finding optimal solution of a min network flow problem












0












$begingroup$


enter image description here



We have the following the min cost network flow problem. Notice that the arcs $(1,2), (2,3), (4,2), (3,6), (5,6)$ give a feasible basis. We easily find a feasible solution:



$$ (x_{12}, x_{23}, x_{36}, x_{42}, x_{56}) = (-6,-1,-1,-3,-1) $$



where we have assumed $b_i < 0$ is supply . When trying to find dual variables, we obtain



begin{align*}
y_1-y_2 = c_{12} = 2 \
y_2-y_3 = c_{23} = 3 \
y_4-y_2 = c_{42} = 5 \
y_3-y_6= c_{36} = 3 \
y_5-y_6 = c_{56} = 7 \
end{align*}



My question is, in solving this system, can we assume $y_1$ or $y_5$ to be $0$ since we can for a root node from them?






linear-programming







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    enter image description here



    We have the following the min cost network flow problem. Notice that the arcs $(1,2), (2,3), (4,2), (3,6), (5,6)$ give a feasible basis. We easily find a feasible solution:



    $$ (x_{12}, x_{23}, x_{36}, x_{42}, x_{56}) = (-6,-1,-1,-3,-1) $$



    where we have assumed $b_i < 0$ is supply . When trying to find dual variables, we obtain



    begin{align*}
    y_1-y_2 = c_{12} = 2 \
    y_2-y_3 = c_{23} = 3 \
    y_4-y_2 = c_{42} = 5 \
    y_3-y_6= c_{36} = 3 \
    y_5-y_6 = c_{56} = 7 \
    end{align*}



    My question is, in solving this system, can we assume $y_1$ or $y_5$ to be $0$ since we can for a root node from them?






    linear-programming







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      enter image description here



      We have the following the min cost network flow problem. Notice that the arcs $(1,2), (2,3), (4,2), (3,6), (5,6)$ give a feasible basis. We easily find a feasible solution:



      $$ (x_{12}, x_{23}, x_{36}, x_{42}, x_{56}) = (-6,-1,-1,-3,-1) $$



      where we have assumed $b_i < 0$ is supply . When trying to find dual variables, we obtain



      begin{align*}
      y_1-y_2 = c_{12} = 2 \
      y_2-y_3 = c_{23} = 3 \
      y_4-y_2 = c_{42} = 5 \
      y_3-y_6= c_{36} = 3 \
      y_5-y_6 = c_{56} = 7 \
      end{align*}



      My question is, in solving this system, can we assume $y_1$ or $y_5$ to be $0$ since we can for a root node from them?






      linear-programming







      share|cite|improve this question









      $endgroup$




      enter image description here



      We have the following the min cost network flow problem. Notice that the arcs $(1,2), (2,3), (4,2), (3,6), (5,6)$ give a feasible basis. We easily find a feasible solution:



      $$ (x_{12}, x_{23}, x_{36}, x_{42}, x_{56}) = (-6,-1,-1,-3,-1) $$



      where we have assumed $b_i < 0$ is supply . When trying to find dual variables, we obtain



      begin{align*}
      y_1-y_2 = c_{12} = 2 \
      y_2-y_3 = c_{23} = 3 \
      y_4-y_2 = c_{42} = 5 \
      y_3-y_6= c_{36} = 3 \
      y_5-y_6 = c_{56} = 7 \
      end{align*}



      My question is, in solving this system, can we assume $y_1$ or $y_5$ to be $0$ since we can for a root node from them?






      linear-programming




      linear-programming






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 3:23









      Jimmy SabaterJimmy Sabater

      2,649321




      2,649321






















          1 Answer
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          $begingroup$

          Yup, that's fine, the notion of a root node is arbitrary, just grab the node that you want it to be the root node, shake it and with the help of gravity, it becomes the root node.



          To see why we can do that in terms of the algorithm. What really matters is the difference, as only the difference of the dual variables/ potential values are used in computing the optimality conditons. We have $n$ potential values and $n-1$ arcs connecting them to form the tree, fixing a value would fully determined the rest. Fixing it differently would adjust two solutions by a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Question: Can we find another Feasible tree ? or the one I found is unique /
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 4:04










          • $begingroup$
            I could have taken path connecting node $2$ and $5$ right?
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 4:16










          • $begingroup$
            How can I see that without doing the computations?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:07










          • $begingroup$
            I let node $2$, get $9$ units, then it passes the extra $1$ unit to node $5$. Then from node $5$, we pass $3$ units to node $6$. I did it by observation. To get a tree, pass $0$ units through one of the arc avoiding forming a cycle.
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 8:10












          • $begingroup$
            but the big-M method can be used to find it as well , correct ?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:12











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Yup, that's fine, the notion of a root node is arbitrary, just grab the node that you want it to be the root node, shake it and with the help of gravity, it becomes the root node.



          To see why we can do that in terms of the algorithm. What really matters is the difference, as only the difference of the dual variables/ potential values are used in computing the optimality conditons. We have $n$ potential values and $n-1$ arcs connecting them to form the tree, fixing a value would fully determined the rest. Fixing it differently would adjust two solutions by a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Question: Can we find another Feasible tree ? or the one I found is unique /
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 4:04










          • $begingroup$
            I could have taken path connecting node $2$ and $5$ right?
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 4:16










          • $begingroup$
            How can I see that without doing the computations?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:07










          • $begingroup$
            I let node $2$, get $9$ units, then it passes the extra $1$ unit to node $5$. Then from node $5$, we pass $3$ units to node $6$. I did it by observation. To get a tree, pass $0$ units through one of the arc avoiding forming a cycle.
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 8:10












          • $begingroup$
            but the big-M method can be used to find it as well , correct ?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:12
















          1












          $begingroup$

          Yup, that's fine, the notion of a root node is arbitrary, just grab the node that you want it to be the root node, shake it and with the help of gravity, it becomes the root node.



