KNAPSACK ALGORITHM FULL

Similarly, we will load the objects in decreasing p/w ratio and we will get the following results-: The maximum profit/weight ratio is of the fourth object, therefore we will load it in the bag. Let us take the below example with the bag capacity of 12. We will calculate the profit/weight value for each object and then we will include the solution from decreasing order of p/w ratio. Therefore, we will calculate the profit/weight ratio to find the optimal solution. Both approaches do not guarantee the optimal solution. If you look at this example, you might think that our second approach is good and will provide us with optimal results every time, but it is not. Now, the bag is completely filled, let us see the profit in this case-: When we load this object into the bag, our bag would be completely filled. Now the capacity of the bag is 5kg but the next minimum weight is вЂ6вЂ™, therefore we will take 5 out of that 6 and the profit will be calculated accordingly. After loading this object, the bag is left with-: The second object with the minimum weight is the 4th object with weight вЂ5вЂ™.The profit associated with this object is вЂ15вЂ™. Therefore, after loading this object to the bag, the remaining bag capacity is-: The weight associated with 3rd object = 2 The first object with the minimum weight = 2nd object Selecting the solution with respect to the minimum weight.Now, let us calculate the profit while selecting the solutions on the basis of minimum weight.

KNAPSACK ALGORITHM FULL

Now, the bag is completely full and we cannot add any more objects to the bag, we shall calculate the profit-:

After loading this object to the bag, the capacity of the bag remains-: The next object with maximum profit is the 3rd object, where the profit is вЂ14вЂ™ and the weight is вЂ7вЂ™. Therefore, if we load this object to the bag we are left with = 12-5 = 7 The weight associated with the 4th object = 5 If we select maximum objects with the maximum profit to fill the bag, then the solution will be-: Selecting the solution with respect to the maximum profit.(Read blog: What Does Probabilistic Programming Work?) Now, what should be our approach to fill the bag? We can either choose the object with a maximum profit or maximum weight, letвЂ™s try out both approaches. We can see that the total weight of the objects is exceeding the capacity of the bag( total weight: 6 + 2 + 7 + 5 = 20). We can see in the above example that there are 4 objects and in the example, we can also see the profit and weights of the object. Now let us take objects, object weights, and profit associated with that weight. The weight of the objects should be less than or equal to the capacity of the bag.Д¬onsider a bag that has a capacity of 12 kg, let's take the maximum capacity of a bag as вЂmвЂ™, here m = 12. The constraint is to fill the objects such that the weight of the objects doesnвЂ™t exceed the capacity of a bag. As we are trying to maximize the profit, this problem is optimization as well as maximization problem.Д®very optimization problem has some sort of constraint associated with it, what is the constraint here? The knapsack examples help in real-world such as resource allocation problems.ДЄ knapsack can also be considered as a bag and the problem is to fill the bag with the objects in such a way that the profit is maximized. The Knapsack problem is used in logistics, mathematics, cryptography, computer science, and more. The optimization problem needs to find an optimal solution and hence no exhaustive search approach could be applied to it.

The knapsack problem is one of the famous and important problems that come under the greedy method.ДЄs this problem is solved using a greedy method, this problem is one of the optimization problems, more precisely a combinatorial optimization.

Knapsack algorithm