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

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

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

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 0.0000351085. 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!

468 thoughts on “A Step by Step Backpropagation Example”

1. Buzz says:

Quick question to anyone who can answer… I know that the Bias inputs remain 1 at all times. But do b1 and b2 change throughout the algorithm as w1-w8 do? Or do we keep them as the initial guess values?

• Buzz says:

Thank you in advance.

• jizele says:

It stays the same throughout

• mfernezir says:

Yes, in practice biases get updated as well. Moreover, each of the neurons should have its own bias. For example, this means that in practice there would not be just one b1 for both h1 and h2. Each of those 2 neurons would have a separate bias that also gets updated in the backward pass.

• Satrajit Kar says:

I had the same question and from available online literature that I read up on, it seems that the bias weights should also get adjusted. For example, what if one started with a bias weight b1 of 100 million – it might never converge since the net(h1) and net(h2) values would hardly be influenced by i1 and i2. Am not totally sure, but it seems only logical that bias weights should also be adjusted the same way w1-w8 are being currently done. If they shouldn’t get adjusted, then it would really be great if someone could explain why.

2. filippo says:

Ok, everything is clear! But at the moment when I choose for the output layer not a sigmoidal activation function, but linear, what changes in the algorithm?

• zq lee says:

The derivatives, for example \partial out_h1 / \partial new_h1 will be changed.

3. Is this same as gradient descent, or that is different from this?

• zq lee says:

I believe this is gradient descent.

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5. Daniel Philpott says:

What is your suggestion if you have a huge number of input nodes and therefore the sum e.g. neth1 is huge, and so your activation function turns literally every neth1 into 1 because there is a (1 / e^-nethx) in the activation function?
It feels primitive to simply change the function by dividing for example

• zq lee says:

This is a very insightful question! There are alternative activation functions such as tanh, ReLU, Leaky ReLU etc.. Dead ReLU problem is usually referred as your question. One remedy is to use Leaky ReLU.

• George Saman says:

I am having the same problem, Your article is very helpful and i designed my neural network based on your article. The problem arises when number of inputs is high and that causes the activation function at the hidden layer to always output a 1. Any suggestions?

6. Minseok kim says:

Hello, I’m university student in korea.
I study neural netwrok, so I like your writing because it is easy to understand.
Now i make program back propagation in c#.
I have a question one thing. AND gate, OR gate is understand, but
How do I handle actual data?? for example, weather data, fonding data..

• Kostyan1996 says:

It’s too complicated question for such topic, in fact, this is what professionals use neural networks for. Main and the hardest task for making neural network is choosing parameters and form of input for neural network. You should read some articles about it, but basically you should represent input data in numbers that neural network can work with, choose certain number of hidden and output neurons and then experimentate. Experiments is what neural networks are all about.

• Minseok kim says:

Thank you. I wanted solution, but new thing leaned to your reply.

7. Mak says:

Outstanding explanation Matt ! Appreciate the excellent and detailed walk-through. This will help anybody trying to put together NN’s to get a complete understanding of the underlying math.

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9. katch says:

thanks!

10. After this line :
“
Here’s how we calculate the total net input for h_1:
“

instead of w2 there should be w3. means w1*i1 + w3*i2 + b1*1

• greenFr0g says:

yes, this part doesn’t make sense with the diagram, and I think should be corrected.

• greenFr0g says:

I stand corrected, w2 is the weight from i2 to h1, and w3 is the weight from i1 to h2

11. You forgot to include how bias is updated.

12. Melo says:

Great!

13. Edmund Gyasi says:

Good job !, I like the way you explain stuff…..

14. DangHoang says:

Thank you for your writing. It’s really interested.

15. Dimitris K says:

Dear Sir Mazur, I cant thank you enough for your exquisite tutorial. My only question is the following.
If we had more hidden layers, lets say 2, how can I calculate the derivative θΕ(tot)/θout(h-1), in order to correct the weights linking my input neurons with the first h.l. neurons? In the second hidden layer we wouldn’t have that kind of problem since we can link θΕ(tot)/θout(h1)=θΕ1/θout(h1)+θΕ2/θout(h1) as you wrote, but for a previews layer there is not “Error association”, as far as I can perceive. I apologise if my question is not well stated.

Thank you in advance

16. Mark M says:

Looks like you have an error in one of your formulas:
net_{h1} = w_{1} * i_{1} + w_{2} * i_{2} + b_{1} * 1

should be:
net_{h1} = w_{1} * i_{1} + w_{3} * i_{2} + b_{1} * 1

• hallberg87 says:

The diagram should have better spacing to eliminate the confusion, but,
w2 is the weight from i2 to h1, and w3 is the weight from i1 to h2
So his formula is good.

