Mixing
Changes to LP if a new constraint is added
Applying simplex, I found out the optimal solution to be $z^*=16$ and $x^{*} = (8,0,0,0,12)$ where $x_4,x_5$ are slack variables.
Now, suppose we add constraint $x_2+2x_3 = 3$
So, If I were to add this constraint to the optimal tableau, we see that this does ${bf not}$ affect the values of $z_j-c_j$ and so optimality is not changed. This suggests that we have to consider $x_2+2x_3 leq 3 $ and $x_2+2x_3 geq 3$ which is same as $-x_2-2x_3 leq -3$ and so after adding slack variables $x_6,x_7$,then we see that the feasibility is changed (primal feasibility). In this situation, do we run the dual simplex?
or the way to handle this situation is different than the way I thought?
linear-programming
asked Nov 21 '18 at 5:34
Neymar
374114
add a comment |
Applying simplex, I found out the optimal solution to be $z^*=16$ and $x^{*} = (8,0,0,0,12)$ where $x_4,x_5$ are slack variables.
Now, suppose we add constraint $x_2+2x_3 = 3$
So, If I were to add this constraint to the optimal tableau, we see that this does ${bf not}$ affect the values of $z_j-c_j$ and so optimality is not changed. This suggests that we have to consider $x_2+2x_3 leq 3 $ and $x_2+2x_3 geq 3$ which is same as $-x_2-2x_3 leq -3$ and so after adding slack variables $x_6,x_7$,then we see that the feasibility is changed (primal feasibility). In this situation, do we run the dual simplex?
or the way to handle this situation is different than the way I thought?
linear-programming
asked Nov 21 '18 at 5:34
Neymar
374114
add a comment |
Applying simplex, I found out the optimal solution to be $z^*=16$ and $x^{*} = (8,0,0,0,12)$ where $x_4,x_5$ are slack variables.
Now, suppose we add constraint $x_2+2x_3 = 3$
So, If I were to add this constraint to the optimal tableau, we see that this does ${bf not}$ affect the values of $z_j-c_j$ and so optimality is not changed. This suggests that we have to consider $x_2+2x_3 leq 3 $ and $x_2+2x_3 geq 3$ which is same as $-x_2-2x_3 leq -3$ and so after adding slack variables $x_6,x_7$,then we see that the feasibility is changed (primal feasibility). In this situation, do we run the dual simplex?
or the way to handle this situation is different than the way I thought?
linear-programming
asked Nov 21 '18 at 5:34
Neymar
374114
Applying simplex, I found out the optimal solution to be $z^*=16$ and $x^{*} = (8,0,0,0,12)$ where $x_4,x_5$ are slack variables.
Now, suppose we add constraint $x_2+2x_3 = 3$
So, If I were to add this constraint to the optimal tableau, we see that this does ${bf not}$ affect the values of $z_j-c_j$ and so optimality is not changed. This suggests that we have to consider $x_2+2x_3 leq 3 $ and $x_2+2x_3 geq 3$ which is same as $-x_2-2x_3 leq -3$ and so after adding slack variables $x_6,x_7$,then we see that the feasibility is changed (primal feasibility). In this situation, do we run the dual simplex?
or the way to handle this situation is different than the way I thought?
linear-programming
linear-programming
asked Nov 21 '18 at 5:34
Neymar
374114
asked Nov 21 '18 at 5:34
Neymar
374114
edited Nov 21 '18 at 5:44
asked Nov 21 '18 at 5:34
Neymar
374114
asked Nov 21 '18 at 5:34
Neymar
374114
asked Nov 21 '18 at 5:34
Neymar
374114
374114
add a comment |
add a comment |
1 Answer
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$(8,0,0)$ doesn't satisfy $x_2+2x_3=3.$ The value on the LHS is less than the $RHS$.
$$max 2x_1+x_2-x_3-Mx_6$$
subject to
$$x_1+2x_2+x_3+x_4=8$$
$$-x_1+x_2-2x_3 + x_5=4$$
$$x_2+2x_3+x_6=3$$
$$x ge 0$$
At $(x_1, x_2, x_3, x_4, x_5, x_6)=(8,0,0,0,12,3)$, we have $x_1, x_5, x_6$ being basic variables and we can run primal simplex.
If the previous basis matrix is $B$, now, our basis matrix is $begin{bmatrix} B & 0 \ 0 & 1end{bmatrix}$ with its inverse being $begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}.$ The main entries of the simplex tables are $$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} A & 0\ a_3^T & 1end{bmatrix}=begin{bmatrix} B^{-1}A & 0\ a_3^T & 1end{bmatrix}$$
$$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} b \ b_3end{bmatrix}=begin{bmatrix} B^{-1}b \ b_3 end{bmatrix}$$
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
Why did you add slack to the new constraint if it was already an equiality?
