Last Updated :
21 Dec, 2018
Consider the following instances of knap-sack problem:
Item |
Weight(in Kgs) |
Value(in $) |
1 |
3 |
10 |
2 |
2 |
25 |
3 |
2 |
20 |
4 |
5 |
15 |
5 |
7 |
40 |
6 |
4 |
36 |
Maximum weight of knapsack is 10 Kg. What will be the maximum profit with the optimal solution of the 0/1 knapsack problem?
(A) 101
(B) 96
(C) 92.42
(D) 81
Answer: (D)
Explanation: Knapsack Instance with value / weight:
Item |
Weight |
Value |
Value / weight |
1 |
3 |
10 |
3.33 |
2 |
2 |
25 |
12.5 |
3 |
2 |
20 |
10 |
4 |
5 |
15 |
3 |
5 |
7 |
40 |
5.71 |
6 |
4 |
36 |
9 |
Decrersing order of value/weight:
Item |
Weight |
Value |
Value / weight |
2 |
2 |
25 |
12.5 |
3 |
2 |
20 |
10 |
6 |
4 |
36 |
9 |
5 |
7 |
40 |
5.71 |
1 |
3 |
10 |
3.33 |
4 |
5 |
15 |
3 |
Now start filling the sack-
item 2 fully:
left weight = 10 – 2 = 8
item 3 fully:
left weight = 8 – 2 = 6
item 6 fully:
left weight = 6 – 4 = 2
So, only P2, P3, and P6 are chosen. Now no other item can be filled in because it is a 0/1 knap sack.
Total profit = 25 + 20 + 36 = 81.
So, option (D) is correct.
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