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HW3SolutionKey - DMOR OEM 2007 Homework 3 Solution Key...

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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 , 0 . 5 , - 0 . 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 , 0 , 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 , 0 , - 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). x 1 + x 2 - x 3 = 1 x 1 + 2 x 2 = 3 2 x 2 + x 3 = 3 - x 4 Adding the first and the last equalities yields ( x 1 + 3 x 2 = 4 - x 4 ). Subtracting the second equation from this, we get ( x 2 = 1 - x 4 ). Plugging
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