Network flow model - How can I turn this diagram into a matrix that when converted to RREF solves for max flow?Many-to-many Breadth First SearchFormulation of a constraint in a MIP for an element in different SetsRunning a linear programming model to maximize binned predictionsA variant of the Multiple Traveling Salesman ProblemGenerating all extreme raysFind feasible point in polynomial time in linear programmingHow to linearize a constraint with maxShould I factor in time as a parameter or a variable in a scheduling problem with MILP?How To Linearize $X = maxx_1,x_2$
How safe is the 4% rule if the U.S. goes back to the mean?
Symbolise polygon outline where it doesn't coincide with other feature using geometry generator in QGIS?
Did Terry Pratchett ever explain the inspiration behind the Luggage?
How should I understand FPGA architecture?
In the twin paradox does the returning twin also come back permanently length contracted flatter than the twin on earth?
Could an American state survive nuclear war?
Idiom for a situation or event that makes one poor or even poorer?
What is the design rationale for having armor and magic penetration mechanics?
Mapping string into integers
Does a restocking fee still qualify as a business expense?
A man condemned to serve his sentence in other times
How do I get my boyfriend to remove pictures of his ex girlfriend hanging in his apartment?
What does "stirring tanks" mean?
Fermat's polygonal number theorem
Do more Americans want the Bidens investigated than Trump impeached?
Tokenizing SGML text for NLTK analysis
If we should encrypt the message rather than the method of transfer, why do we care about wifi security? Is this just security theatre?
What does the British parliament hope to achieve by requesting a third Brexit extension?
How to prove that invoices are really unpaid?
Easy way of generating a 50-150W load @12V
Not returning after leave
Network flow model - How can I turn this diagram into a matrix that when converted to RREF solves for max flow?
Does my protagonist need to be the most important character?
Find number 8 with the least number of tries
Network flow model - How can I turn this diagram into a matrix that when converted to RREF solves for max flow?
Many-to-many Breadth First SearchFormulation of a constraint in a MIP for an element in different SetsRunning a linear programming model to maximize binned predictionsA variant of the Multiple Traveling Salesman ProblemGenerating all extreme raysFind feasible point in polynomial time in linear programmingHow to linearize a constraint with maxShould I factor in time as a parameter or a variable in a scheduling problem with MILP?How To Linearize $X = maxx_1,x_2$
$begingroup$
I have the following network flow model diagram and I have already calculated maximum flow using the R package igraph
to be 28. However, what I would like to know how to do is to solve this for maximum flow using the simplex method of linear programming.
I know that I would need to put values into a matrix which I could then pass through the R function rref()
which puts the matrix into row reduced echelon form, but what I am not sure about is what values need to be put into this matrix. What would such a matrix look like?
I have tried to follow Oguz Toragay's advice but I must be doing something wrong because my matrix is clearly not correct. I made a LHS
matrix which was all the inflows for each node, and a RHS
list that was all the outflows for each node, then binded them together to make A
, but the sides clearly do not equal each other. I have written the following code:
#LHS
#from 1 2 3 4 5 6 7
node1 = c( 0, 0, 0, 0, 0, 0, 0)
node2 = c(20, 0, 0, 0, 15, 0, 0)
node3 = c(15, 0, 0, 13, 0, 0, 0)
node4 = c( 0, 10, 13, 0, 0, 0, 0)
node5 = c( 0, 15, 0, 10, 0, 7, 0)
node6 = c( 0, 0, 15, 0, 7, 0, 8)
node7 = c( 0, 0, 3, 12, 0, 8, 0)
node8 = c( 0, 0, 0, 0, 10, 8, 10)
LHS = rbind(node1,node2,node3,node4,node5,node6,node7,node8)
#RHS
#to 1 2 3 4 5 6 7 8
node1o = sum( 0, 20, 15, 0, 0, 0, 0, 0)
node2o = sum( 0, 0, 0, 10, 15, 0, 0, 0)
node3o = sum( 0, 0, 0, 13, 0, 15, 10, 0)
node4o = sum( 0, 0, 13, 0, 10, 0, 12, 0)
node5o = sum( 0, 15, 0, 0, 0, 7, 0, 10)
node6o = sum( 0, 0, 0, 0, 7, 0, 8, 8)
node7o = sum( 0, 0, 0, 0, 0, 8, 0, 10)
node8o = sum( 0, 0, 0, 0, 0, 0, 0, 0)
RHS = c(node1o,node2o,node3o,node4o,node5o,node6o,node7o,node8o)
A = cbind(LHS, RHS)
# Calling A returns:
RHS
node1 0 0 0 0 0 0 0 35
node2 20 0 0 0 15 0 0 25
node3 15 0 0 13 0 0 0 38
node4 0 10 13 0 0 0 0 35
node5 0 15 0 10 0 7 0 32
node6 0 0 15 0 7 0 8 23
node7 0 0 3 12 0 8 0 18
node8 0 0 0 0 10 8 10 0
```
linear-programming linearization graphs simplex network-flow
$endgroup$
add a comment
|
$begingroup$
I have the following network flow model diagram and I have already calculated maximum flow using the R package igraph
to be 28. However, what I would like to know how to do is to solve this for maximum flow using the simplex method of linear programming.