          To see why we can do that in terms of the algorithm. What really matters is the difference, as only the difference of the dual variables/ potential values are used in computing the optimality conditons. We have $n$ potential values and $n-1$ arcs connecting them to form the tree, fixing a value would fully determined the rest. Fixing it differently would adjust two solutions by a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Question: Can we find another Feasible tree ? or the one I found is unique /
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 4:04










          • $begingroup$
            I could have taken path connecting node $2$ and $5$ right?
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 4:16










          • $begingroup$
            How can I see that without doing the computations?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:07










          • $begingroup$
            I let node $2$, get $9$ units, then it passes the extra $1$ unit to node $5$. Then from node $5$, we pass $3$ units to node $6$. I did it by observation. To get a tree, pass $0$ units through one of the arc avoiding forming a cycle.
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 8:10












          • $begingroup$
            but the big-M method can be used to find it as well , correct ?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:12














          1












          1








          1





          $begingroup$

          Yup, that's fine, the notion of a root node is arbitrary, just grab the node that you want it to be the root node, shake it and with the help of gravity, it becomes the root node.



          To see why we can do that in terms of the algorithm. What really matters is the difference, as only the difference of the dual variables/ potential values are used in computing the optimality conditons. We have $n$ potential values and $n-1$ arcs connecting them to form the tree, fixing a value would fully determined the rest. Fixing it differently would adjust two solutions by a constant.






          share|cite|improve this answer











          $endgroup$



          Yup, that's fine, the notion of a root node is arbitrary, just grab the node that you want it to be the root node, shake it and with the help of gravity, it becomes the root node.



          To see why we can do that in terms of the algorithm. What really matters is the difference, as only the difference of the dual variables/ potential values are used in computing the optimality conditons. We have $n$ potential values and $n-1$ arcs connecting them to form the tree, fixing a value would fully determined the rest. Fixing it differently would adjust two solutions by a constant.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 4:23

























          answered Dec 10 '18 at 3:47









          Siong Thye GohSiong Thye Goh

          101k1466117




          101k1466117












          • $begingroup$
            Question: Can we find another Feasible tree ? or the one I found is unique /
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 4:04










          • $begingroup$
            I could have taken path connecting node $2$ and $5$ right?
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 4:16










          • $begingroup$
            How can I see that without doing the computations?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:07










          • $begingroup$
            I let node $2$, get $9$ units, then it passes the extra $1$ unit to node $5$. Then from node $5$, we pass $3$ units to node $6$. I did it by observation. To get a tree, pass $0$ units through one of the arc avoiding forming a cycle.
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 8:10












          • $begingroup$
            but the big-M method can be used to find it as well , correct ?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:12


















          • $begingroup$
            Question: Can we find another Feasible tree ? or the one I found is unique /
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 4:04










          • $begingroup$
            I could have taken path connecting node $2$ and $5$ right?
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 4:16










          • $begingroup$
            How can I see that without doing the computations?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:07










          • $begingroup$
            I let node $2$, get $9$ units, then it passes the extra $1$ unit to node $5$. Then from node $5$, we pass $3$ units to node $6$. I did it by observation. To get a tree, pass $0$ units through one of the arc avoiding forming a cycle.
            $endgroup$
            – Siong Thye Goh
            Dec 10 '18 at 8:10












          • $begingroup$
            but the big-M method can be used to find it as well , correct ?
            $endgroup$
            – Jimmy Sabater
            Dec 10 '18 at 8:12
















          $begingroup$
          Question: Can we find another Feasible tree ? or the one I found is unique /
          $endgroup$
          – Jimmy Sabater
          Dec 10 '18 at 4:04




          $begingroup$
          Question: Can we find another Feasible tree ? or the one I found is unique /
          $endgroup$
          – Jimmy Sabater
          Dec 10 '18 at 4:04












          $begingroup$
          I could have taken path connecting node $2$ and $5$ right?
          $endgroup$
          – Siong Thye Goh
          Dec 10 '18 at 4:16




          $begingroup$
          I could have taken path connecting node $2$ and $5$ right?
          $endgroup$
          – Siong Thye Goh
          Dec 10 '18 at 4:16












          $begingroup$
          How can I see that without doing the computations?
          $endgroup$
          – Jimmy Sabater
          Dec 10 '18 at 8:07




          $begingroup$
          How can I see that without doing the computations?
          $endgroup$
          – Jimmy Sabater
          Dec 10 '18 at 8:07












          $begingroup$
          I let node $2$, get $9$ units, then it passes the extra $1$ unit to node $5$. Then from node $5$, we pass $3$ units to node $6$. I did it by observation. To get a tree, pass $0$ units through one of the arc avoiding forming a cycle.
          $endgroup$
          – Siong Thye Goh
          Dec 10 '18 at 8:10






          $begingroup$
          I let node $2$, get $9$ units, then it passes the extra $1$ unit to node $5$. Then from node $5$, we pass $3$ units to node $6$. I did it by observation. To get a tree, pass $0$ units through one of the arc avoiding forming a cycle.
          $endgroup$
          – Siong Thye Goh
          Dec 10 '18 at 8:10














          $begingroup$
          but the big-M method can be used to find it as well , correct ?
          $endgroup$
          – Jimmy Sabater
          Dec 10 '18 at 8:12




          $begingroup$
          but the big-M method can be used to find it as well , correct ?
          $endgroup$
          – Jimmy Sabater
          Dec 10 '18 at 8:12


















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