– Neymar
Nov 22 '18 at 16:04
so that we can move on from the current position in our simplex implementation.
– Siong Thye Goh
Nov 22 '18 at 16:16
Is it possible to add constraint to the optimal tableau of the “old” problem and take it from there? Maybe using dual simple?
– Neymar
Nov 22 '18 at 16:28
Sure. I added a constraint and the variable $x_6$. With $x_1, x_5, x_6$ as the basis, most entries are identical with the old problem.
– Siong Thye Goh
Nov 22 '18 at 16:36
I have added an explanation why most entries are the same.
– Siong Thye Goh
Nov 22 '18 at 16:43
|
show 2 more comments
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$(8,0,0)$ doesn't satisfy $x_2+2x_3=3.$ The value on the LHS is less than the $RHS$.
$$max 2x_1+x_2-x_3-Mx_6$$
subject to
$$x_1+2x_2+x_3+x_4=8$$
$$-x_1+x_2-2x_3 + x_5=4$$
$$x_2+2x_3+x_6=3$$
$$x ge 0$$
At $(x_1, x_2, x_3, x_4, x_5, x_6)=(8,0,0,0,12,3)$, we have $x_1, x_5, x_6$ being basic variables and we can run primal simplex.
If the previous basis matrix is $B$, now, our basis matrix is $begin{bmatrix} B & 0 \ 0 & 1end{bmatrix}$ with its inverse being $begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}.$ The main entries of the simplex tables are $$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} A & 0\ a_3^T & 1end{bmatrix}=begin{bmatrix} B^{-1}A & 0\ a_3^T & 1end{bmatrix}$$
$$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} b \ b_3end{bmatrix}=begin{bmatrix} B^{-1}b \ b_3 end{bmatrix}$$
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
Why did you add slack to the new constraint if it was already an equiality?
– Neymar
Nov 22 '18 at 16:04
so that we can move on from the current position in our simplex implementation.
– Siong Thye Goh
Nov 22 '18 at 16:16
Is it possible to add constraint to the optimal tableau of the “old” problem and take it from there? Maybe using dual simple?
– Neymar
Nov 22 '18 at 16:28
Sure. I added a constraint and the variable $x_6$. With $x_1, x_5, x_6$ as the basis, most entries are identical with the old problem.
– Siong Thye Goh
Nov 22 '18 at 16:36
I have added an explanation why most entries are the same.
– Siong Thye Goh
Nov 22 '18 at 16:43
|
show 2 more comments
$(8,0,0)$ doesn't satisfy $x_2+2x_3=3.$ The value on the LHS is less than the $RHS$.
$$max 2x_1+x_2-x_3-Mx_6$$
subject to
$$x_1+2x_2+x_3+x_4=8$$
$$-x_1+x_2-2x_3 + x_5=4$$
$$x_2+2x_3+x_6=3$$
$$x ge 0$$
At $(x_1, x_2, x_3, x_4, x_5, x_6)=(8,0,0,0,12,3)$, we have $x_1, x_5, x_6$ being basic variables and we can run primal simplex.
If the previous basis matrix is $B$, now, our basis matrix is $begin{bmatrix} B & 0 \ 0 & 1end{bmatrix}$ with its inverse being $begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}.$ The main entries of the simplex tables are $$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} A & 0\ a_3^T & 1end{bmatrix}=begin{bmatrix} B^{-1}A & 0\ a_3^T & 1end{bmatrix}$$
$$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} b \ b_3end{bmatrix}=begin{bmatrix} B^{-1}b \ b_3 end{bmatrix}$$
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
Why did you add slack to the new constraint if it was already an equiality?
– Neymar
Nov 22 '18 at 16:04
so that we can move on from the current position in our simplex implementation.
– Siong Thye Goh
Nov 22 '18 at 16:16
Is it possible to add constraint to the optimal tableau of the “old” problem and take it from there? Maybe using dual simple?
– Neymar
Nov 22 '18 at 16:28
Sure. I added a constraint and the variable $x_6$. With $x_1, x_5, x_6$ as the basis, most entries are identical with the old problem.
– Siong Thye Goh
Nov 22 '18 at 16:36
I have added an explanation why most entries are the same.
– Siong Thye Goh
Nov 22 '18 at 16:43
|
show 2 more comments
$(8,0,0)$ doesn't satisfy $x_2+2x_3=3.$ The value on the LHS is less than the $RHS$.
$$max 2x_1+x_2-x_3-Mx_6$$
subject to
$$x_1+2x_2+x_3+x_4=8$$
$$-x_1+x_2-2x_3 + x_5=4$$
$$x_2+2x_3+x_6=3$$
$$x ge 0$$
At $(x_1, x_2, x_3, x_4, x_5, x_6)=(8,0,0,0,12,3)$, we have $x_1, x_5, x_6$ being basic variables and we can run primal simplex.
