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please provide an example of a neural network with 2 input nodes 2 hidden layer nodes and one output node. Calibrate it with inputs 2,3 and output 4 twice. using ReLU (Rectified Linear Unit) show calibration calculations and all forward and backward propagation formulas and calculations and show how output gets closer to 4 and how loss and Gradient Calculations are performed during each calibration.
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To demonstrate a simple neural network with 2 input nodes, 2 hidden layer nodes, and 1 output node, we will go through the process of forward and backward propagation using ReLU activation, and calibration using the inputs (2, 3) and the target output 4. We will illustrate how the network learns by adjusting weights through gradient descent over two iterations.

Network Structure

  1. Input Layer: 2 nodes
  2. Hidden Layer: 2 nodes with ReLU activation
  3. Output Layer: 1 node (linear activation)

Notations

  • Inputs: x1,x2x_1, x_2x1​,x2​
  • Weights between Input and Hidden Layer: w11,w12,w21,w22w_{11}, w_{12}, w_{21}, w_{22}w11​,w12​,w21​,w22​
  • Weights between Hidden and Output Layer: w1,w2w_{1}, w_{2}w1​,w2​
  • Biases for Hidden Layer: b1,b2b_1, b_2b1​,b2​
  • Bias for Output Layer: bob_obo​
  • Learning Rate: η\etaη

Initial Parameters

Let's initialize the weights and biases arbitrarily (small random values for simplicity):

  • w11=0.1,w12=0.2,w21=0.3,w22=0.4w_{11} = 0.1, w_{12} = 0.2, w_{21} = 0.3, w_{22} = 0.4w11​=0.1,w12​=0.2,w21​=0.3,w22​=0.4
  • w1=0.5,w2=0.6w_1 = 0.5, w_2 = 0.6w1​=0.5,w2​=0.6
  • b1=0.1,b2=0.1,bo=0.1b_1 = 0.1, b_2 = 0.1, b_o = 0.1b1​=0.1,b2​=0.1,bo​=0.1
  • η=0.01\eta = 0.01η=0.01 (learning rate)

Forward Propagation

  1. Hidden Layer Calculation:

    z1=w11x1+w21x2+b1z_1 = w_{11}x_1 + w_{21}x_2 + b_1z1​=w11​x1​+w21​x2​+b1​ z2=w12x1+w22x2+b2z_2 = w_{12}x_1 + w_{22}x_2 + b_2z2​=w12​x1​+w22​x2​+b2​

    Apply ReLU activation:

    a1=max⁡(0,z1)a_1 = \max(0, z_1)a1​=max(0,z1​) a2=max⁡(0,z2)a_2 = \max(0, z_2)a2​=max(0,z2​)

  2. Output Layer Calculation:

    y^=w1a1+w2a2+bo\hat{y} = w_1a_1 + w_2a_2 + b_oy^​=w1​a1​+w2​a2​+bo​

Loss Calculation

Using Mean Squared Error (MSE) as the loss function:

L=12(y^−y)2L = \frac{1}{2}(\hat{y} - y)^2L=21​(y^​−y)2

Where yyy is the target output.

Backward Propagation

  1. Output Layer Gradients:

    ∂L∂y^=y^−y\frac{\partial L}{\partial \hat{y}} = \hat{y} - y∂y^​∂L​=y^​−y

    For weights and bias:

    ∂L∂w1=∂L∂y^⋅a1\frac{\partial L}{\partial w_1} = \frac{\partial L}{\partial \hat{y}} \cdot a_1∂w1​∂L​=∂y^​∂L​⋅a1​ ∂L∂w2=∂L∂y^⋅a2\frac{\partial L}{\partial w_2} = \frac{\partial L}{\partial \hat{y}} \cdot a_2∂w2​∂L​=∂y^​∂L​⋅a2​ ∂L∂bo=∂L∂y^\frac{\partial L}{\partial b_o} = \frac{\partial L}{\partial \hat{y}}∂bo​∂L​=∂y^​∂L​

