# 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.

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:

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 $\frac{\partial E_{o2}}{\partial out_{h1}}$, 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!

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

1. yoba says:

Very comprehensive and useful example. Many thanks !

2. Himanshu says:

While calculating w1, shoul i use updated w5 or original w5?

• Hitesh says:

original weight

• No, 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).

• Nehi says:

Original

• Sushovan Haldar says:

The forward pass and backward pass consists of one epoch.In this one epoch the weights are being updated using the just immediate previously calculated weights.
Since in this example,the author has updated the weights only once,so the original value of w5 is to be used here.

3. Really good tutorial Matt. I finally had something like a “aha” moment ;-)

With the inspiration of this tutorial I translated this real number example into matrix multiplication operations.

4. jerpint says:

Hello, thanks for your explanation. Can you please clarify why the biases b1 and b2 are the same for their entire
respective layers? Shouldnt we have a bias vector b of the same length as the number of neurons per layer?

5. Like to print things out and read over. Seems like the print out starts omitting detail once it hits the hidden layers. Odd. All in all this is quite helpful to understand the back propagation.

6. Ab says:

Really appreciate this, thanks!

7. KG says:

thank you very much best explaining i have ever seen

8. mahdi says:

that was very helpful < thanx from a fighter in

9. Nitzan Wilnai says:

This is really great. Like the others I have 2 questions. Why aren’t you updating the bias weights, and when moving to w1-w4, should we use the original w5-w8 or the new updated ones?

Thanks for this article….But i would like to know how performing backpropagation using deep learning? if you can give me an example of how training a DB of n 2D images.

11. TOMAZ STIH says:

Great work. I was able to develop my first net using your tutorial as a test case. One question: using Etotal one can update weights for the output layer. Then using both E1 and E2 one can update weights for the hidden layer. What would you use for errors if you had more then one hidden layer (same E1 and E2?)

• Messy but brilliant. Was about to do it in Excel so you saved me a weekend. Thank you. Well done.

12. thank you very much for best example

13. Wow. could not ask for much granular level explanation. Thanks so much.

14. M Karthikeyan says:

Thanks for this useful explanation. I have a doubt if you have the time. Can the same back propagation be modified slightly to work with a convolution neural net. Any comments would be very helpful.

• MsMC says:

At the start of backward propgation, why is the E_total being considered for partial derivative with respect to w5. Shouldn’t it just be E_01? (Since E_02 doesn’t have any effect of changing w5).

15. Satya says:

Matt good explanation. One quick question is that you pointed out after 10k iteration the error should drop. Does that mean to repeat with same instance (ie i1 is 0.05 and i2 is 0.10) or do you randomly generate i1=2*i2 where 0<i1<0.45 of 10k instances

16. Suchir Bhatnagar says:

Hi,
While calculating H1 should we not use the weight 0.25 and not 0.20

• matze says:

Nope, I had the same issue, but the sketch of the net is a little bit confusing.

17. Abdul Waheed says:

“Maa Shaa Allah”, Great Skills. First time i understood completely what is Back propagation. I am PhD student @ bsauniv.ac.in

18. At the end I have understood backpropagation. Thank you

19. bob sam says:

Thanks to your simple explanation of backpropagation I was able to successfully build a neural network in java. You have the best explanation of this stuff on the internet!

20. Aram says:

Sir, first of all, it is very clear. However; I think you should use the updated weights for the next training input not the same input.
Can you kindly clarify that?

21. 反向传播算法入门资源索引 | 机器虫

22. Sushant says:

Thanks Matt. This is very helpful. I a newbie question. How that the NN is trained how can i use this code to give me an output of 0.01,0.99 when the input is 0.05,0.1

23. dataminor says:

it is cool! thx!

24. Ryan says:

What are the weights after the 10,000 iteration?

25. Fernando says:

Hi, first I´d like to thank you. this tutorial made everything clear to me and helped me building my neural network.
But, I have a question, What book or referencee did you use to learn to develope this tutorial ? I really liked the way you exposed the topic, so I wanted to learn a little bit more.

Hoping for hearing from you soon.

26. This truly is an amazing explanation of backpropogation!

I’m stuck on one thing though, and I could really use some help. Just after this statement:

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

There’s a negative one (-1) in the formula of the partial derivative that I can’t for the life of me figure out where it’s coming from.

Any help is greatly appreciated, thanks so much!

• Timothy Kanarsky says:

A bit late, but hope this helps:

The -1 is a result of the chain rule. Let target[o1] be x and out[o1] be y. Then the expression we want to differentiate (ignoring the output and target of o2) is (1/2)*(x – y)^2. We first find the derivative of the outer expression (using the power rule): 2*(1/2)*(x – y), which simplifies to (x – y). We need to continue differentiating, however, because x is constant and does not vary with y. So the derivative of -y is -1. Then, by the chain rule, we multiply the derivatives of inner and outer expressions: (x-y)*-1. That’s where the -1 comes from! Hope this helps!

• I had this confusion too. Thanks for clarifying.

• Matthew says:

Yes, perfect, thanks so much!

• maashu says:

Yes, thanks so much!

27. Farid Haziyev says:

Thank you so much. It was very helpful and clear.

28. Thanks much – this is very helpful. But one question: Shouldn’t the b1 and b2 values also get updated during the back propagation? Or did I miss that?

29. LSTMs in even more excruciating detail – statisticalinterference

30. Leo says:

Thanks a lot. You clearly explained everything.

31. John says:

Thanks so much. Finally I understand backprop.

32. Yuumnam Kirani Singh says:

Nice tutorial, well explained!

33. Breezy says:

In backpropagation how is E_o2/out_h1 = -0.019049119 is calculated? What is the chain rule equation for this? i am not getting how to relate error E_02 and h_1. Kindly help me

34. your the man brooo thanks ….

35. Minh says:

Very good example, thank you very much!

It is very detail of Backpropagation NN, Thanks!

and, I has a question to more detail:
In this example, you had only one pattern {0.5, 0.1}, so you repeated the training many times on the pattern. If I have a patterns vector (many pattern to input), then at each repeating, we will use next pattern (of the vector)?

one more, Thanks!

37. Sean Walton says:

I want to be able to create a neural net that can recognize voiced phonemes. This means that I will not know how many A-to-D samples I will get until the pattern completes. I guess a network could check sample by sample, returning a FAIL, until the phoneme finishes. Do you have any ideas?

38. Jacob Johnson says:

Where you’re calculating the derivative of netO1 with respect to w5, you’re taking w5 to the power of (1-1) which should be to the power of 0, which should be equal to just 1. Instead, you put .59 (the value of w5) there. What gives?

39. Thanks a ton!!

40. ThangPM says:

Thank you so much for your post!!! I really appreciate it!

41. Homer Simpson Guide to Backpropagation | Sefik Ilkin Serengil

42. Matt
I am very new to ML but I am ok with Math. Thank so much for a very good illustration. Made so much more sense to me. To many tutorials remove the rigorous treatment you put in.
Thank you.
Mike, Cape Town, South Africa
Ps: Have you thought of doing this with an Excel as well?

43. Thank you! This help me find a mistake in my neural network implementation. I added a unittest making sure a backpropagation step produces just these outputs.

44. Julia says:

Thank you very much! It helps a lot! Very useful article!

45. Anuj Sharma says:

what is the use of bias in neural networks?
it would be great Matt sir if you elaborate my doubt with another brilliant example.