Lecture 5

3thesimplexalgorithmmaxlps theresultis canonical form

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 8 4 1 x1 8 4 1 3 z 1 x1 4 1 4 z 1 x1 1 x2 -30 6 2 0.75 1 x2 15 6 2 0.75 1 x2 15 2 0.75 1 x2 15 -1 0.75 1 x3 -20 1 1.5 0.25 x3 -5 1 1.5 0.25 x3 -5 -1 1.5 0.25 x3 -5 -1 0.5 0.25 s1 s2 s3 s4 1 1 0.5 s1 s2 s3 30 1 s4 1 1 0.5 s1 s2 1 s3 30 -4 1 s4 1 0.5 s1 s2 1 1 s3 30 -4 -2 0.5 1 s4 1 rhs 48 20 4 5 rhs 240 48 20 4 5 rhs 240 16 20 4 5 rhs 240 16 4 4 5 ero row 3 divided by 1/2 60 times row 3 added to row 0 - 8 times row 3 added to row 1 - 4 times row 3 added to row 2 4.3 – The Simplex Algorithm (max LPs) The result is: Canonical Form 1 Row 0 z + 15x2 - Row 1 Row 2 - Basic Variable 5x3 x3 + s1 x2 + 0.5 x3 Row 3 x1 + 0.75x2 + 0.25x3 Row 4 x2 + 30s3 = 240 z = 240 - 4s3 = 16 s1 = 16 2 s3 =4 s2 = 4 + 0.5s3 =4 x1 = 4 =5 s4 = 5 + s2 - + s4 In canonical form 1, BV = {z, s1, s2, x1, s4} and NBV = {s3, x2, x3 }. yielding the bfs z = 240, s1 = 16, s2 = 4, x1 = 4, s4 = 5, s3 = x2 = x3...
View Full Document

This note was uploaded on 09/20/2011 for the course ENG 300 taught by Professor Albin during the Fall '11 term at Rutgers.

Ask a homework question - tutors are online