lc 13 dual simplex
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3-Variable Problem by Simplex Method
Minimize: z = 51 2 83 subject to: 2x1 + 5x2 3 1
-2x1 - 122 + 3 9
-3x1 - 82 + 23 4
xi 0, i = 1 , ... , 3.
Starting with Basic Feasible Solution
Introduce slack variables as basic variables and the actual variables
as non-basic (set to 0), then the right-hand side of each constraint
represents the basic solution..
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Canonical Form2x1 + 5x2 - x3 + x4 = 1
-2x1 - 12x2 + 3x3 + x5 = 9
-3x1 - 8x2 + 2x3 + x6 = 4
Treating the slack variables as basic and the others as non-basic,
the starting basic feasible solution is
Basic: x4 = 1, x5 = 9, x6 = 4
Non-basic: x1 = x2 = x3 = 0
z=0 is a basic feasible solution
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We have the following situation:
Basic: x4 = 1, x5 = 9, x6 = 4 ;
Non basic: x1 = x2 = x3 = 0 ;
z= 0: 5x1 3x2 8x3 = 0.
The largest negative coefficient in the f equation is that of x3.
Thus, our next basic feasible solution should use x3 as one of
its basic variables.
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The variable x3 is to be brought into the basic set as entering
variable implying that it will have a value greater than or equal to
zero in the next basic feasible solution. Therefore the columncontaining x3 is now pivotal column. For constraint 1, the
coefficient of x3 is negative and therefore, the choice of Row 1 aspivotal row is ruled out. constraint will remain positive as aresult of making x3 basic.
Basic x1 x2 x3 x4 x5 x6 zRHS (bi)
x4 2 5 -1 1 0 0 0 1
x5 -2 -12 3 0 1 0 0 9
x6 -3 -8 2 0 0 1 0 4
Obj. 5 -4 -8 0 0 0 -1 0
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Basic x1 x2 x3 x4 x5 x6 zRHS (bi)
x4 1/2 1 0 1 0 0 3 R1+1/2R3x5 5/2 0 0 0 1 -3/2 0 3 R2-3/2R3x6 -3/2 -4 1 0 0 1/2 0 2
Obj. -7 -35 0 0 0 4 -1 16
Basic x1 x2 x3 x4 x5 x6 zRHS (bi)
x4 2 5 -1 1 0 0 0 1
x5 -2 -12 3 0 1 0 0 9
x6 -3 -8 2 0 0 1 0 4
Obj. 5 -4 -8 0 0 0 -1 0
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Basic x1 x2 x3 x4 x5 x6 zRHS (bi)
x4 1/2 1 0 1 0 0 3
x5 5/2 0 0 0 1 -3/2 0 3
x6 -3/2 -4 1 0 0 1/2 0 2
Obj. -7 -35 0 0 0 4 -1 16
Basic x1 x2 x3 x4 x5 x6 zRHS (bi)
x4 1/2 1 0 1 0 0 3 Already 1
x5 5/2 0 0 0 1 -3/2 0 3 Already 0
x6 0 1 4 0 3/2 0 14 R3+4R1Obj. 21/2 0 0 35 0 43/2 -1 121
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General Notation and Computer input
Matrix form
Input A B and CT
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Now let us see what are
Surplus Variables
1 1
n m
ji i j
i j
a x b
1 1
1.n m
ji i n j j
i j
a x x b
But then another limitation arises when the coefficient of slack variables becomes
negative. To take care of such situation, Artificial Variables are added to the
system. The LPP Transforms to minimizing
Minimize '
1
n
i
i
z y
Subject to
1 1
n m
ji i i j
i j
a x y b
,i ix y 0
Try Examples Two phase Method on Page 137, 141, 145
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DEGENERACY AND CYCLING
Let a LPP in Canonical form be given by
If there are several candidates for departing variables (tie among many) by criteria.
Then the value of basic variable not selected turns out to be zero.
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A basic feasible solution in which some basic variables are zero is called
degenerate.
Example
2
4
6
1 3
1 2
1 2
1 2
1 2
z = 5x +3x
x - x
2x + x
-3x +2x
x 0 x 0
Examples
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CHECK EXAMPLE PAGE 127
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DUALITY
Maximize Minimize
Tc
z = x
Ax b
x 0 (13)
Primal
T
T
z' = b w
A x c
w 0(13)
Dual
2
5
1
1 2
1 2
1 2
1 2
1 2
z = 2x +3x
3x +2x
-x +2x
4x + x
x 0 x 0
4 2
3
1 2 3
1 2 3
1 2 3
1 2 3
z' = 2w +5w w
3w - w w
2w +2w w
w 0, w 0, w 0
Important Result: What is dual of the Dual problem? Prove it.
Check Statements of Theorems 3.4, 3.5, 3.6
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Imp: Duality Theorem
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Dual Simplex Method
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With initial Tableau
Basic x1=x x2 =y x3=u x4 =v z =f bj
x3 2 2 1 0 0 8
x4 5 3 0 1 0 15
Obj.R -120 -100 0 0 1 0
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Basic x1=x x2 =y x3=u x4 =v z =f bj
x3 0 4/5 1 -1/5 0 2
x1 1 3/5 0 1/5 0 3
Obj.R 0 -28 0 24 1 360
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Basic x1=x x2 =y x3=u x4 =v z =f Xb
x2 0 1 5/4 -1/2 0 5/2
x1 1 0 -3/4 1/2 0 3/2
Obj.R 0 0 35 10 1 430
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