  2. Hidden Layer Gradients:

    ∂L∂a1=∂L∂y^⋅w1\frac{\partial L}{\partial a_1} = \frac{\partial L}{\partial \hat{y}} \cdot w_1∂a1​∂L​=∂y^​∂L​⋅w1​ ∂L∂a2=∂L∂y^⋅w2\frac{\partial L}{\partial a_2} = \frac{\partial L}{\partial \hat{y}} \cdot w_2∂a2​∂L​=∂y^​∂L​⋅w2​

    ReLU derivative:

    ∂L∂z1=∂L∂a1⋅(1z1>0)\frac{\partial L}{\partial z_1} = \frac{\partial L}{\partial a_1} \cdot (1_{z_1 > 0})∂z1​∂L​=∂a1​∂L​⋅(1z1​>0​) ∂L∂z2=∂L∂a2⋅(1z2>0)\frac{\partial L}{\partial z_2} = \frac{\partial L}{\partial a_2} \cdot (1_{z_2 > 0})∂z2​∂L​=∂a2​∂L​⋅(1z2​>0​)

    For weights and biases:

    ∂L∂w11=∂L∂z1⋅x1\frac{\partial L}{\partial w_{11}} = \frac{\partial L}{\partial z_1} \cdot x_1∂w11​∂L​=∂z1​∂L​⋅x1​ ∂L∂w21=∂L∂z1⋅x2\frac{\partial L}{\partial w_{21}} = \frac{\partial L}{\partial z_1} \cdot x_2∂w21​∂L​=∂z1​∂L​⋅x2​ ∂L∂w12=∂L∂z2⋅x1\frac{\partial L}{\partial w_{12}} = \frac{\partial L}{\partial z_2} \cdot x_1∂w12​∂L​=∂z2​∂L​⋅x1​ ∂L∂w22=∂L∂z2⋅x2\frac{\partial L}{\partial w_{22}} = \frac{\partial L}{\partial z_2} \cdot x_2∂w22​∂L​=∂z2​∂L​⋅x2​ ∂L∂b1=∂L∂z1\frac{\partial L}{\partial b_1} = \frac{\partial L}{\partial z_1}∂b1​∂L​=∂z1​∂L​ ∂L∂b2=∂L∂z2\frac{\partial L}{\partial b_2} = \frac{\partial L}{\partial z_2}∂b2​∂L​=∂z2​∂L​

Weight and Bias Update

wij=wij−η∂L∂wijw_{ij} = w_{ij} - \eta \frac{\partial L}{\partial w_{ij}}wij​=wij​−η∂wij​∂L​ bi=bi−η∂L∂bib_i = b_i - \eta \frac{\partial L}{\partial b_i}bi​=bi​−η∂bi​∂L​ wi=wi−η∂L∂wiw_{i} = w_{i} - \eta \frac{\partial L}{\partial w_{i}}wi​=wi​−η∂wi​∂L​ bo=bo−η∂L∂bob_o = b_o - \eta \frac{\partial L}{\partial b_o}bo​=bo​−η∂bo​∂L​

Iteration 1 Calculations

Let's perform the calculations for the first iteration.

Forward Propagation

  1. Hidden Layer Calculation:

    z1=(0.1⋅2)+(0.3⋅3)+0.1=1.2z_1 = (0.1 \cdot 2) + (0.3 \cdot 3) + 0.1 = 1.2z1​=(0.1⋅2)+(0.3⋅3)+0.1=1.2 z2=(0.2⋅2)+(0.4⋅3)+0.1=2.1z_2 = (0.2 \cdot 2) + (0.4 \cdot 3) + 0.1 = 2.1z2​=(0.2⋅2)+(0.4⋅3)+0.1=2.1 a1=max⁡(0,1.2)=1.2a_1 = \max(0, 1.2) = 1.2a1​=max(0,1.2)=1.2 a2=max⁡(0,2.1)=2.1a_2 = \max(0, 2.1) = 2.1a2​=max(0,2.1)=2.1

