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Network flow model - How can I turn this diagram into a matrix that when converted to RREF solves for max flow?

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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 $X = maxx_1,x_2$













4














$begingroup$


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?



enter image description here



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
```









share|improve this question












$endgroup$




















    4














    $begingroup$


    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?



    enter image description here



    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
    ```









    share|improve this question












    $endgroup$


















      4












      4








      4





      $begingroup$


      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?



      enter image description here



      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
      ```









      share|improve this question












      $endgroup$




      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?



      enter image description here



      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






      share|improve this question
















      share|improve this question













      share|improve this question




      share|improve this question








      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























          1 Answer
          1






          active

          oldest

          votes


















          4
















          $begingroup$

          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.






          share|improve this answer












          $endgroup$














          • $begingroup$
            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 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












          Your Answer








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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4
















          $begingroup$

          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.






          share|improve this answer












          $endgroup$














          • $begingroup$
            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 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















          4
















          $begingroup$

          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.






          share|improve this answer












          $endgroup$














          • $begingroup$
            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 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













          4














          4










          4







          $begingroup$

          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.






          share|improve this answer












          $endgroup$



          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.







          share|improve this answer















          share|improve this answer




          share|improve this answer








          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 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
















          • $begingroup$
            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 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















          $begingroup$
          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


















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