# Eight to Late

Sensemaking and Analytics for Organizations

## A gentle introduction to Monte Carlo simulation for project managers

This article covers the why, what and how of Monte Carlo simulation using a canonical example from project management –  estimating the duration of a small project. Before starting, however, I’d like say a few words about the tool I’m going to use.

Despite the bad rap spreadsheets get from tech types – and I have to admit that many of their complaints are justified – the fact is, Excel remains one of the most ubiquitous “computational” tools in the corporate world. Most business professionals would have used it at one time or another. So, if you you’re a project manager and want the rationale behind your estimates to be accessible to the widest possible audience, you are probably better off presenting them in Excel than in SPSS, SAS, Python, R or pretty much anything else. Consequently, the tool I’ll use in this article is Microsoft Excel. For those who know about Monte Carlo and want to cut to the chase, here’s the Excel workbook containing all the calculations detailed in later sections. However, if you’re unfamiliar with the technique, you may want to have a read of the article before playing with the spreadsheet.

In keeping with the format of the tutorials on this blog, I’ve assumed very little prior knowledge about probability, let alone Monte Carlo simulation. Consequently, the article is verbose and the tone somewhat didactic.

### Introduction

Estimation is key part of a project manager’s role. The most frequent (and consequential) estimates they are asked deliver relate to time and cost.  Often these are calculated and presented as point estimates: i.e. single numbers – as in, this task will take 3 days. Or, a little better, as two-point ranges – as in, this task will take between 2 and 5 days.  Better still, many use a PERT-like approach wherein estimates are based on 3 points: best, most likely and worst case scenarios – as in, this task will take between 2 and 5 days, but it’s most likely that we’ll finish on day 3.  We’ll use three-point estimates as a starting point for Monte Carlo simulation, but first, some relevant background.

It is a truism, well borne out by experience, that it is easier to estimate small, simple tasks than large, complex ones. Indeed, this is why one of the early to-dos in a project is the construction of a work breakdown structure. However, a problem arises when one combines the estimates for individual elements into an overall estimate for a project or a phase thereof. It is that a straightforward addition of individual estimates or bounds will almost always lead to a grossly incorrect estimation of overall time or cost. The reason for this is simple: estimates are necessarily based on probabilities and probabilities do not combine additively. Monte Carlo simulation provides a principled and intuitive way to obtain probabilistic estimates at the level of an entire project based on estimates of the individual tasks that comprise it.

### The problem

The best way to explain Monte Carlo is through a simple worked example. So, let’s consider a 4 task project shown in Figure 1. In the project, the second task is dependent on the first, and third and fourth are dependent on the second but not on each other. The upshot of this is that the first two tasks have to be performed sequentially and the last two can be done at the same time, but can only be started after the second task is completed.

To summarise: the first two tasks must be done in series and the last two can be done in parallel.

Figure 1; A project with 4 tasks.

Figure 1 also shows the three point estimates for each task – that is the minimum, maximum and most likely completion times. For completeness I’ve listed them below:

• Task 1 – Min: 2 days; Most Likely: 4 days; Max: 8 days
• Task 2 – Min: 3 days; Most Likely: 5 days; Max: 10 days
• Task 3 – Min: 3 days; Most Likely: 6 days; Max: 9 days
• Task 4 – Min: 2 days; Most Likely: 4 days; Max: 7 days

OK, so that’s the situation as it is given to us. The first step to  developing  an estimate is to formulate the problem in a way that it can be tackled using Monte Carlo simulation. This bring us to the important topic of the shape of uncertainty aka probability distributions.

### The shape of uncertainty

Consider the data for Task 1. You have been told that it most often finishes on day 4.  However, if things go well, it could take as little as 2 days; but if things go badly it could take as long as 8 days.  Therefore, your range of possible finish times (outcomes) is between 2 to 8 days.

Clearly, each of these outcomes is not equally likely.  The most likely outcome is that you will finish the task in 4 days (from what your team member has told you). Moreover, the likelihood of finishing in less than 2 days or more than 8 days is zero. If we plot the likelihood of completion against completion time, it would look something like Figure 2.

Figure 2: Likelihood of finishing on day 2, day 4 and day 8.

Figure 2 begs a couple of questions:

1. What are the relative likelihoods of completion for all intermediate times – i.e. those between 2 to 4 days and 4 to 8 days?
2. How can one quantify the likelihood of intermediate times? In other words, how can one get a numerical value of the likelihood for all times between 2 to 8 days?  Note that we know from the earlier discussion that this must be zero for any time less than 2 or greater than 8 days.

The two questions are actually related. As we shall soon see, once we know the relative likelihood of completion at all times (compared to the maximum), we can work out its numerical value.

Since we don’t know anything about intermediate times (I’m assuming there is no other historical data available), the simplest thing to do is to assume that the likelihood increases linearly (as a straight line) from 2 to 4 days and decreases in the same way from 4 to 8 days as shown in Figure 3. This gives us the well-known triangular distribution.

Jargon Buster: The term distribution is simply a fancy word for a plot of likelihood vs. time.

Figure 3: Triangular distribution fitted to points in Figure 1

Of course, this isn’t the only possibility; there are an infinite number of others. Figure 4 is another (admittedly weird) example.

Figure 4: Another distribution that fits the points in Figure 2.

Further, it is quite possible that the upper limit (8 days) is not a hard one. It may be that in exceptional cases the task could take much longer (for example, if your team member calls in sick for two weeks) or even not be completed at all (for example, if she then leaves for that mythical greener pasture).  Catering for the latter possibility, the shape of the likelihood might resemble Figure 5.