If the previous basis matrix is $B$, now, our basis matrix is $begin{bmatrix} B & 0 \ 0 & 1end{bmatrix}$ with its inverse being $begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}.$ The main entries of the simplex tables are $$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} A & 0\ a_3^T & 1end{bmatrix}=begin{bmatrix} B^{-1}A & 0\ a_3^T & 1end{bmatrix}$$
$$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} b \ b_3end{bmatrix}=begin{bmatrix} B^{-1}b \ b_3 end{bmatrix}$$
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
$(8,0,0)$ doesn't satisfy $x_2+2x_3=3.$ The value on the LHS is less than the $RHS$.
$$max 2x_1+x_2-x_3-Mx_6$$
subject to
$$x_1+2x_2+x_3+x_4=8$$
$$-x_1+x_2-2x_3 + x_5=4$$
$$x_2+2x_3+x_6=3$$
$$x ge 0$$
At $(x_1, x_2, x_3, x_4, x_5, x_6)=(8,0,0,0,12,3)$, we have $x_1, x_5, x_6$ being basic variables and we can run primal simplex.
If the previous basis matrix is $B$, now, our basis matrix is $begin{bmatrix} B & 0 \ 0 & 1end{bmatrix}$ with its inverse being $begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}.$ The main entries of the simplex tables are $$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} A & 0\ a_3^T & 1end{bmatrix}=begin{bmatrix} B^{-1}A & 0\ a_3^T & 1end{bmatrix}$$
$$begin{bmatrix} B^{-1} & 0 \ 0 & 1end{bmatrix}begin{bmatrix} b \ b_3end{bmatrix}=begin{bmatrix} B^{-1}b \ b_3 end{bmatrix}$$
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
edited Nov 22 '18 at 16:50
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
answered Nov 22 '18 at 13:02
Siong Thye Goh
99.5k1464117
99.5k1464117
Why did you add slack to the new constraint if it was already an equiality?
– Neymar
Nov 22 '18 at 16:04
so that we can move on from the current position in our simplex implementation.
– Siong Thye Goh
Nov 22 '18 at 16:16
Is it possible to add constraint to the optimal tableau of the “old” problem and take it from there? Maybe using dual simple?
– Neymar
Nov 22 '18 at 16:28
Sure. I added a constraint and the variable $x_6$. With $x_1, x_5, x_6$ as the basis, most entries are identical with the old problem.
– Siong Thye Goh
Nov 22 '18 at 16:36
I have added an explanation why most entries are the same.
– Siong Thye Goh
Nov 22 '18 at 16:43
|
show 2 more comments
Why did you add slack to the new constraint if it was already an equiality?
– Neymar
Nov 22 '18 at 16:04
so that we can move on from the current position in our simplex implementation.
– Siong Thye Goh
Nov 22 '18 at 16:16
Is it possible to add constraint to the optimal tableau of the “old” problem and take it from there? Maybe using dual simple?
– Neymar
Nov 22 '18 at 16:28
Sure. I added a constraint and the variable $x_6$. With $x_1, x_5, x_6$ as the basis, most entries are identical with the old problem.
– Siong Thye Goh
Nov 22 '18 at 16:36
I have added an explanation why most entries are the same.
– Siong Thye Goh
Nov 22 '18 at 16:43
Why did you add slack to the new constraint if it was already an equiality?
– Neymar
Nov 22 '18 at 16:04
Why did you add slack to the new constraint if it was already an equiality?
– Neymar
Nov 22 '18 at 16:04
so that we can move on from the current position in our simplex implementation.
– Siong Thye Goh
Nov 22 '18 at 16:16
so that we can move on from the current position in our simplex implementation.
– Siong Thye Goh
Nov 22 '18 at 16:16
Is it possible to add constraint to the optimal tableau of the “old” problem and take it from there? Maybe using dual simple?
– Neymar
Nov 22 '18 at 16:28
Is it possible to add constraint to the optimal tableau of the “old” problem and take it from there? Maybe using dual simple?
– Neymar
Nov 22 '18 at 16:28
Sure. I added a constraint and the variable $x_6$. With $x_1, x_5, x_6$ as the basis, most entries are identical with the old problem.
– Siong Thye Goh
Nov 22 '18 at 16:36
Sure. I added a constraint and the variable $x_6$. With $x_1, x_5, x_6$ as the basis, most entries are identical with the old problem.
– Siong Thye Goh
Nov 22 '18 at 16:36
I have added an explanation why most entries are the same.
– Siong Thye Goh
Nov 22 '18 at 16:43
I have added an explanation why most entries are the same.
– Siong Thye Goh
Nov 22 '18 at 16:43
|
show 2 more comments
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