD

2. Introduction
 With only two decision variables it is possible
to use graphical methods to solve LP problems
 But most real life LP problems are too complex
for simple graphical procedures
 We need a more powerful procedure called the
simplex method
 The simplex method examines the corner
points in a systematic fashion using basic
algebraic concepts
 It does this in an iterative manner until an
optimal solution is found
 Each iteration moves us closer to the optimal
solution

3. Basic Terms
 Slack Variables – A variable added to
convert an inequality of equal to or less than
type (≤) to equality.
 Surplus Variable – A variable subtracted to
convert an equality of equal to or greater
than type (≥) to equality.
 Artificial Variable – A fictitious variable
included in the case of inequality of (≥) type
and equality (=), to satisfy the non negativity
condition of a basic feasible solution.

4. Particulars Slack Variable Surplus
Variable
Artificial
Variable
Meaning Unused
resources or
idle resources
Excess amount
of resources
utilized
No physical or
economic
meaning. It is
fictitious
When to be
used
(≤) inequality (≥) inequality (≥) and (=)
constraints
Co-efficient in
the constraint
+1 -1 +1
Co-efficient in
the objective
function Z
0 0 +M for
minimization
and –M for
maximization
Use as Initial
Program
Variable
Used as starting
point (Initial
Table)
Cannot be used
since unit
matrix condition
is not satisfied
Initially used
but later on
eliminated
Presence in
the Optimal
Table
Helps to
interpret idle &
key resources
- - - Indicates
infeasible
solution

5. Procedure for Maximization Problems
1. Set up the inequalities describing the
problem constraints.
2. Introduce slack variables and convert
inequalities into equations.
3. Enter the equalities into the simplex
table.
4. Calculate the Zj and Cj – Zj value for
this solution.
5. Determine the entering variable by
choosing the highest Cj – Zj.

6. 6. Determine the row to be replaced from
the minimum ratio column (only
compute the ratios for rows whose
element is greater than zero).
7. Compute the value of the new key row.
8. Compute the values for remaining rows.
9. Calculate Cj and Zj value for this
solution.
10. If there is non-negative Cj – Zj value
and return to step 5 above.
11. If there is no non – negative Cj – Zj
values the final solution has been
obtained.

7. Procedure for Minimization Problems
1. Set up the inequalities describing the
problem constraints.
2. Convert any inequalities to equalities by
introducing surplus variables.
3. Add artificial variable to all equalities
involving surplus variables.
4. Enter the resulting equalities in the
simplex table.
5. Calculate Cj – Zj value for this solution.

8. 6. Determine the entering variable selecting
the one with the largest negative Cj – Zj
values.
7. Determine the departing row by choosing
the smallest non – negative ratio ignoring
infinity and negative. Zero is positive for
the purpose.
8. Compute the values for the new key row.
9. Calculate the values for the remaining rows.
10. Calculate the Cj – Zj value for this solution.
11. If there is negative element in the Cj – Zj
row then return to step 6. If there is no
negative Cj – Zj then final solution has been
obtained.

9. Simplex Problem (Mixed Constraints)
1. A situation may arise when the
constraints are ≥ type or ≤ type or =
type i.e. mixed constraints.
2. If any constraint has –ve constant at
the RHS, then multiply both sides with -
1 and it will reverse the directions of
inequalities.
3. Introduce slack variable, in case of ≤
type constraint.
4. Introduce surplus variables and artificial
variables in case of ≥ type constraints.

10. 5. Introduce artificial variables if the
constraint is ‘=‘ type.
6. Assign zero co-efficient to the slack
variables and surplus variables in the
objective function.
7. Assign +M in case of minimization and –M
in case of maximization problems to the
artificial variables in the objective
function.
8. In case of maximization, stop where Cj –
Zj row contains zero or negative elements.
9. In case of minimization, stop where Cj – Zj
row contains all elements zero or positive
values.

11. Special Cases
1. Tie for key column
2. Tie for key row (Degeneracy)
3. Unbounded problems
4. Multiple optimal solutions
5. Infeasible problems
6. Redundant constraints
7. Unrestricted variables

12. 1. Tie for key column
 Problem can arise in case of tie
between identical Cj – Zj values,
i.e., two or more columns.
 In such a case, selection for key
column can be made arbitrarily.
 There is no wrong choice, although
selection of one variable may result
in more iteration.

13. 2. Tie for key row (Degeneracy)
Degeneracy occurs when there is tie
for minimum ratio for choosing the
departing variable.
Methods to select the key row:
1. Choose arbitrarily.
2. Use alternate method.

14. Alternate Procedure for Degeneracy
i. Locate the rows in which smallest non-negative
ratio are tied (equal).
ii. Find the coefficient of the slack variables and divide
each coefficient by the corresponding positive
numbers of the key column in the row, starting
from left to right in order to break the tie.
iii. If the ratios do not break the tie, find the similar
ratios for the coefficients of decision variables.
iv. Compare the resulting ratio, column by column.
v. Select the row which contains smallest ratio. This
row becomes the key row.
vi. After the resolving of this tie, simplex method is
applied to obtain the optimum tie.

15. 3. Unbounded problems
 It is a situation when key row
cannot be selected because
minimum ratio column contains all
negative or infinity, the solution is
unbounded.
 The table does not indicate optimal
solution, yet the simplex process is
prohibited from continuing.

16. 4. Multiple optimal solutions
 In the final simplex table, if the
index row indicates the value of Cj
– Zj for a non-basic variable to
be zero, there exists an
alternative optimum solution.

17. 5. Infeasible problems
 It is a situation when an artificial
variable is present as a basic
variable in the final solution.
 This condition occurs when the
problem has incompatible
constraints.

18. 6. Redundant constraints
 When on constraint is less
restrictive than the other, it is
redundant constraint.
For example
4x1 + 3x2 < 12
4x1 + 3x2 < 16 (Redundant)

19. 7. Unrestricted variables
 Unrestricted variable is that decision
variable which does not carry any
value.
 This variable can take positive,
negative and zero values.
 To solve the problem, this variable
can take two values one +ve and
one –ve because difference between
these two is zero.