mthsc810-lecture09-1x2

mthsc810-lecture09-1x2 - MthSc 810: Mathematical...

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MthSc 810: Mathematical Programming Lecture 9 Pietro Belotti Dept. of Mathematical Sciences Clemson University September 22, 2011 Reading for today: Sections 3.4-3.5, textbook Seminar at 4:45pm @M-103 by Banu Soylu Midterm next Tuesday. Open or Closed books/notes? Homework #5 is out – due Tuesday, October 4, 2011. Reading for Sep. 29: Section 3.3, textbook Recap: Dictionaries and LPs Consider the LP problem: min 5 x 1 4 x 2 3 x 3 s . t . 2 x 1 + 3 x 2 + x 3 5 4 x 1 + x 2 + 2 x 3 11 3 x 1 + 4 x 2 + 2 x 3 8 x 1 , x 2 , x 3 0 , and rewrite it in standard form: min 5 x 1 4 x 2 3 x 3 s . t . 2 x 1 + 3 x 2 + x 3 + x 4 = 5 4 x 1 + x 2 + 2 x 3 + x 5 = 11 3 x 1 + 4 x 2 + 2 x 3 + x 6 = 8 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 0 .
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Recap: Dictionaries and LPs A dictionary is simply another way to express the LP: min 5 x 1 4 x 2 3 x 3 s . t . x 4 = 5 2 x 1 3 x 2 x 3 x 5 = 11 4 x 1 x 2 2 x 3 x 6 = 8 3 x 1 4 x 2 2 x 3 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 0 in many 1 equivalent ways—this is the second dictionary: min 25 2 + 7 2 x 2 1 2 x 3 + 5 2 x 4 s . t . x 1 = 5 2 3 2 x 2 1 2 x 3 1 2 x 4 x 5 = 1 + 5 x 2 + 2 x 4 x 6 = 1 2 + 1 2 x 2 1 2 x 3 + 3 2 x 4 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 0 . 1 To be precise, ( m + n m ) = ( 3 + 3 3 ) = 20 ways. Possible troubles with the simplex methods The example we’ve seen last time is ideal in that none of the following problems occurs: Unboundedness : There is no exiting variable; Initialization : For the initial basis, B 1 b 6≥ 0 ; Degeneracy : Two or more (potential) exiting variables. The latter generates a lot of other problems.
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Unbounded problems What if we had x 4 = 5 2 x 1 3 x 2 + x 3 x 5 = 11 4 x 1 x 2 + 2 x 3 x 6 = 8 3 x 1 4 x 2 + 2 x 3 z = 5 x 1 4 x 2 3 x 3 For any x 3 = ϑ > 0, x + ϑ d = ( 0 , 1 , ϑ, 5 + ϑ, 11 + 2 ϑ, 8 + 2 ϑ ) 0 because of the positive coef±cients 1 , 2 , 2 of x 3 ’s column The objective function tends to + for ϑ → ∞ . The problem is unbounded, i.e., it admits no optimal solution of ±nite value. Initialization For the problem min 5 x 1 4 x 2 3 x 3 s . t . 2 x 1 + 3 x 2 + x 3 5 4 x 1 + x 2 2 x 3 11 x 1 , x 2 , x 3 0 , or, equivalently, min 5 x 1 4 x 2 3 x 3 s . t . 2 x 1 + 3 x 2 + x 3 + x 4 = 5 4 x 1 + x 2 2 x 3 + x 5 = 11 x 1 , x 2 , x 3 , x 4 , x 5 0 , the initial dictionary x 4 = 5 2 x 1 3 x 2 x 3 x 5 = 11 4 x 1 x 2 + 2 x 3 z = 0 5 x 1 4 x 2 3 x 3 ± ± ± corresponds to no feasible solution.
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Initialization: the two phase method Multiply by 1 all constraints a
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mthsc810-lecture09-1x2 - MthSc 810: Mathematical...

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