Eight step procedures: Difference between revisions

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Remaining capacity s= 1,2,3,4,5  
Remaining capacity s= 1,2,3,4,5  
Boundary Conditions:
<math>f^{*}_{n+1}(s) = 0</math>,    ''s=0,1,2,3,4,5''      ''C=5''


<math>  
<math>  
Line 56: Line 60:
</math>= '''{0,1,2}'''
</math>= '''{0,1,2}'''


 
<math> f^{*}_{2}(5)= max\left \{ b[2,j]+ f^{*}_{3}(5-j*w[2]) \right \} </math>=
Boundary Conditions: <math>f^{*}_{n+1}(s) = 0</math>,    ''s=0,1,2,3,4,5''      ''C=5''
 
 
 
<math> f^{*}_{n}(s)= \overset{max}{j=0,1,...,min\left \{ a[n],\left \lfloor \frac{s}{w[n]} \right \rfloor \right \}} \left \{ b[n,j]+ f^{*}_{n+1}(s-j*w[n]) \right \} </math>
{| class="wikitable"
{| class="wikitable"
|+
|+
!Unused Capacity s  
!Unused Capacity s
!<math>f^{*}_{1}(s)</math>
!<math>f^{*}_{1}(s)</math>
!Type 1 opt <math>U^{*}_{1}(s)</math>
!Type 1 opt <math>U^{*}_{1}(s)</math>

Revision as of 02:18, 22 November 2020

Author: Eljona Pushaj, Diana Bogdanowich, Stephanie Keomany
Steward: Fengqi You

Introduction

Theory, Methodology, and/or Algorithmic Discussion

Methodology

To solve a problem using the 8-step procedure, one must use the following steps:

Step 1: Specify the stages of the problem
The stages of a dynamic programming problem can be defined as points where decisions are made. These are often denoted with the variable $ n $.

Step 2: Specify the states for each stage
The states of a problem are defined as the knowledge necessary to make a decision, or $ s $. We set $ C $ equal to the maximum value of $ s $.

Step 3: Specify the allowable actions for each state in each stage
This can be defined as:
$ U_{n}(s)\, or\, j\, =\, 0,1,...,min\left \{ a[n], \left \lfloor \frac{s}{w[n]} \right \rfloor \right \} $

Step 4: Describe the optimization function using an English-language description.
In this sentence, we describe the optimization function for each state, or $ s $, and each stage, or $ n $. This can also be called $ f^{*}_{n}(s) $

Step 5: Define the boundary conditions
This helps create a starting point to finding a solution to the problem. First, we set $ f^{*}_{n+1}(s) = 0 $ for all values of $ s $. Here, we can note that $ s=0,...,C $

Step 6: Define the recurrence relation
During this step, we make an allowable decision involving $ j $ items for the remaining capacity $ s $ for items $ n $. We can write this statement as:
$ f^{*}_{n}(s)= \overset{max}{j=0,1,...,min\left \{ a[n],\left \lfloor \frac{s}{w[n]} \right \rfloor \right \}} \left \{ b[n,j]+ f^{*}_{n+1}(s-j*w[n]) \right \} $

Step 7: Compute the optimal value from the bottom-up
In this step, a table is made containing all $ s $, $ f^{*}_{n}(s) $, and optimal values for all $ n $ variables. This step can be done manually or by using programming.

Step 8: Arrive at the optimal solution
Once the value for $ f^{*}_{n}(s) $ is computed, we would look at the optimal decision that corresponds to the table entry for that value. We start with the optimal value for our first $ n $, calculate our remaining space $ s $, and use that value to arrive at an optimal value for all $ n $.

Numerical Example

Weight capacity of C=5 and N=2

Item types are stages: n=1,2

Remaining capacity s= 1,2,3,4,5

Boundary Conditions:

$ f^{*}_{n+1}(s) = 0 $, s=0,1,2,3,4,5 C=5

$ U_{2}(5)\, =\, 0,1,...,min\left \{ a[2], \left \lfloor \frac{5}{w[2]}\right \rfloor \right \} $= {0,1,2}

$ f^{*}_{2}(5)= max\left \{ b[2,j]+ f^{*}_{3}(5-j*w[2]) \right \} $=

Unused Capacity s $ f^{*}_{1}(s) $ Type 1 opt $ U^{*}_{1}(s) $ $ f^{*}_{2}(s) $ Type 2 opt $ U^{*}_{2}(s) $ $ f^{*}_{3}(s) $
5 0
4 0
3 0
2 0
1 0
0 0

Applications

Conclusion

References