Question
Download Solution PDFIn a feed forward neural network with the following specifications:
Input layer has 4 neurons, hidden layer has 3 neurons and output layer has 2 neurons using the sigmoid activation function for given input values [0.5, 0.8, 0.2, 0.6] as well as the initial weights for the connections.
W1: [0.1, 0.3, 0.5, 0.2]
W2: [0.2, 0.4, 0.6, 0.2] Input layer to hidden layer weights
W3: [0.3, 0.5, 0.7, 0.2]
W4: [0.4, 0.1, 0.3]
W5: [0.5, 0.2, 0.4] Hidden layer to output layer weights
What is the output of the output layer when the given input values are passed through neural network? Round the answer to two decimal places :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is [0.62, 0.68]
Key Points
To compute the output of the neural network, we need to follow these steps:
- Compute the input to the hidden layer by taking the dot product of the input values and the weights (adding a bias term, which is 1, for each neuron in the hidden layer).
- Apply the sigmoid activation function to each element of the hidden layer.
- Compute the input to the output layer by taking the dot product of the hidden layer outputs and the weights (adding a bias term, which is 1, for each neuron in the output layer).
- Apply the sigmoid activation function to each element of the output layer.
Calculations:
1. Compute the input to the hidden layer:
- \(\text{Input\_Hidden} = \sigma(W1 \cdot X)\)
- \( \text{Input\_Hidden} = \sigma(0.1 \cdot 0.5 + 0.3 \cdot 0.8 + 0.5 \cdot 0.2 + 0.2 \cdot 0.6) \)
- \(\text{Input\_Hidden} \approx \sigma(0.05 + 0.24 + 0.1 + 0.12)\)
- \(\text{Input\_Hidden} \approx \sigma(0.51)\)
- \(\text{Input\_Hidden} \approx 0.6246\)
2. W2:
- \( \text{Input\_Hidden} = \sigma(0.2 \cdot 0.5 + 0.4 \cdot 0.8 + 0.6 \cdot 0.2 + 0.2 \cdot 0.6) \)
- \(\text{Input\_Hidden} \approx \sigma(0.10 + 0.32 + 0.12 + 0.12)\)
- \(\text{Hidden\_Output} = \sigma(0.66)\)
- \(\text{Hidden\_Output} \approx 0.6592\)
3. W3:
- \( \text{Input\_Hidden} = \sigma(0.3 \cdot 0.5 + 0.5 \cdot 0.8 + 0.7 \cdot 0.2 + 0.2 \cdot 0.6) \)
- \(\text{Input\_Hidden} \approx \sigma(0.15 + 0.40 + 0.14 + 0.12)\)
- \(\text{Hidden\_Output} = \sigma(0.81)\)
- \(\text{Hidden\_Output} \approx 0.6921\)
4. W4:
- \(\text{Input\_Hidden} = \sigma(W4 \cdot Y)\)
- \( \text{Input\_Hidden} = \sigma(0.4 \cdot 0.6248 + 0.1 \cdot 0.6592 + 0.2 \cdot 0.6921) \)
- \(\text{Input\_Hidden} \approx \sigma(0.2499 + 0.0659 + 0.2076)\)
- \(\text{Hidden\_Output} = \sigma(0.5234)\)
- \(\text{Output} \approx 0.6279\)
5. W5:
- \(\text{Input\_Hidden} = \sigma(W5 \cdot Y)\)
- \( \text{Input\_Hidden} = \sigma(0.5 \cdot 0.6248 + 0.2 \cdot 0.6592 + 0.4 \cdot 0.6921) \)
- \(\text{Input\_Hidden} \approx \sigma(0.3124 + 0.1318 + 0.2768)\)
- \(\text{Hidden\_Output} = \sigma(0.721)\)
- \(\text{Hidden\_Output} \approx 0.6728\)
Rounding to two decimal places, the correct output of the output layer is approximately \([0.62, 0.68]\).
Last updated on Jun 6, 2025
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