A Step by Step Backpropagation Example

Background

Backpropagation is a common method for training a neural network. There is no shortage of papers online that attempt to explain how backpropagation works, but few that include an example with actual numbers. This post is my attempt to explain how it works with a concrete example that folks can compare their own calculations to in order to ensure they understand backpropagation correctly.

If this kind of thing interests you, you should sign up for my newsletter where I post about AI-related projects that I’m working on.

Backpropagation in Python

You can play around with a Python script that I wrote that implements the backpropagation algorithm in this Github repo.

Backpropagation Visualization

For an interactive visualization showing a neural network as it learns, check out my Neural Network visualization.

Additional Resources

If you find this tutorial useful and want to continue learning about neural networks, machine learning, and deep learning, I highly recommend checking out Adrian Rosebrock’s new book, Deep Learning for Computer Vision with Python. I really enjoyed the book and will have a full review up soon.

Overview

For this tutorial, we’re going to use a neural network with two inputs, two hidden neurons, two output neurons. Additionally, the hidden and output neurons will include a bias.

Here’s the basic structure:

neural_network (7)

In order to have some numbers to work with, here are the initial weights, the biases, and training inputs/outputs:

neural_network (9)

The goal of backpropagation is to optimize the weights so that the neural network can learn how to correctly map arbitrary inputs to outputs.

For the rest of this tutorial we’re going to work with a single training set: given inputs 0.05 and 0.10, we want the neural network to output 0.01 and 0.99.

The Forward Pass

To begin, lets see what the neural network currently predicts given the weights and biases above and inputs of 0.05 and 0.10. To do this we’ll feed those inputs forward though the network.

We figure out the total net input to each hidden layer neuron, squash the total net input using an activation function (here we use the logistic function), then repeat the process with the output layer neurons.

Total net input is also referred to as just net input by some sources.

Here’s how we calculate the total net input for $h_1$ :

$net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1$

$net_{h1} = 0.15 * 0.05 + 0.2 * 0.1 + 0.35 * 1 = 0.3775$

We then squash it using the logistic function to get the output of $h_1$ :

$out_{h1} = \frac{1}{1+e^{-net_{h1}}} = \frac{1}{1+e^{-0.3775}} = 0.593269992$

Carrying out the same process for $h_2$ we get:

$out_{h2} = 0.596884378$

We repeat this process for the output layer neurons, using the output from the hidden layer neurons as inputs.

Here’s the output for $o_1$ :

$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$

$net_{o1} = 0.4 * 0.593269992 + 0.45 * 0.596884378 + 0.6 * 1 = 1.105905967$

$out_{o1} = \frac{1}{1+e^{-net_{o1}}} = \frac{1}{1+e^{-1.105905967}} = 0.75136507$

And carrying out the same process for $o_2$ we get:

$out_{o2} = 0.772928465$

Calculating the Total Error

We can now calculate the error for each output neuron using the squared error function and sum them to get the total error:

$E_{total} = \sum \frac{1}{2}(target - output)^{2}$

Some sources refer to the target as the ideal and the output as the actual.

The $\frac{1}{2}$ is included so that exponent is cancelled when we differentiate later on. The result is eventually multiplied by a learning rate anyway so it doesn’t matter that we introduce a constant here [1].

For example, the target output for $o_1$ is 0.01 but the neural network output 0.75136507, therefore its error is:

$E_{o1} = \frac{1}{2}(target_{o1} - out_{o1})^{2} = \frac{1}{2}(0.01 - 0.75136507)^{2} = 0.274811083$

Repeating this process for $o_2$ (remembering that the target is 0.99) we get:

$E_{o2} = 0.023560026$

The total error for the neural network is the sum of these errors:

$E_{total} = E_{o1} + E_{o2} = 0.274811083 + 0.023560026 = 0.298371109$

The Backwards Pass

Our goal with backpropagation is to update each of the weights in the network so that they cause the actual output to be closer the target output, thereby minimizing the error for each output neuron and the network as a whole.

Output Layer

Consider $w_5$ . We want to know how much a change in $w_5$ affects the total error, aka $\frac{\partial E_{total}}{\partial w_{5}}$ .

$\frac{\partial E_{total}}{\partial w_{5}}$ is read as “the partial derivative of $E_{total}$ with respect to $w_{5}$ “. You can also say “the gradient with respect to $w_{5}$ “.

By applying the chain rule we know that:

$\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}$

Visually, here’s what we’re doing:

output_1_backprop (4)

We need to figure out each piece in this equation.

First, how much does the total error change with respect to the output?

