EJEMPLO : Min 3x 1 +2x 2 +x 3 +2x 4 +2x 5 / x 1 -x 2 +2x 3 -x 4 +x 5 +2x 6 = 1

  • Published on
    13-Jan-2016

  • View
    32

  • Download
    3

DESCRIPTION

EJEMPLO : Min 3x 1 +2x 2 +x 3 +2x 4 +2x 5 / x 1 -x 2 +2x 3 -x 4 +x 5 +2x 6 = 1 - x 1 +2x 2 +x 3 -2x 4 -x 5 +x 6 = 3 2 x 1 +x 2 -x 3 +x 4 -2x 5 +x 6 = 2 x i 0 i=1..6 Solucin dual factible inicial: T =(0, 0, 0) ya que: - PowerPoint PPT Presentation

Transcript

  • EJEMPLO:

    Min 3x1+2x2+x3+2x4+2x5 / x1-x2+2x3-x4+x5+2x6 = 1-x1+2x2+x3-2x4-x5+x6 = 3 2x1+x2-x3+x4-2x5+x6 = 2 xi0 i=1..6

    Solucin dual factible inicial: T=(0, 0, 0) ya que: TA=(0, 0, 0, 0, 0, 0) cT=(3, 2, 1, 2, 2, 0)

    Sea P= {i / TAi=ci }= ndices de holguras duales nulas P = {6}, la prmera variable a entrar es la x6:

  • 1-12-1121-121-2-11321-11-2123212200cB x6 t1 t2 t3121001110113110002-40006

  • w = 4 0 no es ptimo

    1 = cTBA-1B = [0 1 1]. = [-1 1 1]

    cB x6 t1 t2 t301001/210-105/210-013/202004

  • Construccin de nueva solucin dual factible:2 = 1 + .1 / cT - 2TA 0 cT - 1TA - .1T.A 0

    d = 1T*A.=

    c = [3 2 1 2 2 0], 1T=[0 0 0] d = [0 4 -2 0 -4 0]= Min {ci/di , di >0} = 2 = [-1 1 1]TcR = cT - 2TA = [3 0 2 2 4 0]

    En el prximo paso entran x2 y x6 en el problema reducido: P = {2, 6},

  • cB x2 x6 t1 t2 t30-1/21001/215/20-105/213/20-013/2-402004

    cB x2 x6 t1 t2 t30011/301/311001/31-5/30010-1/302/31002/308/30

  • w=0 es el ptimo, solucin: [0 1 0 0 0 1]

Recommended

View more >