Topics to be covered
 Introduction
 Flow
to simplex method
chart
 Computational procedure of simplex method
 Examples
Introduction to Simplex Method
It has not been possible to obtain the graphical solution of more than
two variables. In such cases, a simplex method is used which was
developed by G. Dantzig in 1947.
The simplex method provides a systematic algorithm which consists of
moving from one basic feasible solution to another in a prescribed
manner so that the value of the objective function is improved.
The simplex algorithm is an step by step procedure for solving LP
problems. It consists of:
(i) Having a trial basic feasible solutions to constraint equations.
(ii) Testing whether it is an optimal solution.
(iii) Improving the first trial solution by a set of rules and repeating the
process till an optimum solution is obtained
In simplex method, One of the non-basic variables at one iteration
becomes basic at the following iteration, and it is called an entering
variable. One of the basic variables at one iteration becomes non-basic at
the following iteration, and is called a departing variable.
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Put the LPP
standard form
Find the initial
basic
feasible
solution
in
Construct
the
starting
simplex
method
Update the new
simplex table so
obtained
Compute ∆j=ZjCj=CBXj-cj
and
examine?
Find key element
and obtain the
new solution by
matrix
transformation
Is any
∆j<0 ?
Select
the
row
corresponding to
min ratio to find
the vector leaving
the basic B
Solution can be
improved.
Compute the ratio
(XB/XK,XK>0)
Solution under
test is optimal.
Stop
No
Find the vector Xk
entering the basisB
sothat ∆k=min∆j
No
Corresponding to
any j, are all the
elements of
entering vector
Xk<=0
Yes
Solution is
unbounded.
Stop
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Computational procedure of simplex method
Step 1: If the problem is one of minimization, convert it to a
maximization problem by considering –z, instead of z, use the min z=max (-z)
Step 2: We checkup all bi’s for non-negativity. If some of the bi’s are
negative, multiply the corresponding constraints through by -1 in order to
ensure all bi>=0
Step 3: We change the inequalities to equation by adding slack and
surplus variable, if necessary.
Step 4: We now construct the starting simplex table. From this table,
basic feasible solution is obtained.
Step 5: We obtain the values of ∆j=Zj-Cj=CBXj-cj and examine the
values of ∆j.
There will be three exhaustive possibilities:
(i) All ∆j>=0. In this cases, the basic feasible solution under test will be
optimal.
(ii) Some ∆j<0 and for at least one of the corresponding xj all xrj<=0. In
this case , solution will be unbounded.
(iii) Some ∆j<=0 and all the corresponding xj have at least one xij>0. In
this case, further improvement is possible.
Step 6: Further improvement is done by replacing the vectors. Rules are:
(i) To select “incoming vector”. We find such value of k for which
△k=min △j. Then the vector coming into basis matrix will be Xk.
(ii) To select “outgoing vector”. The vector going out of the basis matrix
will be βr, if we determine the suffix r by the minimum ratio rule.
(iii) Xβr / Xrk = min[XBi/Xik,Xik>0]
Step 7: We now construct the next improvement table by using the
simplex matrix transformation rules.
Step 8: Now return to step 6, then go to steps 8 and 9, if necessary. This
process is repeated till we reach the desired conclusion.
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Example 1: Consider the linear programming problem
Maximize Z=3x1+2x2
X1+x2<=4, X1-x2<=2 and x1,x2>=0
Solution:
Basic var
S1
S2
CB
0
0
XB
4
2
3
2
0
0
X1
X2
S1
S2
1
1
1
-1
1
0
0
1
0
0
4/1
2/1
△j=zj-cj
X1=X2=0
z=CBXB=0
-3
-2
S1
X1
0
3
0
1
2
-1
1
0
-1
1
X2=S2=0
Z=CBXB=6
0
-5
0
3
X2
2
1
0
1
½
-1/2
X1
3
3
1
0
½
1/2
S1=S2=0
Z=CB=XB=11
0
5/2
½
2
2
0
Min Ratio(XB/XK)
2/2
-△j
All △j>=0
Optimum solution: X1=3, X2=1, Max Z=11
Important points:
1. In the first iteration only, since △j are the same as –cj’s, so there is no
need to calculate it separately.
2. Mark min(△j)by ’ ‘ which at once indicates the columns Xk needed for
computing the minimum ratio(XB/XK)
3. ‘Key element’ is found at the place where the upward directed arrow ‘
‘ of min △j and the left directed arrow (
) of minimum ratio
intersect each other in the simplex table.
4 ‘Key element’ indicates that the current table must be transformed in
such a way that the key element becomes 1 and all other elements in
that column become 0
5. Since △j’s corresponding to unit column vectors are always zero,
there is no need to calculating them.
6. While transforming the table by row operations, the value of z and
corresponding △j’s are also computed at the same time.
Example 2: Min z=x1-3x2+2x3
3X1-X2+3X3<=7,-2X1+4X2<=12,-4X1+3X2+8X3<=10 and
X1,X2,X3>=0
Solution: Objective function is: Max –z= -X1+3X2-2X3
3 -1 3
-2 4 0
-4 3 8
X1
X2
X3
7
= 12
10
-1
3
-2
0
0
0
CB
XB
X1
X2
X3
X4
X5
X6 Min Ratio(XB/XK)
0
0
7
12
3
-2
-1
4
3
0
1
0
0
1
0
0
-12/4
X6
0
10
-4
X1=X2=X3=0 z=CBXB=0 1
3
8
0
0
1
10/3
-3
2
0
0
0
Basic var
X4
X5
△j=zj-cj
X4
0
10
5/2
0
3
1
¼
0
X2
X6
3
0
3
1
-1/2
-5/2
1
0
0
8
0
0
¼
3/4
0
1
4
5
11
-1/2
1
0
0
0
0
1
0
2
6/5
3/5
11
0
0
13/5
X1=x3=x5=0
X1
X2
X6
X3=X4=X2=0
z’= 9
-1
3
0
Z’= 11
0
2/5
1/5
1
1/5
Min
¾
1/10
3/10
-1/2
0
0
0
1
8/10
0
10/2.5
Min
---
△j>=0
Optimal solution is: X1=4,X2=5,X3=0,Min z=-11
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Thanks