$E_{total} = \frac{1}{2}(target_{o1} - out_{o1})^{2} + \frac{1}{2}(target_{o2} - out_{o2})^{2}$

$\frac{\partial E_{total}}{\partial out_{o1}} = 2 * \frac{1}{2}(target_{o1} - out_{o1})^{2 - 1} * -1 + 0$

$\frac{\partial E_{total}}{\partial out_{o1}} = -(target_{o1} - out_{o1}) = -(0.01 - 0.75136507) = 0.74136507$

$-(target - out)$ is sometimes expressed as $out - target$

When we take the partial derivative of the total error with respect to $out_{o1}$ , the quantity $\frac{1}{2}(target_{o2} - out_{o2})^{2}$ becomes zero because $out_{o1}$ does not affect it which means we’re taking the derivative of a constant which is zero.

Next, how much does the output of $o_1$ change with respect to its total net input?

The partial derivative of the logistic function is the output multiplied by 1 minus the output:

$out_{o1} = \frac{1}{1+e^{-net_{o1}}}$

$\frac{\partial out_{o1}}{\partial net_{o1}} = out_{o1}(1 - out_{o1}) = 0.75136507(1 - 0.75136507) = 0.186815602$

Finally, how much does the total net input of $o1$ change with respect to $w_5$ ?

$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$

$\frac{\partial net_{o1}}{\partial w_{5}} = 1 * out_{h1} * w_5^{(1 - 1)} + 0 + 0 = out_{h1} = 0.593269992$

Putting it all together:

$\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}$

$\frac{\partial E_{total}}{\partial w_{5}} = 0.74136507 * 0.186815602 * 0.593269992 = 0.082167041$

You’ll often see this calculation combined in the form of the delta rule:

$\frac{\partial E_{total}}{\partial w_{5}} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1}) * out_{h1}$

Alternatively, we have $\frac{\partial E_{total}}{\partial out_{o1}}$ and $\frac{\partial out_{o1}}{\partial net_{o1}}$ which can be written as $\frac{\partial E_{total}}{\partial net_{o1}}$ , aka $\delta_{o1}$ (the Greek letter delta) aka the node delta. We can use this to rewrite the calculation above:

$\delta_{o1} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = \frac{\partial E_{total}}{\partial net_{o1}}$

$\delta_{o1} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1})$

Therefore:

$\frac{\partial E_{total}}{\partial w_{5}} = \delta_{o1} out_{h1}$

Some sources extract the negative sign from $\delta$ so it would be written as:

$\frac{\partial E_{total}}{\partial w_{5}} = -\delta_{o1} out_{h1}$

To decrease the error, we then subtract this value from the current weight (optionally multiplied by some learning rate, eta, which we’ll set to 0.5):

$w_5^{+} = w_5 - \eta * \frac{\partial E_{total}}{\partial w_{5}} = 0.4 - 0.5 * 0.082167041 = 0.35891648$

Some sources use $\alpha$ (alpha) to represent the learning rate, others use $\eta$ (eta), and others even use $\epsilon$ (epsilon).

We can repeat this process to get the new weights $w_6$ , $w_7$ , and $w_8$ :

$w_6^{+} = 0.408666186$

$w_7^{+} = 0.511301270$

$w_8^{+} = 0.561370121$

We perform the actual updates in the neural network after we have the new weights leading into the hidden layer neurons (ie, we use the original weights, not the updated weights, when we continue the backpropagation algorithm below).

Hidden Layer

Next, we’ll continue the backwards pass by calculating new values for $w_1$ , $w_2$ , $w_3$ , and $w_4$ .

Big picture, here’s what we need to figure out:

$\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$

Visually:

We’re going to use a similar process as we did for the output layer, but slightly different to account for the fact that the output of each hidden layer neuron contributes to the output (and therefore error) of multiple output neurons. We know that $out_{h1}$ affects both $out_{o1}$ and $out_{o2}$ therefore the $\frac{\partial E_{total}}{\partial out_{h1}}$ needs to take into consideration its effect on the both output neurons:

$\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}}$

Starting with $\frac{\partial E_{o1}}{\partial out_{h1}}$ :

$\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}}$

We can calculate $\frac{\partial E_{o1}}{\partial net_{o1}}$ using values we calculated earlier:

$\frac{\partial E_{o1}}{\partial net_{o1}} = \frac{\partial E_{o1}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = 0.74136507 * 0.186815602 = 0.138498562$

And $\frac{\partial net_{o1}}{\partial out_{h1}}$ is equal to $w_5$ :

$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$

$\frac{\partial net_{o1}}{\partial out_{h1}} = w_5 = 0.40$

Plugging them in:

$\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}} = 0.138498562 * 0.40 = 0.055399425$

Following the same process for $\frac{\partial E_{o2}}{\partial out_{h1}}$ , we get:

$\frac{\partial E_{o2}}{\partial out_{h1}} = -0.019049119$

Therefore:

$\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}} = 0.055399425 + -0.019049119 = 0.036350306$

