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Unformatted text preview: ORIE 3310/5310 Optimization II Spring 2012 Homework # 1 Solutions posted Monday 2/13 at 4:00p.m. Homework quiz in recitation Tuesday, 2/14 Friday, 2/17. 1. Three representations from a (max) LP solved in the LP notes for ORIE 3310: ( A ) ( z ) + 5 3 x 2 4 3 x 3 = 12 1 3 x 2 + 1 3 x 3 + x 1 = 3 x 2 x 3 + x 4 = 1 2 3 x 2 1 3 x 3 + x 5 = 1 x 2 , x 3 , x 1 , x 4 , x 5 ( B ) ( z ) + 4 x 1 + 3 x 2 = 3 x 1 + x 2 + x 3 = 9 3 x 1 + 2 x 2 + x 4 = 10 x 1 + x 2 + x 5 = 4 x 1 , x 2 , x 3 , x 4 , x 5 ( C ) ( z ) + 1 3 x 3 5 3 x 4 = 41 3 2 3 x 3 1 3 x 4 + x 1 = 8 3 x 3 + x 4 + x 2 = 1 1 3 x 3 2 3 x 4 + x 5 = 1 3 x 3 , x 4 , x 1 , x 2 , x 5 (a) Which representation has the highest objective value? Is it optimal? (b) Which variables are basic in representation (C)? What are the values taken by these (basic) variables in the corresponding basic solution? (c) Which variables are nonbasic in representation (A)? What are the values taken by these (nonbasic) variables in the corresponding basic solution? (d) Do variables x 2 ,x 4 ,x 5 define a basis for this problem? Why/why not? Is this basis feasible? Why/why not? (e) Determine by inspection from the above data inverses of the following matrices: 3 1 0 3 2 0 1 1 1 1 1 3 1 0 2 3 1 Explain how you arrive at your answers. (f) If the problems objective function is changed to max 5 x 1 + 3 x 2 , which representation is optimal? Why? Are there alternative optimal solutions for the new objective? Explain. (g) If the righthand side coefficients in representation (B) are changed from (9 , 10 , 4) to (3 , 2 , 5) , does representation (A) remain feasible? Explain. 1 2. We would like to prove the following result, known as the Farkas Theorem . For any m n real matrix A and mvector b exactly one of the two following systems is feasible: (I) Ax = b , x (II) yA 0 , yb < 0 . (a) Show that (I) and (II) cannot both be feasible. (b) If system (I) is infeasible then at the end of Phase I (maximizing z = m i =1 w i , the negative sum of artificials) of the simplex algorithm we have a tableau of the following form: x 1 x n w 1 w m RHS ( z ) > . . . Show how to determine a vector y which is feasible in system (II) directly from the data in the above tableau. (Thus when (I) fails, (II) must be feasible, completing the proof of the theorem.) 3. The LP feasible region F = { x : Ax = b,x } 6 = is bounded provided F { x : x j j } for some positive constant ; i.e., we can place the LP feasible region in a hypercube of side length 2 ....
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 Spring '08
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