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Optimization Methods: Linear Programming- Revised Simplex Method
1
Module – 3 Lecture Notes – 5
Revised Simplex Method, Duality and Sensitivity analysis
Introduction
In the previous class, the simplex method was discussed where the simplex tableau at each
iteration needs to be computed entirely. However, revised simplex method is an improvement
over simplex method. Revised simplex method is computationally more efficient and accurate.
Duality of LP problem is a useful property that makes the problem easier in some cases and
leads to dual simplex method. This is also helpful in sensitivity or post optimality analysis of
decision variables.
In this lecture, revised simplex method, duality of LP, dual simplex method and sensitivity or
post optimality analysis will be discussed.
Revised Simplex method
Benefit of revised simplex method is clearly comprehended in case of large LP problems. In
simplex method the entire simplex tableau is updated while a small part of it is used. The
revised simplex method uses exactly the same steps as those in simplex method. The only
difference occurs in the details of computing the entering variables and departing variable as
explained below.
Let us consider the following LP problem, with general notations, after transforming it to its
standard form and incorporating all required slack, surplus and artificial variables.
Z 
 xi 
x 
j
 xl 
c1 x1  c2 x2  c3 x3 
 cn xn  Z  0
c11 x1  c12 x2  c13 x3 
 c1n xn
 b1
c21 x1  c22 x2  c23 x3 
 c2 n xn
 b2
cm1 x1  cm 2 x2  cm3 x3 
 cmn xn
 bm
As the revised simplex method is mostly beneficial for large LP problems, it will be
discussed in the context of matrix notation. Matrix notation of above LP problem can be
expressed as follows:
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
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Minimize z  C T X
subject to : AX  B
with : X  0
 b1 
 x1 
 c1 
 c11
0 
 
x 
c
c 
0 
 b2 
2
2

 21





X

C

A

B

where
,
,
, 0
,


 

 
 

 
 


0
 xn 
c m1
c n 

bm 
c12
c 22

cm2
 c1n 
 c 2 n 
  

 c mn 
It can be noted for subsequent discussion that column vector corresponding to a decision
 c1k 
c 
2k
variable x k is   .
  
 
c mk 
Let X S is the column vector of basic variables. Also let C S is the row vector of costs
coefficients corresponding to X S and S is the basis matrix corresponding to X S .
1. Selection of entering variable
For each of the nonbasic variables, calculate the coefficient WP  c  , where, P is the
corresponding column vector associated with the nonbasic variable at hand, c is the cost
coefficient associated with that nonbasic variable and W  C S S 1 .
For maximization (minimization) problem, nonbasic variable, having the lowest negative
(highest positive) coefficient, as calculated above, is the entering variable.
2. Selection of departing variable
a. A new column vector U is calculated as U  S 1B .
b. Corresponding to the entering variable, another vector V is calculated as V  S 1 P ,
where P is the column vector corresponding to entering variable.
c. It may be noted that length of both U and V is same (  m ). For i  1,, m , the
ratios,
U i 
, are calculated provided Vi   0 . i  r , for which the ratio is least, is
V i 
noted. The r th basic variable of the current basis is the departing variable.
If it is found that Vi   0 for all i , then further calculation is stopped concluding that
bounded solution does not exist for the LP problem at hand.
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M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
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3. Update to new basis

