Maximum Flow Problem (Ford and Fulkerson, 1956)

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1 Maximum Flow Problem (Ford and Fulkerson, 196) In this problem we find the maximum flow possible in a directed connected network with arc capacities. There is unlimited quantity available in the given source and we wish to maximize the quantity that can be sent to the sink (destination) through the capacitated arcs. Let there be n nodes and m arcs in the network. The formulation of the maximum flow problem is given by Maximize f X ij, f 0. The problem is a Linear Programming problem and can be solved using the Simplex algorithm. It is possible to reduce the number of variables and constraints by substituting for f. This can be used when we solve the maximum flow problem as an LP. This problem usually is solved using flow augmenting path algorithms whose optimality is based on the dual of the given formulation. Hence the given formulation is always used. The above formulation is called node-arc formulation. There is also a path-arc formulation where the flows on all possible paths are defined. In this formulation, X j is the flow in path j. The capacity constraints on the arcs limit the flow. This formulation requires that we enumerate all possible paths, which is difficult as the problem size increases. Illustration 7.10 Consider a network with 6 nodes and 10 arcs given in Figure 7.. The arc capacities are given in brackets: 1-2 (20), 1- (60), 2- (0), 2- (0), 2- (10), - (), -6 (10), - (0), -6 (2) and -6 (0). Find the maximum flow between nodes 1 and 6? Solve using the node arc and path arc formulations?

2 Figure 7. Network for the maximum flow problem The node-arc formulation for the given instance is to Maximize f X 12 + X 1 = f -X 12 + X 2 + X 2 + X 2 = 0 -X 1 -X 2 + X + X 6 = 0 -X 2 -X + X + X 6 = 0 X 6 + X 6 + X 6 = f X X 1 60 X 2 0 X 2 0 X 2 10 X X 6 10 X 0 X 6 2 X 6 0 X ij 0. The optimum solution to the above LP is given by X 12 = 20, X 1 =, X 2 = 20, X =, X 6 = 10, X = 0, X 6 = 2, X 6 = 0 and f = 6. The maximum slow possible is 6 units. Arcs 1-2, -, -6, -, and -6 are full and have been used to maximum capacity. Other arcs have unfilled or remaining capacity. It is to be noted that X ij and f need not be restricted to integer and hence the given problem is a LP.

3 We first enumerate all the paths in the given network. These are , , , , , , , 1---6, There are nine paths and we denote the flow in path j as X j. The objective is to Maximize X 1 + X 2 + X + X + X + X 6 + X 7 + X 8 + X 9 X 1 + X 2 + X + X + X + X 6 20 X 7 + X 8 + X 9 60 X 1 + X 2 + X 0 X + X 6 0 X 10 X 1 + X 2 +X 7 + X 8 X + X 9 10 X 1 + X + X 7 0 X 2 + X 6 + X 8 2 X 1 + X + X X 7 0 X j 0. The objective function maximizes the total flow which is the sum of the flows in each path. There are as many constraints as the number of arcs and the constraints ensure that arc capacities are not violated. The optimum solution is given by X 8 = 1, X 7 = 20, X 9 = 10, X 6 = 10 and X = 10 with total flow = 6. Algorithms considering flows in various paths The path arc formulation requires that we enumerate all the paths and include as many variables as the number of paths. We consider an algorithm where we progressively consider each path and allot the maximum possible flow to the path which is the minimum of the available capacities in the arcs. This algorithm is called the flow augmenting path algorithm. We also consider diverting the flows so as to permit additional flow when a new path is introduced. The flow augmenting path algorithm when implemented by considering the paths in the non decreasing order of arcs in them is called the Shortest augmenting path algorithm (Ahuja et al, 199). This algorithm is optimum and also does not require us to divert the flow when new paths are considered. We explain all the algorithms using an example. Illustration Consider a network with 6 nodes and 10 arcs given in Figure 7.. The arc capacities are given in brackets: 1-2 (20), 1- (60), 2- (0), 2- (0), 2- (10), - (), -6 (10), - (0), -6 (2)

