Operations Research

GRAPHICAL METHOD

machine products table

Maximize X_{1}+1.5X_{2}

Constraints:

  • 2X_{1}+2X_{2}\leq 16
  • X_{1}+2X_{2}\leq 12
  • 4X_{1}+2X_{2}\leq 28
  • X_{1},X_{2}\geq 03

glp linear programming

The optimal solution is X_{1}=4 and X_{2}=4 V(P)=1(4)+1,5(4)=10 

 

SIMPLEX METHOD

Maximize z = 3x1 + 2x2

subject to

-x1 + 2x2 ≤ 4
3x1 + 2x≤ 14
x1 – x2 ≤ 3

x1, x2 ≥ 0

-x1 + 2x2 + x3 = 4
3x1 + 2x2 + x4 = 14
x1 – x2 + x5 = 3
x1, x2, x3, x4, x5 ≥ 0

Maximize z = 3x1 + 2x2 + 0x3 + 0x4 + 0x5

x1 = 0, x2 = 0, z = 0

x3 = 4, x4 = 14, x5 = 3

Simplex Method Examples, Operations Research

a11 = -1, a12 = 2, a13 = 1, a14 = 0, a15 = 0, b1 = 4
a21 = 3, a22 = 2, a23 = 0, a24 = 1, a25 = 0, b2 = 14
a31= 1, a32 = -1, a33 = 0, a34 = 0, a35 = 1, b= 3

Calculating values for the index row (z– cj)

z– c1 = (0 X (-1) + 0 X 3 + 0 X 1) – 3 = -3
z– c2 = (0 X 2 + 0 X 2 + 0 X (-1)) – 2 = -2
z– c3 = (0 X 1 + 0 X 0 + 0 X 0) – 0 = 0
z– c4 = (0 X 0 + 0 X 1 + 0 X 0) – 0 = 0
z– c5 = (0 X 0 + 0 X 0 + 0 X 1) – 0 = 0

x3 row

a11 = -1 – 1 X ((-1)/1) = 0
a12 = 2 – (-1) X ((-1)/1) = 1
a13 = 1 – 0 X ((-1)/1) = 1
a14 = 0 – 0 X ((-1)/1) = 0
a15 = 0 – 1 X ((-1)/1) = 1
b1 = 4 – 3 X ((-1)/1) = 7

x4 row

a21 = 3 – 1 X (3/1) = 0
a22 = 2 – (-1) X (3/1) = 5
a23 = 0 – 0 X (3/1) = 0
a24 = 1 – 0 X (3/1) = 1
a25 = 0 – 1 X (3/1) = -3
b2 = 14 – 3 X (3/1) = 5

x1 row

a31 = 1/1 = 1
a32 = -1/1 = -1
a33 = 0/1 = 0
a34 = 0/1 = 0
a35 = 1/1 = 1
b3 = 3/1 = 3

Table 2

asd.png

Calculating values for the index row (z– cj)

z– c1 = (0 X 0 + 0 X 0 + 3 X 1) – 3 = 0
z– c2 = (0 X 1 + 0 X 5 + 3 X (-1)) – 2 = -5
z– c3 = (0 X 1 + 0 X 0 + 3 X 0) – 0 = 0
z– c4 = (0 X 0 + 0 X 1 + 3 X 0) – 0 = 0
z– c5 = (0 X 1 + 0 X (-3) + 3 X 1) – 0 = 3

Key column = xcolumn
Minimum (7/1, 5/5) = 1
Key row = x4 row
Pivot element = 5
x4 departs and xenters.

Calculating values for table 3

x3 row

a11 = 0 – 0 X (1/5) = 0
a12 = 1 – 5 X (1/5) = 0
a13 = 1 – 0 X (1/5) = 1
a14 = 0 – 1 X (1/5) = -1/5
a15 = 1 – (-3) X (1/5) = 8/5
b1 = 7 – 5 X (1/5) = 6

x2 row

a21 = 0/5 = 0
a22 = 5/5 = 1
a23 = 0/5 = 0
a24 = 1/5
a25 = -3/5
b2 = 5/5 = 1

x1 row

a31 = 1 – 0 X (-1/5) = 1
a32 = -1 – 5 X (-1/5) = 0
a33 = 0 – 0 X (-1/5) = 0
a34 = 0 – 1 X (-1/5) = 1/5
a35 = 1 – (-3) X (-1/5) = 2/5
b3 = 3 – 5 X (-1/5) = 4

asd.png

 

DUALITY AND SENSITIVITY

Forming Dual when Primal is in Canonical Form

Constraint Relationships of the Primal and Dual

Duality in Linear Programme

Duality in Linear Programme

Duality in Linear Programme

clip_image008

 

 

TRANSPORTATION MODEL

NWC/R:

North-West Corner-1a

North-West Corner-1b

 

 

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