Combinatorial optimization: Knapsack problem 2: Unterschied zwischen den Versionen

Version vom 26. Juni 2013, 09:57 Uhr

Knappsack-Problem

1. Introduction

2. Application and definition

3. Example

   A woman wants to go on vacation. But now she has to overcome a serious problem, because the airline prescribe 
   a maximum weight of 25 kg (b = 25 kg)for her luggage (the worst that can happen to a woman).
   She has the choice between six different items which she can pack. 
   Each object has a special benefit c_i for the woman and a weight a_i.
   What items should be selected by her to have the maximum benefit without exceeding the maximum weight of 25 kg?

4. Methods 1 - 3

   To apply the different methods a ranking must be formed first. 
   
   For this purpose, a quotient is formed from the benefit c_i and the weight a_i.
   Afterwards, the values are sorted according to the size.

   Ordered list:
   

   This list is the basis for all applicable methods.

   a) Kolesar

   The goal is to obtain an integer solution. One begins with the object of rank 1, then rank 2 and so on 
   and sets the variable equal 1 or 0. The items can be taken as long as the maximum weight of 25 kg is reached.
   This means that the last possible item mostly can't be taken completely.
   The solution with the higher objective function value is branched. If the objective function value is less 
   than the value that provides the best integer solution, this solution is not branched.
   

   b) Greenberg & Hegerich
   
   In this method, one starts with the non-integer variable and set this equal to 0 and 1 respectively to look
   for a better solution. This procedure should be continued until an optimal solution is found.

   c) Cascading Tree

5. Method 4

6. complexity problem