Figure 5: A distribution that allows for a very long (potentially) infinite completion time

The main takeaway from the above is that uncertainties should be expressed as shapes rather than numbers, a notion popularised by Sam Savage in his book, The Flaw of Averages.

[Aside:  you may have noticed that all the distributions shown above are skewed to the right – that  is they have a long tail. This is a general feature of distributions that describe time (or cost) of project tasks. It would take me too far afield to discuss why this is so, but if you’re interested you may want to check out my post on the inherent uncertainty of project task estimates.

### From likelihood to probability

Thus far, I have used the word “likelihood” without bothering to define it.  It’s time to make the notion more precise.  I’ll begin by asking the question: what common sense properties do we expect a quantitative measure of likelihood to have?

Consider the following:

1. If an event is impossible, its likelihood should be zero.
2. The sum of likelihoods of all possible events should equal complete certainty. That is, it should be a constant. As this constant can be anything, let us define it to be 1.

In terms of the example above, if we denote time by $t$ and the likelihood by $P(t)$  then:

$P(t) = 0$ for $t< 2$ and  $t> 8$

And

$\sum_{t}P(t) = 1$ where $2\leq t< 8$

Where $\sum_{t}$ denotes the sum of all non-zero likelihoods – i.e. those that lie between 2 and 8 days. In simple terms this is the area enclosed by the likelihood curves and the x axis in figures 2 to 5.  (Technical Note:  Since $t$ is a continuous variable, this should be denoted by an integral rather than a simple sum, but this is a technicality that need not concern us here)

$P(t)$ is , in fact, what mathematicians call probability– which explains why I have used the symbol $P$ rather than $L$. Now that I’ve explained what it  is, I’ll use the word “probability” instead of ” likelihood” in the remainder of this article.

With these assumptions in hand, we can now obtain numerical values for the probability of completion for all times between 2 and 8 days. This can be figured out by noting that the area under the probability curve (the triangle in figure 3 and the weird shape in figure 4) must equal 1, and we’ll do this next.  Indeed, for the problem at hand, we’ll assume that all four task durations can be fitted to triangular distributions. This is primarily to keep things  simple. However, I should emphasise that you can use any shape so long as you can express it mathematically, and I’ll say more about this towards the end of this article.

### The triangular distribution

Let’s look at the estimate for Task 1. We have three numbers corresponding to a minimummost likely and maximum time.  To keep the discussion general, we’ll call these $t_{min}$, $t_{ml}$ and $t_{max}$ respectively, (we’ll get back to our estimator’s specific numbers later).

Now, what about the probabilities associated with each of these times?

Since $t_{min}$ and $t_{max}$ correspond to the minimum and maximum times,  the probability associated with these is zero. Why?  Because if it wasn’t zero, then there would be a non-zero probability of completion for a time less than $t_{min}$ or greater than $t_{max}$ – which isn’t possible [Note: this is a consequence of the assumption that the probability varies continuously –  so if it takes on non-zero value, $p_{0}$,  at $t_{min}$ then it must take on a value slightly less than $p_{0}$ – but greater than 0 –  at $t$ slightly smaller than $t_{min}$ ] .   As far as  the most likely time,  $t_{ml}$,  is concerned:  by definition, the probability attains its highest value at time $t_{ml}$.    So, assuming the probability can be described by a triangular function, the distribution must have the form shown in Figure 6 below.

Figure 6: Triangular distribution redux.

For the simulation, we need to know the equation describing the above distribution.  Although Wikipedia will tell us the answer in a mouse-click, it is instructive to figure it out for ourselves. First, note that the area under the triangle must be equal to  1 because the task must finish at some time between $t_{min}$ and $t_{max}$.   As a consequence we have:

$\frac{1}{2}\times{base}\times{altitude}=\frac{1}{2}\times{(t_{max}-t_{min})}\times{p(t_{ml})}=1\ldots\ldots{(1)}$

where $p(t_{ml})$ is the probability corresponding to time $t_{ml}$.  With a bit of rearranging we get,

$p(t_{ml})=\frac{2}{(t_{max}-t_{min})}\ldots\ldots(2)$

To derive the probability for any time $t$ lying between $t_{min}$ and $t_{ml}$, we note that:

$\frac{(t-t_{min})}{p(t)}=\frac{(t_{ml}-t_{min})}{p(t_{ml})}\ldots\ldots(3)$

This is a consequence of the fact that the ratios on either side of equation (3)  are  equal to the slope of the line joining the points $(t_{min},0)$ and $(t_{ml}, p(t_{ml}))$.

Figure 7

Substituting (2) in (3) and simplifying a bit, we obtain:

$p(t)=\frac{2(t-t_{min})}{(t_{ml}-t_{min})(t_{max}-t_{min})}\dots\ldots(4)$ for $t_{min}\leq t \leq t_{ml}$

In a similar fashion one can show that the probability for times lying between $t_{ml}$ and $t_{max}$ is given by:

$p(t)=\frac{2(t_{max}-t)}{(t_{max}-t_{ml})(t_{max}-t_{min})}\dots\ldots(5)$ for $t_{ml}\leq t \leq t_{max}$

Equations 4 and 5 together describe the probability distribution function (or PDF)  for all times between $t_{min}$ and $t_{max}$.

