/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q10E For the following network, with ... [FREE SOLUTION] | 91影视

91影视

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}

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An edge of a flow network is called critical if decreasing the capacity of this edge results in a decrease in the maximum flow. Give an efficient algorithm that finds a critical edge in a network

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.

Hall鈥檚 theorem. Returning to the matchmaking scenario of Section 7.3, suppose we have a bipartite graph with boys on the left and an equal number of girls on the right. Hall鈥檚 theorem says that there is a perfect matching if and only if the following condition holds: any subset sof boys is connected to at least |s|girls.

Prove this theorem. (Hint: The max-flow min-cut theorem should be helpful.)

Consider the following linear program.

maximize 5x+3y

5x-2y0x+y7x5x0y0

Plot the feasible region and identify the optimal solution.

Question: Duckwheat is produced in Kansas and Mexico and consumed in New York and California. Kansas produces 15 shnupells of duckwheat and Mexico 8. Meanwhile, New York consumes 10 shnupells and California 13. The transportation costs per shnupell are \(4 from Mexico to New York, \)1 from Mexico to California, \(2 from Kansas to New York, and \)3 and from Kansas to California. Write a linear program that decides the amounts of duckwheat (in shnupells and fractions of a shnupell) to be transported from each producer to each consumer, so as to minimize the overall transportation cost

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.