Now that we have $\frac{\partial E_{total}}{\partial out_{h1}}$ , we need to figure out $\frac{\partial out_{h1}}{\partial net_{h1}}$ and then $\frac{\partial net_{h1}}{\partial w}$ for each weight:

$out_{h1} = \frac{1}{1+e^{-net_{h1}}}$

$\frac{\partial out_{h1}}{\partial net_{h1}} = out_{h1}(1 - out_{h1}) = 0.59326999(1 - 0.59326999 ) = 0.241300709$

We calculate the partial derivative of the total net input to $h_1$ with respect to $w_1$ the same as we did for the output neuron:

$net_{h1} = w_1 * i_1 + w_3 * i_2 + b_1 * 1$

$\frac{\partial net_{h1}}{\partial w_1} = i_1 = 0.05$

Putting it all together:

$\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$

$\frac{\partial E_{total}}{\partial w_{1}} = 0.036350306 * 0.241300709 * 0.05 = 0.000438568$

You might also see this written as:

$\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\frac{\partial E_{total}}{\partial out_{o}} * \frac{\partial out_{o}}{\partial net_{o}} * \frac{\partial net_{o}}{\partial out_{h1}}}) * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$

$\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\delta_{o} * w_{ho}}) * out_{h1}(1 - out_{h1}) * i_{1}$

$\frac{\partial E_{total}}{\partial w_{1}} = \delta_{h1}i_{1}$

We can now update $w_1$ :

$w_1^{+} = w_1 - \eta * \frac{\partial E_{total}}{\partial w_{1}} = 0.15 - 0.5 * 0.000438568 = 0.149780716$

Repeating this for $w_2$ , $w_3$ , and $w_4$

$w_2^{+} = 0.19956143$

$w_3^{+} = 0.24975114$

$w_4^{+} = 0.29950229$

Finally, we’ve updated all of our weights! When we fed forward the 0.05 and 0.1 inputs originally, the error on the network was 0.298371109. After this first round of backpropagation, the total error is now down to 0.291027924. It might not seem like much, but after repeating this process 10,000 times, for example, the error plummets to 0.0000351085. At this point, when we feed forward 0.05 and 0.1, the two outputs neurons generate 0.015912196 (vs 0.01 target) and 0.984065734 (vs 0.99 target).

If you’ve made it this far and found any errors in any of the above or can think of any ways to make it clearer for future readers, don’t hesitate to drop me a note. Thanks!

849 thoughts on “A Step by Step Backpropagation Example”

bharati says:

December 5, 2018 at 2:04 am

nice explanation.

Reply
H.Taner Unal says:

December 10, 2018 at 1:31 pm

Hey Matt. I’m stuck here: I got dEo2/douth1=-0.017144288. you got -0.019049. Can you check again?

Reply
Arindam says:

December 13, 2018 at 3:43 am

Nicely explained. Thanks :)

Reply
Florian says:

December 16, 2018 at 6:10 am

Very helpful. Thanks

Reply
Tansuluu says:

December 19, 2018 at 7:26 am

Thank you a lot! Didn’t get on lecture.Here is the best explanation!

Reply
Loftur says:

December 19, 2018 at 8:57 am

To get the net input of h1 you need to calculate w1*i1 + w3*i2 + b1*1. The weight w2 does not connect to h1

Reply
- Braydon Burkhardt says:
  
  January 10, 2019 at 10:13 pm
  
  It can look like that in the pictures, however h1 is in fact from w1 and w2. The weight # is created based off of the hidden layer. For example h1 has w1 and w2 while h2 has w3 and w4.
  
  Reply
Mike says:

December 20, 2018 at 6:15 am

If we want to extend this to a batch of training samples, do we calculate the output layer errors simply by averaging the error at each output over the training samples, then obtain the total by summing the averages?

e.g. E[O][1] = sum between 1 and N of E[O][1][n], where n is the training sample index and {1 <= n <= N} (Is that correct?)

Reply
- Mike says:
  
  December 20, 2018 at 6:25 am
  
  EDIT – forgot to divide sum by N
  
  Reply
John says:

December 20, 2018 at 4:35 pm

I want to thank you for this write up. Best one on the net! Your breakdown of the math is incredibly helpful

Reply
Oleg says:

December 23, 2018 at 5:22 am

Don’t we need to update b1 and b2?

Reply
- Braydon Burkhardt says:
  
  January 10, 2019 at 10:07 pm
  
  For really basic NN’s, it is not needed and can be left alone. Think of the biases as shifting your graph while the weights change the slope. If you wish to update the biases, simply do the same thing as the weights but don’t add the previous node as it is not attached to it. For example b2 is the same as w5, but without the hiddens (w5 would be h1) because it is not connected to any hiddens. b1 is the same as w1 but without the hidden stuff like h1(1-h1). Hope this helps, if not just leave another question.
  
