# 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.

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## 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.

If you find this tutorial useful and want to continue learning about neural networks and their applications, I highly recommend checking out Adrian Rosebrock’s excellent tutorial on Getting Started with Deep Learning and Python.

## 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:

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

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:

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 $\frac{\partial E_{o2}}{\partial out_{o1}}$, 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:

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

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

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 0.000035085. 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!

## 184 thoughts on “A Step by Step Backpropagation Example”

1. awesome article !! I have just one doubt…while calculating delta o1 , shouldn’t we multiply the given equation by weight matrix of layer 2 ? I saw that in Andrew N.Gs videos he does this and am hence confused

• alex hao says:

Think you ,papi! This is the best explanation about bp I have ever seen.

2. Machine Learning : Resources | New Technology Information

3. Diego Domenech says:

Thanks a lot, best explanation I’ve seen about backpropagation

4. Lavanya says:

Hey Matt, that was a very neat explanation! I am workin on BPA for SIMO system and tryin to implement using Simulink.

Is there any easy way to do in simulink?

5. 111 says:

Thanks a ton!!!!!!

• my guide is now published – it aims to make understanding how neural networks work as accessible as possible – with lots of diagrams, examples and discussion – there are even fun experiments to “see inside the mind of a neural network” and getting it all working on a Raspberry Pi Zero (£4 or \$5)

• Flipperty says:

I’ll be buying your book–it’s a great idea and well executed; but, I was hoping you could walk me through the differentiation of the total error function on pages 94-95 in the draft pdf version online. A minus that pops up that I can’t account for. The same happens in the above post, and I’m just not sure why.
Any hep you can provide is much appreciated.

• The minus sign pops up because d/dx (a-x) is -1.

In the example we have d/dx (a-x)^2 which is done in two steps. First the power rule to get 2*(a-x) but then the inside of the brackets too which gives you the extra linux sign.

This is actually the chain rule: df/dx = df/dy * dy/dx
so if f=(a-x)^2 and y=(a-x) then df/dx = df/dy * dy/dx = 2y * (-1) = (a-x) * (-1)

=================

Some people also get caught out by the weight updates. The weights are changed in proportion to – dE/dw .. that minus is there so that the weights go in the opposite direction of the gradient. My book illustrates this too with pictures.

• Flipperty says:

Thank you! That’s what I thought, but I just needed confirmation. A lot of people don’t mention the chain rule, so that’s what was throwing me.

Your book is excellent, and I bought a copy.

I hope it does really well.

• Thanks Flipperty – if you like it, please do leave a review on amazon. And if you have suggestions for improvement, do let me know too!

6. Amitesh says:

Hi Matt! Great article,probably the best I’ve come across on the internet for Backpropagation. Just noticed a small error – When you calculate outo1 for the first time in the forward pass, the formula should be 1/1-(e^-neto1) whereas you have written it to be 1/1-(e^neth1). That’s it! Amazing article once again. You’ve got me hooked on to NNets now and earned yourself a worshipper :)

• Thank you (and the others) for pointing this out. Fixed!

7. Hi Mat, why aren’t you putting weights on biases and update them during back prop?
thanks
joseph

8. I still see the typo here, is there a revised version of this page or you just missed it?

• Thanks, that was very helpful. Fixed!

9. Dara says:

Doesn’t it map the training output before calculate the error with target value?. Because “logistic function” outputs are always between 0 and 1. Then what happen if my target is really high (eg:300)? error would be really high?. After finish training, my target is 300 and final output would be somewhere between 0 and 1, since transfer function at the output node is “logistic”. That is the part which I still cannot understand. even after training, doesn’t it map the output to be large (should be taken near to 300 from 0-1)?

• You should map your desired outputs to match the activation function – otherwise you will drive the network to saturation. The same is also true of the inputs – you should try to optimise them to that they match the region of greatest variation of the activation function .. some people call this normalisation. My gentle intro to neural networks has a section on this .. http://www.amazon.com/dp/B01EER4Z4G

10. John says:

Hey Mazur, great tutorial, as said earlier, the best on Internet. Just have a question, when you have more hidden layers, the concept for the feedforward is the same, but the derivatives surely will modify, my question is how to calculate it? It will be something like Dtotal = DTotal/Dout * Dout/Dnet * Dnet/Dweight * Dweight/Douthidden * Douthidden/Dnethidden * Dnethidden/Dweight? Don’t know if you will understand the way i wrote it xD Thanks in advance.

11. roberto brunialti says:

Thanks Matt for the great article. Even me, a math dummy, was able to implement the back prop algorithm in C++…
I’m worring abaout generalization of this algorithm: is it viable for networks with more than an hidden layer? It seems to me that the hidden part of the algorithm
is specific for jus one layer.

• Krisztián says:

Theoretically only one hidden layer is enough for approximating any function, there is no need for more than an hidden layer.

• sar33 says:

True, but the key word here is “theory.” In practice, using one enormous layer is vastly inefficient and prone to runaway numerical errors. In practice, always use multiple (smaller) layers if your models need a large representational capacity.

12. mehmet says:

This is wonderful. Thks a lot.

13. Great article Matt, thanks! How do the bias parameters of each neurone come into play with regards to back propagation?

14. clear explantion,thank u!

