08) # Branch and Bound

8/3/2019 08) # Branch and Bound

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Branch and bound


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Branch and Bound Method

The design technique known as branch andbound is similar to backtracking in that itsearches a tree model of the solution space andis applicable to a wide variety of discretecombinatorial problems.

Backtracking algorithms try to find one or allconfigurations modeled as N-tuples, whichsatisfy certain properties.

Branch and bound are more oriented towardsoptimization.


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Here all the children of the E- node aregenerated before any other live node can

become E- node.Here two state space trees can be formed

BFS (FIFO)and D search (LIFO).


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1

2 34 5

X1=1

X1=4

6 7 8

X2=2X2=4

9 10

11

12 13

X3=3

16

1415

X4=4 BFSSEARCH forsum ofsubsetproblem

X3=4

X1=2

X2=3


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1

2 3

X1=0X1=1

4 5

6 7

8 910 11

12 13

14 1516 17

18 19

20 21

22 23

Nodes are generated in D -search manner for sum ofsubset problem


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Traveling salesman problem

The salesman problem is to find a least cost tourof N cities in his sales region.

The tour is to visit each city exactly once. Salesman has a cost matrix C where the

element cij equals the cost (usually in terms oftime, money, or distance) of direct travel

between city I and city j. Assume cii=infinity forall i. Also cij= infinity if it is not possible to movedirectly from city I to city j.


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Branch and bound algorithms for travelingsalesman problem can be formulated in a

variety of ways.Without loss of generality we can assumethat every tour starts and ends at city one.

So the solution space S is given by{ 1 , , 1 | is a permutation of (2,3,4 n)}

|S|= (n-1)!.


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1

2 3 4

5 6

1112

7 8

13 14

9 10

15 16

A Sate space tree for traveling salesman problem with n= 4

I1=2

I1=3

I1=4

I2=3I2=4

I3=4

I2=3

I3=2

Tour 1 2 3 4 1 1 2 4 3 1


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all tours

------

{3,5} {3,5}

-------

{2,1} {2,1}

A branch and bound state space tree for traveling salesmanproblem


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What is meant by bounding?

With each vertex in the tree we associate alower bound on the cost of any tour in theset represented by the vertex.

The computation of these lower bounds ismajor labor saving device in any branchand bound algorithm.

There fore much thought should be given to

obtain tight bounds.


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Assume that we have constructed a specificcomplete tour with cost m.

If the lower bound associated with the set oftours represented by a vertex v is M.

And

M>= m

Then no need to search further fordescendants of v for the optimum tour.


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Basic steps for the computation of

lower bounds

The basic step in the computation of lower bound is knownas reduction. It is based on following observations:

1- In the cost matrix C every full tour contains exactly one

element from each row and each column.Note: converse need not be true e.g {(1,5),(5,1),(2,3),(

3,4),(4,2)}.

2- If a constant h is subtracted from every entry in any rowor column of C , the cost of any tour under the newmatrix C is exactly h less than the cost of the same tourunder matrix C.This subtraction is called a row(column)reduction


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3- By a reduction of the entire cost matrix Cwe mean the following: Sequentially go

down the rows of C and subtract the valueof each rows smallest element hi fromevery element in the row. Then do thesame for each column.

Let h = hi summation over all rows andcolumns

The resulting cost matrix will be called the

reduction of C.h is a lower bound on the cost of any

tour.


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Let A be the reduced cost matrix for a nodeR. Let S be a child of R such that edge

(R,S) corresponds to including edge (i,j) inthe tour.

1- change all entries in row i and column j ofA to .(so that no edge from this row

(column)leaving from I(coming to j), maybe included in the tour in future).

2- set A(j,1) =

This prevents A(j,1) since node 1 should bethe last node of the tour.


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4-Reduce all rows and columns in theresulting matrix except for rows and

columns containing only .

Let the resulting matrix be B.

Let r be the total amount subtracted thenlower bound on S is

lower bound for (R) + A(i,j) + r


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example 20 30 10 1115 16 2 2

3 5 2 4

19 6 18 316 4 7 16

10 17 0 112 11 2 0

0 3 0 2

15 3 12 011 0 0 12

Reduced cost matrix lower bound =

25(subtracting 10,2,2,3,4) and 1,3 fromcolumn 1 and 3

So all tours in the given graph havelength atleast 25.

Cost matrix


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1 25

2 3 4535 53 25 31

6 78

9 10

11

28

50

36

5228

28

I1=2

I1=3

I1=5

I2=2i2

= 3


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1 25

2 3 4535 53 25 31

6 78

9 10

11

50

36

I1=2

I1=3

I1=5

I2=2i2

= 3


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11 2 0

0 0 215 12 011 0 12

1 2 0

3 0 2

4 3 0

0 0 12

12

9 0 0 3 0 12 0 9

0 0

12

Path (1,2) node 2 (25 + 10=35) path(1,3) node 3

path(1,5) node 5

12

11 00 3 2

3 12 011 0 0 Path 1,4

node 4


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Least cost (LC) Search

In both LIFO and FIFO branch and bound theselection rule for the next E- node is rather rigidand in a sense blind.The selection rule for the

next E- node does not give any preference to anode that has a very good chance of getting thesearch to an answer node quickly.

The search foe an answer node can often be

speeded by using an intelligent ranking C*(.)for live nodes.The next E- node is selected onthe basis of this ranking function.


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08) # Branch and Bound