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.

Additional Resources

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:

neural_network (7)

In order to have some numbers to work with, here are the initial weights, the biases, and training inputs/outputs:

neural_network (9)

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:

output_1_backprop (4)

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 $\frac{\partial E_{o2}}{\partial out_{o1}}$ , 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 0.000035085. 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!

224 thoughts on “A Step by Step Backpropagation Example”

leafartz says:

August 18, 2016 at 10:55 am

Thanks dude! finally, i could understand very well this issue.

Reply
Ronald says:

August 18, 2016 at 1:12 pm

THANK’S I helped a lot
Greetings from Peru

Reply
mayankr says:

August 19, 2016 at 8:20 am

Great explaination, Thanks!

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mayankr says:

August 19, 2016 at 8:25 am

Great post, clear concepts! Thanks!!

Reply
max putilov (@baio1980) says:

August 19, 2016 at 3:23 pm

Helps a lot, thanks

Reply
kanu says:

August 20, 2016 at 8:53 am

nice one

Reply
A P says:

August 20, 2016 at 2:41 pm

It was a revelation. Equations with many Greek letters does not explain much alone. Keep going!

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max putilov (@baio1980) says:

August 21, 2016 at 6:58 am

Btw, why dont you calculate gradients for biases ?

Reply
- Mazur says:
  
  August 22, 2016 at 9:34 am
  
  Some guides do that, others don’t. I didn’t for this tutorial.
  
  Reply
  - Sjoerd Redeker says:
    
    September 9, 2016 at 2:45 am
    
    Is there a possibility you can show us how to calculate the gradients for the biases? Love this tutorial by the way it is very informative and clear!
    
    Reply
Sabyasachi Mohanty says:

August 30, 2016 at 1:51 pm

Can you please do a tutorial for back propagation in Elmann recurrent neural networks!!…. It would be of a lot of help…

Reply
Dion says:

September 4, 2016 at 8:59 am

Great article! I struggle with one aspect though and that is calculating the partial derivatives from out/net, using (output * (1.0 – output). In cases where output is 0 or 1, it effectively kills the pass through of error. How do you handle this when, especially at startup, the network can devolve (or even start) in this state?
Thanks!

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Ann says:

September 6, 2016 at 5:24 am

should be the first stop for anyone to understand backward propagation, very well explained…thanks a lot

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Reza says:

September 6, 2016 at 9:49 am

Very informative, Thank you

Reply
Philip Rhoades says:

September 7, 2016 at 2:34 pm

In trying to understand NNs better I have produced a Google Docs spreadsheet that almost does what this link talks about:

I was hoping that going through that exercise will give me a better mental picture of what a NN is doing when it is working. I should be able to see easily by direct comparison what happens to the set of numbers with each iteration (ie each new pair of FP and BP sheets). I think I have got it nearly working except for the stuff in the dashed purple box. I want to confirm that the purple arrow is pointing to the wrong blue arrow? I also want someone to tell me how to implement the purple box . . If I could work that out I think I could then repeat the FR and BP sheets and see how the Diff column evolves . . here is the current SS:

https://docs.google.com/spreadsheets/d/1-YxT_PuzDt3VXrOucOBHzxBpSiB5USiy2ULaqE75Wcg/pubhtml

I still found your description above heavy going – could you help me finish my spreadsheet or turn your example into a spreadsheet?

Thanks,
Phil.

Reply
Hari Prasad says:

September 12, 2016 at 4:51 am

This is how any teaching or learning has to happen! Great work buddy!

Reply
rambo says:

September 12, 2016 at 6:22 pm

Many thanks. The best tutorial I have ever seen

Reply
Erhard M. Dinhobl says:

September 13, 2016 at 7:30 am

It seems that the learningrate of “0.5” in this example here is too high. because it diverges with the given values. Maybe 0.05?

Reply
- Erhard M. Dinhobl says:
  
  September 13, 2016 at 10:39 am
  
  sorry, forgot to set the energy variable in the neuron back to zero after training
  
  Reply
Michael Prager says:

September 17, 2016 at 7:17 pm

I found this post very helpful but there is one thing that is confusing me. You apply the logistic function on the output nodes to compute out_o1 and out_o2 as the sigmoid function applied to the weighted sum of the hidden nodes. My question is, what if you are predicting an output that has a range wider than 0 to 1. I had thought the output layer was simply a weighted sum of the outputs of the final hidden layer. In that case you could output a value in any range, but this seems very limiting.

Reply
- Mazur says:
  
  September 18, 2016 at 10:39 am
  
  Yeah, this only works on a range of 0 to 1. If you have a range from say 0 to 100 you can divide by 100 to get it down to a range of 0 to 1 so that you can use this neural network.
  
  Reply
Gregory says:

September 20, 2016 at 2:20 am

what does the Who mean in the expression for the hidden layers?
The “ho” is small

Reply
- Mazur says:
  
  September 20, 2016 at 8:19 pm
  
  Weight of the hidden output.
  
  Reply
  - Gregory says:
    
    September 23, 2016 at 9:25 pm
    
    Shouldn’t it be the weights connecting the last hidden layer and the output layer? If the network had an input of 200, hidden of 100, and another hidden of 50, and an output of 10; it wouldn’t work. Because the layer of 1 hundred would add the outputs multiplied by 10 of the 50 layer weights, right? Sorry if I am not very clear. What I’m just trying to ask is this: Is the Who referring to the weight connecting the last layer and output layer or is it connecting the current layer and next layer?
    
    Reply
Anirban says:

September 20, 2016 at 3:27 am

You have used a squared error function. When using that as a cost function along with the sigmoid function, won’t your cost function have a lot of local minima ?

Reply
Hajji Hicham says:

September 20, 2016 at 6:55 am

Thanks, it’s the best and clear step by step tutorial of the backpropagation alg I have ever read

Reply
gsaldanha2 says:

September 21, 2016 at 12:37 am

Thanks so much!! It’s a very clear and thorough explanation :)

Reply
Roopak Neevan says:

September 21, 2016 at 1:17 am

Great Post with the step by step explanation.
Based on your explanation I am able to implement the same for LOGICAL GATE solution where I have used 2 Input Nodes (for logical Inputs), 1 Hidden Layer with 2 Nodes and one Output Node.
Thanks for this detailed explanation.
Regards
Neevan

Reply
Nimesh says:

September 22, 2016 at 2:44 am

Couldn’t agree less with Hajji.
Its one the best and simplified article.

Reply
research scholar says:

September 23, 2016 at 10:27 pm

Thanks for beautiful workout .algorithm has become transparent and very easy

Reply
Apoorva Bansal says:

September 25, 2016 at 12:52 am

Amazingly simple explanation to something everyone tends to complicate! :) Thanks a lot!!

Reply
H. Paudel says:

September 25, 2016 at 7:38 am

It was a great explanation, Mazur. Thank you very much for the effort.
It would be great if you could response my following query.
I have been trying to train a model using neuralnet package. My model contains five input one output and a hidden layer with 10 nodes. logistic transfer function is selected as an activation function. All data are normalized in 0,1 range. As the logistic function is bounded between 0,1 the results are expected to be in the range 0,1. But some of the training results are negative. What is causing the model to compute negative output?
Regards
H. Paudel

Reply
gsaldanha2 says:

September 25, 2016 at 11:40 pm

How would you compute the weight for a hidden layer in a multi-layer network?

Reply
反向传播算法入门资源索引 | 我爱自然语言处理

Matt Mazur

A Step by Step Backpropagation Example

Background

Backpropagation in Python

Backpropagation Visualization

Additional Resources

Overview