As it turns out, in Monte Carlo simulations, we don’t directly work with the probability distribution function. Instead we work with the cumulative distribution function (or CDF) which is the probability, $P$,  that the task is completed by time $t$. To reiterate, the PDF, $p(t)$, is the probability of the task finishing at time $t$ whereas the CDF, $P(t)$, is the probability of the task completing by time $t$. The CDF, $P(t)$,  is essentially a sum of all probabilities between $t_{min}$ and $t$. For $t < t_{min}$ this is the area under the triangle with apexes at   ($t_{min}$, 0), (t, 0) and (t, p(t)).  Using the formula for the area of a triangle (1/2 base times height) and equation (4) we get:

$P(t)=\frac{(t-t_{min})^2}{(t_{ml}-t_{min})(t_{max}-t_{min})}\ldots\ldots(6)$ for $t_{min}\leq t \leq t_{ml}$

Noting that for $t \geq t_{ml}$, the area under the curve equals the total area minus the area enclosed by the triangle with base between t and $t_{max}$, we have:

$P(t)=1- \frac{(t_{max}-t)^2}{(t_{max}-t_{ml})(t_{max}-t_{min})}\ldots\ldots(7)$ for $t_{ml}\leq t \leq t_{max}$

As expected,  $P(t)$  starts out with a value 0 at $t_{min}$ and then increases monotonically, attaining a value of 1 at $t_{max}$.

To end this section let’s plug in the numbers quoted by our estimator at the start of this section: $t_{min}=2$, $t_{ml}=4$ and $t_{max}=8$.  The resulting PDF and CDF are shown in figures 8 and 9.

Figure 8: PDF for triangular distribution (tmin=2, tml=4, tmax=8)

Figure 9 – CDF for triangular distribution (tmin=2, tml=4, tmax=8)

### Monte Carlo in a minute

Now with all that conceptual work done, we can get to the main topic of this post:  Monte Carlo estimation. The basic idea behind Monte Carlo is to simulate the entire project (all 4 tasks in this case) a large number N (say 10,000) times and thus obtain N overall completion times.  In each of the N trials, we simulate each of the tasks in the project and add them up appropriately to give us an overall project completion time for the trial.  The resulting N overall completion times will all be different, ranging from the sum of the minimum completion times to the sum of the maximum completion times.  In other words, we will obtain the PDF and CDF for the overall completion time, which will enable us to answer questions such as:

• How likely is it that the project will be completed within 17 days?
• What’s the estimated time for which I can be 90% certain that the project will be completed? For brevity, I’ll call this the 90% completion time in the rest of this piece.

“OK, that sounds great”, you say, “but how exactly do we simulate a single task”?

Good question, and I was just about to get to that…

### Simulating a single task using the CDF

As we saw earlier, the CDF for the triangular has a S shape and ranges from 0 to 1 in value. It turns out that the S shape is characteristic of all CDFs, regardless of the details underlying PDF. Why? Because, the cumulative probability must lie between 0 and 1 (remember, probabilities can never exceed 1, nor can they be negative).

OK, so to simulate a task, we:

• generate a random number between 0 and 1, this corresponds to the probability that the task will finish at time t.
• find the time, t, that this corresponds to this value of probability. This is the completion time for the task for this trial.

Incidentally, this method is called inverse transform sampling.

An example might help clarify how inverse transform sampling works.  Assume that the random number generated is 0.4905. From the CDF for the first task, we see that this value of probability corresponds to a completion time of 4.503 days, which is the completion for this trial (see Figure 10). Simple!

Figure 10: Illustrating inverse transform sampling

In this case we found the time directly from the computed CDF. That’s not too convenient when you’re simulating the project 10,000 times. Instead, we need a programmable math expression that gives us the time corresponding to the probability directly. This can be obtained by solving equations (6) and (7) for $t$. Some straightforward algebra, yields the following two expressions for $t$:

$t = t_{min} + \sqrt{P(t)(t_{ml} - t_{min})(t_{max} - t_{min})} \ldots\ldots(8)$ for $t_{min}\leq t \leq t_{ml}$

And

$t = t_{max} - \sqrt{[1-P(t)](t_{max} - t_{ml})(t_{max} - t_{min})} \ldots\ldots(9)$ for $t_{ml}\leq t \leq t_{max}$

These can be easily combined in a single Excel formula using an IF function, and I’ll show you exactly how in a minute. Yes, we can now finally get down to the Excel simulation proper and you may want to download the workbook if you haven’t done so already.

### The simulation

Open up the workbook and focus on the first three columns of the first sheet to begin with. These simulate the first task in Figure 1, which also happens to be the task we have used to illustrate the construction of the triangular distribution as well as the mechanics of Monte Carlo.

Rows 2 to 4 in columns A and B list the min, most likely and max completion times while the same rows in column C list the probabilities associated with each of the times. For $t_{min}$ the probability is 0 and for $t_{max}$ it is 1.  The probability at $t_{ml}$ can be calculated using equation (6) which, for $t=t_{max}$, reduces to

$P(t_{ml}) =\frac{(t_{ml}-t_{min})}{t_{max}-t_{min}}\ldots\ldots(10)$

Rows 6 through 10005 in column A are simulated probabilities of completion for Task 1. These are obtained via the Excel RAND() function, which generates uniformly distributed random numbers lying between 0 and 1.  This gives us a list of probabilities corresponding to 10,000 independent simulations of Task 1.

The 10,000 probabilities need to be translated into completion times for the task. This is done using equations (8) or (9) depending on whether the simulated probability is less or greater than $P(t_{ml})$, which is in cell C3 (and given by Equation (10) above). The conditional statement can be coded in an Excel formula using the IF() function.