4 and -6 (0). Find the maximum flow between nodes 1 and 6? Solve using the flow augmenting path algorithm and the shortest augmenting path algorithm? The nine paths that we listed are , , , , , , , 1---6, We consider the path The maximum possible flow in this path is 20, which is the capacity of arc 1-2. We allow a flow of 20 and update the flow in the arcs. 1-2 (20), 2- (20), - (20), - (20), -6 (20) and f = We cannot increase the flow using paths 2 to 6 because they involve the arc 1-2 whose capacity is fully used.. We consider the path The available capacities are 1- (60), - (1), - (10), -6 (0). We increase the flow by 10 and f = 0. The flows in the various arcs are 1-2 (20), 2- (20), - (0), - (0), -6 (0), 1- (10).. We consider the path The available capacities are 1- (0), - (), -6 (2). The maximum possible flow is which is the minimum of the capacities. We increase the flow b y and f =. The updated arc flows are 1-2 (20), 2- (20), - (), - (0), -6 (0), 1- (1), -6 ().. We consider the path The available capacities are 1- () and -6 (10). The maximum possible increase is 10 and f =. The updates arc flows are 1-2 (20), 2- (20), - (), - (0), -6 (0), 1- (2), -6 (), - (1) 6. We have exhausted all the paths and the maximum flow possible at this point is. We observe that the maximum flow using the flow augmenting path algorithm is at the moment while the optimum is 6. We now have to divert flows in existing paths and effectively consider additional paths (which we could not consider earlier). 1. We consider the path This appears as an invalid path because the network does not contain an arc -2. We allow this as a valid path when there is existing flow from 2- and there are arc capacities in the other arcs. The available capacities in the arcs are 1- (), -2 (20), which is the flow of 20 from 2-, 2- (10), -6 (20). The additional flow is 10 which is the minimum of the possible flows and f =. We update the flows in the arcs by adding 10 to all arcs. The updated flows in the arcs are 1-2 (20), 2- (10), - (), - (0), -6 (0), 1- (), -6 (), - (1), 2- (10). We observe that the flow in 2- has reduced because we have added 10 in the opposite direction (-2) which is the same as subtracting 10 from the existing flow. This was possible because we considered the opposite arc -2 only when the actual arc 2- has a positive flow. We also observe that in the process we have effectively considered the path (which we could not consider earlier) by diverting flow using

5 2. We next consider the path This also has an arc -2 because the actual arc 2- has a positive flow. The available capacities are 1- (2), -2 (10 which is the flow in 2-), 2- (0), -6 (2). The minimum possible addition is 10 and f = 6. The updated flows are 1-2 (20), - (), - (0), -6 (0), 1- (), -6 (1), - (1), 2- (10), 2- (10). Now there is no flow in 2- (since a flow of 10 has been subtracted).. We now observe that we cannot divert any more flow and conclude that we have reached the optimum. The flow augmenting path algorithm is optimal when we also consider diverting existing flows to accommodate further paths and flows. We apply the shortest augmenting path algorithm for our example. The nine paths are arranged in increasing (or non decreasing) order of the number of arcs in the path. The order is 1--6, , , , 1---6, , , , Consider The available capacities are 1- (60) and -6 (10). The maximum flow possible is 10 and f = 10. We update the flows as 1- (10) and -6 (10). 2. We cannot use since the capacity of -6 is used fully. Consider The available capacities are 1-2 (20), 2- (0), -6 (2). The flow can be increased by 20 and f = 0. The updated flows are 1- (10), 1-2 (20), 2- (20), -6 (20) and -6 (10).. We cannot use because the capacity of 1-2 is used fully. Consider The available capacities are 1- (0), - () and -6 (). The maximum flow permissible is and f =. The updated flows are 1- (1), 1-2 (20), 2- (20), -6 (2), -6 (10), - ().. We cannot use because the capacities of 1-2 and -6 are used fully. We cannot use because the capacity of 1-2 is used fully.. We consider The available capacities are 1- (), - (0), - (0), -6 (0). The maximum permissible flow is 0 and f = 6. The updated flows are 1- (), 1-2 (20), 2- (20), -6 (2), -6 (10), - (), - (0), -6 (0). 6. We cannot increase the flow by considering because capacity of 1-2 is utilized fully. The algorithm terminates giving a maximum flow of 6. The shortest augmenting path algorithm gives the optimum solution. There is no need to consider paths with negative arcs that divert flows. The flow augmenting path algorithm effectively used the same paths that the shortest augmenting path algorithm used when we considered diverting the flows. Because we considered path that diverted flows, we considered more paths while applying the flow augmenting path algorithm. Maximum flows and Minimum cuts (Ford and Fulkerson 1962) A cut or cut set is a set of arcs which when removed from the network, permits no flow from the source and the sink. It also divides the nodes into two sets where the source and sink

