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Unformatted text preview: DMOR OEM 2007 Homework 3 Solution Key Solution for problem 1: (questions, comments, suggestions please to Zeki Caner Taskin, [email protected] ) a) Since there are 4 variables and 3 equality constraints, there is a single nonbasic variable and 3 basic variables. First let x 1 be the nonbasic variable. Setting x 1 = 0 essentially eliminates it from the system, and we are left with: x 2 x 3 = 1 2 x 2 = 3 2 x 2 + x 3 + x 4 = 3 This is a linear system of 3 variables and 3 equalities with a unique solution. Therefore our first basic solution is ( x 1 , x 2 , x 3 , x 4 ) = (0 , 1 . 5 , . 5 , . 5). Since x 4 < 0, this is not a basic feasible solution. Similarly we set x 2 = 0 and solve for the basic variables to get ( x 1 , x 2 , x 3 , x 4 ) = (3 , , 2 , 1). Since all variables are nonnegative, this corresponds to a basic feasible solution and hence an extreme point. Setting x 3 = 0 leads to ( x 1 , x 2 , x 3 , x 4 ) = ( 1 , 2 , , 1), which is infea sible, and finally setting x 4 = 0 leads to ( x 1 , x 2 , x 3 , x 4 ) = (1 , 1 , 1 , 0), which is another extreme point. b) The underlying idea is the same as part a) except the way we treat x 4 . x 4 is still a nonbasic variable, but here we treat it as a constant and move to the right hand side of the equations (instead of setting it equal to 0 and removing from the system altogether)....
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This note was uploaded on 02/25/2010 for the course ESI 6314 taught by Professor Vladimirlboginski during the Fall '09 term at University of Florida.
 Fall '09
 VLADIMIRLBOGINSKI

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