Tasks 2-4 are coded in exactly the same way, with distribution parameters in rows 2 through 4 and simulation details in rows 6 through 10005 in the columns listed below:

• Task 2 – probabilities in column D; times in column F
• Task 3 – probabilities in column H; times in column I
• Task 4 – probabilities in column K; times in column L

That’s basically it for the simulation of individual tasks. Now let’s see how to combine them.

For tasks in series (Tasks 1 and 2), we simply sum the completion times for each task to get the overall completion times for the two tasks.  This is what’s shown in rows 6 through 10005 of column G.

For tasks in parallel (Tasks 3 and 4), the overall completion time is the maximum of the completion times for the two tasks. This is computed and stored in rows 6 through 10005 of column N.

Finally, the overall project completion time for each simulation is then simply the sum of columns G and N (shown in column O)

Sheets 2 and 3 are plots of the probability and cumulative probability distributions for overall project completion times. I’ll cover these in the next section.

### Discussion – probabilities and estimates

The figure on Sheet 2 of the Excel workbook (reproduced in Figure 11 below) is the probability distribution function (PDF) of completion times. The x-axis shows the elapsed time in days and the y-axis the number of Monte Carlo trials that have a completion time that lie in the relevant time bin (of width 0.5 days). As an example, for the simulation shown in the Figure 11, there were 882 trials (out of 10,000) that had a completion time that lie between 16.25 and 16.75 days. Your numbers will vary, of course, but you should have a maximum in the 16 to 17 day range and a trial number that is reasonably close to the one I got.

Figure 11: Probability distribution of completion times (N=10,000)

I’ll say a bit more about Figure 11 in the next section. For now, let’s move on to Sheet 3 of workbook which shows the cumulative probability of completion by a particular day (Figure 12 below).  The figure shows the cumulative probability function (CDF), which is the sum of all completion times from the earliest possible completion day to the particular day.

Figure 12: Probability of completion by a particular day (N=10,000)

To reiterate a point made earlier,  the reason we work with the CDF  rather than the PDF is that we are interested in knowing the probability of completion by a particular date (e.g. it is 90% likely that we will finish by April 20th) rather than the probability of completion on a particular date (e.g. there’s a 10% chance we’ll finish on April 17th). We can now answer the two questions we posed earlier. As a reminder, they are:

• How likely is it that the project will be completed within 17 days?
• What’s the 90% likely completion time?

Both questions are easily answered by using the cumulative distribution chart on Sheet 3 (or Fig 12).  Reading the relevant numbers from the chart, I see that:

• There’s a 60% chance that the project will be completed in 17 days.
• The 90% likely completion time is 19.5 days.

How does the latter compare to the sum of the 90% likely completion times for the individual tasks? The 90% likely completion time for a given task can be calculated by solving Equation 9 for $t$, with appropriate values for the parameters $t_{min}$, $t_{max}$ and $t_{ml}$ plugged in, and $P(t)$ set to 0.9. This gives the following values for the 90% likely completion times:

• Task 1 – 6.5 days
• Task 2 – 8.1 days
• Task 3 – 7.7 days
• Task 4 – 5.8 days

Summing up the first three tasks (remember, Tasks 3 and 4 are in parallel) we get a total of 22.3 days, which is clearly an overestimation. Now, with the benefit of having gone through the simulation, it is easy to see that the sum of 90% likely completion times for individual tasks does not equal the 90% likely completion time for the sum of the relevant individual tasks – the first three tasks in this particular case. Why? Essentially because a Monte Carlo run in which the first three tasks tasks take as long as their (individual) 90% likely completion times is highly unlikely. Exercise:  use the worksheet to estimate how likely this is.

There’s much more that can be learnt from the CDF. For example, it also tells us that the greatest uncertainty in the estimate is in the 5 day period from ~14 to 19 days because that’s the region in which the probability changes most rapidly as a function of elapsed time. Of course, the exact numbers are dependent on the assumed form of the distribution. I’ll say more about this in the final section.

To close this section, I’d like to reprise a point I mentioned earlier: that uncertainty is a shape, not a number. Monte Carlo simulations make the uncertainty in estimates explicit and can help you frame your estimates in the language of probability…and using a tool like Excel can help you explain these to non-technical people like your manager.

### Closing remarks

We’ve covered a fair bit of ground: starting from general observations about how long a task takes, saw how to construct simple probability distributions and then combine these using Monte Carlo simulation.  Before I close, there are a few general points I should mention for completeness…and as warning.

First up, it should be clear that the estimates one obtains from a simulation depend critically on the form and parameters of the distribution used. The parameters are essentially an empirical matter; they should be determined using historical data. The form of the function, is another matter altogether: as pointed out in an earlier section, one cannot determine the shape of a function from a finite number of data points. Instead, one has to focus on the properties that are important. For example, is there a small but finite chance that a task can take an unreasonably long time? If so, you may want to use a lognormal distribution…but remember, you will need to find a sensible way to estimate the distribution parameters from your historical data.

Second, you may have noted from the probability distribution curve (Figure 11)  that despite the skewed distributions of the individual tasks, the distribution of the overall completion time is somewhat symmetric with a minimum of ~9 days, most likely time of ~16 days and maximum of 24 days.  It turns out that this is a general property of distributions that are generated by adding a large number of independent probabilistic variables. As the number of variables increases, the overall distribution will tend to the ubiquitous Normal distribution.

