# 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.

## 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!

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

1. Как устроена нейросеть / Блог компании BCS FinTech / Хабр

2. Matt says:

Thanks for this nice illustration of backpropagation!
I am wondering how the calculations must be modified if we have more than 1 training sample data (e.g. 2 samples).

• Matt says:

I reply to myself… I forgot to apply the chainrule

3. Shahrum says:

It seems that you have totally forgotten to update b1 and b2! They are part of the weights (parameters) of the network. Or am I missing something here?

4. Seong says:

Thanks to your nice illustration, now I’ve understood backpropagation.

5. deva says:

This is exactly what i was needed , great job sir, super easy explanation.

6. pawandtu@gmail.com says:

Just wondering about the range of the learning rate. Why can’t it be greater than 1?

7. ANIL BABU B says:

Hi Matt
Thanks for giving the link, but i have following queries, can you please clarify
1. Why bias weights are not updated anywhere
2.Outputs at hidden and Output layers are not independent of the initial weights chosen at the input layer. So for calculated optimal weights at input layer (w1 to w4) why final Etot is again differentiated w.r.t w1, instead should we not calculate the errors at the hidden layer using the revised weights of w5 to w8 and then use the same method for calculating revised weights w1 to w4 by differentiating this error at hidden layer w.r.t w1.
3.Error at hidden layer can be calculated as follows: We already know the out puts at the hidden layer in forward propagation , these we will take as initial values, then using the revised weights of w5 to w8 we will back calculate the revised outputs at hidden layer, the difference we can take as errors
4. i calculated the errors as mentioned in step 3, i got the outputs at h1 and h2 are -3.8326165 and 4.6039905. Since these are outputs at hidden layer , these are outputs of sigmoid function so values should always be between 0 and 1, but the values here are outside the outputs of sigmoid function range

Please clarify why and where the flaw is

8. Clara Liu says:

Awesome tutorial!

But are there possibly calculation errors for the undemonstrated weights? I kept getting slightly different updated weight values for the hidden layer…

But let’s take a simpler one for example:
For dEtotal/dw7, the calculation should be very similar to dEtotal/dw5, by just changing the last partial derivative to dnet o1/dw7, which is essentially out h2.So dEtotal/dw7 = 0.74136507*0.186815602*0.596884378 = 0.08266763

new w7 = 0.5-(0.5*0.08266763)= 0.458666185.

Perhaps I made a mistake in my calculation? Some clarification would be great!

• Albrecht Ehlert says:

0.5113012702387375 is right …

• AJ says:

The number you have there, 0.08266763, is actually dEtotal/dw6. You will see that applying it to the original w6 yields the value he gave: 0.45 – (0.5*0.08266763) = 0.40866186.

Maybe you confused w7 and w6? The diagram makes it easy to confuse them. W7 is the weight between h1 and o2.

To find dEtotal/dw7 you would have to find:
dEtotal/dout_{o_2} * dout_{o_2}/dnet_{o_2} * dnet_{o_2}/dw7.

so dEtotal/dw7 = -0.21707153 * 0.17551005 * 0.59326999 = -0.02260254

new w7 = 0.5 – (0.5 * -0.02260254) = 0.511301270
hope this helped

9. Vectorization of Neural Nets | My Universal NK

10. Luke says:

After many hours of looking for a resource that can efficiently and clearly explain math behind backprop, I finally found it! Fantastic work!

11. Joshua Laferriere says:

Heaton in his book on neural networks math say
node deltas are based on [sum] “sum is for derivatives, output is for gradient, else your applying the activation function twice?”

but I’m starting to question his book because he also applies derivatives to the sum

“Ii is important to note that in the above equation, we are multiplying by the output of hidden I. not the sum. When dealing directly with a derivative you should supply the sum Otherwise, you would be indirectly applying the activation function twice.”

but I see your example and one more where that’s not the case
https://github.com/thistleknot/Ann-v2/blob/master/myNueralNet.cpp

I see two examples where the derivative is applied to the output

12. Anam Anwar says:

Very well explained…… Really helped alot in my final exams….. Thanks

13. Neural Network – Bagus's digital notes

14. Albrecht Ehlert says:

Dear Matt,
thank you for the nice illustration!
I built the network and get exactly your outputs:

Weights and Bias of Hidden Layer:
Neuron 1: 0.1497807161327628 0.19956143226552567 0.35
Neuron 2: 0.24975114363236958 0.29950228726473915 0.35

