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, 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:

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:

nn-calculation

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!

805 thoughts on “A Step by Step Backpropagation Example

    • 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.

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

  2. 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.

  3. 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

  4. 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.

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  6. 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!

  7. 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.

      • I got the same problem and I don’t understand the answer.
        Why is dE/dout i1 =dE/dnet h1 * dnet h1/dout i1?
        This ignores the contribution of out i1 to the error through h2!?
        So, I assume the term dE/dnet h2 * dnet h2/dout i1 needs to be added!? But then, why is it OK to just use the sum, mathematically?

  8. 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!

    • In reality each bias would be adjusted too, but in the example seems like they are only focusing on weights, and specifically the hidden layers.

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  11. 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

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  14. 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 ?

  15. 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?

  16. 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?

    • Training sets can have as many inputs and outputs as you want. What the “1 row with 2 features” means is that there is only 1 training set in the entire list. For example if you input a (0,1), it will output a (0.5,0.5). For 2 ‘rows’ or training sets, you can have (0,1)->(0.5,0.5) OR if it is for example a (1,1)->(0.6,0.2). The row means a new ‘if’ statement. I am not sure what you mean by ‘feature’ if you’re talking about the amount of hidden neurons, the average between input/output neurons, or the amount of biases.

  17. 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)

  18. 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

  19. This article was really very helpful for me as i was very confused and stuck with tha backprop calculations and u just explained it with an example in a very nice way thanks a lot

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  21. Final wieghts after 10,000 iterations are
    0.378509988350132
    0.657019976700263
    0.477300705684786
    0.754601411369569
    -3.892977198865724
    -3.868416687703616
    2.861810948620059
    2.925688966771616

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  23. This is an awesome post, I was confused about how the error propagates through the hidden layer. Working through each step made the equations seem intuitive. Someone should really make an animation for this.

  24. [D] Is there a book or resource about machine learning that teaches different algorithms with step by step calculations for simple examples? Like how a mathematics or physics book would. | Premium Blog! | Development code, Android, Ios anh Tranning IT

  25. Hi Experts, I tried this approach, for some reason my weights and error keeps increasing, doesn’t seem to reduce for certain inputs, any idea whats happening ?

  26. it is my answer,
    0 0.291027774
    1 0.283547133
    2 0.275943289
    3 0.268232761
    4 0.260434393
    5 0.252569176
    6 0.244659999
    7 0.236731316
    8 0.228808741
    9 0.220918592
    10 0.213087389
    11 0.205341328
    12 0.197705769
    13 0.190204742
    14 0.182860503
    15 0.175693166
    16 0.168720403
    17 0.16195725
    18 0.155415989
    19 0.149106135
    20 0.14303449
    21 0.137205276
    22 0.131620316
    23 0.126279262
    24 0.121179847
    25 0.116318143
    26 0.111688831
    27 0.107285459
    28 0.103100677
    29 0.099126464
    30 0.095354325
    31 0.091775464
    32 0.088380932
    33 0.085161757
    34 0.082109043
    I am not sure that it is right.

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  28. Great explanation. Thanks. Many authors who write books consider themselves as being smart just because they provide some complicated formulas from wikipedia with 10,000 different index symbols. But I do not think they are smart at all. They do not know their audience. This simple example is better than all books that I have which try to explain back propagation with fomrulas… I did not realize it is that simple. It is just taking derivative over and over again. Thank you very much!

  29. Thank you very much but I have a question.
    At the beginning w1 = 0.15, w2 = 0.2, b1 = 0.35 => How to initialize these weight, bias values.
    => How to nit matrix of Height, Bias ?

  30. Great explanation. I had no problem following your description, but I am missing a critical piece. In reality, we always have more than 1 row of inputs and outputs for training data (we could have hundreds of thousands of rows). How does this then affect the computations of the derivatives? I’m sure all training rows play into the total error, but I am missing what is probably a simple connection on calculating new weights in light of all the data. Can you explain how you would get the updated weights if you had two rows of training data? Do you process it one row at a time and then average the new computed weights together to determine the updated values? Is there a fast way to do this?

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