The assumption of independence merits a closer look.  In the case it hand,  it implies that the completion times for each task are independent of each other. As most project managers will know from experience, this is rarely the case: in real life,  a task that is delayed will usually have knock-on effects on subsequent tasks. One can easily incorporate such dependencies in a Monte Carlo simulation. A formal way to do this is to introduce a non-zero correlation coefficient between tasks as I have done here. A simpler and more realistic approach is to introduce conditional inter-task dependencies As an example, one could have an inter-task delay that kicks in only if the predecessor task takes more than 80%  of its maximum time.

Thirdly, you may have wondered why I used 10,000 trials: why not 100, or 1000 or 20,000. This has to do with the tricky issue of convergence. In a nutshell, the estimates we obtain should not depend on the number of trials used.  Why? Because if they did, they’d be meaningless.

Operationally, convergence means that any predicted quantity based on aggregates should not vary with number of trials.  So, if our Monte Carlo simulation has converged, our prediction of 19.5 days for the 90% likely completion time should not change substantially if I increase the number of trials from ten to twenty thousand. I did this and obtained almost the same value of 19.5 days. The average and median completion times (shown in cell Q3 and Q4 of Sheet 1) also remained much the same (16.8 days). If you wish to repeat the calculation, be sure to change the formulas on all three sheets appropriately. I was lazy and hardcoded the number of trials. Sorry!

Finally, I should mention that simulations can be usefully performed at a higher level than individual tasks. In their highly-readable book,  Waltzing With Bears: Managing Risk on Software Projects, Tom De Marco and Timothy Lister show how Monte Carlo methods can be used for variables such as  velocity, time, cost etc.  at the project level as opposed to the task level. I believe it is better to perform simulations at the lowest possible level, the main reason being that it is easier, and less error-prone, to estimate individual tasks than entire projects. Nevertheless, high level simulations can be very useful if one has reliable data to base these on.

There are a few more things I could say about the usefulness of the generated distribution functions and Monte Carlo in general, but they are best relegated to a future article. This one is much too long already and I think I’ve tested your patience enough. Thanks so much for reading, I really do appreciate it and hope that you found it useful.

Acknowledgement: My thanks to Peter Holberton for pointing out a few typographical and coding errors in an earlier version of this article. These have now been fixed. I’d be grateful if readers could bring any errors they find to my attention.

Written by K

March 27, 2018 at 4:11 pm

Tagged with

## A gentle introduction to logistic regression and lasso regularisation using R

In this day and age of artificial intelligence and deep learning, it is easy to forget that simple algorithms can work well for a surprisingly large range of practical business problems.  And the simplest place to start is with the granddaddy of data science algorithms: linear regression and its close cousin, logistic regression. Indeed, in his acclaimed MOOC and accompanying textbook, Yaser Abu-Mostafa spends a good portion of his time talking about linear methods, and with good reason too: linear methods are not only a good way to learn the key principles of machine learning, they can also be remarkably helpful in zeroing in on the most important predictors.

My main aim in this post is to provide a beginner level introduction to logistic regression using R and also introduce LASSO (Least Absolute Shrinkage and Selection Operator), a powerful feature selection technique that is very useful for regression problems. Lasso is essentially a regularization method. If you’re unfamiliar with the term, think of it as a way to reduce overfitting using less complicated functions (and if that means nothing to you, check out my prelude to machine learning).  One way to do this is to toss out less important variables, after checking that they aren’t important.  As we’ll discuss later, this can be done manually by examining p-values of coefficients and discarding those variables whose coefficients are not significant. However, this can become tedious for classification problems with many independent variables.  In such situations, lasso offers a neat way to model the dependent variable while automagically selecting significant variables by shrinking the coefficients of unimportant predictors to zero.  All this without having to mess around with p-values or obscure information criteria. How good is that?

### Why not linear regression?

In linear regression one attempts to model a dependent variable (i.e. the one being predicted) using the best straight line fit to a set of predictor variables.  The best fit is usually taken to be one that minimises the root mean square error,  which is the sum of square of the differences between the actual and predicted values of the dependent variable. One can think of logistic regression as the equivalent of linear regression for a classification problem.  In what follows we’ll look at binary classification – i.e. a situation where the dependent variable takes on one of two possible values (Yes/No, True/False, 0/1 etc.).

First up, you might be wondering why one can’t use linear regression for such problems. The main reason is that classification problems are about determining class membership rather than predicting variable values, and linear regression is more naturally suited to the latter than the former. One could, in principle, use linear regression for situations where there is a natural ordering of categories like High, Medium and Low for example. However, one then has to map sub-ranges of the predicted values to categories. Moreover, since predicted values are potentially unbounded (in data as yet unseen) there remains a degree of arbitrariness associated with such a mapping.

Logistic regression sidesteps the aforementioned issues by modelling class probabilities instead.  Any input to the model yields a number lying between 0 and 1, representing the probability of class membership. One is still left with the problem of determining the threshold probability, i.e. the probability at which the category flips from one to the other.  By default this is set to p=0.5, but in reality it should be settled based on how the model will be used.  For example, for a marketing model that identifies potentially responsive customers, the threshold for a positive event might be set low (much less than 0.5) because the client does not really care about mailouts going to a non-responsive customer (the negative event). Indeed they may be more than OK with it as there’s always a chance – however small – that a non-responsive customer will actually respond.  As an opposing example, the cost of a false positive would be high in a machine learning application that grants access to sensitive information. In this case, one might want to set the threshold probability to a value closer to 1, say 0.9 or even higher. The point is, the setting an appropriate threshold probability is a business issue, not a technical one.

