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Unformatted text preview: Setting x 1 =0 we get e 1 =2 from (1), x 2 =3 from (3) and s 2 = 7 from (2). Basic solution 1: x 1 =0, x 2 = 3, e 1 = 2, s 2 = 7 Solution 2: Select x 2 to be the nonbasic variable. Setting x 2 =0, we get : x 1 =3/2 from (3), e 1 =1/2 from (1), and s 2 = 5/2 from (2) Basic solution 2: x 1 =3/2, x 2 =0, e 1 = 1/2, s 2 = 5/2 Solution 3: Select e 1 to be the nonbasic variable. Setting e 1 =0, we get : x 1 =2 from (1) , x 2 =1 from (3), and s 2 = 1 from (2) Basic solution 3: x 1 =2, x 2 =1, e 1 =0, s 2 = 1 Solution 4: Select s 2 to be the nonbasic variable. Setting s 2 =0, we get : x 1 =7/3 and x 2 =5/3 from (2) and (3), and e 1 =1/3 from (1). Basic solution 4: x 1 =7/3, x 2 =5/3, e 1 =1/3, s 2 =0 c. Solutions 3 and 4 are feasible. d. For solution 3, z=3(2)+ 1 =7 For solution 4, z=3(7/3) + 5/3 = 8 2/3 Thus solution 3 is optimal since the objective is to minimize z. e. See graph on Excel chart....
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This note was uploaded on 04/17/2008 for the course CSA 273 taught by Professor Patton during the Spring '08 term at Miami University.
 Spring '08
 Patton

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