6 are in different sets and each of the remaining nodes is in one of the sets. For example, the arcs that leave the source node represent a cut set that has the source in one set and all other vertices in the other set. The capacity of the arcs in the cut set gives an upper bound (or upper estimate) of the maximum flow possible. An important theorem that relates the maximum flow and the cut set is the max-flow min cut theorem that states that the maximum flow is equal to the capacity of the minimum cut set (Ford and Fulkerson, 1962). Illustration Consider the network with 6 nodes and 10 arcs given in Figure 7.. The arc capacities are given in brackets: 1-2 (20), 1- (60), 2- (0), 2- (0), 2- (10), - (), -6 (10), - (0), - 6 (2) and -6 (0). Find the minimal cut set from the optimum solution given by 1- (), 1-2 (20), 2- (20), -6 (2), -6 (10), - (), - (0), -6 (0). From the optimum solution we observe that arcs 1-2, - and -6 with capacities 20, and 10 are consumed fully and removing them gives the total capacity of 6 and separated the network into two parts with no flow possible. The arcs 1-2, - and -6 form the minimum cut set that divide the nodes into two sets {1,} and {2,,, 6}. Minimum cost flow problem (Ahuja et al, 199) In the minimum cost flow problem or transhipment problem, we minimize the total cost of transporting a single item from multiple sources to multiple destinations. This is similar to the transportation problem but with two differences. In this problem some nodes act as source nodes, some nodes act as destination nodes and some act as intermediate nodes (or transit points). In the transportation problem a given node can be only a source or destination. In the minimum cost flow problem, items can be also transported from one source node to another and from one destination node to another. Arcs exist amongst the source nodes and amongst the destination nodes. The transportation problem is bipartite and there are no arcs between source nodes or destination nodes. The formulation of the minimum cost flow problem is as follows: Let X ij be the quantity transported between nodes i and j. Let C ij be the unit transportion cost between nodes i and j. Let b j be the supply at node j (demand is shown as negative supply and transit as zero supply). The objective is to Minimize X ij 0.

7 There can be capacity constraints of the form X ij u ij. In this case the problem is called the capacitated minimum cost flow problem. This problem can also be modelled as a transportation problem with each node becoming a potential source as well as destination. Solving this as a transportation problem is advantageous when we use special algorithms for the transportation problem. Otherwise the LP formulation can be used. Illustration 7.1 Consider the network with 6 nodes and 10 arcs given in Figure 7.6. The node supplies are b 1 = 60, b 2 = 0, b = b = 0, b = -0 and b 6 = -80. The network is given below where the unit transportation costs are shown as arc weights. The arc capacities are given in brackets: 1-2 (0), 1- (60), 2- (0), 2- (0), 2- (10), - (), -6 (10), - (0), -6 (0) and -6 (0). b=0 2 9 b= b= b= b=-80 b=0 Figure 7.6 Network for minimum cost flow problem The formulation is to Minimize 10X X 1 + X 2 + 6X 2 + 9X 2 + 7X + 1X 6 + X + 6X 6 + X 6 X 12 + X 1 = 60 -X 12 + X 2 + X 2 + X 2 = 0 -X 1 X 2 + X + X 6 = 0 -X 2 X + X + X 6 = -0 -X 2 X + X 6 = 0 -X 6 - X 6 - X 6 = -80 X ij 0. The optimum solution to this LP is given by X 1 = 60, X 2 = 0, X = 60, X 6 = 80 with Z = 160.