15. qwerty says:

I have a question on calculating ∂Eo1/∂ω1 using the chain rule. Eo1 is associated with both of outh1 and outh2. In your calculation, however, only the ∂neto1/∂outh1 is taken into account without consideration for ∂neto1/∂outh2. Could you explain why ∂neto1/∂outh2 can be ignored in computation?

• sigmoid says:

hi qwerty, im not 100% sure with this but im guessing that w1 is not directly related to H2 that’s why. And since youre only computing for the rate of change of Eo1 with respect to change in w1, dNeto1/dOuth2 would just like be a constant/negligible.

16. Thank you. Finally something putting the theory into figures for all to understand! Next to understand layering ;-)

17. silvio says:

Thank you. It help me a lot.

18. Thank you so much for awesome tutorial. Is there such a tutorial for the matrix form ?

19. ZZD says:

does b1,b2 need to be updated ?

20. Anup says:

Yeah, same query. Don’t we need to update the biases?

21. mobaptest says:

Excellent explanation

Excellent article. Can you please explain a little bit “pattern detection in EEG signals” where at the output node one would classify “YES” (if certain pattern detected) and “NO” (if certain pattern not detected). Only thing I’m unable to understand is that how an actual pattern can be represented with a single Label & where do we use the actual pattern in NN.

23. thor says:

Hello Matt, great tutorial !

When you calculate w5 in the backpropagatino pass, you got that Derivative(Etotal) / Derivative(w5) = 0,082167041.

I simulated the NN here and when I changed w5 from 0,4 to 1,4 , the Etotal didn´t changed by 0,082167041. What does it means the Derivative(Etotal) / Derivative(w5) ? How can we interpret this ?

Thanks

Thor

24. what if we have a deep network, like 1+ hidden layers ? then how this error propogation is done ? ( I would like to know how the equation will get updated ).

25. Excellent explanation . Thank you so much .

26. ana says:

A lot of steps are missing in this tutorial, but I’m pretty sure the numbers w6+ and w7+ are wrong.

27. ana says:

Further to my previous comment, pretty sure you got w6+ and w7+ wrong. You have mixed up the two deltas that go into their calculations.

28. jesse says:

nice work

29. Alexandra says:

Great explanation! It helped me so much, thank you!

30. Totoro says:

This is great! However I feel that the OUTh1 is incorrect. (0.25*0.1)+(0.3*0.1)+(0.35)=0.405, and 1/(1+e^(-0.405))=0.599888368864, not 0.596884378 as stated above.
But this explanation is great, thank you for helping me learn!

31. Dear Matt,

Thanks for this lovely post. I have tried to implement this using C#, source uploaded at github https://github.com/animesh/ann.

I could reproduce values after first iteration, however after 10000 iterations, you report

Error 0.000035085
Output1 0.015912196 (vs 0.01 target)
Output2 0.984065734 (vs 0.99 target).

while i am getting

3.51018778297886E-05
0.0159136204435507
0.984064273514624

changes are minor, but I am just wondering if we are actually diverging the values due to the way double type is handled in Python and C#?

Best regards,

Ani

32. Satheesh says:

The point that I canNOT relate or understand clearly is,
a) why should we use derivative in neural network, how exactly does it help
b) Why should we activation function, in most cases its Sigmoid function.
c) I could not get a complete picture of how derivatives helps neural network.
d) What’s actually happening with all those calculations and derivatives

33. Naysan says:

Hi Matt,

thank you so much for this tutorial. I really appreciated the numerical values you provided, they helped me check that my own computations were correct.

Your tutorial inspired me to write a python code that would replicate the neural network from your tutorial. I made the same neural net (with the same initial values as in your tutorial) run for 1000 steps and displayed the evolution of the outputs and errors in a plot.

Here’s how the plot looks like :

The python code is on gitlab here : https://gitlab.com/NaysanSaran/simple-python-neural-net-example

It’s my very first time writing a neural net so I would love to have your input on how I can make the computations more efficient.

34. So as far as I can understand, the biases are constants so we don’t need to bother changing them but why don’t we just get rid of them?

35. Los mundos de Carchenilla

36. TND says:

the best tutorial on the web.

37. Backpropagation Example | Audio & Speech Signal Processing using Machine Learning

38. neoli365 says:

I have a clear intuition now, thanks for you work. I want to translate it into Chinese, if you don’t mind. But, to be honest, what I should do is only to edit some conjunctions, since you have made so many pictures which are easy to understand! :)

• It’s fine if you want to translate to Chinese :)

• neoli365 says:

Thanks again!

39. johhny applesed says:

Great tutorial, thanks a bunch!

Why is it that w1 can be readjusted without including w7? I ask because w5 is included, and both w5 and w7 connect to outputs so I can’t see the difference.

40. AKat says:

Thank you very much for your article. It is very well explained. I am reading neuralnetworksanddeeplearning by Michael Nielsen and had trouble understanding backpropagation. Your explanation is very intuitive and extremely useful in understanding how to do it by hand without reading lots of lots of text several times.

41. This is a really clear explanation about backpropagation, it has been a pleasant reading. Thank you very much for sharing!

42. Backpropagation | Neural Networks

43. Pawel says:

Thanks for this article. I think I found one typo, correct me if I’m wrong. In the last blue frame, first formula. Shouldn’t it be d_Eo / d_out_h1 instead of d_Etotal / d_out_ o ?

44. Alexey Balashov says:

Thank you for a good example.
I have a question.
Is it possible to implement the algorithm in Matlab?