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Lecture4

# Lecture4 - MS&E211 ondoingtheproject WorkonyourHW1 MS&E211...

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MS&E  211

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4 units no longer requires a paper Project is only required for project credit Reply to the announcement if you intend  on doing the project Discussion Section starts tomorrow Work on your HW1 MS&E 211
Basic Solutions in Standard Form Are basic solutions feasible? Fundamental Theorem of LP’s (proof sketch) Implications Simplex Method The History The Tableau Pivoting Optimality Check Initialization MS&E 211

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MS&E 211
MS&E 211

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MS&E 211
MS&E 211

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MS&E 211
MS&E 211 X B X Non-trivial  Intersection X* If the Linear Program is feasible and bounded, then  there exists an optimal basic solution.

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MS&E 211 X B X Non-trivial  Intersection
MS&E 211 Take x  ϵ X Rearrange A & x: If B has less than m columns, you are done. Otherwise, the p > m columns of B must not be linearly independent. Therefore, there exists a non-trivial w ϵ R p : Bw = 0 Add zeros to the end of w to make it n-dimensional. Take x new = x – ε w, for the value ε such that the number of zero components of x increases by one, and no components of x are made negative.

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MS&E 211
MS&E 211 X B X X* Non-trivial  Intersection

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MS&E 211 Take x*  ϵ X* Count number of zero components If there are not enough of them, then find w as before. Since both (x* – ε w) AND (x* + ε w) ϵ X for small ε , and have objective values c’ x* ± ε c’w, respectively, then c’w must equal zero. Else x* is not optimal. Thus take x* new = x* – ε w, as before. It is also optimal. It has one more zero component. It is also still feasible.
MS&E 211 c w

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MS&E 211 It is sufficient to search X B  for an x B* X B X Non-trivial  Intersection X*
Find an x ϵ X B Use it to find an x ϵ X B X

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