  2. Output Layer Calculation:

    y^=(0.5⋅1.2)+(0.6⋅2.1)+0.1=1.92\hat{y} = (0.5 \cdot 1.2) + (0.6 \cdot 2.1) + 0.1 = 1.92y^​=(0.5⋅1.2)+(0.6⋅2.1)+0.1=1.92

Loss Calculation

L=12(1.92−4)2=4.3264L = \frac{1}{2}(1.92 - 4)^2 = 4.3264L=21​(1.92−4)2=4.3264

Backward Propagation

  1. Output Layer Gradients:

    ∂L∂y^=1.92−4=−2.08\frac{\partial L}{\partial \hat{y}} = 1.92 - 4 = -2.08∂y^​∂L​=1.92−4=−2.08 ∂L∂w1=−2.08⋅1.2=−2.496\frac{\partial L}{\partial w_1} = -2.08 \cdot 1.2 = -2.496∂w1​∂L​=−2.08⋅1.2=−2.496 ∂L∂w2=−2.08⋅2.1=−4.368\frac{\partial L}{\partial w_2} = -2.08 \cdot 2.1 = -4.368∂w2​∂L​=−2.08⋅2.1=−4.368 ∂L∂bo=−2.08\frac{\partial L}{\partial b_o} = -2.08∂bo​∂L​=−2.08

  2. Hidden Layer Gradients:

    ∂L∂a1=−2.08⋅0.5=−1.04\frac{\partial L}{\partial a_1} = -2.08 \cdot 0.5 = -1.04∂a1​∂L​=−2.08⋅0.5=−1.04 ∂L∂a2=−2.08⋅0.6=−1.248\frac{\partial L}{\partial a_2} = -2.08 \cdot 0.6 = -1.248∂a2​∂L​=−2.08⋅0.6=−1.248 ∂L∂z1=−1.04⋅1=−1.04\frac{\partial L}{\partial z_1} = -1.04 \cdot 1 = -1.04∂z1​∂L​=−1.04⋅1=−1.04 ∂L∂z2=−1.248⋅1=−1.248\frac{\partial L}{\partial z_2} = -1.248 \cdot 1 = -1.248∂z2​∂L​=−1.248⋅1=−1.248 ∂L∂w11=−1.04⋅2=−2.08\frac{\partial L}{\partial w_{11}} = -1.04 \cdot 2 = -2.08∂w11​∂L​=−1.04⋅2=−2.08 ∂L∂w21=−1.04⋅3=−3.12\frac{\partial L}{\partial w_{21}} = -1.04 \cdot 3 = -3.12∂w21​∂L​=−1.04⋅3=−3.12 ∂L∂w12=−1.248⋅2=−2.496\frac{\partial L}{\partial w_{12}} = -1.248 \cdot 2 = -2.496∂w12​∂L​=−1.248⋅2=−2.496 ∂L∂w22=−1.248⋅3=−3.744\frac{\partial L}{\partial w_{22}} = -1.248 \cdot 3 = -3.744∂w22​∂L​=−1.248⋅3=−3.744 ∂L∂b1=−1.04\frac{\partial L}{\partial b_1} = -1.04∂b1​∂L​=−1.04 ∂L∂b2=−1.248\frac{\partial L}{\partial b_2} = -1.248∂b2​∂L​=−1.248

