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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

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 Solución dual factible inicial:  T =(0, 0, 0) ya que:

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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

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  1. 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 xi0 i=1..6 Solución 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 }= índices de holguras duales nulas  P = {6}, la prmera variable a entrar es la x6:

  2. x1 x2 x3 x4 x5 x6

  3. w = 4  0  no es óptimo 1 = cTBA-1B = [0 1 1]. = [-1 1 1]

  4. Construcción de nueva solución 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} = ½  2 = ½ [-1 1 1]T cR = cT - 2TA = [3 0 2 2 4 0] En el próximo paso entran x2 y x6 en el problema reducido: P = {2, 6},

  5. w=0  es el óptimo, solución: [0 1 0 0 0 1]

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