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homework-5-solutions

homework-5-solutions - MthSc810 – Mathematical...

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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 ϑ ⋆ = 3 (we need x 1 = 3 → 3- 1 3 ϑ ≥ 0 and x 4 = 5 → 5- 5 3 ϑ ≥ 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 ϑ ⋆ = 5 (we need x 1 = 2 → 2- 2 5 ϑ ≥ and x 4 = 3 → 3 + 1 5 ϑ ≥ 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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homework-5-solutions - MthSc810 – Mathematical...

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