The 0/1 Knapsack Problem Solved
by the Branch-and-Bound Strategy
The 0/1 knapsack problem
P1, P2, …, Pn (profit)
W1,W2, …,Wn (weight)
M (capacity)
 Positive integers
n
Maximize  Pi X i
i 1
n
subject to  Wi X i  M Xi = 0 or 1, i =1, …, n.
i 1
The problem is modified:
n
Minimize   Pi X i
i 1
5-2
 e.g. n = 6, M = 34
i
1
2
3
4
5
6
Pi
6
10
4
5
6
4
Wi
10
19
8
10
12
8
(Pi/Wi  Pi+1/Wi+1)
 A feasible solution:
X1 = 1, X2 = 1, X3 = 0, X4 = 0,
X5 = 0, X6 = 0
-(P1+P2) = -16 (upper bound)
Any solution higher than -16 can not be an optimal solution.
5-3
Relax the Restriction
 Relax our restriction from Xi = 0 or 1 to 0  Xi  1 (knapsack
problem)
n
Let   Pi X i be an optimal solution for 0/1
i 1
n
knapsack problem and   Pi X i be an optimal
i 1
solution for knapsack problem.
n
Let Y=   Pi X i , Y’ =
i 1
 Y’  Y
5-4
n
  Pi X i .
i 1
Upper Bound and Lower Bound
 We can use the greedy method to find an optimal solution for
knapsack problem:
X1 = 1, X2 =1, X3 = 5/8, X4 = 0, X5 = 0, X6 =0
-(P1+P2+5/8P3) = -18.5 (lower bound)
-18 is our lower bound. (only consider integers)
 -18  optimal solution  -16
optimal solution: X1 = 1, X2 = 0, X3 = 0, X4 = 1, X5 = 1, X6 = 0
-(P1+P4+P5) = -17
5-5
U.B.=-16
L.B.=-18
0
X1=1
1
U.B.=-16
L.B.=-18
X3=1
5
X2=0
U.B.=-16
L.B.=-18
infeasible
X5=1
9
X2=1
U.B.=-16
L.B.=-18
6
13
8
U.B.=-16
L.B.=-18
10
U.B.=-15
L.B.=-18
22
X5=0
U.B.=-16
L.B.=-18
X5=1
27
28
U.B.=-15
L.B.=-18
11
X6=0
infeasible
infeasible
12
X6=1
U.B.=-16
L.B.=-16
29
30
infeasible
U.B.=-16
L.B.=-18
24
X5=1
23
X6=1
U.B.=-14
L.B.=-14
17
X4=0
16
X6=1
U.B.=-16
L.B.=-16
19
U.B.=-16
L.B.=-16
U.B.=-17 X =0
5
L.B.=-18
18
26
infeasible
U.B.=-17
L.B.=-18
X5=1
X6=0
25
15
U.B.=-16
L.B.=-18
U.B.=-15
L.B.=-15
U.B.=-17
L.B.=-18
X4=1
X5=0
X6=0
X6=1
14
X4=0
U.B.=-15
L.B.=-18
21
X5=0
infeasible
X3=0
X4=1
X4=0
infeasible
U.B.=-15
L.B.=-18
4
X3=0
X4=1
7
U.B.=-14
L.B.=-17
2
X2=1
3
X1=0
infeasible
U.B.=-15
L.B.=-15
X6=0
20
U.B.=-17
L.B.=-17
A Job Scheduling Problem Solved by
the Branch-and Boubd Approach
Job Scheduling
Job Scheduling
A
B
F
G
C
E
D
H
I
K
L
J
T1 T2 T3 T4 T5 T6 T7 T8 T9
3 2 2 2 4 5 3 2 3
M
N
P
O
A Time Profile
 T1 T2 T3 T4 T5 T6 T7 T8 T9
3 2 2 2 4 5 3 2 3
This time profile indicates that at time t=1,only three precessors
can be used
Solution
Part of a Solution Tree
Rule1：
Common Successors Effect
A
C
D
I
E
P
J
(C,D)分享同一個子結點，(E,P)也是相同的狀況
在可能解當中，(C,E)與(D,P)是相同的狀況
有相同子代和父代的連線，互換不受影響
Rule2
Internal Node First Strategy
第三種答案
選擇(B,E,P)展開之後會發現(C,D)(F,J)可以使用
Rule3：Maximizing the Number of
Processed Job
Rule4：Accumulated Idle Processors
Strategy
Accumulated Processors Effect
The Experimental Result
Conclude