Simplex Method- Revise part-1

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• Revised Simplex Method
• Prepared by
• Boina Anil Kumar
• Asst.Prof(Maths)
• MITS, Rayagada.

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Revised Simplex Method.

• Streamlined versions of the simplex method for
  computer implementations do not follow the
  original simplex algorithm (algebraic or tabular
  form)
• ? The Revised Simplex Algorithm explicitly uses
  matrix manipulations
• ?it computes and stores only the information
  required for each iteration, namely the
  coefficients of the non-basic variables in
  Equation 0(zero), the coefficients of the basic
  entering variable in the other equations, and the
  RHSs of the equations.

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• To explain revised simplex Procedure we will take
  the following example.
• Example-
• Solve the following LPP by revised simplex
  method.
• Note- If the problem is given in minimization,
  then convert it in to Maximization.

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Solution-

• Step-1
• 1)If the problem is in minimization, then convert
  it in to maximization(i.e. objective function.)
• 2) Make all bi 0 (if not).
• 3) Use Slack variables as per need to make the
  constraints to equations.
• 4)Objective function also considered as the first
  constraint equation.
• Thus the given problem can be re-written in
  standard form for revised simplex as follow.

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• Standard form in Revised simplex Method.
• Here z is as large as possible and unrestricted
  in sign.

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• Step-2
• Now write the Matrix form of above standard form.
• Here B is basis matrix and intially it is
  identity matrix.
• So B-1B.

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• Step-3
• In this step we construct the Revised Simplex
  table as follow..

Non basic variable table
Basic Variable ß0 (z) ß1 ß2 XB Xk
Basic Variable
z


A1 A2



z is one of the basic variable for every revised
simplex table.
ß0 always contains the coefficents of z in the
system of equations in standard form.
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• 1st column Basic variables is filled with the
  variables corresponding to the columns forming
  identity matrix in the Matrix A.
• Here ß0,ß1,ß2 are the columns from the Matrix
  B-1.
• Non basic variable table is filled with the
  column vectors in Matrix A are not forming
  identity matrix.
• First we will fill all the data in Revised
  simplex table as well as Non basic variable table
  except XB and Xk column.

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Table-1
Basic Variable ß0 (z) ß1 (X3) ß2 (X4)
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) XB Xk
z 1 0 0
X3 0 1 0
X4 0 0 1
Non basic variable table.
A1 (X1) A2 (X2)
-2 -1
3 4
6 1
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• Now XB is filled after calculating it by the
  formula.
• Now the table is

Non basic variable table
Table-1
Basic Variable ß0 (z) ß1 (X3) ß2 (X4)
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) XB Xk
z 1 0 0 0
X3 0 1 0 6
X4 0 0 1 3
A1 (X1) A2 (X2)
-2 -1
3 4
6 1
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• Now compute ?js by following formula.
• ?j (first row of B-1 )(Aj from Non basic
  variable
  table)
• So here ?1(first row of B-1)(A1)(1,0,0)(-2,3,6
  ) 1(-2)0306-2
• Similarly?2(first row of B-1)(A2)(1,0,0)(-1,4,
  1) 1(-1)0401-1
• Hint- we may also calculate ?js as follow

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• Step-4
• Now test for optimality.
• If all the ?j0 then the optimal solution is
  obtained and XB gives the corresponding values of
  z and other basic variables. Value of non basic
  variables are zero.
• In our example both the ?js are negative we need
  to minimum.Min?1,?2-2,-1-2?1. i.e. k1So
  the corresponding vector A1 of ?1 will enter in
  to the basis.

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• Step-5
• Now calculate the Xk by following formula and
  fill it in the table.
• XkB-1Ak.
• Here k1 so calculate X1B-1A1.
• After filling the Xk column the table becomes

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Table-1
A1 (X1) A2 (X2)
-2 -1
3 4
6 1
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) X1
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) XB Xk Min (XB/Xk)
z 1 0 0 0 -2 Min (XB/Xk)
X3 0 1 0 6 3 6/3
X4 0 0 1 3 6 3/6
6
Here calculate min(XB/Xk). The vector
corresponding to min ratio will leave the basis.
In our example X4 will leave the basis and Xk
i.e. X1 will enter to the basis. So 6 is the
pivot element (which is highlighted.).
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• Step-6
• To get the improve table we need to do the
  following operations.
• Now we need to make pivot element become 1 and
  other elements of that column as zero by matrix
  operations.
• So do the following operations

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• So we will get
• Now replace ß1,ß2 in the previous table by new
  ß1,ß2 in
• Entering vector Ak in non basic variable table
  will replaced by Xk in
• In improved table X4 is removed and X1 will
  enter.
• In the same way Ak is now X4. (i.e. A1 is X4)
• So the improved table is as follow

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Table-2
Non basic variable table
Basic Variable ß0 (z) ß1 (X3) ß2 (X4)
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) XB Xk Min (XB/Xk)
z 1 0 1/3 Min (XB/Xk)
X3 0 1 -1/2
X1 0 0 1/6
A1 (X4) A2 (X2)
0 -1
0 4
1 1
Here again B-1 is
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• Again XB is calculated by XBB-1b.
• So
• Then table become

Table-2
Non basic variable table
Basic Variable ß0 (z) ß1 (X3) ß2 (X4)
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) XB Xk Min (XB/Xk)
z 1 0 1/3 1 Min (XB/Xk)
X3 0 1 -1/2 9/2
X1 0 0 1/6 1/2
A1 (X4) A2 (X2)
0 -1
0 4
1 1
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• Again test for optimal as follow.
• Here ?2-2/3, so optimal solution is not
  obtained.
• So A2 will enter. i.e. vectorX2 will enter to the
  basis.
• So again we obtain Xk by XkB-1A2.

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• On repeating the above process we will get the
  following revised simplex tables.

Table-2
Non basic variable Table.
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) X2
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) XB Xk Min (XB/Xk)
z 1 0 1/3 1 -2/3 Min (XB/Xk)
X3 0 1 -1/2 9/2 7/2 9/7
X1 0 0 1/6 1/2 1/6 3
A1 (X4) A2 (X2)
0 -1
0 4
1 1
7/2
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• Next simplex table is
• On testing all ?j0. So optimal solution is
  obtained.
• Optimal solution is z13/7, x12/7 and x29/7.
• Note In entire process the 1st column ß0 is not
  changed. It is applicable for all the problems.

Non basic variable table
Table-3
Basic Variable ß0 (z) ß1 (X3) ß2 (X4)
Basic Variable ß0 (z) ß1 (X3) ß2 (X4) XB Xk
z 1 4/21 5/21 13/7
X2 0 2/7 -1/7 9/7
X1 0 -1/21 4/21 2/7
A1 (X4) A2 (X3)
0 0
0 1
1 0
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• Thank you all