I know that I would need to put values into a matrix which I could then pass through the R function rref()
which puts the matrix into row reduced echelon form, but what I am not sure about is what values need to be put into this matrix. What would such a matrix look like?
I have tried to follow Oguz Toragay's advice but I must be doing something wrong because my matrix is clearly not correct. I made a LHS
matrix which was all the inflows for each node, and a RHS
list that was all the outflows for each node, then binded them together to make A
, but the sides clearly do not equal each other. I have written the following code:
#LHS
#from 1 2 3 4 5 6 7
node1 = c( 0, 0, 0, 0, 0, 0, 0)
node2 = c(20, 0, 0, 0, 15, 0, 0)
node3 = c(15, 0, 0, 13, 0, 0, 0)
node4 = c( 0, 10, 13, 0, 0, 0, 0)
node5 = c( 0, 15, 0, 10, 0, 7, 0)
node6 = c( 0, 0, 15, 0, 7, 0, 8)
node7 = c( 0, 0, 3, 12, 0, 8, 0)
node8 = c( 0, 0, 0, 0, 10, 8, 10)
LHS = rbind(node1,node2,node3,node4,node5,node6,node7,node8)
#RHS
#to 1 2 3 4 5 6 7 8
node1o = sum( 0, 20, 15, 0, 0, 0, 0, 0)
node2o = sum( 0, 0, 0, 10, 15, 0, 0, 0)
node3o = sum( 0, 0, 0, 13, 0, 15, 10, 0)
node4o = sum( 0, 0, 13, 0, 10, 0, 12, 0)
node5o = sum( 0, 15, 0, 0, 0, 7, 0, 10)
node6o = sum( 0, 0, 0, 0, 7, 0, 8, 8)
node7o = sum( 0, 0, 0, 0, 0, 8, 0, 10)
node8o = sum( 0, 0, 0, 0, 0, 0, 0, 0)
RHS = c(node1o,node2o,node3o,node4o,node5o,node6o,node7o,node8o)
A = cbind(LHS, RHS)
# Calling A returns:
RHS
node1 0 0 0 0 0 0 0 35
node2 20 0 0 0 15 0 0 25
node3 15 0 0 13 0 0 0 38
node4 0 10 13 0 0 0 0 35
node5 0 15 0 10 0 7 0 32
node6 0 0 15 0 7 0 8 23
node7 0 0 3 12 0 8 0 18
node8 0 0 0 0 10 8 10 0
```
linear-programming linearization graphs simplex network-flow
$endgroup$
add a comment
|
$begingroup$
I have the following network flow model diagram and I have already calculated maximum flow using the R package igraph
to be 28. However, what I would like to know how to do is to solve this for maximum flow using the simplex method of linear programming.
I know that I would need to put values into a matrix which I could then pass through the R function rref()
which puts the matrix into row reduced echelon form, but what I am not sure about is what values need to be put into this matrix. What would such a matrix look like?