8 We include the arc capacities in the form of constraints X 12 0; X 1 60; X 2 0; X 2 0; X 2 10; X ; X 6 10; X 0; X 6 0; X 6 0. The optimum solution is X 1 = 60, X 2 = 0, X 2 = 10, X = 0, X 6 = 10, X = 10, X 6 = 0, X 6 = 20 with Z = The capacitated problem understandably has a higher cost. Circulatory flow problem (Bazaara et al, 1990) In this problem, the flow in each arc is constrained by a lower bound as well as an upper bound. The network is such that there is circulatory flow. We wish to minimize the total cost associate with the circulatory flow. The formulation is the same as that of the minimum cost flow problem except that the demand and supply are zero at all nodes and there is an additional lower bound on the flow. We use the same notation for the minimum cost flow problem. The formulation is to Minimize u ij X ij l ij Illustration 7.1 Consider the network with nodes and 8 arcs given in Figure 7.7. The values of C ij, l ij and u ij for the arcs are given in brackets: 1-2 (, 1, 8), -1(, 1, 10), 2- (6,, 12), 2- (, 0, ), 2- (2, 0, 10), - (7, 1, 6), - (8, 2, 12), - (10, 2, 9). 2 2,0,10,1, 1 6,,12,0, 7,1,6 10,2,9,1,10 8,2,12 Figure 7.7 Network for circulatory flow problem The formulation of the circulatory flow problem is to Minimize X X 2 + X 2 + 2X 2 + X 1 + 7X + 10X + 8X

9 subject to X 12 - X 1 = 0 -X 12 + X 2 + X 2 + X 2 = 0 -X 2 - X - X + X 1 = 0 -X 2 + X + X = 0 -X 2 - X + X = 0 X 12 8; X 2 12; X 2 ; X 2 10; X 1 10; X 6; X 9; X 12; X 12 1; X 2 ; X 2 0; X 2 0; X 1 1; X 1; X 2; X 2. The optimum solution to the LP is given by X 12 = 6, X 2 =, X 2 =, X 1 = 6, X = 1, X = 2and X = 2 with Z = 12. Multi commodity Flows (Hu, 196) The multi commodity flow problem addresses the flow of different types of commodities on a given network. We assume that there are K different commodities. Let C ijk represent the unit cost of transporting commodity k in arc i-j. Let u ijk be the capacity of arc i-j for commodity k. Let b lk be the supply of commodity k in node l. The demands are shown as negative supplies. Let U ij be the maximum capacity of arc i-j for all commodities put together. The problem is to Minimize X ijk 0. The above formulation is an LP formulation. Illustration 7. 1 Consider the network with nodes and 8 arcs given in Figure 7.8. There are two commodities. The values of C ij1, C ij2, u ij1, u ij2 and U ij for the arcs i-j are shown in the figure. The five values are given in brackets for arc i-j as follows: 1-2 (,,,2,), 1- (7,,,6,), 2- (6,1,,2,6), 2- (,8,,2,), 2- (6,,,,6), - (2,,,,) - (,,,,) and - (1,1,,1,6). Six units of commodity 1 is available in node 1 and the demands are and in nodes and. Four units of commodity 2 is available in node 2and one unit in node 1 and the demands are 1 and in nodes and. Solve the multi commodity flow formulation?

10 b2= 2 6,,,,6 b1=6 b2=1,,,2,,8,,2, 6,1,,2,6 1 7,,,6, 2,,,, 1,1,,,6 b1=-,,,, b1=- b2=- b2=-1 Figure 7.8 Network for multi commodity flow problem Let X ij1 and X ij2 represent the flow of the two commodities in arc i-j. There are 16 variables. The multi commodity flow formulation is to Minimize X X X 11 + X X 21 + X 22 + X X X 21 + X X 1 + X 2 + X 1 + X 2 + X 1 + X 2 X X 11 = 6 -X X 21 + X 21 + X 21 = 0 -X 11 X 21 + X 1 + X 1 = 0 -X 21 X 1 + X 1 = - -X 2 X X = - X X 12 = 1 -X X 22 + X 22 + X 22 = -X 12 X 22 + X 2 + X 2 = -1 -X 22 X 2 + X 2 = 0 -X 22 X 2 X 2 = - X 121 ; X 122 2: X 11 ; X 12 6; X 21 ; X 22 2; X 21 ; X 22 2; X 21 ; X 22 ; X 1 ; X 2 ; X 1 ; X 2 ; X 1 ; X 2 X X 122 ; X 11 + X 12 ; X 21 + X 22 6; X 21 + X 22 ; X 21 + X 22 6; X 1 + X 2 ; X 1 + X 2 ; X 1 + X 2 6 X ijk 0. The optimum solution to the LP is given by X 121 =, X 122 = 1, X 11 =, X 22 = 2, X 21 =, X 22 =, X 1 =, X 2 = 1 with cost = 76. This solution is given in Figure 7.9

