Flow with Lower Bounds

Feasible Flow Problem with Lower Bounds:

Find a flow v that satisfies the following conditions:

          $$\Sigma_{\{j:(i,j) \in A\}} x_{ij} - \Sigma_{\{j:(j,i) \in A\}... ... all i \in N -\{s,t\} \\ -v & for i = t\\ \end{array}\right.$$

          $l_{ij} \leq x_{ij} \leq u_{ij}$ for each $(i, j) \in A$, where $A$ is the set of arcs of the network.

To solve this problem we first add an arc $(t,s)$ of infinite capacity to the network. Then solve the following problem:

Transformed Feasible Flow Problem:

Find a flow that satisfies the following conditions:

          $\Sigma_{\{j:(i,j) \in A\}} x'_{ij} - \Sigma_{\{j:(j,i) \in A\}} x'_{ji} = b(i)$ for all $i \in N$,

          $0 \leq x'_{ij} \leq u_{ij} - l_{ij}$ for all $(i, j) \in A$,

          where $A$ is he set of arcs and $N$ is the set of vertices of the network and $b( )$ is defined as

          $b(i) = \Sigma_{\{j:(j,i) \in A\}} l_{ji} - \Sigma_{\{j:(i,j) \in A\}} l_{ij}$.

This is the same as the feasible flow problem with no lower bounds.

A feasible flow $x_{ij}$'s for the original Feasible Flow Problem with Lower Bounds is obtained from this solution $x'_{ij}$'s of Transformed Feasible Flow Problem by $x_{ij} = x'_{ij} + l_{ij}$.
Note that Transformed Feasible Flow Problem is obtained from Feasible Flow Problem with Lower Bounds by $x_{ij} = x'_{ij} + l_{ij}$.

Maximum Flow Problem with Lower Bounds:

Maximize flow v

subject to

          $$\Sigma_{\{j:(i,j) \in A\}} x_{ij} - \Sigma_{\{j:(j,i) \in A\}... ... all i \in N -\{s,t\} \\ -v & for i = t\\ \end{array}\right.$$

          $l_{ij} \leq x_{ij} \leq u_{ij}$ for each $(i, j) \in A$, where $A$ is the set of arcs of the network.

To solve this problem we first find a feasible flow through the network. Then starting with that feasible flow, we solve the maximum flow problem with the residue capacity $r_{ij} = (u_{ij} - x_{ij}) + (x_{ji} - l_{ji})$. The first term shows the remaining capacity through arc $(i, j)$ and the second term shows how much the flow from $i$ to $j$ can be increased by canceling the existing flow through arc $(j,i)$. We can use any of the known max flow algorithms to find this maximum flow.