2016_09_29(1).pdf - Step 1 Determine an index k for which...

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Step 1: Determine an index k for which the reduced cost ¯ c k is negative . min c T x s.t. Ax = b x 0 Revised Simplex Method Let J be the current basic set, and assume A - 1 J has been computed. (If all reduced costs are non-negative, then optimality has been reached, and the algorithm terminates.) To compute reduced costs, first calculate the vector ¯ y T = c T J A - 1 J . vector of original costs c j for the basic variables Then begin computing reduced costs one at a time: ¯ c k = c k - ¯ y T A k original cost for x k original column for x k There is no point in computing reduced costs of basic variables. (Why? What will be the values of those reduced costs?) The entering variable can be any x k for which ¯ c k < 0, so we might as well stop computing reduced costs as soon as we find one that is negative.
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Let x k be the entering variable. To apply the minimum-ratio test we first need to compute the vectors ¯ b = A - 1 J b and ¯ A k = A - 1 J A k | {z } . original RHS vector original column for x k this needs to be computed only for the entering variable x k , not for any of the other variables min i ¯ b i ¯ a i,k : ¯ a i,k > 0 ¯ A k = 2 4 ¯ a 1 ,k ¯ a 2 ,k . . . ¯ a m,k 3 5 ¯ b = 2 6 4 ¯ b 1 ¯ b 2 . . . ¯ b m 3 7 5 notation for individual coe ffi cients of ¯ b and ¯ A k 9 = ; Else, compute the minimum-ratio: If ¯ a i,k 0 for all i , then the LP has unbounded optimal value and the algorithm now terminates. For whichever index i the minimum occurs, the i th basic variable is the blocking variable. This is the variable x j i where j i is the i th index in the (ordered) set J = { j 1 , j 2 , . . . , j m } (If the minimum occurs for multiple indices i , choose any one of them to provide the blocking variable x j i ) Step 2: Use the minimum-ratio test to find the blocking variable .
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Thus, according to lecture on September 8, computing the inverse for ˜ B = A ˜ J can be done e ffi ciently by relying on the inverse for B = A J (which already has been computed). Let x k be the entering variable, and let x j i denote the blocking variable. The new basic set – call it ˜ J – is the same as J except that k replaces (whatever was) the index j i in the i th position. Thus, A ˜ J di ers from A J only in the i th column. For A ˜ J , the i th column is A k (the original column for the incoming variable x k ). B - 1 ˜ B i | B - 1 Pivot on the i th entry in this column vector so as to make the column become the i th column of the identity matrix I , at which time the matrix on the right will be ˜ B - 1 (the desired inverse). | {z } A - 1 J A k | A - 1 J | {z } This is the vector ¯ A k you computed for the ratio test. Thus, it’s precisely the vector ¯ A k on which you want to pivot, to make it into the i th column of I . substitute for B and ˜ B Step 3: Update the basic set, and update the inverse .
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For each k , the computation requires approximately m arithmetic operations (if A k has lots of zeros, far less work is required ).
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