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
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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
$begingroup$
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?
linear-programming simplex
$endgroup$
add a comment |
$begingroup$
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?
linear-programming simplex
$endgroup$
add a comment |
$begingroup$
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?
linear-programming simplex
$endgroup$
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
linear-programming simplex
edited Mar 25 at 6:53
Prachi
asked Mar 25 at 6:30
PrachiPrachi
12
12
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