Weight and Bias Update

w11=0.1−0.01×(−2.08)=0.1208w_{11} = 0.1 - 0.01 \times (-2.08) = 0.1208w11​=0.1−0.01×(−2.08)=0.1208 w21=0.3−0.01×(−3.12)=0.3312w_{21} = 0.3 - 0.01 \times (-3.12) = 0.3312w21​=0.3−0.01×(−3.12)=0.3312 w12=0.2−0.01×(−2.496)=0.22496w_{12} = 0.2 - 0.01 \times (-2.496) = 0.22496w12​=0.2−0.01×(−2.496)=0.22496 w22=0.4−0.01×(−3.744)=0.43744w_{22} = 0.4 - 0.01 \times (-3.744) = 0.43744w22​=0.4−0.01×(−3.744)=0.43744 w1=0.5−0.01×(−2.496)=0.52496w_1 = 0.5 - 0.01 \times (-2.496) = 0.52496w1​=0.5−0.01×(−2.496)=0.52496 w2=0.6−0.01×(−4.368)=0.64368w_2 = 0.6 - 0.01 \times (-4.368) = 0.64368w2​=0.6−0.01×(−4.368)=0.64368 b1=0.1−0.01×(−1.04)=0.1104b_1 = 0.1 - 0.01 \times (-1.04) = 0.1104b1​=0.1−0.01×(−1.04)=0.1104 b2=0.1−0.01×(−1.248)=0.11248b_2 = 0.1 - 0.01 \times (-1.248) = 0.11248b2​=0.1−0.01×(−1.248)=0.11248 bo=0.1−0.01×(−2.08)=0.1208b_o = 0.1 - 0.01 \times (-2.08) = 0.1208bo​=0.1−0.01×(−2.08)=0.1208

Iteration 2 Calculations

Let's perform the calculations for the second iteration with the updated weights and biases.

Forward Propagation

  1. Hidden Layer Calculation:

    z1=(0.1208⋅2)+(0.3312⋅3)+0.1104=1.4456z_1 = (0.1208 \cdot 2) + (0.3312 \cdot 3) + 0.1104 = 1.4456z1​=(0.1208⋅2)+(0.3312⋅3)+0.1104=1.4456 z2=(0.22496⋅2)+(0.43744⋅3)+0.11248=2.3192z_2 = (0.22496 \cdot 2) + (0.43744 \cdot 3) + 0.11248 = 2.3192z2​=(0.22496⋅2)+(0.43744⋅3)+0.11248=2.3192 a1=max⁡(0,1.4456)=1.4456a_1 = \max(0, 1.4456) = 1.4456a1​=max(0,1.4456)=1.4456 a2=max⁡(0,2.3192)=2.3192a_2 = \max(0, 2.3192) = 2.3192a2​=max(0,2.3192)=2.3192

  2. Output Layer Calculation:

    y^=(0.52496⋅1.4456)+(0.64368⋅2.3192)+0.1208=2.492704\hat{y} = (0.52496 \cdot 1.4456) + (0.64368 \cdot 2.3192) + 0.1208 = 2.492704y^​=(0.52496⋅1.4456)+(0.64368⋅2.3192)+0.1208=2.492704

Loss Calculation

L=12(2.492704−4)2=1.136195L = \frac{1}{2}(2.492704 - 4)^2 = 1.136195L=21​(2.492704−4)2=1.136195

Backward Propagation

  1. Output Layer Gradients:

    ∂L∂y^=2.492704−4=−1.507296\frac{\partial L}{\partial \hat{y}} = 2.492704 - 4 = -1.507296∂y^​∂L​=2.492704−4=−1.507296 ∂L∂w1=−1.507296⋅1.4456=−2.178768\frac{\partial L}{\partial w_1} = -1.507296 \cdot 1.4456 = -2.178768∂w1​∂L​=−1.507296⋅1.4456=−2.178768 ∂L∂w2=−1.507296⋅2.3192=−3.496984\frac{\partial L}{\partial w_2} = -1.507296 \cdot 2.3192 = -3.496984∂w2​∂L​=−1.507296⋅2.3192=−3.496984 ∂L∂bo=−1.507296\frac{\partial L}{\partial b_o} = -1.507296∂bo​∂L​=−1.507296

