Page 54 - DCOM303_DMGT504_OPERATION_RESEARCH

Unit 3: Linear Programming Problem – Simplex Method

(c) The same multiple should be used to other elements in the row to adjust the rest of Notes
the elements. But, the adjusted key row elements should be used for deducting out
of the earlier iteration row.
(d) The same iteration is continued until the values of Z – C become either '0' or
j j
positive.
5. Find the 'Z' value given by C , X .
B B
3.1.1 Maximisation Cases

Example:

Maximise ‘Z’ = 5x + 3x [Subject to constraints]
1 2
x + x  2
1 2
5x + 2x  10
1 2
3x + 8x  12
1 2
Where, x , x  0 [Non-negativity constraints]
1 2
Solution:

Step 1: Conversion of inequalities into equalities adding slack variables
x + x + x = 2
1 2 3
5x + 2x + x = 10
1 2 4
3x + 8x + x = 12
1 2 5
Where, x , x and x are slack variables.
3 4 5
Step 2: Fit the data into the matrix form AX = B

 Y 1 Y 2 S 1 S 2 S 3   x 1 
 x x x x x   x 
 1 2 3 4 5   2   2 

A   1 1 1 0 0  X   x 3   B  10 
       
 5 2 0 1 0   x 4   12 
   
 3 8 0 0 1   x 5 
Step 3: Fit the data into first iteration of Simplex Method
BV CB XB Y1 Y2 S1 S2 S3 Min. Ratio
S1 0 2 1 1 1 0 0 2/1 = 2(KR) 
S2 0 10 5 2 0 1 0 10/5 = 2
S3 0 12 3 8 0 0 1 12/3 = 4
Zj 0 0
Cj 5 3
Zj – Cj -5 -3
(KC)
Therefore, Z = C X
B B
= (0×2) + (0×10) + (0×12)
= 0

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