I have tried to follow Oguz Toragay's advice but I must be doing something wrong because my matrix is clearly not correct. I made a LHS
matrix which was all the inflows for each node, and a RHS
list that was all the outflows for each node, then binded them together to make A
, but the sides clearly do not equal each other. I have written the following code:
#LHS
#from 1 2 3 4 5 6 7
node1 = c( 0, 0, 0, 0, 0, 0, 0)
node2 = c(20, 0, 0, 0, 15, 0, 0)
node3 = c(15, 0, 0, 13, 0, 0, 0)
node4 = c( 0, 10, 13, 0, 0, 0, 0)
node5 = c( 0, 15, 0, 10, 0, 7, 0)
node6 = c( 0, 0, 15, 0, 7, 0, 8)
node7 = c( 0, 0, 3, 12, 0, 8, 0)
node8 = c( 0, 0, 0, 0, 10, 8, 10)
LHS = rbind(node1,node2,node3,node4,node5,node6,node7,node8)
#RHS
#to 1 2 3 4 5 6 7 8
node1o = sum( 0, 20, 15, 0, 0, 0, 0, 0)
node2o = sum( 0, 0, 0, 10, 15, 0, 0, 0)
node3o = sum( 0, 0, 0, 13, 0, 15, 10, 0)
node4o = sum( 0, 0, 13, 0, 10, 0, 12, 0)
node5o = sum( 0, 15, 0, 0, 0, 7, 0, 10)
node6o = sum( 0, 0, 0, 0, 7, 0, 8, 8)
node7o = sum( 0, 0, 0, 0, 0, 8, 0, 10)
node8o = sum( 0, 0, 0, 0, 0, 0, 0, 0)
RHS = c(node1o,node2o,node3o,node4o,node5o,node6o,node7o,node8o)
A = cbind(LHS, RHS)
# Calling A returns:
RHS
node1 0 0 0 0 0 0 0 35
node2 20 0 0 0 15 0 0 25
node3 15 0 0 13 0 0 0 38
node4 0 10 13 0 0 0 0 35
node5 0 15 0 10 0 7 0 32
node6 0 0 15 0 7 0 8 23
node7 0 0 3 12 0 8 0 18
node8 0 0 0 0 10 8 10 0
```
linear-programming linearization graphs simplex network-flow
$endgroup$
I have the following network flow model diagram and I have already calculated maximum flow using the R package igraph
to be 28. However, what I would like to know how to do is to solve this for maximum flow using the simplex method of linear programming.
I know that I would need to put values into a matrix which I could then pass through the R function rref()
which puts the matrix into row reduced echelon form, but what I am not sure about is what values need to be put into this matrix. What would such a matrix look like?
I have tried to follow Oguz Toragay's advice but I must be doing something wrong because my matrix is clearly not correct. I made a LHS
matrix which was all the inflows for each node, and a RHS
list that was all the outflows for each node, then binded them together to make A
, but the sides clearly do not equal each other. I have written the following code:
#LHS
#from 1 2 3 4 5 6 7
node1 = c( 0, 0, 0, 0, 0, 0, 0)
node2 = c(20, 0, 0, 0, 15, 0, 0)
node3 = c(15, 0, 0, 13, 0, 0, 0)
node4 = c( 0, 10, 13, 0, 0, 0, 0)
node5 = c( 0, 15, 0, 10, 0, 7, 0)
node6 = c( 0, 0, 15, 0, 7, 0, 8)
node7 = c( 0, 0, 3, 12, 0, 8, 0)
node8 = c( 0, 0, 0, 0, 10, 8, 10)
LHS = rbind(node1,node2,node3,node4,node5,node6,node7,node8)
#RHS
#to 1 2 3 4 5 6 7 8
node1o = sum( 0, 20, 15, 0, 0, 0, 0, 0)
node2o = sum( 0, 0, 0, 10, 15, 0, 0, 0)
node3o = sum( 0, 0, 0, 13, 0, 15, 10, 0)
node4o = sum( 0, 0, 13, 0, 10, 0, 12, 0)
node5o = sum( 0, 15, 0, 0, 0, 7, 0, 10)
node6o = sum( 0, 0, 0, 0, 7, 0, 8, 8)
node7o = sum( 0, 0, 0, 0, 0, 8, 0, 10)
node8o = sum( 0, 0, 0, 0, 0, 0, 0, 0)
RHS = c(node1o,node2o,node3o,node4o,node5o,node6o,node7o,node8o)
A = cbind(LHS, RHS)
# Calling A returns:
RHS
node1 0 0 0 0 0 0 0 35
node2 20 0 0 0 15 0 0 25
node3 15 0 0 13 0 0 0 38
node4 0 10 13 0 0 0 0 35
node5 0 15 0 10 0 7 0 32
node6 0 0 15 0 7 0 8 23
node7 0 0 3 12 0 8 0 18
node8 0 0 0 0 10 8 10 0
```
linear-programming linearization graphs simplex network-flow
linear-programming linearization graphs simplex network-flow
edited 3 hours ago
TheSimpliFire♦
2,6547 silver badges39 bronze badges