### Logistic regression in brief

So how does logistic regression work?

For the discussion let’s assume that the outcome (predicted variable) and predictors are denoted by Y and X respectively and the two classes of interest are denoted by + and – respectively.  We wish to model the conditional probability that the outcome Y is +, given that the input variables (predictors) are X. The conditional probability is denoted by p(Y=+|X)   which we’ll abbreviate as p(X) since we know we are referring to the positive outcome Y=+.

As mentioned earlier, we are after the probability of class membership so we must ensure that the hypothesis function (a fancy word for the model) always lies between 0 and 1. The function assumed in logistic regression is:

$p(X) = \dfrac{\exp^{\beta_0+\beta_1 X}}{1+\exp^{\beta_0 + \beta_1 X}} .....(1)$

You can verify that $p(X)$ does indeed lie between 0 and  1 as $X$ varies from $-\infty$ to $\infty$.  Typically, however, the values of $X$ that make sense are bounded as shown in the example (stolen from Wikipedia) shown in Figure 1. The figure also illustrates the typical S-shaped  curve characteristic of logistic regression.

Figure 1: Logistic function

As an aside, you might be wondering where the name logistic comes from. An equivalent way of expressing the above equation is:

$\log(\dfrac{p(X)}{1-p(X)}) = \beta_0+\beta_1 X .....(2)$

The quantity on the left is the logarithm of the odds. So, the model is a linear regression of the log-odds, sometimes called logit, and hence the name logistic.

The problem is to find the values of $\beta_0$  and $\beta_1$ that results in a $p(X)$ that most accurately classifies all the observed data points – that is, those that belong to the positive class have a probability as close as possible to 1 and those that belong to the negative class have a probability as close as possible to 0. One way to frame this problem is to say that we wish to maximise the product of these probabilities, often referred to as the likelihood:

$\displaystyle\log ( {\prod_{i:Y_i=+} p(X_{i}) \prod_{j:Y_j=-}(1-p(X_{j}))})$

Where $\prod$ represents the products over i and j, which run over the +ve and –ve classed points respectively. This approach, called maximum likelihood estimation, is quite common in many machine learning settings, especially those involving probabilities.

It should be noted that in practice one works with the log likelihood because it is easier to work with mathematically. Moreover, one minimises the negative  log likelihood which, of course, is the same as maximising the log likelihood.  The quantity one minimises is thus:

$L = - \displaystyle\log ( {\prod_{i:Y_i=+} p(X_{i}) \prod_{j:Y_j=-}(1-p(X_{j}))}).....(3)$

However, these are technical details that I mention only for completeness. As you will see next, they have little bearing on the practical use of logistic regression.

### Logistic regression in R – an example

In this example, we’ll use the logistic regression option implemented within the glm function that comes with the base R installation. This function fits a class of models collectively known as generalized linear models. We’ll apply the function to the Pima Indian Diabetes dataset that comes with the mlbench package. The code is quite straightforward – particularly if you’ve read earlier articles in my “gentle introduction” series – so I’ll just list the code below  noting that the logistic regression option is invoked by setting family=”binomial”  in the glm function call.

Here we go:

