Lecture4

# Program the constraints in standard form become

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Unformatted text preview: , + .r, subject to 3x, +x, = 3 -Ix, + 3x, 2 6 I, +2x, 5 -I X, ..K. > 0 Minimize z = 4.x, + x, subject to 3x, + x1 = 3 (no changes made) 4.q + 3.5 - .q = 6 (surplus is subtracted) .Y, + 2x, + .x, = 4 (slack is added) .x, .SI ,.x1, x, 2 0 M-Technique M-Technique (Step 2 add Artiicialr) (Step 3 - Modify Objective Function) Augment all constraints lacking a Slack variable by adding an Artiftcial Variable Augment the Objective Function to include the two Artificial Variables Minimize z = -Ix, + .x, subpxt to 3.~, + rZ + R, = 3 (Artlticlal added) 4r, + 7~: - Y, + R1 = 6 (Artificial added) v, + 2.~. +,x1 =-I (no further changes) .x, .x.. .s, _.I., 2 0 Minmiize z = -I; + .v, 1s aug111enteLl to Mulmnze z = 1.s, +.x2 + R, + Rz \\lnch when penalized by + M becomes Minunize z = 1.x, +x1 + .\IR, + .\fR_ EM-602 I QM-710 (NJ) Lecture 4 Page 4-3 M-Technique M-Technique (Summarizing) (Step 4 - Solving for R’s) The Artificial variables are solved for in terms of the Veal” variables The complete problem becomes: Mitknize z=4x,+x,+MR,+MR, 3.5 + xl + R, = 3 a R, = 3 - 3x, - .r: 4.u,+3.~2-.~,+R,=6=>R=6-4x,-3x,+x, M-Technique M-Technique (Step (Step 5 -Conditioning Objective Function) The Basic Feasible Solutiof~ for the problem with 6 variables and 3 equations can now be wrttten: The Artificials are removed from the Objective Function by substitution and the Objective Function is put into Standard Form: z = 4x, +x2 + .\fR, + l/R2 Z = -bX, + XI + .\ f( 3 - 3X, - .KI 6 - Sating Solution) R, = 3 R,=6 x4 = 4 )+ .\/(6-4.5 -3-r: +.r,) 2=(4-7J/).r, +(I--l.\f).r, +.\l.r,+9.\1 z-(-I-7.\f).r, -(l--l.\f).rz - .\f.r, = 9.\f M-Techruque lterabon 0 Basic -- x. x, x, M-Technique lterabon 1 R. R. x, Soln Basic x. x, x, EM-602 I QM-710 (NJ) Lecture 4‘ Page 4-4 R. R, x, Soln M-Technque ItcrattOn 3 H-Technqx lteratm 2 Basic x. x1 x, R, R, X, Soln Basic x. x_ x, R, R, x, Soln M-Technique l l l M-Technique Drawbacks Observations Numerical instability Artiftcial Variables should be ftrst to leave basis since their elimination is essential Once an Artificial Variable has left the basis, it can never return - Eliminate it! If an optimal tableau contains an Artificial variable in the basic column, two conclusions are possible If the Artific...
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## This document was uploaded on 03/31/2014 for the course MS 602 at NJIT.

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