11 b2= 2 X 121 =, X 122 = 1 X 21 = X 22 = b1=6 b2=1 1, X 22 = 2 b1=- b1=- b2=- X 11 = X 1 =, X 2 = 1 b2=-1 Figure 7.9 Solution to multi commodity flow problem Minimum Spanning Trees (MST) Consider a network with n nodes and m arcs. We assume that the network is connected, which means that it is possible to go from every node to every other node on the network by traversing across arcs. Arc j has a weight C j. Any network that has n nodes and exactly n-1 arcs and is connected is called a tree. Given a network with n nodes and m arcs (m > n) such that it is connected, it is possible to create a sub network (whose node set and arc sets are that of the given network and from the given network) that is connected and has all the n nodes and exactly n-1 arcs. Such a tree is called a spanning tree of the given network. The spanning tree with minimum sum of arc weights is called the minimum spanning tree. The binary integer programming formulation of the minimum spanning tree problem is a follows: Let X j = 1 if arc is in the MST; = 0 otherwise. Let V be the set of nodes. The objective is to Minimize subject to

12 ( ) X j = 0, 1 The formulation has a large number of constraints as we will see in the example. This happens because we have to evaluate all the subsets of the node set V. Illustration 7.16 Consider the network given in Figure 7.10 with nodes and eight arcs. The arc weights are given in the network. Find the length of the minimum spanning tree using the formulation? Figure 7.10 Network for Illustration 7.16 The mathematical programming formulation is as follows: Let X j = 1 if arc j is in the MST; = 0 otherwise. We have to label the arcs. This is done as follows: Arc 1-2 is denoted by variable X 1 ; 1- by X 2, 2- by X, 2- by X, 2- by X, - by X 6, - by X 7 and - by X 8. Minimize X 1 + X 2 + X + X + 6X + 8X 6 + X X 8 X 1 + X 2 + X + X + X + X 6 + X 7 + X 8 = X 1 1; X 2 1; X 1; X 1; X 1; X 6 1; X 7 1; X 8 1; X 1 + X 2 + X 2; X 1 + X 2; X 1 + X 2; X 2 + X 6 2; X 2 + X 7 2; X + X + X 6 2; X + X 7 2; X + X 8 2; X 6 + X 7 + X 8 2 X 1 + X 2 + X + X + X 6 ; X 1 + X 2 + X + X, X 1 + X + X + X 8 ; X 2 + X 6 + X 7 + X 8 ; X + X + X + X 6 + X 7 + X 8 X j = 0,1 The second set of constraints are written considering the node sets {1,2}, {1,}, {2,}, {2,}, {2,}, {,}, {,}, {,}, {1,2,}, {1,2,}, {1,2,}, {1,,}, {1,,}, {1,,}, {2,,}, {2,,}, {2,,},

