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For the following network, with edge capacities as shown, find the maximum flow from S to T, along with a matching cut.

Short Answer

Expert verified

The maximum flow is obtained with the residual capacity 13 and the matching cut is{S,C,F} and {A,B,D,E,G,T}

Step by step solution

01

Step 1:

Choose the augmenting path as S -> A -> D -> G -> T and the edge, which has the lowest capacity along the path, is A -> D andit has the capacity 4.

The following diagram represents it:

In the above diagram,

Left side is the current path and right side is the residual path.

The residual capacity is 4.

Subtract 4 from the capacity of the forward edge. Therefore, its capacity 6 now becomes 2 and the capacity of the reverse edge becomes 4.

Subtract 4 from the capacity of the forward edge. Therefore, its capacity 4 now becomes 0 and the capacity of the reverse edge becomes 4.

Subtract 4 from the capacity of the forward edge. Therefore, its capacity 5 now becomes 1 and the capacity of the reverse edge becomes 4.

Subtract 4 from the capacity of the forward edge. Therefore, its capacity 12 now becomes 8 and the capacity of the reverse edge becomes 4.

02

Step 2:

Choose the augmenting path as S -> A -> B -> E -> G -> T and the edge, which has the lowest capacity along the path, and it has the capacity 2.

The following diagram represents it:

In the above diagram,

Left side is the current path and right side is the residual path.

The residual capacity is 2.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 2 now becomes 0 and the capacity of the reverse edge becomes 6 from 4.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 2 i now becomes 0 and the capacity of the reverse edge becomes 2.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 20 now becomes 18 and the capacity of the reverse edge becomes 2.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 10 now becomes 8 and the capacity of the reverse edge becomes 2.

03

Step 3:

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 8 now becomes 6 and the capacity of the reverse edge becomes 6 from 4.

Choose the augmenting path as S -> B -> E -> G -> T and the edge, which has the lowest capacity along the path, and it has the capacity 1.

The following diagram represents it:

In the above diagram,

Left side is the current path and right side is the residual path.

The residual capacity is 1.

Subtract 1 from the capacity of the forward edge. Therefore, its capacity 1 now becomes 0 and the capacity of the reverse edge becomes 1.

Subtract 1 from the capacity of the forward edge. Therefore, its capacity 18 now becomes 17 and the capacity of the reverse edge becomes 3 from 2.

Subtract 1 from the capacity of the forward edge. Therefore, its capacity 8 now becomes 7 and the capacity of the reverse edge becomes 3 from 2.

Subtract 1 from the capacity of the forward edge. Therefore, its capacity 6 now becomes 5 and the capacity of the reverse edge becomes 7 from 6.

04

Step 4:

Choose the augmenting path as S-> C -> F -> T and the edge, which has the lowest capacity along the path, and it has the capacity 4.

The following diagram represents it:

In the above diagram,

Left side is the current path and right side is the residual path.

The residual capacity is 4.

Subtract 4 from the capacity of the forward edge. Therefore, its capacity 10 now becomes 6 and the capacity of the reverse edge becomes 4.

Subtract 4 from the capacity of the forward edge. Therefore, its capacity 5 now becomes 1 and the capacity of the reverse edge becomes 4.

Subtract 4 from the capacity of the forward edge. Therefore, its capacity 4 now becomes 0 and the capacity of the reverse edge becomes 4.

05

Step 5:

Choose the augmenting path as S -> C -> B -> E -> G -> T and the edge, which has the lowest capacity along the path, and it has the capacity 2.

The following diagram represents it:

In the above diagram,

Left side is the current path and right side is the residual path.

The residual capacity is 2.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 6 now becomes 4 and the capacity of the reverse edge becomes 6 from 4.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 2 now becomes 0 and the capacity of the reverse edge becomes 2.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 17 now becomes 15 and the capacity of the reverse edge becomes 5 from 3.

06

Step 6:

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 7 now becomes 5 and the capacity of the reverse edge becomes 5 from 3.

Subtract 2 from the capacity of the forward edge. Therefore, its capacity 5 now becomes 3 and the capacity of the reverse edge becomes 9 from 7.

The maximum flow is obtained with the residual capacity 4+2+1+4+2=13

Then cut the partition into two, let it be 鈥淢鈥 and 鈥淣鈥. Both should be a disjoint group where 鈥淪鈥 should be in 鈥淢鈥 and 鈥淭鈥 should be in 鈥淣鈥.

LetM={S,C,F}andN={A,B,D,E,G,T}

Cut 鈥淢鈥 is shown below:

Cut 鈥淣鈥 is shown below:

Therefore, the maximum flow is obtained with the residual capacity 13 and the matching cut is {S,C,F} and {A,B,D,E,G,T}

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Most popular questions from this chapter

Find necessary and sufficient conditions on the reals a and b under which the linear program

maxx+yax+by1x,y0

(a) Is infeasible.

(b) Is unbounded.

(c) Has a unique optimal solution.

Suppose someone presents you with a solution to the max-flow problem on some network. Give a linear-time algorithm to determine whether the solution does indeed give a maximum flow.

Question: A linear program for shortest path. Suppose we want to compute the shortest path from node s to node t in a directed graph with edge lengths le>0.

a) Show that this is equivalent to finding an s - tflow fthat minimizes elefesubject to size (f) = 1. There are no capacity constraints.

b) Write the shortest path problem as a linear program.

c) Show that the dual LP can be written as

role="math" localid="1659250472483" maxxs-xtxu-xvluvforall(u,v)E

d) An interpretation for the dual is given in the box on page 223. Why isn鈥檛 our dual LP identical to the one on that page?

There are many common variations of the maximum flow problem. Here are four of them.

(a) There are many sources and many sinks, and we wish to maximize the total flow from all sources to all sinks.

(b) Each vertex also has a capacity on the maximum flow that can enter it.

(c) Each edge has not only a capacity, but also a lower bound on the flow it must carry.

(d) The outgoing flow from each node u is not the same as the incoming flow, but is smaller by a factor of (1-U), whererole="math" localid="1659789093525" u is a loss coefficient associated with node u.

Each of these can be solved efficiently. Show this by reducing (a) and (b) to the original max-flow problem, and reducing (c) and (d) to linear programming.

Consider the following network (the numbers are edge capacities).

(a)Find the maximum flow fand a minimum cut.

(b)Draw the residual graphGf (along with its edge capacities). In this residual network, mark the vertices reachable fromS and the vertices from whichT is reachable.

(c)An edge of a network is called a bottleneck edge if increasing its capacity results in an increase in the maximum flow. List all bottleneck edges in the above network.

(d)Give a very simple example (containing at most four nodes) of a network which has no bottleneck edges.

(e)Give an efficient algorithm to identify all bottleneck edges in a network.

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