Embed Size (px)
Citation preview
1
Branch and bound
2
Branch and Bound Method
• The design technique known as branch and bound is similar to backtracking in that it searches a tree model of the solution space and is applicable to a wide variety of discrete combinatorial problems.
• Backtracking algorithms try to find one or all configurations modeled as N-tuples, which satisfy certain properties.
• Branch and bound are more oriented towards optimization.
3
Here all the children of the E- node are generated before any other live node can become E- node.
Here two state space trees can be formed BFS (FIFO)and D search (LIFO).
4
1
2 34 5
X1=1X1=4
6 7 8
X2=2 X2=4
9 10 11
12 13
X3=3
16
1415
X4=4 BFS SEARCH for sum of subset problem
X3=4
X1=2
X2=3
Variable size tuple
5
Variable size tuple 1
2 1014 16
X1=1X1=4
3 7 9
X2=2 X2=4
11 13 15
4 6
X3=3
5
812
X4=4 DFS SEARCH for sum of subset problem
X3=4
X1=2
X2=3
6
Variable size tuple 1
2 34 5
X1=1X1=4
10 11 12
X2=2 X2=4
7 8 6
14 15
X3=3
16
139
X4=4 D- SEARCH for sum of subset problem
X3=4
X1=2
X2=3
7
Fixed size tuple
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 of subset problem
8
Traveling salesman problem
• The salesman problem is to find a least cost tour of 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 of time, money, or distance) of direct travel between city I and city j.
• Assume cii=infinity for all i. • Also cij= infinity if it is not possible to move
directly from city I to city j.
9
Branch and bound algorithms for traveling salesman problem can be formulated in a variety of ways.
Without loss of generality we can assume that every tour starts and ends at city one.
So the solution space S is given by
{ 1 , π , 1 | π is a permutation of (2,3,4… n)}
|S|= (n-1)!.
10
1
2 3 4
5 6
1112
7 8
13 14
9 10
15 16
A State space tree for traveling salesman problem with n= 4
I1=2I1=3
I1=4
I2=3I2=4
I3=4
I2=3
I3=2
Tour 1 2 3 4 1 1 2 4 3 1
11
all tours
------{3,5} {3,5}
-------{2,1} {2,1}
A branch and bound state space tree for traveling salesman problem
12
What is meant by bounding?
With each vertex in the tree we associate a lower bound on the cost of any tour in the set represented by the vertex.
The computation of these lower bounds is major labor saving device in any branch and bound algorithm.
There fore much thought should be given to obtain tight bounds.
13
Assume that we have constructed a specific complete tour with cost = m.
If the lower bound associated with the set of tours represented by a vertex v is M.
And
M>= m
Then no need to search further for descendants of v for the optimum tour.
14
Basic steps for the computation of lower bounds
The basic step in the computation of lower bound is known as 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)}.
15
{(1,5),(5,1),(2,3),( 3,4),(4,2)}.
16
• 1 5
• 2 3
» 4
17
Row Reduction
2- If a constant h is subtracted from every entry in any row or column of C , the cost of any tour under the new matrix C’ is exactly h less than the cost of the same tour under matrix C. This subtraction is called a row (column) reduction
18
3- By a reduction of the entire cost matrix C we mean the following: Sequentially go down the rows of C and subtract the value of each row’s smallest element hi from every element in the row. Then do the same for each column.
Let h = ∑ hi summation over all rows and columns
The resulting cost matrix will be called the reduction of C.
h is a lower bound on the cost of any tour.
19
Let A be the reduced cost matrix for a node R. Let S be a child of R such that edge (R,S) corresponds to including edge (i,j) in the tour.
1- change all entries in row i and column j of A to .(so that no edge from this row (column)leaving from I(coming to j), may be included in the tour in future).
2- set A(j,1) = ∞
This prevents A(j,1) since node 1 should be the last node of the tour.
20
4-Reduce all rows and columns in the resulting matrix except for rows and columns containing only ∞
.
Let the resulting matrix be B.
Let r be the total amount subtracted then lower bound on S is
lower bound for (R) + A(i,j) + r
21
example
∞ 20 30 10 11
15 ∞ 16 4 2
3 5 ∞ 2 4
19 6 18 ∞ 3
16 4 7 16 ∞
∞ 10 17 0 1
12 ∞ 11 2 0
0 3 ∞ 0 2
15 3 12 ∞ 0
11 0 0 12 ∞
Reduced cost matrix lower bound = 25(subtracting from rows 10,2,2,3,4) and 1,3 from column 1 and 3.
So all tours in the given graph have length at least 25.
Cost matrix Reduced Cost Matrix
22
∞ 20 30 10 11
15 ∞ 16 4 2
3 5 ∞ 2 4
19 6 18 ∞ 3
16 4 7 16 ∞
∞ 10 20 00 1
15 ∞ 16 4 2
3 5 ∞ 2 4
19 6 18 ∞ 3
16 4 7 16 ∞
2 3 1 4 5 2 h1 h2 h3 h4 h5
C1 C2 C3 C4 C5
23
52
1 25
2 3 4535 53 25 31
6 7 8
9 10
11
5036
I1=2
I1=3I1=4
I1=5
I2=2 i2= 3
28
52 28
28
I3=3I3=5
I4=3
24
∞ 10 17 0 1
12 ∞ 11 2 0
0 3 ∞ 0 2
15 3 12 ∞ 0
11 0 0 12 ∞
Reduced Cost Matrix
∞ ∞ ∞ ∞ ∞
∞ ∞ 11 2 0
0 ∞ ∞ 0 215 ∞ 12 ∞ 011 ∞ 0 12 ∞
Path (1,2) node 2(25 + 10=35)
25
∞ ∞ ∞ ∞ ∞
∞ ∞ 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 (25+17+11=53
=
path(1,5) node 5
∞ ∞ ∞ ∞ ∞12 ∞ 11 ∞ 0
0 3 ∞ ∞ 2
∞ 3 12 ∞ 011 0 0 ∞ ∞ Path 1,4
node 4
26
∞ ∞ ∞ ∞ ∞
12 ∞ 9 0 ∞
0 3 ∞ 0 ∞
12 0 9 ∞ ∞
∞ 0 0 12 ∞
∞ 10 17 0 1
12 ∞ 11 2 0
0 3 ∞ 0 2
15 3 12 ∞ 0
11 0 0 12 ∞
Reduced Cost Matrixpath(1,5) node 5
25 + +2 + 3=31
27
∞ 10 17 0 1
12 ∞ 11 2 0
0 3 ∞ 0 2
15 3 12 ∞ 0
11 0 0 12 ∞
Reduced Cost Matrix
∞ ∞ ∞ ∞ ∞
1 ∞ ∞ 2 0
∞ 3 ∞ 0 2
4 3 ∞ ∞ 0
0 0 ∞ 12 ∞
path(1,3) node 3 (25+17+11=53
28
∞ 10 17 0 1
12 ∞ 11 2 0
0 3 ∞ 0 2
15 3 12 ∞ 0
11 0 0 12 ∞
∞ ∞ ∞ ∞ ∞
12 ∞ 11 ∞ 0
0 3 ∞ ∞ 2
∞ 3 12 ∞ 0
11 0 0 ∞ ∞
Path 1,4 node 4Reduced Cost Matrix
29
∞ ∞ ∞ ∞ ∞
12 ∞ 11 ∞ 0
0 3 ∞ ∞ 2
∞ 3 12 ∞ 0
11 0 0 ∞ ∞
∞ ∞ ∞ ∞ ∞
∞ ∞ 11 ∞ 0
0 ∞ ∞ ∞ 2
∞ ∞ ∞ ∞ ∞
11 ∞ 0 ∞ ∞
Path 1-4-2 node 6
Bound 28
Path 1,4 node 4 bound 25
30
∞ ∞ ∞ ∞ ∞
∞ ∞ 11 ∞ 0
0 ∞ ∞ ∞ 2
∞ ∞ ∞ ∞ ∞
11 ∞ 0 ∞ ∞
∞ ∞ ∞ ∞ ∞
∞ ∞ ∞ ∞ ∞
∞ ∞ ∞ ∞ 0
∞ ∞ ∞ ∞ ∞
0 ∞ ∞ ∞ ∞
Path 1-4-2-3 node 9
Bound 52
Path 1-4-2 node 6
Bound 28
31
∞ ∞ ∞ ∞ ∞
∞ ∞ 11 ∞ 0
0 ∞ ∞ ∞ 2
∞ ∞ ∞ ∞ ∞11 ∞ 0 ∞ ∞
∞ ∞ ∞ ∞ ∞
1 ∞ ∞ ∞ 0
∞ 1 ∞ ∞ 0
∞ ∞ ∞ ∞ ∞
0 0 ∞ ∞ ∞
∞ ∞ ∞ ∞ ∞
∞ ∞ ∞ ∞ ∞
∞ ∞ ∞ ∞ 0
∞ ∞ ∞ ∞ ∞
0 ∞ ∞ ∞ ∞
Path 1-4-2 npde 6Path 1 4 3 : node 7
Path 1-4-2-3 node 9
∞ ∞ ∞ ∞ ∞
∞ ∞ ∞ ∞ ∞
0 ∞ ∞ ∞ ∞
∞ ∞ ∞ ∞ ∞
∞ ∞ 0 ∞ ∞
Path 1-4-2-5 node 10
32
Least cost (LC) Search
In both LIFO and FIFO branch and bound the selection rule for the next E- node is rather rigid and in a sense blind.The selection rule for the next E- node does not give any preference to a node that has a very good chance of getting the search to an answer node quickly.