13 {,,}, {1,2,,}, {1,2,,}, {1,2,,}, {1,,,} and {2,,,}. The node set {1,, } gives us only one arc X 8 and there is already a constraint X 8 1 for the node set {,}. The constraint X 8 2 for the node set {1,,} is left out. The optimum solution to the formulation is given by X 1 = X = X = X 7 = 1 with distance = 17. It is also observed that the LP relaxation where all X j 0 also gives us the same solution with X j taking values zero or 1. Algorithms to find the MST The MST formulation is interesting because it has a large number of constraints. If there are n nodes, we have + 1 constraints. In spite of this large number of constraints (that increase exponentially with n) the LP relaxation gives solutions with binary values. Two very popular algorithms exist that find the MST. These are the Prim s algorithm and Kruskal s algorithm: Prim s algorithm (Prim, 197) 1. Find the arc k-l with the smallest weight. Arc i-j is in the MST. Create node set S 1 = {k, l} and node set S 2 = {rest of the nodes} 2. For every node i in S 1 and every node j in S 2 find the arc i-j such that d ij is minimum. Add node j to S 1 and delete j from S 2.. Repeat step 2 such that S 2 = { } or when exactly n-1 arcs have been considered. These arcs form the MST. Kruskal s algorithm (Kruskal, 196) 1. Sort the arcs according to increasing (non decreasing) order of weights. 2. The first arc k-l is in the MST. Create set S 1 = {k,l}.. Consider the next arc k-l in the sorted list.. If inclusion of arc k-l will create a cycle, go to Step.. If one of the nodes is in S 1 add the other node to S 1. If both are not in S 1 add both to S 1. Include the arc into the MST. If both nodes are in S 1 go to step. 6. Repeat steps to till exactly n-1 arcs have been added. These arcs form the MST. Illustration 7.17 Consider the network given in Figure 7.10 with nodes and eight arcs. The arc weights are given in the network. Find the length of the minimum spanning tree using Prim s and Kruskal s algorithms? Prim s algorithm: Arc - has minimum weight. S 1 = {, } and S 2 = {1,2,}. Consider d 1, d 2, d, d 2, d. The minimum distance is for 1- (also for 2- but we consider one of them). S 1 = {1,, } and S 2 = {2, }. Consider d 12, d 2, d 2, d 2, d, d. The minimum distance is for 1-2.

14 S 1 = {1, 2,, } and S 2 = {}. Consider d 2, d and d. The minimum is d 2. Now S 1 = {1, 2,,, }. The algorithm terminates with arcs -, 1-, 1-2 and 2- in the MST. The reader may observe that -, 1-2, 2- and 2- is also an MST with length = 17. Kruskal s algorithm: The sorted list is -, 1-2, 1-, 2-, 2-, 2-, -, -. Consider -. S 1 = {, }. Consider 1-2 and update S 1 = {1, 2,, }. 1- is considered and added because it does not create a cycle. 2- is not considered because adding this arc would create a cycle is added and we have added arcs. The MST has arcs -, 1-2, 1- and 2- with length = 17. Degree constrained MST (Narula and Ho, 1980) A degree constrained MST places a restriction on the number of arcs incident from a vertex. There is an additional constraint restricting the number of arcs in the minimum spanning tree from every vertex to be lesser than or equal to a given constant. This is of the form for all nodes. The formulation can be solved as a binary IP to get the optimum solution. The LP relaxation may result in fractional values for the variables in some instances. The reader may observe that the minimum cost (distance) travelling salesman path problem between a given start node and a given end node can be modelled as a degree constrained minimum spanning tree where the start and end nodes have a degree of 1 and every other node has a degree of 2. Illustration 7.18 Consider the network given in Figure 7.11 with nodes and eight arcs. The arc weights are given in the network. Solve the degree constrained MST problem where the degree of each vertex is not more than 2? Figure 7.11 Network for Illustration 7.18

15 It is customary to first solve the MST using Prim s or Kruskal s algorithm. The solution obtained using Kruskal s algorithm has arcs -, 1-2, 2- and 2- in the MST with length = 17. In the X j notation, we observe that X 1, X, X and X 7 are in the solution. The degree of each vertex represents the number of arcs in the spanning tree that are incident to the vertex. We compute d(1) = 1; d(2) =, d() = 2, d() = 1 and d() = 1. This solution violates the restriction because node 2 has degree of. We add the following constraints X 1 + X 2 2; X 1 + X + X + X 2; X + X 6 + X 7 2; X + X + X 6 2 and X + X 7 + X 8 2. The optimum solution is X 1 = X 2 = X = X 7 = 1 with length = 17. This satisfies the degree constraints. This solution is the other optimum solution to the MST (which also satisfies the degree constraint). If we were to solve the shortest distance between 1 and passing through every other node once and only once the path is with length = 17.

Discrete Optimization 23

Discrete Optimization 23 2 Total Unimodularity (TU) and Its Applications In this section we will discuss the total unimodularity theory and its applications to flows in networks. 2.1 Total Unimodularity: