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Dual simplex method when initial reduced costs are negative



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Dual simplex doubt (unrestricted)solving minimum linear programming with simplex methodWhat variable to choose with the dual algorithm of the Simplex for a $max$ problem with negative variablesSimplex method: Facing with negative RHS?Using simplex method to show that a linear program has no finite optimal solutionMinimize using simplex method with mixed constraintsSolving Linear program by the dual simplex methodTransportation problem into initial simplex tableauMaximization using dual simplex method - problemSolving this LP with Dual simplex method










0












$begingroup$


I have the following problem which I'm trying to solve by dual simplex method:



$$min -6x_1-14x_2-13x_3$$
s.t $$0.5x_1+2x_2+x_3 le 24$$
$$x_1+2x_2+4x_3 le 60$$
$$x_1+x_2 ge 40$$
$$x_1, x_2, x_3 ge 0$$



I change the constraints into equality as:
$$0.5x_1+2x_2+x_3+s_1 = 24$$
$$x_1+2x_2+4x_3+s_2 = 60$$
$$-x_1-x_2+s_3=-40$$
The problem is that in the initial tableau the reduced costs are $-6, -14, -13, 0, 0, 0$ which violate the non-negativity clause for the reduced costs in dual simplex method.



Similarly there could be another type of problem with atleast 1 initial reduced cost as negative, such as:



$$min x_1-8x_2$$
s.t $$x_1+x_2 ge 1$$
$$-x_1+6x_2 le 3$$
$$x_1 le 2$$
$$x_1, x_2, x_3 ge 0$$



Again, I change the constraints into equality as:
$$-x_1-x_2+s_1 = -1$$
$$-x_1+6x_2+s_2 = 3$$
$$x_1+s_3= 2$$
Here my initial reduced cost is $(1, -8, 0, 0)$.
If I try solving the above two questions without taking the negativity of reduced costs into consideration, I stop at some intermediate solution when my $ B^-1 b$ becomes $ge 0$ which is not optimal solution and my reduced vector is still not $ge 0$.
How should I then proceed with such problem if my reduced costs are negative?










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I have the following problem which I'm trying to solve by dual simplex method:



    $$min -6x_1-14x_2-13x_3$$
    s.t $$0.5x_1+2x_2+x_3 le 24$$
    $$x_1+2x_2+4x_3 le 60$$
    $$x_1+x_2 ge 40$$
    $$x_1, x_2, x_3 ge 0$$



    I change the constraints into equality as:
    $$0.5x_1+2x_2+x_3+s_1 = 24$$
    $$x_1+2x_2+4x_3+s_2 = 60$$
    $$-x_1-x_2+s_3=-40$$
    The problem is that in the initial tableau the reduced costs are $-6, -14, -13, 0, 0, 0$ which violate the non-negativity clause for the reduced costs in dual simplex method.



    Similarly there could be another type of problem with atleast 1 initial reduced cost as negative, such as:



    $$min x_1-8x_2$$
    s.t $$x_1+x_2 ge 1$$
    $$-x_1+6x_2 le 3$$
    $$x_1 le 2$$
    $$x_1, x_2, x_3 ge 0$$



    Again, I change the constraints into equality as:
    $$-x_1-x_2+s_1 = -1$$
    $$-x_1+6x_2+s_2 = 3$$
    $$x_1+s_3= 2$$
    Here my initial reduced cost is $(1, -8, 0, 0)$.
    If I try solving the above two questions without taking the negativity of reduced costs into consideration, I stop at some intermediate solution when my $ B^-1 b$ becomes $ge 0$ which is not optimal solution and my reduced vector is still not $ge 0$.
    How should I then proceed with such problem if my reduced costs are negative?










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I have the following problem which I'm trying to solve by dual simplex method:



      $$min -6x_1-14x_2-13x_3$$
      s.t $$0.5x_1+2x_2+x_3 le 24$$
      $$x_1+2x_2+4x_3 le 60$$
      $$x_1+x_2 ge 40$$
      $$x_1, x_2, x_3 ge 0$$



      I change the constraints into equality as:
      $$0.5x_1+2x_2+x_3+s_1 = 24$$
      $$x_1+2x_2+4x_3+s_2 = 60$$
      $$-x_1-x_2+s_3=-40$$
      The problem is that in the initial tableau the reduced costs are $-6, -14, -13, 0, 0, 0$ which violate the non-negativity clause for the reduced costs in dual simplex method.



      Similarly there could be another type of problem with atleast 1 initial reduced cost as negative, such as:



      $$min x_1-8x_2$$
      s.t $$x_1+x_2 ge 1$$
      $$-x_1+6x_2 le 3$$
      $$x_1 le 2$$
      $$x_1, x_2, x_3 ge 0$$



      Again, I change the constraints into equality as:
      $$-x_1-x_2+s_1 = -1$$
      $$-x_1+6x_2+s_2 = 3$$
      $$x_1+s_3= 2$$
      Here my initial reduced cost is $(1, -8, 0, 0)$.
      If I try solving the above two questions without taking the negativity of reduced costs into consideration, I stop at some intermediate solution when my $ B^-1 b$ becomes $ge 0$ which is not optimal solution and my reduced vector is still not $ge 0$.
      How should I then proceed with such problem if my reduced costs are negative?










      share|cite|improve this question











      $endgroup$




      I have the following problem which I'm trying to solve by dual simplex method:



      $$min -6x_1-14x_2-13x_3$$
      s.t $$0.5x_1+2x_2+x_3 le 24$$
      $$x_1+2x_2+4x_3 le 60$$
      $$x_1+x_2 ge 40$$
      $$x_1, x_2, x_3 ge 0$$



      I change the constraints into equality as:
      $$0.5x_1+2x_2+x_3+s_1 = 24$$
      $$x_1+2x_2+4x_3+s_2 = 60$$
      $$-x_1-x_2+s_3=-40$$
      The problem is that in the initial tableau the reduced costs are $-6, -14, -13, 0, 0, 0$ which violate the non-negativity clause for the reduced costs in dual simplex method.



      Similarly there could be another type of problem with atleast 1 initial reduced cost as negative, such as:



      $$min x_1-8x_2$$
      s.t $$x_1+x_2 ge 1$$
      $$-x_1+6x_2 le 3$$
      $$x_1 le 2$$
      $$x_1, x_2, x_3 ge 0$$



      Again, I change the constraints into equality as:
      $$-x_1-x_2+s_1 = -1$$
      $$-x_1+6x_2+s_2 = 3$$
      $$x_1+s_3= 2$$
      Here my initial reduced cost is $(1, -8, 0, 0)$.
      If I try solving the above two questions without taking the negativity of reduced costs into consideration, I stop at some intermediate solution when my $ B^-1 b$ becomes $ge 0$ which is not optimal solution and my reduced vector is still not $ge 0$.
      How should I then proceed with such problem if my reduced costs are negative?







      linear-programming simplex






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Mar 25 at 6:53







      Prachi

















      asked Mar 25 at 6:30









      PrachiPrachi

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