Hello dear experts;
I'm going to implement the ABC algorithm in order to optimize the performance of a trained MLP neural network.
Now the point is that ABC algorithm tries to minimize the MSE error during a descending way. Right, but how can I get its outputs (produced for each data) to be compared with the MLP outputs?
What is the outcome of this algorithm? Does it produce just the amended weights and biases?
I'm so thankful for your responds.
Dear Mohammad Mehrabi,
I suggest you to see links and attached files on topic.
-(PDF) A Comparison of Artificial Bee Colony algorithm and Genetic ...
https://www.researchgate.net/.../275536465_A_Comparison_of_Ar...
-Psi angle outputs of the quadrocopter. ABC: artificial bee colony ...
https://www.researchgate.net/.../Psi-angle-outputs-of-the-quadroco...
-An Artificial Bee Colony Algorithm Based on Problem ... - Science Direct
www.sciencedirect.com/science/article/pii/.../pdf?md5...
-Training ANFIS by using the artificial bee colony algorithm
journals.tubitak.gov.tr/elektrik/issues/.../elk-25-3-6-1601-240.pdf
-A novel artificial bee colony detection algorithm for massive MIMO ...
https://onlinelibrary.wiley.com/doi/full/10.1002/wcm.2754
-Artificial Bee Colony (ABC) Algorithm for Vehicle Routing ... - CiteSeerX
citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.649...
-Artificial bee colony algorithm - Scholarpedia
www.scholarpedia.org/article/Artificial_bee_colony_algorithm
-A modified Artificial Bee Colony algorithm to solve Clustering problems
https://towardsdatascience.com/a-modified-artificial-bee-colony-al...
Best regards
You can see the position of bees after your optimization algorithm terminates..
To be honest I couldn't get your question. However, I will try to help you ..
The objective function of such problem is to find the optimal values for wights and biases which produces the maximum rate of classification accuracy "A" (or) the minimum rate of mean square error "MSE".
Each food source represents a solution, which consists of the weights & bias.
The size of each solution is equal to the total number of weights & biases. Then, when the execution of the algorithm is finished, you should print these values, which represent the optimal weights & biases generated by ABC.
Hope this helps ..
Regards..
Sinan Salih
Dear Sinan,
Your answer is exactly what I need. Thank you.
well, with this in mind, there are some questions please:
1- How to distinguish the obtained parameter for weights and biases? Because they are just in a row. I want to use them to produce new outputs.
2- If we assume
NP = The number of colony size
D = The number of parameters of the problem to be optimized;
In my code, number of food sources = "NP / 2" and finally the obtained foods matrix has "NP/2" rows and "D" columns. How to figure out which row is the desired parameters?
3- This is the MLP equation (with 8 neurons, 4 inputs and 1 output) that is introduced as the Cost function.
Is this right?
function ObjVal = nntrain(x)
x=[]
targets = [];
outputs = zeros (size(x,1),1);
for i=1:size(x,1)
Y1(i) = tansig (-0.7695 * x(i,1) + 0.1115 * x(i,2) - 0.0004 * x(i,3) + 3.5861 * x(i,4) + 4.8819);
Y2(i) = tansig (-0.6921 * x(i,1) + 0.4576 * x(i,2) + 0.0575 * x(i,3) + 0.4881 * x(i,4) + 1.2720);
Y3(i) = tansig (0.5813 * x(i,1) + 0.3150 * x(i,2) - 0.0435 * x(i,3) - 0.5634 * x(i,4) - 0.2423);
Y4(i) = tansig (0.8457 * x(i,1) + 0.4627 * x(i,2) - 0.0133 * x(i,3) - 1.8860 * x(i,4) - 2.0092);
Y5(i) = tansig (1.3264 * x(i,1) + 0.1077 * x(i,2) + 0.0319 * x(i,3) + 1.5664 * x(i,4) - 0.7707);
Y6(i) = tansig (-0.3763 * x(i,1) - 0.1752 * x(i,2) + 0.6218 * x(i,3) + 0.3677 * x(i,4) + 2.0737);
Y7(i)= tansig (-0.7317 * x(i,1) - 0.1839 * x(i,2) + 0.0026 * x(i,3) + 4.0301 * x(i,4) + 5.1392);
Y8(i)= tansig (1.1470 * x(i,1) - 0.3794 * x(i,2) - 0.0121 * x(i,3) + 0.6488 * x(i,4) + 3.4396);
outputs(i,1) = 2.3830 * Y1(i) + 0.2196 * Y2(i) + 0.1933 * Y3(i) + 0.2433 * Y4(i) + 0.0130 * Y5(i) - 0.1561 * Y6(i) - 2.7611 * Y7(i) + 1.8678 * Y8(i) - 2.0766;
ObjVal(i)=mse(targets,outputs);
end;
I'm really grateful.
First of all ..
In my previous comment, I said that each food source represents a solution.
Based on your structure of the network ( 4 input, 3 hidden, 1 output = 8 nodes). Then we have 2 biases for hidden layer and output layer.
The main structure of the solution should be like this :
Size = 4 * 3 + 3 + 1 + 1 = 17
| W1-1 | W1-2 | W1-3 | W1-4| .... | W1-12| W2-1 | W2-2 | W2-3 | B1 | B2 |
The first 12 values represent the weights between the input layer and the hidden layer, while the next 3 values represent the weights between the hidden layer and the output layer. B1 and B2 represent the first and the second biases for both layers respectively.
Suppose you represent the solutions in a 2-D matrix ( in the code ), then the size of this matrix is equal to np/2 * 17. In other words, there are np/2 rows, each one consists of 17 values.
Therefore, each solution is a neural network, which has different weights and biases (17 values). Thus, there are np/2 neural networks, each one has a different accuracy or MSE. The neural network with the maximum accuracy (or) the minimum MSE represents the best solution.
Code Section:
1- You can write a function which calculate the Y value for you, the input is the index of the solution and the data values).
You have used in your example specific values for weights and biases, however, in the code, you just need to add a for loop and get any valye you need from the matrix explained above.
2- the last line in your code, ObjVal(i)=mse(targets,outputs);
If the mse function contains the neural network inside, this is correct, otherwise, this is wrong code. Because the neural network should receive the input data (X) and the weights and biases in order to produce the output, then you can calculate the MSE value by comparing the predicted results with the targeted results.
3- The structure of the neural network should be defined at the bigging of the code. How many input nodes ? how many nodes in the hidden layer? and how many nodes in the output layer ?
Hope this helps,
Regards ..
Temitayo Bankole
To be honest, I haven't work with ANT before but I guess they must be different; because all other optimization algorithms work with separate mechanisms.
In such conditions, the weights are the desicion variables. Therefore, the ABC method is the training algorithm that uses the error of the net as objective function.
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