Old basis S , is updated to new basis S new , as S new  ES 1






where E  







1
0
1
0
0
1
2
0
r
0
0
 m 1
1
0
0
m
0

1
0 

0 


 V i 


 and  i  V r 
1



V r 

0 

1 
for
ir
for
ir
r th column
S is replaced by S new and steps 1 through 3 are repeated. If all the coefficients calculated in
step 1, i.e., WP  c  is positive (negative) in case of maximization (minimization) problem,
then optimum solution is reached and the optimal solution is,
X S  S 1B and z  CX S
Duality of LP problems
Each LP problem (called as Primal in this context) is associated with its counterpart known
as Dual LP problem. Instead of primal, solving the dual LP problem is sometimes easier
when a) the dual has fewer constraints than primal (time required for solving LP problems is
directly affected by the number of constraints, i.e., number of iterations necessary to
converge to an optimum solution which in Simplex method usually ranges from 1.5 to 3
times the number of structural constraints in the problem) and b) the dual involves
maximization of an objective function (it may be possible to avoid artificial variables that
otherwise would be used in a primal minimization problem).
The dual LP problem can be constructed by defining a new decision variable for each
constraint in the primal problem and a new constraint for each variable in the primal. The
coefficients of the j th variable in the dual’s objective function is the i th component of the
primal’s requirements vector (right hand side values of the constraints in the Primal). The
dual’s requirements vector consists of coefficients of decision variables in the primal
objective function. Coefficients of each constraint in the dual (i.e., row vectors) are the
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
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column vectors associated with each decision variable in the coefficients matrix of the primal
problem. In other words, the coefficients matrix of the dual is the transpose of the primal’s
coefficient matrix. Finally, maximizing the primal problem is equivalent to minimizing the
dual and their respective values will be exactly equal.
When a primal constraint is less than equal to in equality, the corresponding variable in the
dual is non-negative. And equality constraint in the primal problem means that the
corresponding dual variable is unrestricted in sign. Obviously, dual’s dual is primal. In
summary the following relationships exists between primal and dual.
Primal
Dual
Maximization
Minimization
Minimization
Maximization
i th variable
i th constraint
j th constraint
j th variable
Inequality sign of i th Constraint:
xi  0
 if dual is maximization
 if dual is minimization
i th variable unrestricted
i th constraint with = sign
j th constraint with = sign
j th variable unrestricted
th
RHS of j constraint
Cost coefficient associated with j th
variable in the objective function
Cost coefficient associated with
i th variable in the objective
RHS of i th constraint
function
See the pictorial representation in the next page for better understanding and quick reference:
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
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Mark the corresponding
decision variables in the dual
Cost coefficients for
the Objective Function
Opposite for the Dual,
i.e., Minimize
Maximize Z  c1 x1  c2 x2 
Subject to
 cn xn
c11 x1  c12 x2 
 c1n xn  b1 
 y1
c21 x1  c22 x2 
 c2 n xn  b2 
 y2
cm1 x1  cm 2 x2 
 cmn xn  bm 
 ym
x1  0, x2 unrestricted,
Coefficients
of the 1st
constraint
, xn  0
Corresponding
sign of the 2nd
constraint is 
Determine the
sign of y1
Thus, the 1st constraint,
c11 y1  c21 y2   cm1 ym  c1
Right hand
side of the 1st
constraint
Corresponding
sign of the 1st
constraint is 
Coefficients
of the 2nd
constraint
Thus the Objective Function,
Minimize b1 y1  b2 y2   bm ym
Right hand
side of the 2nd
constraint
Determine the
sign of y 2
Thus, the 2nd constraint,
c12 y1  c22 y2   cm 2 ym  c2
Determine the
sign of ym
Dual Problem
Minimize Z  b1 y1  b2 y2 
Subject to
 bm ym
c11 y1  c21 y2 
 cm1 ym  c1
c12 y1  c22 y2 
 cm 2 ym  c2
c1n y1  c2 n y2 
 cmn ym  cn
y1 unrestricted, y2  0,
D Nagesh Kumar, IISc, Bangalore
, ym  0
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
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It may be noted that, before finding its dual, all the constraints should be transformed to ‘lessthan-equal-to’ or ‘equal-to’ type for maximization problem and to ‘greater-than-equal-to’ or
‘equal-to’ type for minimization problem. It can be done by multiplying with 1 both sides
of the constraints, so that inequality sign gets reversed.
An example of finding dual problem is illustrated with the following example.
Primal
Z  4 x1  3x2
Maximize
Minimize
Z   6000 y1  2000 y 2  4000 y3
Subject to
Subject to
x1 
Dual
2
x 2  6000
3
y1  y 2  y3  4
x1  x2  2000
2
y1  y 2  3
3
x1  4000
y1  0
x1 unrestricted
y2  0
x2  0
y3  0
It may be noted that second constraint in the primal is transformed to  x1  x2  2000
before constructing the dual.
Primal-Dual relationships
Following points are important to be noted regarding primal-dual relationship:
1. If one problem (either primal or dual) has an optimal feasible solution, other problem
also has an optimal feasible solution. The optimal objective function value is same for
both primal and dual.
2. If one problem has no solution (infeasible), the other problem is either infeasible or
unbounded.
3. If one problem is unbounded the other problem is infeasible.
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
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Dual Simplex Method
Computationally, dual simplex method is same as simplex method. However, their
approaches are different from each other. Simplex method starts with a nonoptimal but
feasible solution where as dual simplex method starts with an optimal but infeasible solution.
Simplex method maintains the feasibility during successive iterations where as dual simplex
method maintains the optimality. Steps involved in the dual simplex method are:
1. All the constraints (except those with equality (=) sign) are modified to ‘less-thanequal-to’ (  ) sign. Constraints with greater-than-equal-to’ (  ) sign are multiplied by
1 through out so that inequality sign gets reversed. Finally, all these constraints are
transformed to equality (=) sign by introducing required slack variables.
2. Modified problem, as in step one, is expressed in the form of a simplex tableau. If all
the cost coefficients are positive (i.e., optimality condition is satisfied) and one or
more basic variables have negative values (i.e., non-feasible solution), then dual
simplex method is applicable.
3. Selection of exiting variable: The basic variable with the highest negative value is
the exiting variable. If there are two candidates for exiting variable, any one is
selected. The row of the selected exiting variable is marked as pivotal row.
4. Selection of entering variable: Cost coefficients, corresponding to all the negative
elements of the pivotal row, are identified. Their ratios are calculated after changing


