Knapsack problem
V=knapsack volume, weight limit
n- elements number
w(1),w(2),...,w(n) – values
v(1),v(2),...,v(n) – volumes, weights
I.
Heuristic (greedy procedure): value/ volume, in the increasing order of this indicator:
Element
Value
volume
V= 10.
A
75
7
B
150
8
C
250
6
Element
Value
volume
V= 5.
A
100
2
B
10
2
C
120
3
Element
Value
volume
V= 30
A
50
5
B
140
20
C
60
10
Element
A
B
C
D
Value
Volume
V= 6
2
1
3
2
3
4
4
5
Element
A
B
C
D
200
50
155
40
115
30
90
25
Value
Volume
V= 95
D
35
4
E
10
3
F
100
9
Linear programming problem:
𝑛
∑ 𝑤(𝑖) 𝑥(𝑖) → 𝑚𝑎𝑥
𝑖=1
𝑛
∑ 𝑣(𝑖) 𝑥(𝑖) ≤ 𝑉
𝑖=1
𝑥(𝑖) = 0,1
II.
K-optimal algorithm (heuristic)
A.
k=1
B.
For all the subsets of the set of elements with cardinality k, which fit into the
knapsack: we put them into the knapsack and complete the knapsack like in the
heuristics I
C.
III.
While k<n-1, k:=k+1 and go to B. Otherwise choose the best solution
Dynamic algorithm (exact)
Table with dimensions „knapsack volume +1” x „elements number +1”
1.
First row and column fill with „0”s
2.
i:=1
3.
IF i=n+1, STOP, interpret the solution
4.
Consider column i and its individual elements s, s=0,1,,,,V. In each
element s draw a horizontal arrow for column i-1, if it is not possible or not
profitable to put element i into the knapsack of capacity s, and draw an arrow
going down from column i-1, with the height equal to the volume of element i,
if it is possible and profitable element i into the knapsack of capacity s. In the
first case copy the value from the beginning of the arrow at the end of the arrow,
in the second case add to the value from the beginning of the arrow the value of
element i and put it at the end of the arrow.
5.
i:=i+1, go to 3.
Multiple knapsack problem – the same element can be available in more than one copy.
Dynamic algorithm – the same, but additionally consider arrows with heights equal
to multiplications of elements weight.
Element
Value
Volume
Numer of
copies
V= 16
A
15
7
4
B
10
5
3
C
4
2
4
Knapsack problem
1.
decision variables: x(i)= 0-1 (put – not put) for each element i=1,..,n
2.
objective function: w(1)x(1)+w(2)x(2)+...+w(n)x(n)--> max
3.
constraint: w(1)x(1)+w(2)x(2)+...+w(n)x(n)<=V