There are basically Branch and bound the 0/1 knapsack for a least cost Branch and Bound , FIFO Branch and Bound

1. NADAR SARASWATHI COLLEGE OF ARTS&SCIENCE,THENI
DEPARTMENT OF COMPUTER
SCIENCE&INFORMATION TECHNOLOGY
V.VANMATHY
I-MSC(CS)

2. DATA STRUCTURE
ALGORITHMS
0/1 knapsack least cost
branch and bound&FIFO
branch and bound
TOPIC:

3. 0/1 knapsack least cost branch and bound
Given N items with weight W[0…..n-1],values
V[0…..n-1] and a knapsack with capacity C select the
items such that:
1. The sum of weight taken into the knapsack I less
than or equal to c.
2. The sum of values of the items in the knapsack I
maximum among all the possible combinations

4. 0/1 knapsack least cost branch and bound
Input: N=4,C=15,V[]={10,10,12,18},W[]={2,4,6,9}
Output:
Items taken into the knapsack are
1101
Maximum profit is 38
Explanation:
1 in the output indicates that the item is included in
the knapsack while 0 indicates that the item I
excluded.

5. 0/1 knapsack least cost branch and bound
Since the maximum possible cost all owed is 15, the
ways to select items are:
(1101)->cost=2+4+9=15,profit=10+10+18=38
(0011)->cost=6+9=15,profit=12+18=30
(1110)->cost=2+4+6=12,profit=32
Hence,maximum profit possible wit hin a cost of 15
is 38.

6. 0/1 knapsack least cost branch and bound
Input: N=4,C=21,V[]={18,20,14,18},w[]={6,3,5,9}
Output:
Items taken into the knapsack are
1101
Maximum profit is 56
Explanation:
Cost=6+3+9=18
Profit=18+20+18=56

7. Least cost
Upper= node:1 profit: 10 10 12 18
M=15 weight:2 4 6 9
c=-38u=-32 c=10+10+12+18/9*3
u=10+10+12 2 + 4 + 6
u=-32
upper value find the value remove 12
the value 18/9*3 find the answer 15-12=3
u=-32 c=10+10+12+2*3
c=-38
1
c

8. Least cost
18/9*5 remove value c=10+10+12+18/9*5
find the value u answer 2+ 4+6
u=-32
x=1 c=-38 10
u=-32 x=0 15-10=5
c=-38 c=-32 u=-22 c=10+12+10
c=-32
u=10+12
node: 2,3 u=-22
1
2 3

9. Least cost
c=10+10+12+18/9*7
u=-32 2+ 4+6
c=-38
u=-32 x=1 x=0 8
c=-38 c=-32 u=-22 15-8=7
x=0 x=1 c=10+12+14
u=-32 u=-22 c=-36
c=-38 c=-36 u=10+12
u=-22
Node:4,5
1
2 3
4 5

10. Least cost
c=10+10+12+18
u=-32 2+ 4+6 +9
x=1 c=-38
u=-32 x=0
c=-38 c=-32 u=-22
x=1 x=0 c=10+10+18
u=-32 u=-22 c=-38
c=-38 c=-36 u=10+10+18
x=1 x=0 u=-38
Node:7 u=-38
c=-38
1
2 3
4 5
6
7
k

11. Least cost
u=-32 lower bound>upper bound
x=1 x=0 c=-38 -20>-38
u=-32 u=-22 kill value node 9
c=-38 c=-32
x=1 x=0
u=-32 u=-22
x=1 x=0c=-38 c=-36 c=10+10
u=-32 u=-38 2+4
c=-38 c=-38 c=20
x=1 x=0
u=-38 node: 9
c=-38 k c=-20
1
3
2
4 5
6 7
8 9
k
X={1,1,0,1}
10+10+0+18=38
2+4+0+9=15

12. FIFO branch and bound solution
Draw the portion of the state space tree generated by the
FIFO BB technique for the knapsack instance of n=5
m=12
(p1,p2,p3,p4,p5)=(10,1,6,8,4)
(w1,w2,w3,w4,w5)=(4,6,3,4,2)
0/1 knapsack problem is a maximization problem but we
require FIFO BB knapsack problem, I a minimization
problem.
Hence,we convert the +ve profits into –ve profits.
Hence, the problem is converted from maximization
problem to minimization problem.

13. FIFO EXAMPLE

14. FIFO

15. FIFO

16. FIFO

17. FIFO

18. FIFO

19. FIFO

20. FIFO

21. FIFO

22. FIFO

23. FIFO

24. FIFO

25. FIFO