Ch04rp - 1 SOLUTIONS TO CHAPTER 4 MP REVIEW PROBLEMS 1. z...

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SOLUTIONS TO CHAPTER 4 MP REVIEW PROBLEMS 1. z x 1 x 2 x 3 s 1 s 2 RHS ___________________________________ 1 -5 -3 -1 0 0 0 ___________________________________ 0 1 1 3 1 0 6 ___________________________________ 0 5 3 6 0 1 15 ___________________________________ ___________________________________ 1 0 0 5 0 1 15 ___________________________________ 0 0 2/5 9/5 1 -1/5 3 ___________________________________ 0 1 3/5 6/5 0 1/5 3 ___________________________________ This is an optimal tableau with the optimal solution being z = 15, x 1 =3, x 2 =0. Pivoting x 2 into the basis yields the alternative optimal solution z = 15, x 1 =0, x 2 = 5. 2. z x 1 x 2 s 1 s 2 RHS _____________________________________ 1 4 -1 0 0 0 _____________________________________ 0 3 1 1 0 6 _____________________________________ 0 -1 2 0 1 0 _____________________________________ 1 0 -7/3 -4/3 0 -8 _____________________________________ 0 1 1/3 1/3 0 2 _____________________________________ 0 0 7/3 1/3 1 2 _____________________________________ This tableau yields the optimal solution z = -8, x 1 =2, x 2 = 0. 3. max z = 5x 1 - x 2 - Ma 1 = 0 s.t. 2x 1 + x 2 + a 1 = 6 x 1 + x 2 + s 2 = 4 x 1 +2x 2 + s 3 = 5 After eliminating a 1 from row 0, we obtain z -(2M + 5)x 1 + (1 - M)x 2 = -6M. 1
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z x 1 x 2 a 1 s 2 s 3 RHS ----------------------------------------------- 1 -2M-5 1-M 0 0 0 -6M ----------------------------------------------- 0 2 1 1 0 0 6 ----------------------------------------------- 0 1 1 0 1 0 4 ----------------------------------------------- 0 1 2 0 0 1 5 ----------------------------------------------- ----------------------------------------------- 1 0 7/2 (2M+5)/2 0 0 15 ----------------------------------------------- 0 1 1/2 1/2 0 0 3 ----------------------------------------------- 0 0 1/2 -1/2 1 0 1 ----------------------------------------------- 0 0 3/2 -1/2 0 1 2 ----------------------------------------------- This is an optimal tableau with the optimal solution being z = 15, x 1 = 3, x 2 = 0. 4. z x 1 x 2 s 1 s 2 RHS Basic Variable ------------------------------ 1 -5 1 0 0 0 z=0 ------------------------------ 0 1 -3 1 0 1 s 1 =1 ------------------------------ 0 1 -4 0 1 3 s 2 =3 ------------------------------ z x 1 x 2 s 1 s 2 RHS Basic Variable ------------------------------ 1 0 -14 5 0 5 z=5 ------------------------------ 0 1 -3 1 0 1 x 1 =1 ------------------------------ 0 0 -1 -1 1 2 s 2 =2 ------------------------------ Unbounded LP (look at x 2 column). 5. z x 1 x 2 s 1 s 2 RHS Basic Variable ________________________________ 1 1 2 0 0 0 z=0 ________________________________ 0 2 1 1 0 5 s 1 =5 ________________________________ 0 1 1 0 1 3 s 2 =3 ________________________________ 2
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z x 1 x 2 s 1 s 2 RHS Basic Variable ________________________________ 1 -1 0 0 -2 -6 z=-6 ________________________________ 0 1 0 1 -1 2 s 1 =2 ________________________________ 0 1 1 0 1 3 x 2 =3 ________________________________ Optimal solution is z = -6, x 2 = 3, x 1 = 0 6. max z= x 1 + x 2 - Ma 1 s.t. 2x 1 + x 2 - e 1 + a 1 = 3 3x 1 + x
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Ch04rp - 1 SOLUTIONS TO CHAPTER 4 MP REVIEW PROBLEMS 1. z...

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