CA427Simplexbigmexample[1] - Z row – M*R 1 row: X 1 X 2 S...

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Simplex – Big M Method Maximise 3X 1 + 4X 2 Subject to 2X 1 + X 2 <= 600 X 1 + X 2 <= 225 5 X 1 + 4X 2 <= 1000 X 1 + 2X 2 >= 150 X 1 , X 2 >= 0 Solution: Standard form: Maximise 3X 1 + 4X 2 Subject to 2X 1 + 3X 2 + S 1 = 600 X 1 + X 2 + S 2 = 225 5 X 1 + 4X 2 + S 3 = 1000 X 1 + 2X 2 - S 4 = 150 X 1 , X 2 , S 1 , S 2 , S 3 , S 4 >= 0 Not in canonical form because there is no basic variable in the fourth equation. Therefore we add an artificial variable to that equation (R 1 ) and give it a large negative coefficient in the objective function, to penalise it: Maximise 3X 1 + 4X 2 Subject to 2X 1 + X 2 + S 1 = 600 X 1 + X 2 + S 2 = 225 5 X 1 + 4X 2 +S 3 = 1000 X 1 + 2X 2 - S 4 + R 1 = 150 X 1 , X 2 , S 1 , S 2 , S 3 , S 4 , R 1 >= 0
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X 1 X 2 S 4 S 1 S 2 S 3 R 1 B Z -3 -4 0 0 0 0 +M S 1 2 3 0 1 0 0 0 600 S 2 1 1 0 0 1 0 0 225 S 3 5 4 0 0 0 1 0 1000 R 1 1 2 -1 0 0 0 1 150 Not in Canonical form because of +M entry on Z row for one basic variable (R 1 ). Pivot to replace +M on Z row by zero -
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Unformatted text preview: Z row – M*R 1 row: X 1 X 2 S 4 S 1 S 2 S 3 R 1 b Z (-3-M) (-4-2M) M 0 0 0 0 -150M S 1 2 3 0 1 0 0 0 600 S 2 1 1 0 0 1 0 0 225 S 3 5 4 0 0 0 1 0 1000 R 1 1 2 -1 0 0 0 1 150 X 1 X 2 S 4 S 1 S 2 S 3 R 1 b Z -1 0 -2 0 0 0 M 800 S 1 ½ 0 3/2 1 0 0 -3/2 375 S 2 ½ 0 ½ 0 1 0 -½ 150 S 3 3 0 2 0 0 1 -2 700 X 2 ½ 1 -½ 0 0 0 ½ 75 X 1 X 2 S 4 S 1 S 2 S 3 R 1 b Z -1/3 0 0 4/3 0 0 M 800 S 4 1/3 0 1 2/3 0 0 -1 250 S 2 1/3 0 0 -1/3 1 0 0 25 S 3 7/3 0 0 -4/3 0 1 0 200 X 2 2/3 1 0 1/3 0 0 0 200 X 1 X 2 S 4 S 1 S 2 S 3 R 1 b Z 0 0 0 1 1 0 M 825 S 4 0 0 1 1 -1 0 -1 225 X 1 1 0 0 -1 3 0 0 75 S 3 0 0 0 1 -7 1 0 25 X 2 0 1 0 1 -2 0 0 250 Optimal tableau: Solution: X 1 * = 75 X 2 * = 150 Z* = 825...
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CA427Simplexbigmexample[1] - Z row – M*R 1 row: X 1 X 2 S...

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