Question

6. Show the trace of execution of the algorithm of Ford and Fulkerson on the following graph, using A and Z as source and sink. Show the minimum cut on this graph, and compute its capacity. 21 3 В E H 5 20 2 12 1.5 8 A N D G 23 24 1.2 2 18 1.4 13 с F 1.5 3

Answer 1

The execution of Ford Fulkerson algorithm for the given graph is as follows:

0/21 0/3 3/21 B/ 3 B E H B E H 0/5 3/5 0/ 20 3 / 20 0/ 2 0/12 0/ 2 0/12 0/ 1.5 0/ 1.5 0/8 0/8 A A Z D G D G 0/ 23 0/ 23 0,24 0/ 1.2 0/2 0/18 0/24 0/ 1.2 0/2 0/18 0/1.4 0/13 0/1.4 0/13 C F - C F - 0/ 1.5 0/ 1.5 0/3 0/3 Minimum Capacity = 0 Path: A-B-E-H-Z Minimum Capacity = 3 5/21 3/ 3 3/3 E 5/21 B H E B 515 H 515 5/ 20 6.5/ 20 2/2 2/12 0/ 1.5 2/2 3.5/12 0/8 1.5/ 1.5 А 1.5/8 Z А. D G Z G o/ 23 1.5/ 23 0,24 0/ 1.2 0/2 0/18 0/ 24 0/ 1.2 0/2 0/18 0/1.4 0/13 0/1.4 0/13 C F 1 C F 0/ 1.5 0/ 1.5 0/3 0/3 Path: A-B-D-H-G-Z Minimum Capacity = 1.5 Path: A-B-E-D-H-Z Minimum Capacity = 2 5/21 3/3 521 3/ 3 B E H B E H 515 515 6.5/ 20 6.5/ 20 2/ 2 3.5/12 2/2 3.5/12 1.5/ 1.5 1.5/ 1.5 1.5/8 1.5/8 A Z A D G Z D G 2.9/ 23 1.5/ 23 2.9/24 0/ 1.2 0/2 0/18 1.5/24 0/ 1.2 0/2 0/18 1.4/1.4 1.5/13 0/1.4 1.5/13 - C F C F - 1.5/ 1.5 1.5/1.5 2.9/ 3 1.5/3 Path: A-C-F-G-Z Minimum Capacity = 1.4 Path: A-C-F-1-Z Minimum Capacity = 1.5 5/21 313 5/21 3/3 B E H B E H 515 515 6.5/ 20 6.5/ 20 2/2 3.5/12 2/ 2 3.5/12 1.5/ 1.5 1.5/8 1.5/ 1.5 1.5/8 A Z А => D G Z D G 4.9/ 23 4.9/ 23 4.9/24 0/ 1.2 2/2 0/18 6.1/24 1.2/1.2 2/2 1.2/18 1.4/1.4 1.5/13 2.7/ 13 1.4/1.4 с F - C F 1 1.5/1.5 1.5/1.5 2.9/3 2.9/3 Path: A-C-G-Z Minimum Capacity = 2 Path: A-C-D-I-Z Minimum Capacity = 1.2

All the set of vertices from reachable to non-reachable vertex are called minimum cut.

Minimum cut on the graph:

21 3 E I B 5 20 12 1.5 8 A Z D G 23 24 1.2 2 18 1.4 13 с F 1.5 3

The Capacity of given graph = 3 + 2 +1.5 + 1.5 + 1.4 + 2 + 1.2 = 12.6