Page 69 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes Step 3: Bring the objective function into a standard form.
Therefore, Maximise ‘Z’ = –4x – 8x – 3x + 0x + 0x – Mx – Mx
1 2 3 4 5 6 7
Step 4: Find out the matrix form of equalities.

 x 1 
 x 
 Y 1 Y 2 Y 3 S 1 S 2 a 1 a 2   2 
 x x x x x x x   x 3  2
A   1 2 3 4 5 6 7  X   x 4  B   
 1 1 0  1 0 1 0     5
   x 
5
 2 0 3 0  1 0 1   
 x 6 
 
 x 7 
Step 5: First iteration of Simplex Method.
BV CB XB Y1 Y2 Y3 S1 S2 a1 a2 Min. Ratio
a1 –M 2 1 1 0 –1 0 1 0 2/1 = 2(KR)
a2 –M 5 2 0 1 0 –1 0 1 5/2 = 2.5
Zj –3M –M –M
Cj –4 8 –3
Zj – Cj –3M+4 –M+8 –M+3
( KC)
Therefore, Z = C X
B B
= (–M × 2) + (–M × 5)

= –2M – 5M
= –7M
Step 6: Second iteration of Simplex Method.

BV CB XB Y1 Y2 Y3 S1 S2 a1 a2 Min. Ratio
y1 –4 2/1 = 2 1/1 = 1 1 0 –1 0 1 0 2/0 = –
a2 –M 5 – 2(2) = 1 2 –1 (2) = –2 0–1 (2) = –2 1–0 (2) = 1 – – – – 1/1 = 1 (KR)
Zj –4 –4 + 2M –M
Cj –4 –8 –3
Zj – Cj 0 4 + 2M M+3
( KC)
Therefore, Z = (–4 × 2) + (–M × 1)
= –8 – M

Step 7: Second iteration of Simplex Method.

BV CB XB y1 y2 y3 S1 S2 a1 a2 Min.
Ratio
y1 –4 2–1 (0) = 2 1–1 (0) = 1 1 – (–2) (0) = 1 0 – – – – –
y3 –3 1/1 = 1 0 –2/1 = –2 1/1 = 1 – – – – –
Zj –4 2 –3
Cj –4 –8 –3
Zj – Cj 0 10 0

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