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# L15 - CO350 Linear Programming Chapter 6 The Simplex Method...

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CO350 Linear Programming Chapter 6: The Simplex Method 6th June 2005

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Chapter 6: The Simplex Method 1 Recap Last week, we learned the definition of a tableau and as- sociated terminology. Example: max. 5 x 1 + 3 x 2 s.t. 2 x 1 + 3 x 2 + x 3 = 15 2 x 1 + x 2 + x 4 = 9 x 1 - x 2 + x 5 = 3 x 1 , x 2 , x 3 , x 4 , x 5 0 Final tableau (see slides on course page): z + 1 4 x 3 + 9 4 x 4 = 24 3 4 x 3 - 5 4 x 4 + x 5 = 3 x 2 + 1 2 x 3 - 1 2 x 4 = 3 x 1 - 1 4 x 3 + 3 4 x 4 = 3
Chapter 6: The Simplex Method 2 Proof of optimality z + 1 4 x 3 + 9 4 x 4 = 24 3 4 x 3 - 5 4 x 4 + x 5 = 3 x 2 + 1 2 x 3 - 1 2 x 4 = 3 x 1 - 1 4 x 3 + 3 4 x 4 = 3 We can actually prove algebraically that the current bfs is optimal. From the z -row: z + 1 4 x 3 + 9 4 x 4 = 24 , we get z = 24 - 1 4 x 3 - 9 4 x 4 Since x 4 , x 5 0 for any feasible solution, we have z = 24 - 1 4 x 3 - 9 4 x 4 24 for any feasible solution. The solution x * = [3 , 3 , 0 , 0 , 3] T has objective value 24, so it must an optimal solution. We can also prove optimality using the C.S. Theorem. (More on this later.)

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Chapter 6: The Simplex Method 3 Working out the Details ( § 6.3) Each step of the simplex method involves 1.Picking a nonbasic variable x k to increase. 2.Determining how much to increase x k . 3.Forming the next tableau. 1. Choosing Entering Variable (Pg 75) By picking a nonbasic x k with ¯ c k > 0 and increase its value, we can hope to increase the objective value z . So we pick such nonbasic variable to join the next basis. We say that x k enters the basis . 2. Choosing Leaving Variable (Pg 75) When we increase x k from 0 to t while holding the other nonbasics at 0 , the basic variables changes as follows x i ( t ) = ¯ b i - ¯ a ik t, ( i
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