Page 63 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes Step 4: First iteration of Simplex Method.
BV CB XB Y1 Y2 Y3 S1 S2 S3 Min. Ratio
S1 0 7 3 –1 3 1 0 0 7/–1 = –7
S2 0 12 –2 4 0 0 0 12/4 = 3 (KR)
S3 0 10 –4 3 8 0 0 1 10/3 = 3.33
Zj 0 0 0
Cj –1 3 –2
Zj – Cj 1 –3 2
( KC)
Therefore, Z = C X
B B
= 0 + 0 + 0
= 0
Step 5: Second iteration of Simplex Method.
BV CB XB Y1 Y2 Y3 S1 S2 S3 Min. Ratio
S1 0 7–3 (–1) = 3 + 0.5 (–1) = –1+1 (–1) 3–0 (–1) – – – 10/2.5 = 4
10 2.5 = 0 = 3 (KR)
Y2 3 12/4 = 3 –2/4 = –0.5 4/4 = 1 0 – – – 3/–0.5 = -
S3 0 10 – 3 (3) = –4 + (–0.5) (3) 3–1 (3) = 0 8–0 (3) = – – – 1/2.5 = -
1 = –5.5 8
Zj –1.5 3 0
Cj –1 –3 –2
Zj – Cj –0.5 –0 2
( KC)
Therefore, Z = C X
B B
= (0 × 0) + (3 × 3) + (0 × 1)
= 0 + 9 + 0

= 9
Step 6: Third iteration of Simplex Method.

BV CB XB Y1 Y2 Y3 S1 S2 S3 Min.
Ratio
y1 –1 10/2.5 = 4 2.5/2.5 = 1 0 3/2.5 = 1.2 – – – –
S2 3 3–4 (–0.5)= 5 –0.5–1 (–0.5)= 0 1 0 – – – –
S3 0 1–4 (–2.5)= 11 –2.5–1 (–2.5)= 0 0 8 – 1 (–2.5) = 10.5 – – – –
Zj –1 3 –1.2
Cj –1 3 –2
Zj–Cj 0 0 0.8

Therefore, Maximise ‘Z’ = C X
B B
= – 4 + 15 + 0
= 11
Therefore, Minimise Z = – 11

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