A Step by Step Backpropagation Example

March 17, 2015February 23, 2024Mazur

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.

Backpropagation in Python

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

Continue learning with Emergent Mind

If you find this tutorial useful and want to continue learning about AI/ML, I encourage you to check out Emergent Mind, a new website I’m working on that uses GPT-4 to surface and explain cutting-edge AI/ML papers:

In time, I hope to use AI to explain complex AI/ML topics on Emergent Mind in a style similar to what you’ll find in the tutorial below.

Now, on with the backpropagation tutorial…

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!

And while I have you…

Again, if you liked this tutorial, please check out Emergent Mind, a site I’m building with an end goal of explaining AI/ML concepts in a similar style as this post. Feedback very much welcome!

1,111 thoughts on “A Step by Step Backpropagation Example”

Asaf Zebulon says:

October 9, 2017 at 10:39 pm

this is the best explanation about backpropagation that I’ve ever found. thanks!

Reply
christineshen says:

October 10, 2017 at 3:29 am

Actually why is that? I don’t see the explanations here… to me the formula here is (output – target) * g'(z). And I still don’t know why I read in some other materials that the error term of the output layer is just (output – target).

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Edu says:

October 10, 2017 at 3:43 pm

Very nice explanation !
I hope you could clarify one thing. In your diagram, the same bias is applied to all perceptrons in a given layer (b1 is applied to h1 and h2 , b2 is applied to o1 and o2).
Is that the normal approach? Shouldn’t we have a bias per perceptron?
Thanks a lot !

Reply
1. Deadletter Grad says:
  
  October 18, 2017 at 3:16 am
  
  It appears there are several ways of doing it. One way is a single bias setting, no weights, to all nodes.
  
  Another is, as above, using one setting for each layer. Equal to that would be ONE bias (of say 1) but different weights to each layer.
  
  Yet another is one setting, but unique weights to each node.
  
  Since the only goal is to help the logistic or sigmoid function not pass through the origin, they all work – but I think the reason to have them be different is to avoid shifting the entire function to the right or left simultaneously at all nodes, forcing adjustments to be very miniscule. With different weights, it offers the chance to adjust different nodes in different ways.
  
  This example doesn’t include it, but some people also update the weights on the bias settings.
  
  Reply
Kaushik says:

October 10, 2017 at 10:05 pm

Awesome tutorial!

Reply
Dorin says:

October 12, 2017 at 4:20 pm

best resource on understanding back propagation I have came across.

Reply
Priya says:

October 14, 2017 at 6:34 am

Thank u for such a needful explanation

Reply
Waasala says:

October 14, 2017 at 1:19 pm

This valuable explanation saved a lot of time…Thanks a lot for the very clear explanation of BP function.

Reply
henry says:

October 14, 2017 at 7:21 pm

Thank you for the great tutorial!
When you do:
d(E_total) / d(out_o1) = 2 * 1 / 2 * (target_o1 – out_o1)^(2-1) * -1 + 0

Where does the part >> * -1<< come from ???

Reply
1. Alex says:
  
  October 18, 2017 at 2:22 am
  
  @henry
  
  (Fog)’ = g’ * (f’og)
  With g being a-x g’ is -1
  F being (a-x)^2
  F’og is 2 (a-x)
  
  Reply
2. Mustafa Murat ARAT says:
  
  October 29, 2017 at 10:22 pm
  
  You are getting the derivative with respect to output_o1. It is a variable with a negative sign, meaning (-1 * output_o1). If you take the derivative of a variable whose degree is 1 (a variable without an
  exponent actually has an exponent of 1), you will have a constant which is -1 here.
  
  Reply
Long le says:

October 17, 2017 at 11:27 pm

Very clear tutorial. Thank you very much

Reply
Jakes says:

October 21, 2017 at 3:58 pm

All examples I can find online shows an explanation of a single data set. How do you train an ANN with let’s say two data sets, as such the weights for both data sets are the same, but it yields the correct results for each data set?

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Jonathan Woodbury says:

October 22, 2017 at 9:50 pm

Thank you very much.

Reply
Emmanuel Tetteh says:

October 24, 2017 at 6:46 am

Very insightful. Great work.

Reply
Nishant says:

October 25, 2017 at 2:28 pm

Thank you so much ! Couldn’t be better :))

Reply
Amit says:

October 26, 2017 at 10:35 am

Awesome stuff. Exactly what beginners need.

Reply
Miguel MG says:

October 27, 2017 at 6:41 pm

Very good explanation

Reply
Shashi Dhungel says:

October 30, 2017 at 2:06 pm

This is a great tutorial. One thing that will help is to explain why the error rate only goes down and why the error does not increase. What forces the error to go down in every iteration. Why should we expect it to go down.

I believe it has to do with the slope. When the slope is negative then it causes the learning rate to add and when the slope is negative it cases the learning rate to subtract from the original rate.

