Simplex method, Operational ResearchLevel 4

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Simplex method, Operational Research-Level 4

• Prepared by T.M.J.A.Cooray
• Department of Mathematics

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• General LP problem is to find the values of
  X1,X2,..Xn which maximizes (or minimizes) the
  objective function
• Max (or Min) Z C1X1C2X2CnXn
• While satisfying the constraints
• a11X1.a1nXn b1 or b1
• a21X1.a2nXn b2 or b2
• ..
• am1X1.amnXn bm or bm
• and X1, .Xn 0

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• This problem can be put in the canonical form as
  follows.
• Inequalities can be converted to equalities.
• a11X1.a1nXn S1 b1
• a21X1.a2nXn S2 b2
• ..
• am1X1.amnXn Sm bm
• ( by adding a slack or a surplus variable)

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Minimization or maximization

• Max (min) Z C1X1C2X2CnXn
• can be converted to a
• min (max) Z- C1X1C2X2CnXn
• by multiplying the objective function by -1.

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Variables unrestricted in sign

• Xi can take any value, either positive or
  negative.
• In such a case it can be replaced by
• Xi Xi-Xi

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DEFINITIONS

Slack variables are
defined, when there are inequalities. In the
example discussed , the available capacities of
the three machines M1,M2 and M3 are 40,40 and 40
respectively.. Unused amounts of the three
machines are denoted by X3,X4 and X5. (They are
0.) Some books denote them by Sj .
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• Surplus variables
• are defined when there are inequalities.
• In the diet planning problem of the dog,
• The minimum protein requirements are specified
  on the r.h.s.. If you feed more, the dog gets
  more than what is required.
• The excess amount of protein is denoted by Xj or
  Sj .(they are 0) The amount overfed is the
  surplus variable.
• By subtracting that amount we get the equality.

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• Non basic variables
• The variables which have the value zero are
  called non basic variables.
• Basic variables
• Variables which are positive are called basic
  variables. However some times the basic variables
  can have zero values and then the solution is
  said to be degenerate .
• .

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Simplex method.

• Consider the same problem solved using the
  graphical method.
• This procedure is equivalent to find the cdts of
  the corner points of the f. region. This method
  consists of changing set of basic variables one
  at a time until Z (or f) is maximized.

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• The first step is to determine an initial basic
  feasible solution an obvious solution is
  x10,x20, giving x340,x440 and x540.
• This is equivalent to corner point O in the
  graphical solution.
• Step 2
• Solve for the basic variables in terms of the non
  basics and express f in terms of non basics.

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• X340-.25x1-.5x2
• X440-.4x1-.2x2
• X5 40 -.8x2
• f2x13x2
• .

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PIVOT COLUMN
40/.5 80 40/.2 200 40/.8 50
PIVOT ROW
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PIVOT COLUMN
Row- Pivot row 3 pivot elt
2 0 0 0 -3.75 -150
0 .25 0 1 0 -.625 15
40/.5 80 40/.2 200 40/.8 50
Row- Pivot row .5 pivot elt
Row- Pivot row .2 pivot elt
0 4 0 0 1 -.25 30
0 0 .8/.8 0 0 1/.8 50
PIVOT ROW PIVOT ELEMENT
X2
PIVOT ROW
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PIVOT COLUMN
Row- Pivot row 3 pivot elt
2 0 0 0 -3.75 -150
0 .25 0 1 0 -.625 15
40/.5 80 40/.2 200 40/.8 50
Row- Pivot row .5 pivot elt
Row- Pivot row .2 pivot elt
0 4 0 0 1 -.25 30
PIVOT ROW PIVOT ELEMENT
0 0 .8/.8 0 0 1/.8 50
PIVOT ROW