  Reply
  - Braydon Burkhardt says:
    
    January 10, 2019 at 10:09 pm
    
    Sorry but I messed this up, the h1(1-h1) would be removed for b2 NOT b1, as b1 connects to hidden layer and the b2 does not. For b1 you would remove the input (i1) stuff.
    
    Reply
Raghavender rao says:

December 26, 2018 at 2:04 pm

This is really amazing.thankyou for the clear explanation.

Reply
Neuronal networking in VBA+Excel |
Raymond Zhang says:

December 29, 2018 at 7:50 am

Excuse me. If I have 1000 training data, In bckpropagation, the loss function would be E=sum((target1-out1)^2+(target2-out2)^2+……(target1000-out1000)^2), should partial derivative of the weights to E be divided by 1000(the number of data) when update the weights

Reply
Ghulamuddin Ansari says:

December 29, 2018 at 10:50 pm

Wonderful explaination!, btw how to reduce biases at input and hidden layer?

Reply
Deep Learning Resources – Machine Learning
Fabio says:

December 31, 2018 at 7:38 am

Bravo Matt, la migliore spiegazione!

Reply
A quick introduction to derivatives for machine learning people – Data Science Austria
prometheusgr says:

January 6, 2019 at 5:49 pm

Hey Matt, I’ve been trying to duplicate your results in a network I am building, and running into an issue. Using your examples as my test data, I have passed all tests except for the weight values in w6 and w7. My values are not matching.

From a sniff test perspective, w6 appears it should be increasing since the output is lower than the expected (0.7 -> ~0.9) , but your calculation has the weight being reduced (0.45 -> ~0.41). With w7 your example has the weight increasing (0.5 -> ~0.51), but it appears it should be decreasing since the output is higher than expected (~0.7 -> 0.1).

I know this is an older post, but if you could double check those values and let me know, that would help me a lot.

:)

Reply
Maria says:

January 7, 2019 at 7:11 am

Thank you Matt for your very good explanation! How do you create the figures of the basic structre, … etc. ?

Reply
amin says:

January 10, 2019 at 4:13 am

Hi I need help, please have someone help me

Reply
- Braydon Burkhardt says:
  
  January 10, 2019 at 10:01 pm
  
  What is your problem?
  
  Reply
sadikul haque says:

January 15, 2019 at 8:24 am

that was very nice

Reply
Johannes G. Mooyman says:

January 15, 2019 at 10:08 am

could the problem be solved linear programming ?

Reply
Introduction to Neural Networks – Glass Box Medicine
signalspa says:

January 23, 2019 at 9:41 am

Wonderful explanation, sir!

Reply
Walter says:

January 24, 2019 at 3:18 pm

I’m stuck at calculating w1. The procedure says to calculate dEtotal/dw1 we need dEo1/douth1, proceeds to use the chain rule, and substitutes 0.74136507 for dEo1/douto1. 0.74136507 was earlier given as the value of dEtotal/douto1. What am I missing? Why are the values the same? How is dEo2/douto1 calculated? Thanks in advance!

Reply
- Walter says:
  
  January 26, 2019 at 7:46 am
  
  I actually got it after a few repeat attempts. The key is, of course, in using the chain rule more to go deeper from d/dout to d/dnet. It’s then possible to calculate everything.
  
  Reply
Furkan says:

January 31, 2019 at 10:23 am

why the w1 partial derivative formula is not,

i.hizliresim.com/nQlkpa.jpg

Am I wrong?

Reply
Ali Çelik says:

February 2, 2019 at 2:54 am

I think neth1 will be 0,25*0,10, not 0,2*0,10, because w3=0,25. Am I wrong ?

Reply
- Ali Çelik says:
  
  February 5, 2019 at 3:31 pm
  
  Yes I am wrong :))
  
  Reply
ticky says:

February 3, 2019 at 1:33 pm

Excellent !!! realy great ! thx !!

Reply
Jeremiah says:

February 8, 2019 at 9:56 am

This was one of the best tutorials which helped me have a firm understanding of the dynamics of backpropagation algorithm in NN. Thanks man. Be blessed.

Reply
EM Farih says:

February 8, 2019 at 7:03 pm

Excellent!
But, is bias value need to be updated as weight?

Reply
Talita Anthonio says:

February 9, 2019 at 3:15 pm

One question: in the explanation of the chain rule (the image) why is w6 connected to h2? w6 is connected to h1 in the image, right? Only w7 and w8 are connected to h2.

Reply
nurjamil says:

February 14, 2019 at 6:52 pm

i dunno how that -1 come from cost function derivative, someone help

Reply
0chen says:

February 17, 2019 at 11:11 pm

Thanks!

Reply
Neural Networks Learning the Basics: Backpropagation – Sam's Blog

Matt Mazur

A Step by Step Backpropagation Example

Background

Backpropagation in Python

Backpropagation Visualization

Additional Resources

Overview