  2. Hidden Layer Gradients:

    ∂L∂a1=−1.507296⋅0.52496=−0.791429\frac{\partial L}{\partial a_1} = -1.507296 \cdot 0.52496 = -0.791429∂a1​∂L​=−1.507296⋅0.52496=−0.791429 ∂L∂a2=−1.507296⋅0.64368=−0.970056\frac{\partial L}{\partial a_2} = -1.507296 \cdot 0.64368 = -0.970056∂a2​∂L​=−1.507296⋅0.64368=−0.970056 ∂L∂z1=−0.791429⋅1=−0.791429\frac{\partial L}{\partial z_1} = -0.791429 \cdot 1 = -0.791429∂z1​∂L​=−0.791429⋅1=−0.791429 ∂L∂z2=−0.970056⋅1=−0.970056\frac{\partial L}{\partial z_2} = -0.970056 \cdot 1 = -0.970056∂z2​∂L​=−0.970056⋅1=−0.970056 ∂L∂w11=−0.791429⋅2=−1.582858\frac{\partial L}{\partial w_{11}} = -0.791429 \cdot 2 = -1.582858∂w11​∂L​=−0.791429⋅2=−1.582858 ∂L∂w21=−0.791429⋅3=−2.374287\frac{\partial L}{\partial w_{21}} = -0.791429 \cdot 3 = -2.374287∂w21​∂L​=−0.791429⋅3=−2.374287 ∂L∂w12=−0.970056⋅2=−1.940112\frac{\partial L}{\partial w_{12}} = -0.970056 \cdot 2 = -1.940112∂w12​∂L​=−0.970056⋅2=−1.940112 ∂L∂w22=−0.970056⋅3=−2.910168\frac{\partial L}{\partial w_{22}} = -0.970056 \cdot 3 = -2.910168∂w22​∂L​=−0.970056⋅3=−2.910168 ∂L∂b1=−0.791429\frac{\partial L}{\partial b_1} = -0.791429∂b1​∂L​=−0.791429 ∂L∂b2=−0.970056\frac{\partial L}{\partial b_2} = -0.970056∂b2​∂L​=−0.970056

Weight and Bias Update

w11=0.1208−0.01×(−1.582858)=0.136628w_{11} = 0.1208 - 0.01 \times (-1.582858) = 0.136628w11​=0.1208−0.01×(−1.582858)=0.136628 w21=0.3312−0.01×(−2.374287)=0.354943w_{21} = 0.3312 - 0.01 \times (-2.374287) = 0.354943w21​=0.3312−0.01×(−2.374287)=0.354943 w12=0.22496−0.01×(−1.940112)=0.24436112w_{12} = 0.22496 - 0.01 \times (-1.940112) = 0.24436112w12​=0.22496−0.01×(−1.940112)=0.24436112 w22=0.43744−0.01×(−2.910168)=0.46654168w_{22} = 0.43744 - 0.01 \times (-2.910168) = 0.46654168w22​=0.43744−0.01×(−2.910168)=0.46654168 w1=0.52496−0.01×(−2.178768)=0.54674768w_1 = 0.52496 - 0.01 \times (-2.178768) = 0.54674768w1​=0.52496−0.01×(−2.178768)=0.54674768 w2=0.64368−0.01×(−3.496984)=0.67864984w_2 = 0.64368 - 0.01 \times (-3.496984) = 0.67864984w2​=0.64368−0.01×(−3.496984)=0.67864984 b1=0.1104−0.01×(−0.791429)=0.11831429b_1 = 0.1104 - 0.01 \times (-0.791429) = 0.11831429b1​=0.1104−0.01×(−0.791429)=0.11831429 b2=0.11248−0.01×(−0.970056)=0.12218056b_2 = 0.11248 - 0.01 \times (-0.970056) = 0.12218056b2​=0.11248−0.01×(−0.970056)=0.12218056 bo=0.1208−0.01×(−1.507296)=0.13587296b_o = 0.1208 - 0.01 \times (-1.507296) = 0.13587296bo​=0.1208−0.01×(−1.507296)=0.13587296

Conclusion

After two iterations, the output y^\hat{y}y^​ has moved closer to the target output (4) from 1.92 to 2.492704. The loss has decreased from 4.3264 to 1.136195, indicating that the network is learning and minimizing the error. The weight and bias updates demonstrate the network's adjustment process based on the gradients computed during backward propagation. Further iterations would continue this process, reducing the loss and improving the accuracy of the output.

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