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Unformatted text preview: MthSc810 Mathematical Programming Fall 2011, Solutions to Homework #5 Problem 1. Solve the following problem (after an appropriate transformation) using the simplex method (i.e., do not use dictionaries but rather matrices B , N and vectors c B , c N , d etc.). min 2 x 1 x 2 s.t. 3 x 1 + x 2 9 x 1 + 2 x 2 8 x 1 , x 2 . Use B = { 1 , 4 } as the first basis. Solution. 1. First iteration: B = parenleftbigg 3 0 1 1 parenrightbigg , N = parenleftbigg 1 1 2 0 parenrightbigg , and B 1 = parenleftBigg 1 3 1 3 1 parenrightBigg . The ba sic solution is x B = parenleftbigg x 1 x 4 parenrightbigg = B 1 b = (3 , 5) and x N = . The vector of reduced costs is c N = c N c B B 1 N = ( 1 , 0) (2 , 0) parenleftBigg 1 3 1 3 1 parenrightBigg parenleftbigg 1 1 2 0 parenrightbigg = ( 1 , 0) ( 2 3 , 0) parenleftbigg 1 1 2 0 parenrightbigg = ( 5 3 , 2 3 ). Select x 2 for entering. The feasible di rection is d N = parenleftbigg d 2 d 3 parenrightbigg = parenleftbigg 1 parenrightbigg , while d B = B 1 A 2 = parenleftBigg 1 3 1 3 1 parenrightBigg parenleftbigg 1 2 parenrightbigg = parenleftBigg 1 3 5 3 parenrightBigg . Hence = 3 (we need x 1 = 3 3 1 3 0 and x 4 = 5 5 5 3 0), and x 4 is the exiting variable. 2. Second iteration: B = { 1 , 2 } , N = { 4 , 3 } . Therefore B = parenleftbigg 3 1 1 2 parenrightbigg , N = parenleftbigg 0 1 1 0 parenrightbigg , and B 1 = parenleftBigg 2 5 1 5 1 5 3 5 parenrightBigg . The basic solution is x B = parenleftbigg x 1 x 2 parenrightbigg = B 1 b = (2 , 3) and x N = . The vector of reduced costs is c N = c N c B B 1 N = (0 , 0) (2 , 1) parenleftBigg 2 5 1 5 1 5 3 5 parenrightBigg parenleftbigg 0 1 1 0 parenrightbigg = (1 , 1) parenleftbigg 0 1 1 0 parenrightbigg = (1 , 1). Select x 3 for entering (note that N = { 4 , 3 } from the way we write N ). The feasible direction is d N = parenleftbigg d 4 d 3 parenrightbigg = parenleftbigg 1 parenrightbigg , while d B = B 1 A 3 = parenleftBigg 2 5 1 5 1 5 3 5 parenrightBigg parenleftbigg 1 parenrightbigg = parenleftBigg 2 5 1 5 parenrightBigg . Hence = 5 (we need x 1 = 2 2 2 5 and x 4 = 3 3 + 1 5 0), and x 1 is the exiting variable. 3. Third iteration: B = { 3 , 2 } , N = { 4 , 1 } . Therefore B = parenleftbigg 1 1 0 2 parenrightbigg , N = parenleftbigg 0 3 1 1 parenrightbigg , and B 1 = parenleftBigg 1 1 2 1 2 parenrightBigg . The basic solution is x B = parenleftbigg x 3 x 2 parenrightbigg = B 1 b = (5 , 4) and x N = . The vector of reduced costs is c N = c N c B B 1 N = (0 , 2) (0 , 1) parenleftBigg 1 1 2 1 2 parenrightBigg parenleftbigg 0 3 1 1 parenrightbigg = (0 , 2) (0 , 1 2 ) parenleftbigg 0 3 1 1 parenrightbigg = ( 1 2 , 5 2 ). This solution is optimal....
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This note was uploaded on 03/14/2012 for the course MTHSC 810 taught by Professor Staff during the Fall '08 term at Clemson.
 Fall '08
 Staff
 Math, Vectors, Matrices

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