Mixing
Finding basic feasible solution graphically
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I'm supposed to find the basic solutions of the given LPP graphically. I know what a bfs means, I can find that in other way (I mean by method mentioned in this post), but I don't know how to do it graphically. The system is:-
$$2x + 3y leq 21$$$$ 3x - y leq 15$$$$ x + y geq 5$$$$ y leq 5$$ $$ x, y geq 0$$
Any hint/solution would be appreciated. Thanks!
optimization linear-programming
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
I'm supposed to find the basic solutions of the given LPP graphically. I know what a bfs means, I can find that in other way (I mean by method mentioned in this post), but I don't know how to do it graphically. The system is:-
$$2x + 3y leq 21$$$$ 3x - y leq 15$$$$ x + y geq 5$$$$ y leq 5$$ $$ x, y geq 0$$
Any hint/solution would be appreciated. Thanks!
optimization linear-programming
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
$endgroup$
$begingroup$
Since you only have two variables you can plot each inequality in the plane and take the intersection, maybe this is what they mean by graphically?
$endgroup$
– Tony S.F.
Feb 3 at 13:12
$begingroup$
@TonyS.F. but that'll be a feasible solution. Not bfs if I'm not wrong...
$endgroup$
– Ankit Kumar
Feb 3 at 13:13
$begingroup$
It depends on which one you pick, of course. You will get a polygon of sorts in the plane and if you pick a vertex you will have BFS.
$endgroup$
– Tony S.F.
Feb 3 at 13:14
add a comment |
$begingroup$
I'm supposed to find the basic solutions of the given LPP graphically. I know what a bfs means, I can find that in other way (I mean by method mentioned in this post), but I don't know how to do it graphically. The system is:-
$$2x + 3y leq 21$$$$ 3x - y leq 15$$$$ x + y geq 5$$$$ y leq 5$$ $$ x, y geq 0$$
Any hint/solution would be appreciated. Thanks!
optimization linear-programming
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
$endgroup$
I'm supposed to find the basic solutions of the given LPP graphically. I know what a bfs means, I can find that in other way (I mean by method mentioned in this post), but I don't know how to do it graphically. The system is:-
$$2x + 3y leq 21$$$$ 3x - y leq 15$$$$ x + y geq 5$$$$ y leq 5$$ $$ x, y geq 0$$
Any hint/solution would be appreciated. Thanks!
optimization linear-programming
optimization linear-programming
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
asked Feb 3 at 12:34
Ankit KumarAnkit Kumar
1,542221
1,542221
$begingroup$
Since you only have two variables you can plot each inequality in the plane and take the intersection, maybe this is what they mean by graphically?
$endgroup$
– Tony S.F.
Feb 3 at 13:12
$begingroup$
@TonyS.F. but that'll be a feasible solution. Not bfs if I'm not wrong...
$endgroup$
– Ankit Kumar
Feb 3 at 13:13
$begingroup$
It depends on which one you pick, of course. You will get a polygon of sorts in the plane and if you pick a vertex you will have BFS.
$endgroup$
– Tony S.F.
Feb 3 at 13:14
add a comment |
$begingroup$
Since you only have two variables you can plot each inequality in the plane and take the intersection, maybe this is what they mean by graphically?
$endgroup$
– Tony S.F.
Feb 3 at 13:12
$begingroup$
@TonyS.F. but that'll be a feasible solution. Not bfs if I'm not wrong...
$endgroup$
– Ankit Kumar
Feb 3 at 13:13
$begingroup$
It depends on which one you pick, of course. You will get a polygon of sorts in the plane and if you pick a vertex you will have BFS.
$endgroup$
– Tony S.F.
Feb 3 at 13:14
$begingroup$
Since you only have two variables you can plot each inequality in the plane and take the intersection, maybe this is what they mean by graphically?
$endgroup$
– Tony S.F.
Feb 3 at 13:12
$begingroup$
Since you only have two variables you can plot each inequality in the plane and take the intersection, maybe this is what they mean by graphically?
