A Step by Step Backpropagation Example


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.

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.


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})


\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}}



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


\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!

939 thoughts on “A Step by Step Backpropagation Example

  1. It’s a great tutorial but I think I found an error:
    at forward pass values should be:
    neth1 = 0.15 * 0.05 + 0.25 * 0.1 + 0.35 * 1 = 0.3825
    outh1 = 1/(1 + e^-0.3825) = 0,594475931
    neth2 = 0.20 * 0.05 + 0.30 * 0.1 + 0.35 * 1 = 0.39
    outh2 = 1/(1 + e^-0.39) = 0.596282699

  2. Good stuff ! Professors should learn from you. Most professors make complex things complex. A real good teacher should make complex things simple.

  3. Hey there very helpful indeed, in the line for net01 = w5*outh1 + ‘w6’*outh2+b2*1, is it not meant to be ‘w7’ ??

  4. Can anyway help me explaining manual calculation for testing outputs with trained weights and bias? Seems it does not give the correct answer when I directly substitute my inputs to the equations. Answers are different than I get from MATLAB NN toolbox.

  5. Hello, A Step by Step Backpropagation Example is helpful. I would like to ask……What are the advantages of using the weights and biases in neural network and how the weight and biases are initialized? based on what and how?

  6. Awesome tutorial. It clears my all doubts about Backpropagation learning algorithm. Thanks for such a nice tutorial…

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  9. At the beginning, while calculating neth1, the term 0.2*0.1 should be 0.25*0.1. I do not know this changes all the numbers.

    Even if all the numbers are wrong, the explanation is excellent. Thanks.

  10. awesome article !! I have just one doubt…while calculating delta o1 , shouldn’t we multiply the given equation by weight matrix of layer 2 ? I saw that in Andrew N.Gs videos he does this and am hence confused

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  12. Hey Matt, that was a very neat explanation! I am workin on BPA for SIMO system and tryin to implement using Simulink.

    Is there any easy way to do in simulink?

    • my guide is now published – it aims to make understanding how neural networks work as accessible as possible – with lots of diagrams, examples and discussion – there are even fun experiments to “see inside the mind of a neural network” and getting it all working on a Raspberry Pi Zero (£4 or $5)

        • I’ll be buying your book–it’s a great idea and well executed; but, I was hoping you could walk me through the differentiation of the total error function on pages 94-95 in the draft pdf version online. A minus that pops up that I can’t account for. The same happens in the above post, and I’m just not sure why.
          Any hep you can provide is much appreciated.

          • The minus sign pops up because d/dx (a-x) is -1.

            In the example we have d/dx (a-x)^2 which is done in two steps. First the power rule to get 2*(a-x) but then the inside of the brackets too which gives you the extra linux sign.

            This is actually the chain rule: df/dx = df/dy * dy/dx
            so if f=(a-x)^2 and y=(a-x) then df/dx = df/dy * dy/dx = 2y * (-1) = (a-x) * (-1)


            Some people also get caught out by the weight updates. The weights are changed in proportion to – dE/dw .. that minus is there so that the weights go in the opposite direction of the gradient. My book illustrates this too with pictures.

            • Thank you! That’s what I thought, but I just needed confirmation. A lot of people don’t mention the chain rule, so that’s what was throwing me.

              Your book is excellent, and I bought a copy.

              Everybody should buy a copy!

              I hope it does really well.

            • Thanks Flipperty – if you like it, please do leave a review on amazon. And if you have suggestions for improvement, do let me know too!

  13. Hi Matt! Great article,probably the best I’ve come across on the internet for Backpropagation. Just noticed a small error – When you calculate outo1 for the first time in the forward pass, the formula should be 1/1-(e^-neto1) whereas you have written it to be 1/1-(e^neth1). That’s it! Amazing article once again. You’ve got me hooked on to NNets now and earned yourself a worshipper :)

  14. Thank you Matt for this deep explanation
    I would like to ask about the bias
    1- Why you set it fixed per layer, not per neuron?
    2- why it is kept fixed throughout the operation, i.e., why it is not updated through the back propagation?

    Thanks in advance for reply?

  15. Thanks a lot Matt!!

    The tutorial was really great and It was really nice how you explained the concept so clearly without leaving the math out.

    I had question:

    When the NN trains for Example 1 and good weights are found for it. Afterwards, when the NN starts to learn for the Example 2. Will it move away from the Example 1 or will it keep accommodating it also along with Example 2 etc?

  16. good explanation, and easy for me to understand. thank you for your effort to explain. I appreciate this

  17. Hi mat, Thanks for the clear explanation, one question , why don’t you use weights for the biases and update then during backprop?

  18. Thanks for your nice post!

    I have a question.

    In the forward pass, there are

    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_{h1}}} = \frac{1}{1+e^{-1.105905967}} = 0.75136507

    In my opinion, “out_{o1} = \frac{1}{1+e^{-net_{h1}}}” should be changed to
    “out_{o1} = \frac{1}{1+e^{-net_{o1}}}” .

    Is it right suggestion?

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  20. I’m new to AI and for that reason I have traveled the net for an understandable explanation/tutorial on Back propagation. Found a lot moor or less understandable (I’m not a mathematician). I fail every time when we have to back propagate the signals. My calculus understanding was to weak. Your introduction with the chain rule together with the example with real numbers was a break through in my understanding. Thank you!

  21. Hi Mazur, this is great. I have question about this.
    Doesn’t it map the training output before calculate the error with target value?. Because “logistic function” outputs are always between 0 and 1. Then what happen if my target is really high (eg:300)? error would be really high?. After finish training, my target is 300 and final output would be somewhere between 0 and 1, since transfer function at the output node is “logistic”. That is the part which I still cannot understand. even after training, doesn’t it map the output to be large (should be taken near to 300 from 0-1)?

    • You should map your desired outputs to match the activation function – otherwise you will drive the network to saturation. The same is also true of the inputs – you should try to optimise them to that they match the region of greatest variation of the activation function .. some people call this normalisation. My gentle intro to neural networks has a section on this .. http://www.amazon.com/dp/B01EER4Z4G

  22. Hey Mazur, great tutorial, as said earlier, the best on Internet. Just have a question, when you have more hidden layers, the concept for the feedforward is the same, but the derivatives surely will modify, my question is how to calculate it? It will be something like Dtotal = DTotal/Dout * Dout/Dnet * Dnet/Dweight * Dweight/Douthidden * Douthidden/Dnethidden * Dnethidden/Dweight? Don’t know if you will understand the way i wrote it xD Thanks in advance.

  23. Thanks Matt for the great article. Even me, a math dummy, was able to implement the back prop algorithm in C++…
    I’m worring abaout generalization of this algorithm: is it viable for networks with more than an hidden layer? It seems to me that the hidden part of the algorithm
    is specific for jus one layer.

    • Theoretically only one hidden layer is enough for approximating any function, there is no need for more than an hidden layer.

      • True, but the key word here is “theory.” In practice, using one enormous layer is vastly inefficient and prone to runaway numerical errors. In practice, always use multiple (smaller) layers if your models need a large representational capacity.

  24. I have a question on calculating ∂Eo1/∂ω1 using the chain rule. Eo1 is associated with both of outh1 and outh2. In your calculation, however, only the ∂neto1/∂outh1 is taken into account without consideration for ∂neto1/∂outh2. Could you explain why ∂neto1/∂outh2 can be ignored in computation?

    • hi qwerty, im not 100% sure with this but im guessing that w1 is not directly related to H2 that’s why. And since youre only computing for the rate of change of Eo1 with respect to change in w1, dNeto1/dOuth2 would just like be a constant/negligible.

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