17. Deepak Mandge says:

Excellent examples with simple explanations. Thank you

18. Shinny says:

Awesome!

19. josephkj20 says:

The Second formula under section: Hidden Layer

should be

\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{tot}}{\partial out_{h1}} + \frac{\partial E_{tot}}{\partial out_{h1}} rt?

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

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22. pete says:

how it be used in DL,I mean calculated in image process(2 demension signal process),such as CNN.

23. Hi, I’m new to this topic so pardon me for asking simple questions or having wrong concepts. I understand that backpropagation algorithm uses gradient descent to find the minimized error, but you did not mention in your example that it stops when it reaches the minimum error value? Or rather, how do you know when you can stop the algorithm in this example?

• BhanuKiran Chaluvadi says:

Usually people set epoch. For example epoch = 10000, neural network will run 10000 times(forward and backward propagation).If they are not satisfied with the result they will tweak number of epoch.

Hi Melodytch, lets I explain with an example. You have 1020 data input and you feed your network 1000 data as training and 20 as test data set. anytime you training data, you calculate error with this formula: MAPE=( |output-target| / target ) * 100 , and calculate this for test data separately. any iteration control the MAPE for the test if MAPE for test improves, it is mean your network get a better result to predict. When MAPE for train improved but MAPE for test stop and get worse its mean your network come to overfitting and you should stop training.
We have other methods for stop training such as fix training epoch(iteration), those other friends mention about that.

24. Great Explanation! Thanks a ton!!!

25. alpha says:

26. kris says:

One of the best explanation for back-propagation that I have come across. Thank you!

27. Derrick says:

Dear Matt,
I have been studying Machine Learning and Neural Networks intensely for the past 9 months and this IS the best explanation I have come across (including 2 online MOOC’s – Stanford and Johns Hopkins Universities). Brilliant!

Bet regards,

Derrick Attfield

28. Sur says:

Thank you so much Matt ! You have explained the basic working backpropagation which is very helpful for the beginners of ML ! :)

29. endre says:

Some of the best stuff I’ve read, extremely clear using a concrete example!

However, I would so much have appreciated that you did your calculations using Cross Entropy instead of (or in addition to!) the Squared Error as error function. Also, I’ll also point out that you do not calculate the Bias change (and the nodes should have separate biases..)

Also, while you’re at it (!), how does the derivation change using ReLU’s instead of Standard Logistc?

30. Dimi says:

Great class! Thank you.

31. hgygty says:

very good thing .

32. Sam says:

This was very helpful for me. Thanks!

33. Pinto says:

I would like to thank for this post and I would like to make one question about the partial derivative of the sigmoid function for out_o1 and out_h1.
the partial derivative of the function: out_o1=1/(1+exp(-net_o1) ) with respect to the net_o1 in this post is : out_1*(1-out_o1) why not :net_o1*(1-net_o1)?

34. Jaloliddin says:

Hi there! Really need to clarify this point.
What steps will neural network run in case of a real data set? Let’s say, we run first record and found some error rate – e1. Will we pass forward our first record’s values again after finishing backpropagation with new weights or new weights will be applied to next record?
in other words, should we keep running backpropagation for one record until its weights are unsatisfactory?

35. Suresh says:

Great article indeed. I could not find any better resource than this one that explains backpropagation with an example.. Thanks for taking time to write this. This helped me a lot.

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37. pablo says:

thank you this was very helpful!

38. CESAR says:

Great, thanks

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40. young says:

Thank you for your nice example.
I’m curious to know how weights of biases are updated in this example. This example includes only one starting weight for each bias.

41. Hi, I really like this resource. Just wanted to reiterate other comments that I read which state that each node should have a separate bias weight (b1-b4) not just one bias per layer (e.g. b1,b2). Would be good to update the images to make that clear.

42. Great Topic with a smooth way of explaining. is there is any article for Convolutional neural network ?

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44. Pengshan Cai says:

Very clear presentation and helps a lot. Thanks!

45. Great explanation. Good illustration and to the point.

46. Navy says:

Hi. Great article! I have a problem with computing net input for second hidden neuron in the forward pass.

Basically I put weights in a two-dimensional array, with input index and hidden neuron index. Then for every hidden neuron and for every input neuron, I calculate the net number: netNumber += weights[inputIndex][hiddenindex]*inputVectors[inputIndex];
Next, I add biases and use activation function. The result is wrong (out_h2 = 0.655318370173895), but the out_h1 gets computed correctly. Any hints?