Here, a multi-perceptron network with three layers is given to us. The input layer contains two neurons, the hidden layer contains two neurons, and the output layer contains one neuron. In this case, given this multi-layer perceptron network, we need to calculate the output at the output layer neuron, that is, five, with input vector 7 and 3. The input at neuron 1 is 7, and the input at neuron 2 is 3 here. The activation functions we need to use are the logistic function (Ru) and Ru with matrix operations. Here, so first we will try to calculate the output using the logistic activation function. But to calculate the output at the output layer neuron, first we need to propagate the input from the input layer neuron to the hidden layer and then from the hidden layer to the output layer. Here, so first we will calculate the output at the hidden layer neurons and then we will propagate it to the output layer to calculate the output. At the hidden layer neurons, first we need to calculate the summation term here, or you can say that the weighted sum in this case. So, the weighted sum at neuron 3 and neuron 4. Now, how to calculate the weighted sum? The weighted sum is nothing but the summation of multiplication between weight and input in this case. That is nothing but Z3 is equal to — if you go with the bias, the weight is equal to 6 and input is equal to 1 here. That is nothing but 6 * 1. If you go with this one, that is the input one, the weight is 0.1, and the input over here is 7, so 0.1 into 7, and from neuron 2 here, the weight is 5 and the input is equal to 3 here. And once you solve it, you'll be getting 8 as the weighted sum at neuron 3. Similarly, we will calculate the weighted sum at neuron 4 here, which is equal to 1.06 in this case. So once you calculate the weighted sum at neuron 3 and 4, we need to apply the logistic activation function and calculate the output at 3 and 4. Here, the logistic activation function is given by \( \frac{1}{1 + e^{-Z}} \). In the first case, that is, \( Z_3 = 8 \), that is nothing but \( \frac{1}{1 + e^{-8}} \), which is equivalent to 0.6942 here. Similarly, for the second neuron, that is Neuron 4, \( \frac{1}{1 + e^{-Z_4}} \), that is nothing but 1.06, which is equivalent to 0.7426, and so on. Now we have calculated the output at neuron 3 and neuron 4. We will use these outputs and then we will calculate the output at neuron 5. Here, to calculate the output at neuron 5, first we need to calculate the weighted sum. The weighted sum at output neuron 5, that is nothing but Z5, is equal to the bias weight, which is 0.9, and the input is equal to one here; that is what I have written here. For this particular neuron 3, the weight is equal to 3, and the output at this neuron we have already calculated in the previous slide, we will put it here. The weight for neuron 4 is equal to 7, and the output is already calculated, we will put it here. Once you solve it, Z5 will become 1.62 in this case. Now we have calculated the weighted sum at neuron 5 here. Now, once you calculate the weighted sum, that is 1.62. We need to apply again the logistic activation function, that is nothing but \( \frac{1}{1 + e^{-Z}} \). Here, \( Z \) is 1.62 in this case. Once I put that particular value, \( A5 \) will become 0.8359 in this case. So, we were able to calculate the output at neuron 5, which is equal to 0.8359, using the logistic activation function. Here, now the next activation function is the ReLU (Rectified Linear Unit) activation function. Once you use the ReLU activation function, again the same thing we need to do: we need to calculate the weighted sum at neurons 3 and 4, calculate the output, and then propagate it to the output layer neuron. In this case, we have already calculated the weighted sum at neurons 3 and 4; they are nothing but 8.2 and 1.06 here. Once you calculate the weighted sum, what we need to do is apply the activation function. What is the activation function given to us? ReLU, in this case, which is defined by this function: \(\text{Max}(0, Z)\). Here, \( Z_3 \) in this case is 8.2, so \(\text{Max}(0, 8.2) = 8.2\). For the second one, \( Z_4 \) is 1.06, so \(\text{Max}(0, 1.06) = 1.06\). Thus, the output at neuron 3 is 8.2, and the output at neuron 4 is 1.06 in this case. Now, we have calculated the output at neurons 3 and 4. Now, we need to calculate the weighted sum at the neuron 5. Now, how to calculate the weighted sum? Again, the same function. \( Z_5 \) is equal to the weight multiplied by the input. The bias weight is 0.9, and the input is equal to 1 here. For neuron 3, the weight is equal to 3, and the output of neuron 3 is equal to \( A_2 \). Then, coming back to the last one, the weight is 7, and the output at neuron 4 is equal to 1.06. Once you solve it, you will get 1.888 in this case. Again, we have calculated the weighted sum at neuron 5. Now, we need to apply the ReLU activation function. Again, the same function we will use: \(\text{Max}(0, Z)\). \( Z \) in this case is 1.888, so \(\text{Max}(0, 1.888)\) = 1.888. So, the output at neuron 5 is equal to 1.888 in this case. In this case, we have calculated the output at the output layer neuron using the ReLU activation function. Coming back to the next one, that is the ReLU function with the Matrix operation. We have been given a template. In this case, we need to put the values in this template, and then we need to calculate the output over here. So, the hidden layer weight matrix is nothing but the weights for this particular neuron 3 and weights for this particular neuron 4. So, what is the first weight? 6, so we will put that particular 6 here. Then, the second weight for this particular neuron 3 is 0.1, and the third one is 0.5 here. So, we will put them into the first two here. Coming back to the second neuron, that is neuron 4, for this one, the weight is equal to 8, this weight is equal to 0.2, and the third one is 4. We will put all those things here, and we will be getting the hidden layer weight matrix in this case. So, that is what I have written in this particular template here. Now, once you write this particular template, the next part is to write the activations at the input layer. That is nothing but the inputs over here. So, for neuron 3, the input is equal to 1, because as I said earlier, the bias input is always equal to one. For this, the bias input is equal to one here. Then, the input at neuron 1 is equal to 7, so that is what we have added here. The input at neuron 2 is equal to 3; that is what I have added here. Now, once you multiply these two things, you will be getting 8.2 and 1.06. That is what we have already calculated in the previous example also. Now, once you calculate these things, what is the next thing we need to do? We need to apply the ReLU activation function. How does the ReLU activation function work? Max of 0, comma Z. Here, both of these values are greater than zero, so they are the maximum, and they are the final values in this case. Now, once you calculate these things, the next part is to calculate the weighted sum and the output at neuron 5. For that reason, first, we need to write the output layer weight matrix. The output layer weight matrix contains 0.93 and 7 here; these are the weights in this case. And what are the inputs to this particular neuron? The first input is 1; the second input is the output of neuron 3, which is 8.2, and the third input is the output of neuron 4, which is 1.06. That is what I have written in this column matrix here. Again, once you multiply these two things, we will be getting 1.1888 in this case. Now, once you calculate 1.1888, that is nothing but the weighted sum. The next part is to apply the activation function. Here, again, the activation function is ReLU. The ReLU takes the maximum of 0, comma Z. Here, in this case, the Z value is 1.1888, which is greater than 0. The meaning of this one is the output is nothing but 1.1888.