Reply
shashidhungel says:

October 30, 2017 at 2:09 pm

This is a great tutorial. I think it will be clearer if you can explain why the error rate only goes down with each iteration (assuming there is only one equilibrium). How the slope directions contributes to ensuring that each iteration decreases the cost?

Reply
Mario A Vinasco says:

October 31, 2017 at 1:27 pm

Matt, very good tutorial !!
Q: how do you apply back propagation on batches? lets say I feed forward 50 training examples from my data set, I calculate the errors for each, then do I take average Error and do the back prop? Thanks

Reply
mariovinasco says:

October 31, 2017 at 1:30 pm

Matt, how does BackPropagation work with batches of training data? If I do 50 training samples per batch, do I take the average of the errors and do the back prop ? thanks

Reply
Yiting says:

October 31, 2017 at 1:42 pm

Super clear explanation 👍🏻

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Sonya says:

November 1, 2017 at 7:34 am

Thank you very much for this example, it has made the concept so much more clear!

Reply
adideshp says:

November 2, 2017 at 8:46 am

Great Tutorial. Thanks a lot :)

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Felix says:

November 2, 2017 at 11:45 pm

I got a solution after 2 months learning.. :) Thank you very much..

Reply
Richard says:

November 3, 2017 at 8:52 pm

Very Very good

Reply
Nasim Mahmud says:

November 4, 2017 at 9:08 am

Hey there, I want to have an output value 2.5 but using the equations above I only get output value 1. What should I do now?? I have set the target value 2.5 but also get output value 1.

Reply
Burak Ipek says:

November 4, 2017 at 6:49 pm

Thank you. I was studying for 2 weeks without understanding this. You saved me.

Reply
Sam says:

November 5, 2017 at 6:29 pm

I’m stuck. How do I update the bias?

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Kaezer says:

November 5, 2017 at 9:13 pm

Thank you, Professor.

Reply
julianthefrank says:

November 7, 2017 at 4:11 am

Reblogged this on Julian Frank's Blog and commented:
A Really Good Step by step Explanation of Back Propagation in Neurons

Reply
D-Frame says:

November 9, 2017 at 4:03 am

Ignoring the whole backpropagation thing, I don’t even understand how we could ever wind up with 0.01 as final output. Using the “squash” logistics function above, even when all inputs and weights are zero, we get 0.5 as output. You can’t ever go below 0.5 this way. What am I missing?

Reply
1. Jian Ji Xing says:
  
  November 23, 2017 at 8:19 pm
  
  The weight can be negative. ;)
  
  Reply
  1. D-Frame says:
    
    November 27, 2017 at 2:44 am
    
    Oh, right! Now that you mention that, it seems obvious. Thanks!
    
    Reply
Hür AKDÜLGER says:

November 17, 2017 at 3:11 pm

You give theory and practice. Very very useful.

Thanks, thanks, thanks.

Reply
Hubert says:

November 18, 2017 at 2:01 am

While each forward path of an neuron to neurons of next layer has its own weight, why is there only one bias weight shared by the whole layer?

Reply
k says:

November 21, 2017 at 9:28 pm

awesome explanation

Reply
MUNEESWARAN.G says:

November 25, 2017 at 7:10 am

Thanks a lot for nice Explanation for beginners.
Muneeswaran.G

Reply
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vu dat says:

November 28, 2017 at 11:51 am

Thanks a lot. but i dont’ understand this “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}}.”
Can you explain more?

Reply
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Dimos says:

November 29, 2017 at 5:34 am

many Many thanks. The most simple and also analytic explanation of backpropagation algorithm. I was looking for a tutrial like this for months!!!! Thank you again

Reply
premi says:

November 29, 2017 at 6:15 am

very much thankyou

Reply
Mithilesh Vaidya (@methi1999) says:

December 2, 2017 at 12:46 am

Thanks a lot! Was looking for an in-depth example for a long time!

Reply
Pavel Koryakin says:

December 4, 2017 at 10:25 pm

Great tutorial!
But how can I update biases using back propagation ?

Reply
1. Matus Moravcik says:
  
  January 28, 2018 at 9:52 am
  
  You can imagine biases as weights of a neuron that always outputs 1.
  
  Reply
Jack says:

December 7, 2017 at 4:03 am

Same way. Just take the partial with respect to the bias instead of the weight.

Reply
1. WangLu says:
  
  December 20, 2017 at 9:51 pm
  
  It’s easy to generalize the method accroding to weight update. You just need use the similar way to calculate dEtotal/dbias to get the step for bias and update it.
  
  Reply
shikha purwar says:

December 8, 2017 at 4:08 am

very good

Reply
HansMueller says:

December 10, 2017 at 1:32 pm

The best explanaition of backpropagation I’ve read! Really helped me, thank you!

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mobymotion says:

December 14, 2017 at 6:16 am

Thank you thank you thank you.

Just about to give up trying to understand back prop, before I saw this.

Reply
Mohamad Zeina says:

December 14, 2017 at 6:17 am

Thank you thank you thank you.

Just about to give up trying to understand back prop, before I saw this.

Reply
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