# 1 x 3 4 plant 1 2 x 2 x 4 12 plant 2 3 x 1 2 x 2 r 1

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1 + x 3 = 4 (Plant 1) 2 x 2 + x 4 = 12 (Plant 2) 3 x 1 + 2 x 2 + R 1 = 18 (Plant 3) x 1 , x 2, x 3 , x 4 , R 1 0 (Non-Neg) (C) Copyright 2012 Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 -3 -5 0 0 M 0 x 3 0 1 0 1 0 0 4 x 4 0 0 2 0 1 0 12 R 1 0 3 2 0 0 1 18 R 1 column does not contain all zeros. Must subtract M times the R 1 row from the Z-row to achieve this
IE310 89 Zeroing out the R 1 column (C) Copyright 2012 Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 -3 -5 0 0 M 0 x 3 0 1 0 1 0 0 4 x 4 0 0 2 0 1 0 12 R 1 0 3 2 0 0 1 18 Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 -3-3M -5-2M 0 0 0 -18M x 3 0 1 0 1 0 0 4 x 4 0 0 2 0 1 0 12 R 1 0 3 2 0 0 1 18 All basic variables columns have a one in a single row and zeros in all others We can begin the simplex algorithm!
IE310 90 Big-M Method: Comparing two terms with M Difficult to choose a single value of M; how would we know it is big enough? Instead, we leave the LP with M as a parameter Since M will only appear in the Z-row, we may need to compare two different Z-row coefficients with M in them to select the entering variable When this happens, the one with more negative M is more negative, and if all have the same amount of negative M, choose the one with a more negative constant term (C) Copyright 2012
IE310 91 Big-M Method, Iteration 1 (C) Copyright 2012 Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 0 -5-2M 3+3M 0 0 12-6M x 1 0 1 0 1 0 0 4 x 4 0 0 2 0 1 0 12 R 1 0 0 2 -3 0 1 6 Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 -3-3M -5-2M 0 0 0 -18M x 3 0 1 0 1 0 0 4 x 4 0 0 2 0 1 0 12 R 1 0 3 2 0 0 1 18 Entering basic var is x 1 4 / 1 = 4 (min ratio) Ignore (zero in x 1 col) 18 / 3 = 6 x 2 has a negative coefficient in the Z-row Continue!
(4,0) Z=12-6M IE310 92 x 1 x 2 (0,9) (6,0) (C) Copyright 2012 (0,6) Z=30-6M (0,0) Z= -18M (4,3) Z=27 (2,6) Z=36 Current BF solution (infeasible if we discard the artificial variable)
IE310 93 Big-M Method, Iteration 2 (C) Copyright 2012 Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 0 0 -9 / 2 0 5/2 + M 27 x 1 0 1 0 1 0 0 4 x 4 0 0 0 3 1 -1 6 x 2 0 0 1 -3 / 2 0 1 / 2 3 Entering basic var is x 2 Ignore (zero in x 1 col) 12 / 2 = 6 6 / 2 = 3 (min ratio) x 3 has a negative coefficient in the Z-row Continue! Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 0 -5-2M 3+3M 0 0 12-6M x 1 0 1 0 1 0 0 4 x 4 0 0 2 0 1 0 12 R 1 0 0 2 -3 0 1 6
(4,0) Z=12-6M IE310 94 x 1 x 2 (0,9) (6,0) (C) Copyright 2012 (0,6) Z=30-6M (0,0) Z= -18M (4,3) Z=27 (2,6) Z=36 Current BF solution (feasible in the original problem)
Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 0 0 -9 / 2 0 5/2 + M 27 x 1 0 1 0 1 0 0 4 x 4 0 0 0 3 1 -1 6 x 2 0 0 1 -3 / 2 0 1 / 2 3 IE310 95 Big-M Method, Iteration 3 (C) Copyright 2012 Basic Var Z x 1 x 2 x 3 x 4 R 1 RHS Z 1 0 0 0 3 / 2 1 + M 36 x 1 0 1 0 0 -1 / 3 1 / 3 2 x 3 0 0 0 1 1 / 3 -1 / 3 2 x 2 0 0 1 0 1 / 2 0 6 Entering basic var is x 3 4 / 1 = 4 6 / 3 = 2 (min ratio) Ignore (negative x 3 col) Done!
(4,0) Z=12-6M IE310 96 x 1 x 2 (0,9) (6,0) (C) Copyright 2012 (0,6) Z=30-6M (0,0) Z= -18M (4,3) Z=27 (2,6) Z=36 Current BF solution (feasible in the original problem)
IE310 97 Two-Phase Method Doing all computations in terms of M can be tedious Instead, break the optimization process into two phases 1. Minimize the sum of the artificial variables to find a BF solution 2. Revert back to the original objective to complete the optimization min Z = R 1 (Profit) subject to x 1 + x 3 = 4 (Plant 1) 2 x 2 + x 4 = 12 (Plant 2) 3 x 1 + 2 x 2 + R 1 = 18 (Plant 3) x 1 , x 2, x 3 , x 4 , R 1 0 (Non-Neg) Convert first phase objective to max Z = - R 1 and use simplex If simplex finds a BF solution, revert objective back to Z = 3 x 1 + 5 x 2 and continue the simplex algorithm (C) Copyright 2012
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