How could stochastic gradient descent save time compared to standard gradient descent?

Question

Standard Gradient Descent would compute gradient for the entire training dataset.

for i in range(nb_epochs):
  params_grad = evaluate_gradient(loss_function, data, params)
  params = params - learning_rate * params_grad

For a pre-defined number of epochs, we first compute the gradient vector weights_grad of the loss function for the whole dataset w.r.t. our parameter vector params.

Stochastic Gradient Descent in contrast performs a parameter update for each training example x(i) and label y(i).

for i in range(nb_epochs):
  np.random.shuffle(data)
  for example in data:
    params_grad = evaluate_gradient(loss_function, example, params)
    params = params - learning_rate * params_grad

SGD is said to be much faster. However, I do not understand how it can be much faster if we still have a loop over all data points. Does the computation of the gradient in GD is much slower than computation of GD for each data point separately?

Code comes from here.

Answer 1

Short answer:

• In many big data setting (say several million data points), calculating cost or gradient takes very long time, because we need to sum over all data points.
• We do NOT need to have exact gradient to reduce the cost in a given iteration. Some approximation of gradient would work OK.
• Stochastic gradient decent (SGD) approximate the gradient using only one data point. So, evaluating gradient saves a lot of time compared to summing over all data.
• With "reasonable" number of iterations (this number could be couple of thousands, and much less than the number of data points, which may be millions), stochastic gradient decent may get a reasonable good solution.

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Long answer:

My notation follows Andrew NG's machine learning Coursera course. If you are not familiar with it, you can review the lecture series here.

Let's assume regression on squared loss, the cost function is

\begin{align} J(\theta)= \frac 1 {2m} \sum_{i=1}^m (h_{\theta}(x^{(i)})-y^{(i)})^2 \end{align}

and the gradient is

\begin{align} \frac {d J(\theta)}{d \theta}= \frac 1 {m} \sum_{i=1}^m (h_{\theta}(x^{(i)})-y^{(i)})x^{(i)} \end{align}

for gradient decent (GD), we update the parameter by

\begin{align} \theta_{new} &=\theta_{old} - \alpha \frac 1 {m} \sum_{i=1}^m (h_{\theta}(x^{(i)})-y^{(i)})x^{(i)} \end{align}

For stochastic gradient decent we get rid of the sum and $1/m$ constant, but get the gradient for current data point $x^{(i)},y^{(i)}$, where comes time saving.

\begin{align} \theta_{new}=\theta_{old} - \alpha \cdot (h_{\theta}(x^{(i)})-y^{(i)})x^{(i)} \end{align}

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Here is why we are saving time:

Suppose we have 1 billion data points.

• In GD, in order to update the parameters once, we need to have the (exact) gradient. This requires to sum up these 1 billion data points to perform 1 update.

• In SGD, we can think of it as trying to get an approximated gradient instead of exact gradient. The approximation is coming from one data point (or several data points called mini batch). Therefore, in SGD, we can update the parameters very quickly. In addition, if we "loop" over all data (called one epoch), we actually have 1 billion updates.

The trick is that, in SGD you do not need to have 1 billion iterations/updates, but much less iterations/updates, say 1 million, and you will have "good enough" model to use.

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I am writing a code to demo the idea. We first solve the linear system by normal equation, then solve it with SGD. Then we compare the results in terms of parameter values and final objective function values. In order to visualize it later, we will have 2 parameters to tune.

    set.seed(0);n_data=1e3;n_feature=2;
    A=matrix(runif(n_data*n_feature),ncol=n_feature)
    b=runif(n_data)
    res1=solve(t(A) %*% A, t(A) %*% b)

    sq_loss<-function(A,b,x){
      e=A %*% x -b
      v=crossprod(e)
      return(v[1])
    }

    sq_loss_gr_approx<-function(A,b,x){
      # note, in GD, we need to sum over all data
      # here i is just one random index sample
      i=sample(1:n_data, 1)
      gr=2*(crossprod(A[i,],x)-b[i])*A[i,]
      return(gr)
    }

    x=runif(n_feature)
    alpha=0.01
    N_iter=300
    loss=rep(0,N_iter)

    for (i in 1:N_iter){
      x=x-alpha*sq_loss_gr_approx(A,b,x)
      loss[i]=sq_loss(A,b,x)
    }

The results:

as.vector(res1)
[1] 0.4368427 0.3991028
x
[1] 0.3580121 0.4782659

Note, although the parameters are not too close, the loss values are $124.1343$ and $123.0355$ which are very close.

Here is the cost function values over iterations, we can see it can effectively decrease the loss, which illustrates the idea: we can use a subset of data to approximate the gradient and get "good enough" results.

Now let's check the computational efforts between two approaches. In the experiment, we have $1000$ data points, using SD, evaluate gradient once needs to sum over them data. BUT in SGD, sq_loss_gr_approx function only sum up 1 data point, and overall we see, the algorithm converges less than $300$ iterations (note, not $1000$ iterations.) This is the computational savings.

Answer 2

First of all, if you do the same number of epochs, SGD won't be faster than GD, because the per-epoch computation complexity is same for SGD and GD, as you pointed out.

However, GD only does one iteration of gradient descent per epoch while SGD does n/m iterations. Those n/m iterations brings greater improvement than the single iteration of GD, even though each of the n/m iterations is probably worse than the single iteration of GD due to sampling.

So, SGD is faster in the sense that it reaches the same accuracy quicker, with fewer epochs, than GD.

Answer 3

Not only does SGD iterate gradients much faster, the stochasticity (noise from randomly picking samples) itself can be an asset for generalization: see ex. On the Generalization Benefit of Noise in Stochastic Gradient Descent (Smith, Elsen, De, ICML 2020)