Simplex Algorithm: Solving Degenerate Linear Programming Problems - Prof. David Phillips, Study notes of Linux skills

Download Simplex Algorithm: Solving Degenerate Linear Programming Problems - Prof. David Phillips and more Study notes Linux skills in PDF only on Docsity! An example of the simplex algorithm Original LP maximize 3x1 + x2 + 2x3 (1) subject to x1 + x2 + 3x3 ≤ 30 (2) 2x1 + 2x2 + 5x3 ≤ 24 (3) 4x1 + x2 + 2x3 ≤ 36 (4) x1, x2, x3 ≥ 0 . (5) Standard form. z = 3x1 + x2 + 2x3 (6) x4 = 30 − x1 − x2 − 3x3 (7) x5 = 24 − 2x1 − 2x2 − 5x3 (8) x6 = 36 − 4x1 − x2 − 2x3 . (9) Pivot in x1. Remove x6 from the basis. z = 27 + x2 4 + x3 2 − 3x6 4 (10) x1 = 9 − x2 4 − x3 2 − x6 4 (11) x4 = 21 − 3x2 4 − 5x3 2 + x6 4 (12) x5 = 6 − 3x2 2 − 4x3 + x6 2 . (13) Pivor in x3. Remove x5. z = 111 4 + x2 16 − x5 8 − 11x6 16 (14) x1 = 33 4 − x2 16 + x5 8 − 5x6 16 (15) x3 = 3 2 − 3x2 8 − x5 4 + x6 8 (16) x4 = 69 4 + 3x2 16 + 5x5 8 − x6 16 . (17) Pivot in x2. Remove x3. z = 28 − x3 6 − x5 6 − 2x6 3 (18) x1 = 8 + x3 6 + x5 6 − x6 3 (19) x2 = 4 − 8x3 3 − 2x5 3 + x6 3 (20) x4 = 18 − x3 2 + x5 2 . (21) 1 Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com 2 Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com A Degenerate LP Definition: An LP is degenerate if in a basic feasible solution, one of the basic variables takes on a zero value. Degeneracy is a problem in practice, because it makes the simplex algorithm slower. Original LP maximize x1 + x2 + x3 (1) subject to x1 + x2 ≤ 8 (2) −x2 + x3 ≤ 0 (3) x1, x2, ≥ 0 . (4) Standard form. z = x1 + x2 + x3 (5) s1 = 8 − x1 − x2 (6) s2 = − x2 + x3 (7) Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com Iteration 1 z = x1 + x2 + x3 (8) s1 = 8 − x1 − x2 (9) s2 = − x2 + x3 (10) Note that one of the basic variables is 0. We choose x1 as the entering variable and s1 as the leaving variable. z = 8 + x3 − s1 (11) x1 = 8 − x2 − s1 (12) s2 = x2 − x3 (13) Note again that one of the basic variables is 0. The previous pivot did increase the objective function value from 0 to 8 though. Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com CSCI 628 – Introduction to Linear Programming Cycling max z = 3 4 x1 −150x2 + 1 50 x3 −6x4 s.t. 1 4 x1 −60x2 − 1 25 x3 +9x4 +x5 = 0 1 2 x1 −90x2 − 1 50 x3 +3x4 +x6 = 0 x3 +x7 = 1 x1, x2, x3, x4, x5, x6, x7 ≥ 0 (1) x5 = 0 − 1 4 x1 +60x2 + 1 25 x3 −9x4 x6 = 0 − 1 2 x1 +90x2 + 1 50 x3 −3x4 x7 = 1 −x3 z = 0 +3 4 x1 −150x2 + 1 50 x3 −6x4 (2) x1 = 0 +240x2 + 4 25 x3 −36x4 −4x5 x6 = 0 −30x2 − 3 50 x3 +15x4 +2x5 x7 = 1 −x3 z = 0 30x2 + 7 50 x3 −33x4 −3x5 (3) x1 = 0 − 8 25 x3 +84x4 +12x5 −8x6 x2 = 0 − 1 500 x3 + 1 2 x4 + 1 15 x5 − 1 30 x6 x7 = 1 −x3 z = 0 + 2 25 x3 −18x4 −x5 −x6 (4) 1 Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com x3 = 0 − 25 8 x1 + 525 2 x4 + 75 2 x5 −25x6 x2 = 0 1 160 x1 − 1 40 x4 − 1 120 x5 + 1 60 x6 x7 = 1 + 25 8 x1 − 525 2 x4 − 75 2 x5 +25x6 z = 0 −1 4 x1 +3x4 +2x5 −3x6 (5) x3 = 0 125 2 x1 −10500x2 −50x5 +150x6 x4 = 0 1 4 x1 −40x2 − 1 3 x5 + 2 3 x6 x7 = 1 − 125 2 x1 +10500x2 +50x5 −150x6 z = 0 1 2 x1 −120x2 +x5 −x6 (6) x5 = 0 5 4 x1 −210x2 − 1 50 x3 +3x6 x4 = 0 − 1 6 x1 +30x2 + 1 150 x3 − 1 3 x6 x7 = 1 −x3 z = 0 7 4 x1 −330x2 − 1 50 x3 +2x6 (7) x5 = 0 − 1 4 x1 +60x2 + 1 25 x3 −9x4 x6 = 0 − 1 2 x1 +90x2 + 1 50 x3 −3x4 x7 = 1 −x3 z = 0 +3 4 x1 −150x2 + 1 50 x3 −6x4 (8) 2 Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com