#set working directory if needed (modify path as needed)
#setwd(“C:/Users/Kailash/Documents/logistic”)
library(mlbench)
data(“PimaIndiansDiabetes”)
#set seed to ensure reproducible results
set.seed(42)
#split into training and test sets
PimaIndiansDiabetes[,”train”] <- ifelse(runif(nrow(PimaIndiansDiabetes))<0.8,1,0)
#separate training and test sets
trainset <- PimaIndiansDiabetes[PimaIndiansDiabetes$train==1,] testset <- PimaIndiansDiabetes[PimaIndiansDiabetes$train==0,]
#get column index of train flag
trainColNum <- grep(“train”,names(trainset))
#remove train flag column from train and test sets
trainset <- trainset[,-trainColNum]
testset <- testset[,-trainColNum]
#get column index of predicted variable in dataset
typeColNum <- grep(“diabetes”,names(PimaIndiansDiabetes))
#build model
glm_model <- glm(diabetes~.,data = trainset, family = binomial)
summary(glm_model)
Call:
glm(formula = diabetes ~ ., family = binomial, data = trainset)
<<output edited>>
Coefficients:
Estimate  Std. Error z value Pr(>|z|)
(Intercept)-8.1485021 0.7835869 -10.399  < 2e-16 ***
pregnant    0.1200493 0.0355617   3.376  0.000736 ***
glucose     0.0348440 0.0040744   8.552  < 2e-16 ***
pressure   -0.0118977 0.0057685  -2.063  0.039158 *
triceps     0.0053380 0.0076523   0.698  0.485449
insulin    -0.0010892 0.0009789  -1.113  0.265872
mass        0.0775352 0.0161255   4.808  1.52e-06 ***
pedigree    1.2143139 0.3368454   3.605  0.000312 ***
age         0.0117270 0.0103418   1.134  0.256816
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#predict probabilities on testset
#type=”response” gives probabilities, type=”class” gives class
glm_prob <- predict.glm(glm_model,testset[,-typeColNum],type=”response”)
#which classes do these probabilities refer to? What are 1 and 0?
contrasts(PimaIndiansDiabetes$diabetes) pos neg 0 pos 1 #make predictions ##…first create vector to hold predictions (we know 0 refers to neg now) glm_predict <- rep(“neg”,nrow(testset)) glm_predict[glm_prob>.5] <- “pos” #confusion matrix table(pred=glm_predict,true=testset$diabetes)
glm_predict  neg pos
neg    90 22
pos     8 33
#accuracy
mean(glm_predict==testset$diabetes) [1] 0.8039216 Although this seems pretty good, we aren’t quite done because there is an issue that is lurking under the hood. To see this, let’s examine the information output from the model summary, in particular the coefficient estimates (i.e. estimates for $\beta$) and their significance. Here’s a summary of the information contained in the table: • Column 2 in the table lists coefficient estimates. • Column 3 list s the standard error of the estimates (the larger the standard error, the less confident we are about the estimate) • Column 4 the z statistic (which is the coefficient estimate (column 2) divided by the standard error of the estimate (column 3)) and • The last column (Pr(>|z|) lists the p-value, which is the probability of getting the listed estimate assuming the predictor has no effect. In essence, the smaller the p-value, the more significant the estimate is likely to be. From the table we can conclude that only 4 predictors are significant – pregnant, glucose, mass and pedigree (and possibly a fifth – pressure). The other variables have little predictive power and worse, may contribute to overfitting. They should, therefore, be eliminated and we’ll do that in a minute. However, there’s an important point to note before we do so… In this case we have only 9 variables, so are able to identify the significant ones by a manual inspection of p-values. As you can well imagine, such a process will quickly become tedious as the number of predictors increases. Wouldn’t it be be nice if there were an algorithm that could somehow automatically shrink the coefficients of these variables or (better!) set them to zero altogether? It turns out that this is precisely what lasso and its close cousin, ridge regression, do. ### Ridge and Lasso Recall that the values of the logistic regression coefficients $\beta_0$ and $\beta_1$ are found by minimising the negative log likelihood described in equation (3). Ridge and lasso regularization work by adding a penalty term to the log likelihood function. In the case of ridge regression, the penalty term is $\beta_1^2$ and in the case of lasso, it is $|\beta_1|$ (Remember, $\beta_1$ is a vector, with as many components as there are predictors). The quantity to be minimised in the two cases is thus: $L +\lambda \sum \beta_1^2.....(4)$ – for ridge regression, and $L +\lambda \sum |\beta_1|.....(5)$ – for lasso regression. Where $\lambda$ is a free parameter which is usually selected in such a way that the resulting model minimises the out of sample error. Typically, the optimal value of $\lambda$ is found using grid search with cross-validation, a process akin to the one described in my discussion on cost-complexity parameter estimation in decision trees. Most canned algorithms provide methods to do this; the one we’ll use in the next section is no exception. In the case of ridge regression, the effect of the penalty term is to shrink the coefficients that contribute most to the error. Put another way, it reduces the magnitude of the coefficients that contribute to increasing $L$. In contrast, in the case of lasso regression, the effect of the penalty term is to set the these coefficients exactly to zero! This is cool because what it mean that lasso regression works like a feature selector that picks out the most important coefficients, i.e. those that are most predictive (and have the lowest p-values). Let’s illustrate this through an example. We’ll use the glmnet package which implements a combined version of ridge and lasso (called elastic net). Instead of minimising (4) or (5) above, glmnet minimises: $L +\lambda[ (1-\alpha)\sum [\beta_1^2 + \alpha\sum|\beta_1|]....(6)$ where $\alpha$ controls the “mix” of ridge and lasso regularisation, with $\alpha=0$ being “pure” ridge and $\alpha=1$ being “pure” lasso. ### Lasso regularisation using glmnet Let’s reanalyse the Pima Indian Diabetes dataset using glmnet with $\alpha=1$ (pure lasso). Before diving into code, it is worth noting that glmnet: • does not have a formula interface, so one has to input the predictors as a matrix and the class labels as a vector. • does not accept categorical predictors, so one has to convert these to numeric values before passing them to glmnet. The glmnet function model.matrix creates the matrix and also converts categorical predictors to appropriate dummy variables. Another important point to note is that we’ll use the function cv.glmnet, which automatically performs a grid search to find the optimal value of $\lambda$. OK, enough said, here we go: #load required library library(glmnet) #convert training data to matrix format x <- model.matrix(diabetes~.,trainset) #convert class to numerical variable y <- ifelse(trainset$diabetes==”pos”,1,0)
#perform grid search to find optimal value of lambda
#family= binomial => logistic regression, alpha=1 => lasso
# check docs to explore other type.measure options
cv.out <- cv.glmnet(x,y,alpha=1,family=”binomial”,type.measure = “mse” )
#plot result
plot(cv.out)

The plot is shown in Figure 2 below:

Figure 2: Error as a function of lambda (select lambda that minimises error)

The plot shows that the log of the optimal value of lambda (i.e. the one that minimises the root mean square error) is approximately -5. The exact value can be viewed by examining the variable lambda_min in the code below. In general though, the objective of regularisation is to balance accuracy and simplicity. In the present context, this means a model with the smallest number of coefficients that also gives a good accuracy.  To this end, the cv.glmnet function  finds the value of lambda that gives the simplest model but also lies within one standard error of the optimal value of lambda. This value of lambda (lambda.1se) is what we’ll use in the rest of the computation. Interested readers should have a look at this article for more on lambda.1se vs lambda.min.