2,6547 silver badges39 bronze badges
asked 9 hours ago
Jacob MyerJacob Myer
904 bronze badges
904 bronze badges
add a comment
|
add a comment
|
1 Answer
1
active
oldest
votes
$begingroup$
Here is a link that includes all the information that you need. The matrix should include all the capacity limitations on all the connections between nodes. Actually, for your example, it should be a $8times8$ matrix with all the coefficients. Each row represents one of the constraints in your LP model. In other words for each row, you consider one of the nodes in the network and write a constraint which implies that:
$$textAll incoming flow = textAll outgoing flow$$
Also in this the matrix for a given simple network extracted from the linear model, which I think would be helpful for your problem.
$endgroup$
$begingroup$
Thank you that helps a lot. However I am still unsure about what to do with the values of some of the arcs because this model allows flow in both directions whereas all the ones in that link do not. For instance, would the arc with flow 15 going from 5 to 2 simply be an outflow of 5 and the arc from 2 to 5 with flow 15 be an outflow of 2? @Oguz Toragay
$endgroup$
– Jacob Myer
8 hours ago
$begingroup$
@JacobMyer yes exactly, putting it in that way, the model decided to choose one of the directions, either from 5 to 2 or 2 to 5. If I remember correctly both of the flows can not be positive simultaneously.
$endgroup$
– Oguz Toragay
8 hours ago
$begingroup$
I edited my post because I ran into a problem and must have misunderstood something about what you were saying. Could you take a look at my matrixA
and tell me what I could change?
$endgroup$
– Jacob Myer
7 hours ago
$begingroup$
@JacobMyer look at the link below from Matlab documentation, in which the same Matlab function called rref() has been used to solve a system of equation. In linear programs the procedure is almost the same. Link: mathworks.com/help/matlab/ref/rref.html#description
$endgroup$
– Oguz Toragay
6 hours ago
add a comment
|
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "700"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2for.stackexchange.com%2fquestions%2f2767%2fnetwork-flow-model-how-can-i-turn-this-diagram-into-a-matrix-that-when-convert%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a link that includes all the information that you need. The matrix should include all the capacity limitations on all the connections between nodes. Actually, for your example, it should be a $8times8$ matrix with all the coefficients. Each row represents one of the constraints in your LP model. In other words for each row, you consider one of the nodes in the network and write a constraint which implies that:
$$textAll incoming flow = textAll outgoing flow$$
Also in this the matrix for a given simple network extracted from the linear model, which I think would be helpful for your problem.
$endgroup$
$begingroup$
Thank you that helps a lot. However I am still unsure about what to do with the values of some of the arcs because this model allows flow in both directions whereas all the ones in that link do not. For instance, would the arc with flow 15 going from 5 to 2 simply be an outflow of 5 and the arc from 2 to 5 with flow 15 be an outflow of 2? @Oguz Toragay
$endgroup$
– Jacob Myer
8 hours ago
$begingroup$
@JacobMyer yes exactly, putting it in that way, the model decided to choose one of the directions, either from 5 to 2 or 2 to 5. If I remember correctly both of the flows can not be positive simultaneously.
