2.13 Maximum flow with a strictly positive initial feasible flow

Transcription

1 ex-.-. Foundations of Operations Research Prof. E. Amaldi. Maximum flow and minimum cut iven the following network with capacities on the arcs find a maximum (feasible) flow from node to node, and determine a corresponding minimum (capacity) cut.. Maximum flow with node capacities In maximum flow problems, how can we deal with capacities on both nodes and arcs? Find a maximum flow from node to node in the network of the previous exercise, with a node capacity of on node.. Maximum flow with a strictly positive initial feasible flow iven the following network with capacities on the arcs find a maximum flow from node to node, starting from the feasible flow of value 0 in which 0 units are sent along the path Determine a corresponding minimum cut.. Indirect application of maximum flows A software house has to handle projects, P,P,P, over the next months. P can only begin after month, and must be completed within month. P and P can begin on month, and Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio

2 ex-.-. Foundations of Operations Research Prof. E. Amaldi must be completed, respectively, within month and. The projects require, respectively,, 0, and man-months. Each month, engineers are available. Due to the internal structure of the company, no more than engineers can work, at the same time, on the same project. Determine whether it is possible to complete the three projects within the time constraints and, if it is possible, find a feasible workforce plan. Describe how this problem can be reduced to the problem of finding a maximum flow in an appropriate graph. [Hint: Look for a feasible flow of value 0?] Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio

3 ex-.-. Foundations of Operations Research Prof. E. Amaldi Solution. Maximum flow and minimum cut We apply the Ford-Fulkerson algorithm. In the following figures, the network on the left reports the current feasible flow x in the graph, indicating on each arc the quantity of product x ij flowing through it, and the capacity k ij. The network on the right indicates the incremental graph Ḡ for the current flow. We start from the null flow x = 0 of value ϕ = 0. Since all arcs are empty, Ḡ is equivalent to. 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, We sent δ = units on path δ is given by arc (,), which has the smallest capacity k ij of the path. 0, 0, 0,, 0, 0,, 0, 0,,, We sent δ = units on path ---. δ is given by arc (,), which has the smallest residual capacity k ij of the path. We obtain a flow of value ϕ =. 0, 0,,, 0,,, 0,,,, Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio

4 ex-.-. Foundations of Operations Research Prof. E. Amaldi We sent δ = units on path ---. δ is given by arc (,). We obtain a flow of value ϕ =.,,,, 0,,, 0,,,, Se send δ = units on path δ is given by arc (,) (or (,)). We obtain a flow of value ϕ = 9.,,,, 0,,,,,,, NoaugmentingpathscanbefoundinḠandthealgorithmhalts. ThesetS = {,,,,,}, highlighted in blue, indicates the nodes that can be reached from the source in Ḡ. S induces the minimum cut corresponding to the flow x, which is highlighted in red. Correctness can be checked by observing whether the value of the flow equals the total capacity of the cut. At any optimal solution, the two values must be equivalent.. Maximum flow with node capacities. We substitute, for each node with a capacity, two auxiliary nodes, which are connected with an arc of capacity equivalent to that of the original node, as shown in the figure c i i i i The network of exercise. is modified as follows. Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio

5 ex-.-. Foundations of Operations Research Prof. E. Amaldi 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, We sent δ = units, on path , 0, 0,, 0, 0, 0,, 0, 0,, 0, We sent δ = units, on path , 0, 0, 0, 0, 0,,,,,,, We sent δ = units, on path ---.,,, 0, 0, 0,,,,,,, Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio

6 ex-.-. Foundations of Operations Research Prof. E. Amaldi No more augmenting paths can be found. The minimum cut is highlighted in red, induced by S = {,,,,,}. It has total capacity k(s ) = ϕ =.. Maximum flow with a strictly positive initial flow The original network and the incremental one, for the initial flow of value 0, are as follows 0,0 0, 0, 0,0 0, ,0 0, 0 0,0 0,0 0, 0,0 0, We sent δ = units, on path ---.,0 0,, 0,0,0 0 0,0 0, 0 0,0 0,0 0, 0,0 0, We sent δ = units, on path ---.,0,, 0,0, ,0, 0 0,0 0,0 0, 0,0 0, We sent δ = units, on path ---.,0,, 0,0,0 9 0,0 0,0 0,0,, 0,0, Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio

7 ex-.-. Foundations of Operations Research Prof. E. Amaldi Observe that all augmenting paths for this network use the backward arc (,). From the flow point of view, this amounts to unload arc (,), decreasing the amount of product flowing through it. We sent δ = 0 units, on path ----.,0,, 0,0, ,0, 0 0,0 0,0, 0,0, 0 No more augmenting paths can be found. The minimum cut is highlighted in red, induced by S = {,,,}. It has total capacity k(s ) = ϕ =.. Indirect application of maximum flows We build a network with nodes m,m,m,m associated to the four months and nodes P,P,P associated to the three projects. m P m P m P m For each pair of month-node m i and project-node P j, the arc (m i,p j ) is included in the network if man-hours of month i can be allocated to project P j. For instance, since project P can only begin after month and must be completed within month, there are only two arcs entering in P, namely (m,p ) and (m,p ). Thus we obtain the following bipartite oriented graph: m P m P m P m Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio

8 ex-.-. Foundations of Operations Research Prof. E. Amaldi To turn this graph into a network and reduce the problem under consideration to a maximum flow problem, the idea is to consider a man-month as a unit of product (flow). Then we add a source s and a sink t, and connect them to the other nodes as follows: s m m m P P P 0 t m All arcs outgoing from s have capacity, since there are engineers available. All arcs connecting month-nodes to project-nodes have capacity, as no more than engineers can work on the same project at the same time. All arcs entering in t have a capacity that is equivalent to the number of man-months needed to complete the corresponding project P j. Since all capacities are integer, the maximum flow will be integer as well. To check whether all projects can be completed within the given time limits, it suffices to check whether the network admits a feasible flow of value +0+ = 0. As a further exercise, find the maximum flow in the network and, if it has a value of 0, derive the corresponding work plan. Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio