Lecture Notes on Branch and Bound in Algorithms, Slides of Design and Analysis of Algorithms

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Branch and Bound Paradigm

• Aims at exploring the Tree corresponding to the

problem statement

• Recall : In backtracking feasible function was used

for pruning the search !!

• In addition to pruning following are additional

requirement:

1. For each node a bound (lower or upper ) is

determined on the best possible value of objective

function that can be obtained by exploring the node

further

2. Keep track of a value of best possible solution

obtained so far

Example: Job Assignment Problem

• N people and N Jobs are available

• People are associated with jobs and cost of

association is given (Matrix)

• Aim: To find an assignment which will minimize

the cost.

A

B

C

D

J1 J2 J3 J

Consider the

Assignment Matrix:

• The problem is that of minimization of cost hence, the

lowest bound for the cost can be found by considering

lowest value in each row and adding them i.e. 4 + 3 + 2 + 7

= 16. Optimal solution without solving the problem !!!

• This bound itself may not give the desired solution as

the problem constraints may be violated!!

• However, it is a lower bound on any legitimate solution
• The same process is applied to partially constructed

solutions

• Lower bound obtained when A is assigned

jobs 2 and B is assigned jobs 1,3 and 4

A→ 2

Lb= 4+3+2+9=

B→ 4

Lb= 4+3+2+9=

B→ 3

Lb= 4+6+5+9=

B→ 1

Lb= 4+7+2+14=

Note: Node 8 has a lower bound of 18 (obtained after

assignment of job 2 to A and job 4 to B) which is minimum

among all assignments and hence node 8 is further

expanded.

• Lower bound obtained when A is assigned

jobs 2, B is assigned job 4 and C is assigned

jobs 1and 3:

B→ 4

Lb= 4+3+2+9=

C→ 3

D→ 1

C→ 1

D→ 3

Lb= 4+3+8+10=25 Lb= 4+3+2+9=

//Implied

Note: Value of Lower bound at node 10 is less than that at

node 9. Moreover Lower bounds at nodes 2 , 4 , 5 , 6 and 7

are all greater than 18. Hence, Lower bound is 18 and

assignments are: A→ 2 ; B→ 4 ; C→ 3 and D→ 1

0/1 Knapsack problem

• N objects with profit are given

• Aim: To select subset of objects such that the

total wieght does not exceed M, the capacity

of knapsack, and profit earned is maximum

• Objects are considered according to

decreasing P/W ratios to maximize profit

• Strategy: Construct a binary tree

corresponding to selection and non selection

of objects one by one

• The upper bound, ub, is obtained by adding the

product of remaining capacity of knapsack and

best p/w value among the remaining objects to p,

i.e

ub = p + (M-w)*(best P/W ratio among the

remaining objects)

• Example: n=4; M=11;

(p1,p2,p3,p4) = (50,49,36,20);

(w1,w2,w3,w4) = (5,7,6,4)

Solve the 0/1 knapsack problem using

Branch and Bound

• N=4; M=10; (p1,p2,p3,p4)=(40,49,25,9)

(w1,w2,w3,w4)=(4,7,5,3)

Traveling Salesman Problem

• Well known problem that has been solved using

dynamic Programming

• If B&B is to be used, all possibilities have to be

investigated using State Space Tree

• Given a Weighted Graph, starting at vertex 1, we

have to investigate all possible paths to find the

least cost path starting at vertex 1 and ending at

it

• More importantly, we need to compute a lower

bound for each node!!

Example: Solve TSP for the given

Graph using Branch and Bound

• Root of Tree corresponds to vertex 1 and the lower bound

is:

• Lb = ½[(1+4)+(1+3)+(4+2)+(3+2)+(4+6)]

= ½[5+4+6+5+10] = ciel (15) = 15

• Lb for node 2 of tree:

Lb = ceil{1/2[(1+4)+(1+3)+(2+4)+(2+3)+(4+6)]}

= ceil{ 1/2[5+4+6+5+10]}= 15

• Lb for node 3 of the tree:

Lb = ceil{1/2[(1+5)+(1+3)+(2+4)+(2+5)+(4+6)]}

= ceil {1/2[6+4+6+7+10]} = ceil {16.5} = 17

• Similarly, Lb for node 4 = 15

V

Lb = 15

V1, V

Lb = ??

V1, V

Lb = ??

V1, V

Lb = ??

Solve the TSP using B&B