$endgroup$
– Oguz Toragay
8 hours ago
$begingroup$
I edited my post because I ran into a problem and must have misunderstood something about what you were saying. Could you take a look at my matrixA
and tell me what I could change?
$endgroup$
– Jacob Myer
7 hours ago
$begingroup$
@JacobMyer look at the link below from Matlab documentation, in which the same Matlab function called rref() has been used to solve a system of equation. In linear programs the procedure is almost the same. Link: mathworks.com/help/matlab/ref/rref.html#description
$endgroup$
– Oguz Toragay
6 hours ago
add a comment
|
$begingroup$
Here is a link that includes all the information that you need. The matrix should include all the capacity limitations on all the connections between nodes. Actually, for your example, it should be a $8times8$ matrix with all the coefficients. Each row represents one of the constraints in your LP model. In other words for each row, you consider one of the nodes in the network and write a constraint which implies that:
$$textAll incoming flow = textAll outgoing flow$$
Also in this the matrix for a given simple network extracted from the linear model, which I think would be helpful for your problem.
$endgroup$
$begingroup$
Thank you that helps a lot. However I am still unsure about what to do with the values of some of the arcs because this model allows flow in both directions whereas all the ones in that link do not. For instance, would the arc with flow 15 going from 5 to 2 simply be an outflow of 5 and the arc from 2 to 5 with flow 15 be an outflow of 2? @Oguz Toragay
$endgroup$
– Jacob Myer
8 hours ago
$begingroup$
@JacobMyer yes exactly, putting it in that way, the model decided to choose one of the directions, either from 5 to 2 or 2 to 5. If I remember correctly both of the flows can not be positive simultaneously.
$endgroup$
– Oguz Toragay
8 hours ago
$begingroup$
I edited my post because I ran into a problem and must have misunderstood something about what you were saying. Could you take a look at my matrixA
and tell me what I could change?
$endgroup$
– Jacob Myer
7 hours ago
$begingroup$
@JacobMyer look at the link below from Matlab documentation, in which the same Matlab function called rref() has been used to solve a system of equation. In linear programs the procedure is almost the same. Link: mathworks.com/help/matlab/ref/rref.html#description
$endgroup$
– Oguz Toragay
6 hours ago
add a comment
|
$begingroup$
Here is a link that includes all the information that you need. The matrix should include all the capacity limitations on all the connections between nodes. Actually, for your example, it should be a $8times8$ matrix with all the coefficients. Each row represents one of the constraints in your LP model. In other words for each row, you consider one of the nodes in the network and write a constraint which implies that:
$$textAll incoming flow = textAll outgoing flow$$
Also in this the matrix for a given simple network extracted from the linear model, which I think would be helpful for your problem.
$endgroup$
Here is a link that includes all the information that you need. The matrix should include all the capacity limitations on all the connections between nodes. Actually, for your example, it should be a $8times8$ matrix with all the coefficients. Each row represents one of the constraints in your LP model. In other words for each row, you consider one of the nodes in the network and write a constraint which implies that:
$$textAll incoming flow = textAll outgoing flow$$
Also in this the matrix for a given simple network extracted from the linear model, which I think would be helpful for your problem.
edited 8 hours ago
answered 8 hours ago
Oguz ToragayOguz Toragay
4,2571 gold badge3 silver badges30 bronze badges
4,2571 gold badge3 silver badges30 bronze badges
$begingroup$
Thank you that helps a lot. However I am still unsure about what to do with the values of some of the arcs because this model allows flow in both directions whereas all the ones in that link do not. For instance, would the arc with flow 15 going from 5 to 2 simply be an outflow of 5 and the arc from 2 to 5 with flow 15 be an outflow of 2? @Oguz Toragay
$endgroup$
– Jacob Myer
8 hours ago
$begingroup$
@JacobMyer yes exactly, putting it in that way, the model decided to choose one of the directions, either from 5 to 2 or 2 to 5. If I remember correctly both of the flows can not be positive simultaneously.
$endgroup$
– Oguz Toragay
8 hours ago
$begingroup$
I edited my post because I ran into a problem and must have misunderstood something about what you were saying. Could you take a look at my matrixA
and tell me what I could change?
