## 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.

## Additional Resources

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.

*net input*by some sources.

Here’s how we calculate the total net input for :

We then squash it using the logistic function to get the output of :

Carrying out the same process for we get:

We repeat this process for the output layer neurons, using the output from the hidden layer neurons as inputs.

Here’s the output for :

And carrying out the same process for we get:

### 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:

For example, the target output for is 0.01 but the neural network output 0.75136507, therefore its error is:

Repeating this process for (remembering that the target is 0.99) we get:

The total error for the neural network is the sum of these errors:

## 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 . We want to know how much a change in affects the total error, aka .

By applying the chain rule we know that:

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?

Next, how much does the output of change with respect to its total net input?

The partial derivative of the logistic function is the output multiplied by 1 minus the output:

Finally, how much does the total net input of change with respect to ?

Putting it all together:

You’ll often see this calculation combined in the form of the delta rule:

Alternatively, we have and which can be written as , aka (the Greek letter delta) aka the *node delta*. We can use this to rewrite the calculation above:

Therefore:

Some sources extract the negative sign from so it would be written as:

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):

We can repeat this process to get the new weights , , and :

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 , , , and .

Big picture, here’s what we need to figure out:

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 affects both and therefore the needs to take into consideration its effect on the both output neurons:

Starting with :

We can calculate using values we calculated earlier:

And is equal to :

Plugging them in:

Following the same process for , we get:

Therefore:

Now that we have , we need to figure out and then for each weight:

We calculate the partial derivative of the total net input to with respect to the same as we did for the output neuron:

Putting it all together:

You might also see this written as:

We can now update :

Repeating this for , , and

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!

can you tell me why h1 = w1*x1 + w2*x2 +b ? why not h1 = w1 * x1 + w3* x3 + b

The labeling is slightly ambiguous because the weights were not put right on top of the links/edges. w2 is not going from i1 to h2, it is going from i2 to h1. That is, the weights are labeled to the left of their respective links/edges, not to the right. This is a little bit more obvious in the first graph.

I think the correct formula for H1 is:

netH1 = w1*i1 + w3*i2 + b1*1

i use w3 instead of w2, since I2 is linked to H1 via w3.

Will you plz confirm?

many errors dude mainly the computation of d(h1)/d(o2))=-(0.01 – 0.772928465)*(0.772928465*(1-0.772928465))*0.45

It really helped me to understand how back propagation works. Keep up the good work.

Thanks Mazur for this numerical example. Such an example provides the best way to learn the working of an algorithm. It helped me a lot. Many thanks.

Going through the example, I was wondering whether the biases b1, b2 are being treated as constants. Is it not necessary to adjust their values also?

Yes. In this article, biases were treated as constants to keep things simple. If you understood the math explained in this article, you can easily update the biases as well. In reality, biases are also updated.

Are we updating the bias values by setting up the dE / db1 and dE / db2 chains?

This is great! Thank you for the step by step explanation

Hi matt

Great document, very pedagogic !

Maybe a little mistake in the calculation of the net_h1 and net_h2:

net_h1= w1×i1 + w3×i3 + b1

(Analog mistake for net_h2)

The mistake is communicated to the numerical application also

This explaination and visualization is very well understanding. It helps me so much because it takes me a lot of time to know how really backpropagation does. Thank you very much.

A Step by Step Backpropagation Example – Deep Learning

Outstanding explainer for back propagation. Thanks, Matt!

I was a bit frustrated with this backpropagation topic and was struggling to have a clear mental picture of backpropagation.

Your article radically improved my understanding. Your article was so clear that I was actually able to write my own code to implement backpropagation. Thank you very much.

If you ever plan to expand on this article, I request you to add some details about how weights are updated for all the samples (your article explained the case for one sample.)

Thanks again for this wonderful article!

Reblogged this on conlatio.

Goog job! But this is 3-layer network only. If it is 4-layer then how we calculate the dEtotal/dout(h1)?

dEh2/dout(h1) will not be know. Because we don’t have the value for dEh2 (the error for hidden layer 2). I need an explanation here. Thanks!

You use the value that comes from the previous layer. For instance dE/dout i1 = dE/dnet h1 * dnet h1/ dout i1 = dE/dout h1 * dout h1 / dnet h1 * dnet h1/dout i1, in which dE/dout h1 and dout h1 / dnet h1 have been calculated by the previous layer.

Question: do all neurons in a layer use the same bias weight or is there a individual weight per neuron? I.e. In the example it looks like that b1 is used for both hidden neurons and b2 for both output neurons.

Btw: nice and easy to follow example!

Implementing a flexible neural network with backpropagation from scratch

i think this is best explain backpropagation with detail kudos!

Important Links | Tejalal Choudhary

This was a fantastic write up. I am using it to study the algorithm while currently in an AI course at uni. I was wondering if you could add to this and describe momentum in the same way. Thank you

Thanks a lot Matt for making this. Your blog and the Stanford’s CS231n lectures are the best resources on this.

Deep learning for product managers – part 1 – Kai's notebook

Thank you very much! I’ve been looking exactly for this

【每日AI收集】BP神经网络 - 每日AI

This page is very helpful to solve the problems.

Amazing page !!

Thank you !!

Thanks for the great example. One question though:

During the backpropagation, you find the new values for w5-w8 and continue backpropagating the previous hidden layer. There you need the value of dnet_o1/dout_h1, which equals w5. Do we use the original value of w5, 0.4, as it is done here or its updated value, 0.35891648 ?

Thank you for explanation. I didn’t get one thing. Why we multyply by -1 derivative of totel error? is this because target is constant and out1 is function?

You said in the beginning that we are going to work with only a single training set (1 row with 2 features ). Then how we are getting two outputs. It should be one right.

Can you please clarify?

Very nice explanation.

Thanks for taking the time to do this.

its really very helpful..thanks for such a wonderful doc

Carrying out the same process for h_2 we get:

out_{h2} = 0.596884378

I got 0.666522 for h2 because i did: activation_function(0.25 * 0.05 + 0.3 * 0.1 + 0.65 * 1)

Hi,

It is a good work. I am struck at here, Following the same process for \frac{\partial E_{o2}}{\partial out_{h1}}, we get:

\frac{\partial E_{o2}}{\partial out_{h1}} = -0.019049119 (partial derivative of H1out from E2)… how to calculate out_H1 from E2

Thanks for this awesome explanation.