0-1 Knapsack Problem: Optimizing Value with Constraints

The knapsack problem states that −
given a set of items, holding weights
and profit values, one must
determine the subset of the items to
be added in a knapsack such that, the
total weight of the items must not
exceed the limit of the knapsack and
its total profit value is maximum.

LIST OF
CONTENTS
Introduction
Fractional knapsack
problem using greedy
method
Steps involved to solve
the problem
Example
problem
Pseudocode

INTROduction
Given the weights and profits of N items, in the
form of {profit, weight} put these items in a
knapsack of capacity W to get the maximum total
profit in the knapsack. In Fractional Knapsack, we
can break items for maximizing the total value of
the knapsack.

MAXIMUM
PROFIT/WEIGH
T RATIO
THE THREE POSSIBLE APPROACHES
MAXIMUM
PROFIT
MINIMUM
WEIGHT

AN EFFICIENT
SOLUTION IS TO
USE THE GREEDY
APPROACH.
The basic idea of the greedy approach is to calculate
the ratio profit/weight for each item and sort the item
on the basis of this ratio. Then take the item with the
highest ratio and add them as much as we can (can be
the whole element or a fraction of it).
This will always give the maximum profit because, in
each step it adds an element such that this is the
maximum possible profit for that much weight.

• CALCULATE THE RATIO
(PROFIT/WEIGHT) FOR EACH
ITEM.
• DO THE FOLLOWING FOR EVERY ITEM I IN THE
SORTED ORDER:
1.If the weight of the current item is less than or equal
to the remaining capacity then add the value of that
item into the result
2.Else add the current item as much as we can and
break out of the loop.
Follow the given steps to solve the problem
• SORT ALL THE ITEMS IN
DECREASING ORDER OF
THE RATIO.
• INITIALIZE RES = 0,
CURR_CAP = GIVEN_CAP.
• RETURN RES.
STEPS

GreedyKnapsack(m, n):
//P[1..n]: Profits of items
//W[1..n]: Weights of items
//x[1..n]: Fraction of items to be taken
//m: Capacity of the knapsack
Sort items such that P[i]/W[i] is in decreasing order
x :=0
u := m
p := 0
for i := 1 to n do
if W[i] <= u then
x[i] := 1
u := u - W[i]
p := p + P[i]
else
x[i] := u / W[i]
p := p + P[i] * x[i]
break
return x, p
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0-1 Knapsack Problem: Optimizing Value with Constraints

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