$endgroup$
– Tony S.F.
Feb 3 at 13:12
$begingroup$
@TonyS.F. but that'll be a feasible solution. Not bfs if I'm not wrong...
$endgroup$
– Ankit Kumar
Feb 3 at 13:13
$begingroup$
@TonyS.F. but that'll be a feasible solution. Not bfs if I'm not wrong...
$endgroup$
– Ankit Kumar
Feb 3 at 13:13
$begingroup$
It depends on which one you pick, of course. You will get a polygon of sorts in the plane and if you pick a vertex you will have BFS.
$endgroup$
– Tony S.F.
Feb 3 at 13:14
$begingroup$
It depends on which one you pick, of course. You will get a polygon of sorts in the plane and if you pick a vertex you will have BFS.
$endgroup$
– Tony S.F.
Feb 3 at 13:14
add a comment |
1 Answer
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$begingroup$
For a linear program like this, the set of all feasible solutions is the intersection of half spaces and thus a polyhedron. A BFS corresponds to a vertex of this polyhedron.
Now, since we only have 2 variables we can plot all our half spaces in the plane and take the intersection to get a polygon. Then, the BFS are just the vertices of this polygon.
Here is a graph of your problem (more or less)
From that graph we can see (0,5), (5,0), (3,5), and (6,3) are all BFS.
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
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add a comment |
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1 Answer
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$begingroup$
For a linear program like this, the set of all feasible solutions is the intersection of half spaces and thus a polyhedron. A BFS corresponds to a vertex of this polyhedron.
Now, since we only have 2 variables we can plot all our half spaces in the plane and take the intersection to get a polygon. Then, the BFS are just the vertices of this polygon.
Here is a graph of your problem (more or less)
From that graph we can see (0,5), (5,0), (3,5), and (6,3) are all BFS.
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
$endgroup$
add a comment |
$begingroup$
For a linear program like this, the set of all feasible solutions is the intersection of half spaces and thus a polyhedron. A BFS corresponds to a vertex of this polyhedron.
Now, since we only have 2 variables we can plot all our half spaces in the plane and take the intersection to get a polygon. Then, the BFS are just the vertices of this polygon.
Here is a graph of your problem (more or less)
From that graph we can see (0,5), (5,0), (3,5), and (6,3) are all BFS.
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
$endgroup$
add a comment |
$begingroup$
For a linear program like this, the set of all feasible solutions is the intersection of half spaces and thus a polyhedron. A BFS corresponds to a vertex of this polyhedron.
Now, since we only have 2 variables we can plot all our half spaces in the plane and take the intersection to get a polygon. Then, the BFS are just the vertices of this polygon.
Here is a graph of your problem (more or less)
From that graph we can see (0,5), (5,0), (3,5), and (6,3) are all BFS.
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
$endgroup$
For a linear program like this, the set of all feasible solutions is the intersection of half spaces and thus a polyhedron. A BFS corresponds to a vertex of this polyhedron.
Now, since we only have 2 variables we can plot all our half spaces in the plane and take the intersection to get a polygon. Then, the BFS are just the vertices of this polygon.
Here is a graph of your problem (more or less)
From that graph we can see (0,5), (5,0), (3,5), and (6,3) are all BFS.
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
edited Feb 3 at 13:22
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
answered Feb 3 at 13:16
Tony S.F.Tony S.F.
3,79921031
3,79921031
add a comment |
add a comment |
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$begingroup$
Since you only have two variables you can plot each inequality in the plane and take the intersection, maybe this is what they mean by graphically?
$endgroup$
– Tony S.F.
Feb 3 at 13:12
$begingroup$
@TonyS.F. but that'll be a feasible solution. Not bfs if I'm not wrong...
$endgroup$
– Ankit Kumar
Feb 3 at 13:13
$begingroup$
It depends on which one you pick, of course. You will get a polygon of sorts in the plane and if you pick a vertex you will have BFS.
$endgroup$
– Tony S.F.
Feb 3 at 13:14