tanotes6 - 3 = 1 x 1 + x 3 = 0 x 2 + x 3 = 1 After one...

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CO 350 Linear Optimization TA’s comments for Assignment 6 General: Many students are still confused by the multiple uses of the word “degenerate”. Recall that “degenerate basis” and “degenerate iterations” are two completely different things. A basis B is degenerate if the associated basic solution x B = A - 1 B b, x N = 0 has some basic variable set to zero, i.e., if x i = 0 for some i B . This has absolutely nothing to do with the simplex method. When we say “degenerate iteration”, we mean an iteration of the simplex method for which t = 0. Recall that, in the simplex method, we choose an entering variable and perform the min ratio test to determine the leaving variable. The result of the min ratio test is t , and the row (or rows) that achieves the minimum determines the leaving variable. Q1 [(a) 7 marks; (b) 3 marks]: For part (b), many students argued that, since there are no ties for the choice of leaving variable, the LP is nondegenerate. This is false, as the following tableau shows: z - x
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Unformatted text preview: 3 = 1 x 1 + x 3 = 0 x 2 + x 3 = 1 After one pivot, this tableau becomes optimal, so there are no ties for the choice of leaving variable during the running of simplex, and yet the basis B = { 1 , 2 } is degenerate. Q2 [(a) 5 marks; (b) 5 marks]: This question was very poorly done. For instance, many students didnt realize that they could use the result from part (a) to prove part (b). Please see the solution. Q3 [10 marks]: Many students didnt realize that they could not apply Strong Duality Theorem in SEF to the dual problem min { b T y : A T y c } . Note that, while the primal LP is in SEF, this LP (the dual) is not in SEF. Thus, to use the Strong Duality Theorem for SEF for this LP, you rst had to transform it into SEF, and form its dual, then translate the conclusion back into the notation of the original problem max { c T x : Ax = b, x } . Q4 [(a) 5 marks; (b) 5 marks]: Generally okay, except for arithmetic mistakes. 1...
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