Cost Coefficients
 .
the sign of the elements of pivotal row, i.e., ratio  
  1  Elements of pivotal row 
The column corresponding to minimum ratio is identified as the pivotal column and
associated decision variable is the entering variable.
5. Pivotal operation: Pivotal operation is exactly same as in the case of simplex
method, considering the pivotal element as the element at the intersection of pivotal
row and pivotal column.
6. Check for optimality: If all the basic variables have nonnegative values then the
optimum solution is reached. Otherwise, Steps 3 to 5 are repeated until the optimum is
reached.
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
8
Consider the following problem:
Minimize
Z  2 x1  x 2
subject to
x1  2
3 x1  4 x 2  24
4 x1  3 x 2  12
 x1  2 x 2  1
By introducing the surplus variables, the problem is reformulated with equality constraints as
follows:
Minimize
Z  2 x1  x 2
subject to
 x1
 x3  2
3x1
4 x 2
 x 4  24
4 x1
3x 2
 x5  12
x1
2 x 2
 x6  1
Expressing the problem in the tableau form:
Variables
Iteration
1
Basis
Z
x1
x2
x3
x4
x5
x6
br
Z
1
-2
-1
0
0
0
0
0
x3
0
-1
0
1
0
0
0
-2
x4
0
3
4
0
1
0
0
24
x5
0
-4
-3
0
0
1
0
-12
x6
0
1
-2
0
0
0
1
-1
0.5
1/3
--
--
--
--
Ratios 
Pivotal Element
Pivotal Row
Pivotal Column
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
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Tableaus for successive iterations are shown below. Pivotal Row, Pivotal Column and Pivotal
Element for each tableau are marked as usual.
Iteration
2
Basis
Z
3
4
x5
x6
br
1
-2/3
0
0
0
-1/3
0
4
x3
0
-1
0
1
0
0
0
-2
x4
0
-7/3
0
0
1
4/3
0
8
x2
0
4/3
1
0
0
-1/3
0
4
x6
0
11/3
0
0
0
-2/3
1
7
2/3
--
--
--
--
--
x1
x2
Variables
x3
x4
x5
x6
br
Basis
Z
Z
1
0
0
-2/3
0
-1/3
0
16/3
x1
0
1
0
-1
0
0
0
2
x4
0
0
0
-7/3
1
4/3
0
38/3
x2
0
0
1
4/3
0
-1/3
0
4/3
x6
0
0
0
11/3
0
-2/3
1
-1/3
--
--
--
--
0.5
--
x1
x2
Variables
x3
x4
x5
x6
Ratios 
Iteration
x2
Z
Ratios 
Iteration
x1
Variables
x3
x4
br
Basis
Z
Z
1
0
0
2.5
0
0
-0.5
5.5
x1
0
1
0
-1
0
0
0
2
x4
0
0
0
5
1
0
2
12
x2
0
0
1
-0.5
0
0
-0.5
1.5
x5
0
0
0
-5.5
0
1
-1.5
0.5
Ratios 
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
10
As all the br are positive, optimum solution is reached. Thus, the optimal solution is Z  5.5
with x1  2 and x2  1.5 .
Solution of Dual from Final Simplex Tableau of Primal
Primal
Dual
Maximize
Z  4 x1  x2  2 x3
subject to
2 x1  x2  2 x3  6
x1  4 x2  2 x3  0
5 x1  2 x2  2 x3  4
x1 , x2 , x3  0
Minimize
Z '  6 y1  0 y 2  4 y3
subject to
2 y1  y 2  5 y3  4
y1  4 y 2  2 y3  1
2 y1  2 y 2  2 y3  2
y1 , y 2 , y3  0
Final simplex tableau of primal:
y1
y2
Z’
y3
As illustrated above solution for the dual can be obtained corresponding to the coefficients of
slack variables of respective constraints in the primal, in the Z row as, y1  1 , y 2 
y3 
1
and
3
1
and Z’=Z=22/3.
3
D Nagesh Kumar, IISc, Bangalore
M3L5
Optimization Methods: Linear Programming- Revised Simplex Method
11
Sensitivity or post optimality analysis
A dual variable, associated with a constraint, indicates a change in Z value (optimum) for a
small change in RHS of that constraint. Thus,
Z  y j bi
where y j is the dual variable associated with the i th constraint, bi is the small change in the
RHS of i th constraint, and Z is the change in objective function owing to bi .
Let, for a LP problem, i th constraint be 2 x1  x2  50 and the optimum value of the objective
function be 250. What if the RHS of the i th constraint changes to 55, i.e., i th constraint
changes to 2 x1  x2  55 ? To answer this question, let, dual variable associated with the i th
constraint is y j , optimum value of which is 2.5 (say). Thus, bi  55  50  5 and y j  2.5 .
So, Z  y j bi  2.5  5  12.5 and revised optimum value of the objective function is
 250  12.5  262.5 .
D Nagesh Kumar, IISc, Bangalore
M3L5