The search for an answer node can often be speeded by using an “ intelligent ranking C*(.) for live nodes.The next E- node is selected on the basis of this ranking function.
33
Let g^ (x) be an estimate of the additional effort needed to reach an answer node from x.
H(x) is the cost of reaching x from the root.F(.) is any non decreasing function.Node x is assigned a rank using c^(.) such thatC^(x)= f(h(x)) + g^(x)Using f(.)=0 usually biases the search algorithm to
make deep probes into the search tree.Note: BFS and d-search are special cases of LC
search.If g^(x)= 0 and f(h(x)) = level of node x then a LC
search generates nodes by level.Which is BFS
34
Assignment Problem
35
There are n people who need to be assigned to execute n jobs, one person per job. (i.e. each person is assigned to exactly one job and each job is assigned to exactly one person).
C(i, j) is the cost if the i th person is assigned j th job
for each pair i, j =1,2…nThe problem is to find an assignment
with the smallest total cost.
36
• Hungarian method is much more efficient for this problem.
37
Lower bound
There are many ways to find a lower bound.
We can relax the condition on person,
i.e. one Person may be assigned more than one job
Or
We can relax he condition on jobs
More than one person may be assigned to a job
38
problem
Job1 Job2 Job3 Job4 persons
9 2 7 8 A
6 4 3 7 B
5 8 1 8 C
7 6 9 4 D
39
Person A to job 1Cut first row and first column
then try to assign the remaining persons to cheapest jobs lower bound 9+3+1+4=17
Job1 Job2 Job3 Job4 persons
9 2 7 8 A
6 4 3 7 B
5 8 1 8 C
7 6 9 4 D
40
Start
Lb=10
a 2
Lb=10
a1
Lb=17
a 3
Lb=20
a 4
Lb =18
b4
Lb=17
b 3
Lb=14
b1
Lb=13
c4
d 3
Lb=25
c3
d 4
Lb=13
For a 2 delete 2nd col ,1st row find min of each row and total of it
41
Lower bound when A-2 and B-1Lower bound 2+6+1+4=13
Job1 Job2 Job3 Job4 persons
9 2 7 8 A
6 4 3 7 B
5 8 1 8 C
7 6 9 4 D
42
Lower bound=2+3+1+4=10
Job1 Job2 Job3 Job4 persons
9 2 7 8 A
6 4 3 7 B
5 8 1 8 C
7 6 9 4 D
43
Knapsack Problem
Item
Wt Val Val/ wt
1 4 40 10
2 7 42 6
3 5 25 5
4 3 12 4
Knapsack ’s capacity is 10
44
• A simple way to compute upper bound is
= v + (W – w) (vi+1/wi+1)
Where v = total value of items already added in bag
w = wt of items already selected
i+1 is the best per unit payoff among the remaining items
45
W=0,v=0
Up=100
W=4,v=40
Up=76
W=0,v=0
Up=60
W /o 1With 1
W=4, v=40
Up=70W=11
Not feas
w/ o 2 With 2
W=4, v=40
Up=64
W=9, v=65
Up=69
W= 12
Not feas
W=9, v=65
Up=65
Sate space tree
w/ o 3 With 3
Item
Wt
Val
Val/ wt
1 4 40 10
2 7 42 6
3 5 25 5
4 3 12 4
46
Traveling salesman problem
a
e
c d
b
9
4
1
3
7
3 9
56
8
2
47
Lower Bound
for each city I find the sum si of the distances from city I to the two nearest cities; compute the sum s of these n numbers; divide the result by two and round up if all distances are integers.
Lb = [(1 +3)+ (3 +6) + (1 +2) + (3 +4) + (2+3)]/2 = 14
48
aLb=14
a, eLb=19
a, dLb=16
a, cX
a, bLb=14
a, b, eLb=19
a, b, dLb=16
a, b, cLb=16
a, b, d,e(c, a)
Lb= 16
a, b, d, c(e, a)Lb=24
a, b, c ,e(d, a)Lb=19
a, b, c, d(e,a)
Lb= 24
First tour Better tour Inferior tour Optimal tour
X
X X
