Many practical optimization models contain some network structures. In virtually all cases, these networks are modeled as **directed graphs**. I.e., between nodes \(i\) and \(j\), we consider arcs \(i \rightarrow j\) and \(j \rightarrow i\). In graph theory, it is very customary to talk about **undirected graphs**. In this case, there is just one link between nodes \(i\) and \(j\). Why am I using directed graphs, with two uni-directional arcs for each pair of nodes? I will try to answer this in this post.

As an example, let's consider the shortest path problem. This is really the same as a min-cost problem. I took the following data (from [1]):

---- 36 SET i nodes node0, node1, node2, node3, node4, node5, node6, node7, node8 ---- 36 SET u undirected arcs node1 node2 node3 node4 node5 node6 node7 node8 node0 YES YES node1 YES YES node2 YES YES YES node3 YES YES node4 YES node5 YES node6 YES YES node7 YES ---- 36 PARAMETER ucost undirected cost node1 node2 node3 node4 node5 node6 node7 node8 node0 4.000 8.000 node1 8.000 11.000 node2 7.000 4.000 2.000 node3 9.000 14.000 node4 10.000 node5 2.000 node6 1.000 6.000 node7 7.000 ---- 36 SET a directed arcs node0 node1 node2 node3 node4 node5 node6 node7 node8 node0 YES YES node1 YES YES YES node2 YES YES YES YES node3 YES YES YES node4 YES YES node5 YES YES YES YES node6 YES YES YES node7 YES YES YES YES node8 YES YES YES ---- 36 PARAMETER acost directed cost node0 node1 node2 node3 node4 node5 node6 node7 node8 node0 4.000 8.000 node1 4.000 8.000 11.000 node2 8.000 7.000 4.000 2.000 node3 7.000 9.000 14.000 node4 9.000 10.000 node5 4.000 14.000 10.000 2.000 node6 2.000 1.000 6.000 node7 8.000 11.000 1.000 7.000 node8 2.000 6.000 7.000

From this, it looks like that the undirected version is preferable: half of the arcs and half of the data. As verification:

---- 91 PARAMETER count number of arcs undirected 14.000, directed 28.000

In addition, we need to know the source node, and the sink node. Here we model this as a sparse vector, with an exogenous supply of 1 (for the source node) and demand of 1 (for the sink node).

---- 43 PARAMETER ex exogenous supply/demand node0 1.000, node4 -1.000

With this, a standard shortest path model with directed arcs can be formulated as:

Shortest Path Model: Directed Arcs |
---|

\[\begin{align}\min\>&\sum_{(i,j)\in \color{darkblue}A} \mathit{\color{darkblue}{acost}}_{i,j} \cdot \color{darkred}f_{i,j}\\ & \sum_{(i,k)\in \color{darkblue}A} \color{darkred}f_{i,k} + \mathit{\color{darkblue}{ex}}_{k} = \sum_{(k,j)\in \color{darkblue}A} \color{darkred}f_{k,j} && \forall k\\ &\color{darkred}f_{i,j} \in [0,1] \end{align}\] |

The constraint says: the inflow for node \(k\) is equal to the outflow. The flows are positive, and we have an arc capacity of 1 (no need to ship more around). The results are:

---- 62 VARIABLE f.L flow node4 node5 node6 node7 node0 1.000 node5 1.000 node6 1.000 node7 1.000 ---- 62 VARIABLE z.L = 21.000 total cost (objective)

As expected, we see integer flows, even though we solved the problem as an LP. Pure networks yield integer solutions automatically.

So the question is how do we handle this with an undirected network. The first thing is that we need to allow for negative flows. E.g. a unit flow from node 5 to 4, will be seen as \(\color{darkred}f_{4,5}=-1\). I.e. our capacity restriction for all links with \((i,j)\in \color{darkblue}U\) becomes \(|\color{darkred}f_{i,j}| \le 1\) or \(\color{darkred}f_{i,j} \in [-1,1]\).

The flow conservation constraint does not change much, although the interpretation is a bit different. On the left-hand side, we no longer have pure inflow. We may see a negative flow \(\color{darkred}f_{i,k} \lt 0\). We could call it **net inflow** from nodes \(i\) with \((i,k)\in \color{darkblue}U\). The right-hand side has net-outflow to nodes \(j\) with \((k,j)\in \color{darkblue}U\) (again, we may see negative terms \(\color{darkred}f_{k,j} \lt 0\)).

The objective is more problematic. We can expect flows to be \(-1\), \(0\) or \(+1\). This means the objective becomes \[\min\sum_{(i,j)\in \color{darkblue}U} \mathit{\color{darkblue}{ucost}}_{i,j} \cdot |\color{darkred}f_{i,j}|\] This can be linearized, which leads to:

Shortest Path Model: Undirected Arcs |
---|

\[\begin{align}\min\>&\sum_{(i,j)\in \color{darkblue}U} \mathit{\color{darkblue}{ucost}}_{i,j} \cdot \mathit{\color{darkred}{fabs}}_{i,j} \\ & \color{darkred}f_{i,j} \le \mathit{\color{darkred}{fabs}}_{i,j} && \forall (i,j) \in \color{darkblue}U\\ &\color{darkred}f_{i,j} \ge -\mathit{\color{darkred}{fabs}}_{i,j} && \forall (i,j) \in \color{darkblue}U \\ & \sum_{(i,k)\in \color{darkblue}U} \color{darkred}f_{i,k} + \mathit{\color{darkblue}{ex}}_{k} = \sum_{(k,j)\in \color{darkblue}U} \color{darkred}f_{k,j} && \forall k\\ &\color{darkred}f_{i,j} \in [-1,1] \\ &\mathit{\color{darkred}{fabs}}_{i,j} \in [0,1] \end{align}\] |

The results look like:

---- 85 VARIABLE f.L flow node5 node6 node7 node0 1.000 node4 -1.000 node5 -1.000 node6 -1.000 ---- 85 VARIABLE z.L = 21.000 total cost (objective)

Clearly, in this model, we have lost the advantage of the undirected graph being smaller. The linearization of the absolute values makes this a rather unattractive approach.

#### Conclusion

There is a good reason why I often use directed arcs in my math programming models: they make things easier. The savings you might expect from using undirected arcs may prove to be illusory.

## No comments:

## Post a Comment