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Tutorial 3

# Tutorial 3 - a 2 a RHS Z 1 1/3-4/3 1 8/3-20 1 x 3 2/3 1/3 1...

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IELM202 Exercises 1/6/2011 1. Solve the following LP by simplex method (use tableaux): Maximize Z= 3 2 1 2 4 X X X + + Subject to 5 4 3 2 1 + + X X X 10 2 3 2 1 + + - X X X 0 , 0 3 2 X X Since 1 X is unrestricted, we have to replace 1 X throughout the model by the difference of two new nonnegative variables. = 1 X + 1 X - - 1 X , 0 1 + X 0 1 - X Row Basic z + 1 X - 1 X x 2 x 3 x 4 x 5 RHS 0 z 1 -1 1 -4 -2 0 0 0 1 x 4 0 4 -4 1 1 1 0 5 2 x 5 0 -1 1 1 2 0 1 10 Row Basic z + 1 X - 1 X x 2 x 3 x 4 x 5 RHS 0 z 1 15 -15 0 2 4 0 20 1 x 2 0 4 -4 1 1 1 0 5 2 x 5 0 -5 5 0 1 -1 1 5 Row Basic z + 1 X - 1 X x 2 x 3 x 4 x 5 RHS 0 z 1 0 0 0 5 1 3 35 1 x 2 0 0 0 1 1.8 0.2 0.8 9 2 - 1 X 0 -1 1 0 0.2 -0.2 0.2 1 Basic variables ( x 2, - 1 X )= (9, 1) and non-basic variables ( + 1 X , x 3 , x 4 , x 5 )= (0, 0, 0, 0), where = 1 X -1. Optimal solution ( 1 X * , 2 X * , * 3 X )= (-1, 9, 0) Z 35 * =

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IELM202 Exercises 1/6/2011 2. Solve this problem by Two-Phase method Minimize W=3 3 2 1 4 2 X X X + + Subject to 0 , 0 , 0 120 5 3 3 60 3 2 3 2 1 3 2 1 3 2 1 + + = + + X X X X X X X X X Phase I Maximize Z= 2 1 a a - - Row z x 1 x 2 x 3 x 4 1 a 2 a RHS 0 z 1 0 0 0 0 1 1 0 1 1 a 0 2 1 3 0 1 0 60 2 2 a 0 3 3 5 –1 0 1 120 New Row 0 =Old Row 0-Row 1-Row 2 Row Z x 1 x 2 x 3 x 4 1 a 2 a RHS 0 Z 1 -5 -4 -8 1 0 0 -180 1 1 a 0 2 1 3 0 1 0 60 2 2 a 0 3 3 5 -1 0 1 120 Row Z x 1 x 2
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Unformatted text preview: a 2 a RHS Z 1 1/3-4/3 1 8/3-20 1 x 3 2/3 1/3 1 1/3 20 2 2 a-1/3 4/3-1-5/3 1 20 Row Z x 1 x 2 x 3 x 4 1 a 2 a RHS Z 1 1 1 1 x 3 3/4 1 1/4 3/4-1/4 15 2 x 2-1/4 1-3/4-5/4 3/4 15 IELM202 Exercises 1/6/2011 Phase II Max Z=-W= 3 2 1 4 2 3 X X X---Row z x 1 x 2 x 3 x 4 RHS z 1 3 2 4 1 x 3 3/4 1 1/4 15 2 x 2-1/4 1-3/4 15 New Row (0)=Old Row (0)-4*Row (1)-2*Row(2) Row z x 1 x 2 x 3 x 4 RHS z 1 1/2 1/2-90 1 x 3 3/4 1 1/4 15 2 x 2-1/4 1-3/4 15...
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Tutorial 3 - a 2 a RHS Z 1 1/3-4/3 1 8/3-20 1 x 3 2/3 1/3 1...

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