Weights and Bias of Output Layer:
Neuron 1: 0.35891647971788465 0.4086661860762334 0.6
Neuron 2: 0.5113012702387375 0.5613701211079891 0.6

output:
0.7513650695523157 0.7729284653214625

Than I made a experiment with the bias. Without changing the bias I got after 1000 epoches the following outputs:

Weights and Bias of Hidden Layer:
Neuron 1: 0.2820419392605305 0.4640838785210599 0.35
Neuron 2: 0.3805890849512254 0.5611781699024483 0.35

Weights and Bias of Output Layer:
Neuron 1: -3.0640975297007556 -3.034730378052809 0.6
Neuron 2: 2.0517051904569836 2.110885730396752 0.6

output:
0.044075530730776365 0.9572825838174545

With backpropagation of the bias the outputs getting better:

Weights and Bias of Hidden Layer:
Neuron 1: 0.20668916041682514 0.3133783208336505 1.4753841161727905
Neuron 2: 0.3058492464890622 0.4116984929781265 1.4753841161727905

Weights and Bias of Output Layer:
Neuron 1: -2.0761119815104956 -2.038231681376019 -0.08713942766189575
Neuron 2: 2.137631425033325 2.194909264537856 -0.08713942766189575

output:
0.03031757858059988 0.9698293077608338

Sincerly
Albrecht Ehlert from Germany

15. 한성주 says:

I finally understood BP thanks to you. You are my hero.

16. Matt says:

When you derive E_total for out_o1 could you please explain where the -1 comes from? I get the normal derivative and the 0 for the second error term but I don’t get where the -1 appeared from.

• AJ says:

It’s because of the chain rule. In the original equation (1/2)(target – out_{o1})^2, when you end up taking the derivative of the (…)^2 part, you have to multiply that by the derivative of the inside. The derivative of the inside with respect to out_{o1} is 0 – 1= -1

17. Luqing says:

I read many explanations on back propagation, you are the best in explaining the process. Thank you very much.

18. Neural Networks – Feedforward Math – Shahzina Khan

19. apoola says:

Matt, thanks a lot for the explanation….However, I noticed

net_{h1} = 0.15 * 0.05 + 0.2 * 0.1 + 0.35 * 1 = 0.3775

should it be

net_{h1} = 0.15 * 0.05 + 0.25 * 0.1 + 0.35 * 1 = 0.3825

ie. 0.25 instead of 0.2 (based on the network weights) ?

Again I greatly appreciate all the explanation

• AJ says:

I think you may have misread the second diagram (to be fair its very confusingly labeled). Take a look at the first diagram in the section “The Backwards Pass.” Here we see that neuron o_1 has associated weights w5 & w6. If you look back at the first diagram, w5 & w6 are the top two labeled weights, so it would follow logically that neuron h1 is has the weights w1 & w2.
W2 has a value of .20, which is consistent with the way he performed the other calculations.

20. gellaz says:

Great explanation Matt! I really appreciate your work. I noticed a small mistake:
I noticed a small mistake at the end of the post:
net_{h1} = w_1 * i_1 + w_3 * i_2 + b_1 * 1
should be:
net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1

21. طیب مظہر says:

When calculating for w1, why are you doing it like :

Eo1/OUTh1 = Eo1/NETo1 * NETo1/OUTh1

and not like:

Eo1/OUTh1 = Eo1/OUTo1 * OUTo1/NETo1 * NETo1/OUTh1

Why are you going from Eo1 to NetO1 directly, when there is OUTo1 in the middle.

• Aravinth M says:

Eo1/OUTh1 = Eo1/OUTo1 * OUTo1/NETo1 * NETo1/OUTh1. Thanks for giving this explanation bro.

22. Hemant says:

any reason why back propagation is necessary ? Can we not do this with just forward propagation in a brute force way ? or is the forward propagation is somehow much slower than back propagation.

given that we have a network with weights,
can we not get dE(over all training examples)/dWi as follows:

[1] store current error Ec across all samples as sum of [ Oactual – Odesired } ^2 for all output nodes of all samples
[2] simply change the Wi by say 0.001 and propagate the change through the network and get new error En over all training examples.
[3] then set Wi back to its old value. (so we do not mess it up for another Wi)
[4] and then dE/dWi = (En – Ec) / 0.001
[5] Do this for all weights to get all weight sensitivities.
[6] then change all weights Wi = Wi – dE/DWi * learning rate
[7] propagate through the network get Ec
[8] Repeat 1 to 7 until Ec is lower than acceptable error threshold

23. Great article! Just what I was looking for, thank you.