# 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!

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

1. Niloufar says:

thank you very much
it was so good :D

2. Caspar says:

Really great tutorial. How would the pattern of back propagation carry on with multiple hidden layers instead of just the one? What would the overall algorithm look like where the hidden layer count, weights, etc, are all variable?

• Shakezz says:

I also would be very interested in the answer!

3. Lukas Cerny says:

Thank you an amazing article! Everywhere else the description of the algorithm is extremely theoretically, so I am glad that you got your hands dirty by really describing the actual implementation with numbers and not just by loads of symbols, which particularly in the case of backpropagation was very confusing for me. Cheers from Edinburgh.

4. Caspar says:

• Hello Caspar, did you get information about how to implement it for multiple hidden layer ? I am also looking for this. Thank you.

Thanks for great article!
Question: how do we tune the weights of biases?

Hi there,
I think those are same with same layer’s weight but the output value won’t update.( in example suppose 1 for biases weight )

Sorry Matt please update my post with this:
Hi there,
I think those are same with same layer’s weight but the value of biases won’t update.( in example supposed 1 for biases weight )

6. Exploring Neural Networks… – Cluster Chord

7. ali hesri says:

great

8. Syed says:

Thanks alot for such a wonderful explanation.

9. Nishant says:

Awesome!

10. Tariq says:

This was a pleasure to read with beautiful diagrams and numbers to make it real. Thank you.

11. Josef Kerner says:

Best explanation so far, thanks so much!

12. This is probably the best explanation I’ve seen so far. Thank you so much!

13. Lawine says:

Thank you for your great explanation! but i have some questions about it.

in your post, at [The Backwards Pass // Output Layer] part,
(d E_total/ d out_o1) = 0.74136507.

and at [The Backwards Pass // Hidden Layer] part, you said
‘We can calculate (d E_o1/d out_o1) using values we calculated earlier.’
(d E_o1/d out_o1) = 0.74136507.

(d E_o1/d out_o1) = (d E_total/ d out_o1) ?

• dogan says:

I wonder this too, is there anybody?

• neyim123 says:

Yeah, I wonder this too, anybody?

• Shree Ranga Raju says:

I guess it is because, E_O2 is independent of Out_O1 and is treated as a constant and when you take the derivative it is zero. Could be wrong.

• Dubois LEE says:

maybe:
E_total = E_o1 + E_o2
d(E_total) = d(E_o1 + E_o2) = d(E_o1) + d(E_o2)
so, with respect to d_out1, the second part will be zero.
As @Shree Ranga Raju said.

I think, Yes
those are same:
E_total = E_output1 + E_output2 => E_total = E_output1 + 0 for E_o1 and E_total = 0 + E_output2 for E_o2.
for calculate E_o1, E_output2 can’t effect and is equal 0 then they have same amount.

14. Li says:

A very helpful explanation thank you.

15. amazing article, those diagrams are very clear. thanks.

i have same question, it was asked before. how do we update biases weights?

16. shelton says:

Thanks! Best explanation I have ever read.
Question: Does it make any difference if I choose bias to be -1?

17. richardshandross says:

Hi, thanks for this explanation. What I don’t understand is why the weight adjustment is dEtot/dw instead of (dw/dEtot)*(Etot)*(eta).

Thanks, Rich

18. sridhar reddy says:

19. Best explanation I’ve seen so far on backpropagation. Great job, many thanks!

20. Mojeeb says:

Thanks much…

21. Learning Machine Learning | ebc

22. Sam says:

Why is the weight for the bias the same for a layer? For instance, for the input layer, the bias going into the hidden layer nodes h1, h1 has weight b2.

23. seaofocean says:

amazing! Thanks!

24. 7 Steps to Understanding Deep Learning – What does a Ph.D need?

How can I implement it for multiple hidden layers (more than one) ?

I think, You can do it for any number of hidden layer just you need to repeat process between output layer and hidden layer(Section Hidden Layer) for extra hidden layer you have.

26. Nice tutorial thanks. But how can we make this to work for multiple hidden layers(not a single hidden layer) ?

27. Kanchan says:

Nice Explanation. Thank you.

28. dfdds says:

Awesome!

29. hathal says:

Thanks.

30. Jerzy Kaczorek says:

Very good article, however:
neth1 = 0.15*0.5 + 0.1*0.2 + 1*0.35 is not equal 0.3775 but 0.445
I suggest you examining all the calculations in order to improve your credibility

• Jerzy Kaczorek says:

Sorry, I was wrong. You have 0.05 not 0.5. I made a mistake by wrong copying data.

31. You are explaining that dE/dO1 = 1/2 * 2 * (target1 – output1) * (-1) = 0.7414
But dE/dO2 = 1/2 * 2 * (target2 – output2) * (-1) = -0.217
Above you have considered it as +0.217
By taking it as negative, makes w7+ = 0.48871 and not 0.51130 as you have calculated.
Could you verify it?

32. Davide Quaroni says:

Thank you very much, this was a great explanation and I could check the results of my neural network output by confronting them with your calculations. I also managed to improve the error reduction after 10000 cycles by updating the biases of neurons during backpropagation.

33. How I learn Neural Network (and Deep Learning) | Malioboro;

34. Excellent! Thanks for writing this up.

35. superbo says:

Thank you. This is easy to follow. This helps me a lot.

36. Ognyan Zhelezov says:

As far as I understand, all these calculations are made to get two-dimensional output vector (target) as a result of two-dimensional input vector. I think, one linear system of two equations can give the same result. Am I right?

• Nops. That neural network forms a non linear equation. Note that each node act like a function and each of them have a output to another function. This kind of network can produce equations like f(x,y) = x^1.24 + y^0.59 +2.25x + 1.1345x*0.1y – 0.33y, depending of number of nodes.

37. Murry says:

It was an amazing tutorial. Succint, yet comprehensive. Thanks a million.

38. Nicely done! Thanks for doing this – helped me a lot!

40. Paresh Kamble says:

Very nice explanation with sound examples.

41. Thank you,

Very clear and easy to understand.

It gave me an end-to-end vision.

Other articles only explain math calculus, but not a systematic diagrama.

42. Neural Networks: The mechanics of backpropagation | Giga thoughts …

43. Bikram Gupta says:

Amazing explanation. Thank you!

44. AI Project, Part 2: ANNs – Ethan the Block

45. Russell Dunk says:

Thank you so much!! This made this algorithm so much more clear to me. The one typo I noticed is that the error after 10000 epochs in your github code is 3.5102e-05 and when I created my own network I got the same.(3.510187782978861e-05) but the article says 0.000035085.