$endgroup$
– Jacob Myer
7 hours ago
$begingroup$
@JacobMyer look at the link below from Matlab documentation, in which the same Matlab function called rref() has been used to solve a system of equation. In linear programs the procedure is almost the same. Link: mathworks.com/help/matlab/ref/rref.html#description
$endgroup$
– Oguz Toragay
6 hours ago
add a comment
|
$begingroup$
Thank you that helps a lot. However I am still unsure about what to do with the values of some of the arcs because this model allows flow in both directions whereas all the ones in that link do not. For instance, would the arc with flow 15 going from 5 to 2 simply be an outflow of 5 and the arc from 2 to 5 with flow 15 be an outflow of 2? @Oguz Toragay
$endgroup$
– Jacob Myer
8 hours ago
$begingroup$
@JacobMyer yes exactly, putting it in that way, the model decided to choose one of the directions, either from 5 to 2 or 2 to 5. If I remember correctly both of the flows can not be positive simultaneously.
$endgroup$
– Oguz Toragay
8 hours ago
$begingroup$
I edited my post because I ran into a problem and must have misunderstood something about what you were saying. Could you take a look at my matrixA
and tell me what I could change?
$endgroup$
– Jacob Myer
7 hours ago
$begingroup$
@JacobMyer look at the link below from Matlab documentation, in which the same Matlab function called rref() has been used to solve a system of equation. In linear programs the procedure is almost the same. Link: mathworks.com/help/matlab/ref/rref.html#description
$endgroup$
– Oguz Toragay
6 hours ago
$begingroup$
Thank you that helps a lot. However I am still unsure about what to do with the values of some of the arcs because this model allows flow in both directions whereas all the ones in that link do not. For instance, would the arc with flow 15 going from 5 to 2 simply be an outflow of 5 and the arc from 2 to 5 with flow 15 be an outflow of 2? @Oguz Toragay
$endgroup$
– Jacob Myer
8 hours ago
$begingroup$
Thank you that helps a lot. However I am still unsure about what to do with the values of some of the arcs because this model allows flow in both directions whereas all the ones in that link do not. For instance, would the arc with flow 15 going from 5 to 2 simply be an outflow of 5 and the arc from 2 to 5 with flow 15 be an outflow of 2? @Oguz Toragay
$endgroup$
– Jacob Myer
8 hours ago
$begingroup$
@JacobMyer yes exactly, putting it in that way, the model decided to choose one of the directions, either from 5 to 2 or 2 to 5. If I remember correctly both of the flows can not be positive simultaneously.
$endgroup$
– Oguz Toragay
8 hours ago
$begingroup$
@JacobMyer yes exactly, putting it in that way, the model decided to choose one of the directions, either from 5 to 2 or 2 to 5. If I remember correctly both of the flows can not be positive simultaneously.
$endgroup$
– Oguz Toragay
8 hours ago
$begingroup$
I edited my post because I ran into a problem and must have misunderstood something about what you were saying. Could you take a look at my matrix
A
and tell me what I could change?$endgroup$
– Jacob Myer
7 hours ago
$begingroup$
I edited my post because I ran into a problem and must have misunderstood something about what you were saying. Could you take a look at my matrix
A
and tell me what I could change?$endgroup$
– Jacob Myer
7 hours ago
$begingroup$
@JacobMyer look at the link below from Matlab documentation, in which the same Matlab function called rref() has been used to solve a system of equation. In linear programs the procedure is almost the same. Link: mathworks.com/help/matlab/ref/rref.html#description
$endgroup$
– Oguz Toragay
6 hours ago
$begingroup$
@JacobMyer look at the link below from Matlab documentation, in which the same Matlab function called rref() has been used to solve a system of equation. In linear programs the procedure is almost the same. Link: mathworks.com/help/matlab/ref/rref.html#description
$endgroup$
– Oguz Toragay
6 hours ago
add a comment
|
Thanks for contributing an answer to Operations Research Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2for.stackexchange.com%2fquestions%2f2767%2fnetwork-flow-model-how-can-i-turn-this-diagram-into-a-matrix-that-when-convert%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown