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mthsc810-lecture08-1x2

# mthsc810-lecture08-1x2 - MthSc 810 Mathematical Programming...

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MthSc 810: Mathematical Programming Lecture 8 Pietro Belotti Dept. of Mathematical Sciences Clemson University September 20, 2011 Reading for today: Chapters 3.3 and 3.4, textbook Reading for Sep. 22: Chapters 3.5, textbook Midterm exam : Next Tuesday, 2:00pm-3:15pm, in class 1: procedure S IMPLEX ( A , b , c ) 2: Find an initial basis B , N 3: loop 4: Reduced costs: ¯ c c N - c B B 1 N 5: if ¯ c 0, STOP : optimum found 6: Select j ∈ N : ¯ c j < 0 j is the entering variable 7: d B ← - B 1 A j 8: Ifd B 0, STOP : problem unbounded 9: argmin i ∈N : d i < 0 - x i d i ⊲ ℓ = exiting var. 10: B ← B ∪ { j } \ { } ; N ← N \ { j } ∪ { } 11: endloop 12: endprocedure

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Dictionaries and tableaus Associate the objective function with a new variable, z := n i = 1 c i x i . Once a basis B is chosen, a dictionary describes the basic variables as affine functions of the non-basic variables: x B = B 1 b - B 1 N x N z = c B B 1 b + ( c N - c B B 1 N ) x N Tableaus and dictionaries are equivalent. Example Given the problem in standard form max { c x : A x b , x 0 } , max 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 , 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 . and chose B = { 4 , 5 , 6 } for convenience.
The corresponding dictionary 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 = 0 - 5 x 1 - 4 x 2 - 3 x 3 As long as the three equations defining x 4 , x 5 , x 6 are satisfied and all variables are nonnegative, we have a feasible solution.

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