least_index - OR 3310/5310/5311 Spring 2012 Supplemental...

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OR 3310/5310/5311 Spring 2012 Supplemental Handout II.15 Theorem: With least index rules, the simplex algorithm will not cycle. Proof: Assume least index rules are used, but a cycle occurs; we show this is impossible by deriving a contradiction. During the cycle we pass through a sequence of degenerate pivots, eventually repeating a basis; in these iterations certain variables enter and leave the basis at level zero. Among these variables, let x be the one with the largest index. Denote by A,b,c the data for the representation when x leaves the basis. Suppose x is basic in row r of this representation (hence x = b r = 0) and that x k enters the basis to replace x on this iteration. We then have: c k > 0 ( x k enters) (1) a rk > 0 (pivot on a rk ) . (2) Denote by ¯ A, ¯ b, ¯ c the data for the representation when x enters the basis. At this iteration we have: ¯ c > 0 ( x enters) . (3) And since x k also enters/leaves during the cycle, we have k < ‘ , so the least index choice of
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