#min value of lambda
lambda_min <- cv.out$lambda.min #best value of lambda lambda_1se <- cv.out$lambda.1se
#regression coefficients
coef(cv.out,s=lambda_1se)
10 x 1 sparse Matrix of class “dgCMatrix”
1
(Intercept) -4.61706681
(Intercept)  .
pregnant     0.03077434
glucose      0.02314107
pressure     .
triceps      .
insulin      .
mass         0.02779252
pedigree     0.20999511
age          .

The output shows that only those variables that we had determined to be significant on the basis of p-values have non-zero coefficients. The coefficients of all other variables have been set to zero by the algorithm! Lasso has reduced the complexity of the fitting function massively…and you are no doubt wondering what effect this  has on accuracy. Let’s see by running the model against our test data:

#get test data
x_test <- model.matrix(diabetes~.,testset)
#predict class, type=”class”
lasso_prob <- predict(cv.out,newx = x_test,s=lambda_1se,type=”response”)
#translate probabilities to predictions
lasso_predict <- rep(“neg”,nrow(testset))
lasso_predict[lasso_prob>.5] <- “pos”
#confusion matrix
table(pred=lasso_predict,true=testset$diabetes) pred neg pos neg 94 28 pos 4 27 #accuracy mean(lasso_predict==testset$diabetes)
[1] 0.7908497

Which is a bit less than what we got with the more complex model. So, we get  a similar out-of-sample accuracy as we did before, and we do so using a way simpler function (4 non-zero coefficients) than the original one (9  nonzero coefficients). What this means is that the simpler function does at least as good a job fitting the signal in the data as the more complicated one.  The bias-variance tradeoff tells us that the simpler function should be preferred because it is less likely to overfit the training data.

Paraphrasing William of Ockhamall other things being equal, a simple hypothesis should be preferred over a complex one.

### Wrapping up

In this post I have tried to provide a detailed introduction to logistic regression, one of the simplest (and oldest) classification techniques in the machine learning practitioners arsenal. Despite it’s simplicity (or I should say, because of it!) logistic regression works well for many business applications which often have a simple decision boundary. Moreover, because of its simplicity it is less prone to overfitting than flexible methods such as decision trees. Further, as we have shown, variables that contribute to overfitting can be eliminated using lasso (or ridge) regularisation, without compromising out-of-sample accuracy. Given these advantages and its inherent simplicity, it isn’t surprising that logistic regression remains a workhorse for data scientists.

Written by K

July 11, 2017 at 10:00 pm

## The improbability of success

Anyone who has tidied up after a toddler intuitively understands that making a mess is far easier than creating order. The fundamental reason for this is that the number of messy states in the universe (or a toddler’s room) far outnumbers the ordered ones.  As this point might not be obvious, I’ll demonstrate it via a simple thought experiment involving marbles:

Throw three marbles onto a flat surface.  When the marbles come to rest, you are most likely to end up with a random configuration  as in Figure 1.

Figure 1: A random configuration of 3 marbles

Indeed, you’d be extremely surprised if the three ended up being collinear as in Figure 2.   Note that Figure 2 is just one example of many collinear possibilities, but the point I’m making is that if the marbles are thrown randomly, they are more likely to end up in a random state than a lined-up one.

Figure 2: an unlikely (ordered) configuration

This raises a couple of questions:

Question: On what basis can one claim that the collinear configuration is tidier or more ordered than the non-collinear one?

Naive answer:  It looks more ordered. Yes, tidiness is in the eye of the beholder so it is necessarily subjective. However, I’ll wager that if one took a poll, an overwhelming number of people would say that the configuration in Figure 2 is more ordered than the one in Figure 1.

More sophisticated answer : The “state” of collinear marbles can be described using 2 parameters, the slope and intercept of the straight line that three marbles lie on (in any coordinate system) whereas the description of the nonlinear state requires 3 parameters. The first state is tidier because it requires fewer parameters.  Another way to think about is that the line can be described by two marbles; the third one is redundant as far as the description of the state is concerned.

Question: Why is a tidier configuration less likely than a messy one?

Answer:  May be you see this intuitively and need no proof, but here’s one just in case. Imagine rolling the three marbles one after the other. The first two, regardless of where they end up, will necessarily lie along a line (two points lie on the straight line joining them). Now, I think it is easy to see that if we throw the third marble randomly, it is highly unlikely end up on that line. Indeed, for the third marble to end up exactly on the same straight line requires a coincidence of near cosmic proportions.

I know, I know, this is not a proof, but I trust it makes the point.

Now, although it is near impossible to get to a collinear end state via random throws, it is possible to approximate it by changing the way we throw the marbles. Here’s how:

1. Throw the marbles consecutively rather than in one go.
2. When throwing the third marble, adjust its initial speed and direction in a way that takes into account the positions of the two marbles that are already on the surface. Remember these two already define a straight line.

The third throw is no longer random because it is designed to maximise the chance that the last marble will get as close as possible to the straight line defined by the first two. Done right, you’ll end up with something closer to the configuration in Figure 3 rather than the one in Figure 2.

Figure 3: an “approximately ordered” state

Now you’re probably wondering what this has to do with success. I’ll make the connection via an example that will be familiar to many readers of this blog: an organisation’s strategy. However, as I will reiterate later, the arguments I present are very general and can be applied to just about any initiative or situation.

Typically, a strategy sets out goals for an organisation and a plan to achieve them in a specified timeframe. The goals define a number of desirable outcomes, or states which, by design, are constrained to belong to a (very) small subset of all possible states the organisation can end up in.  In direct analogy with the simple model discussed above it is clear that, left to its own devices, the organisation is more likely to end up in one of the much overwhelmingly larger number of “failed states” than one of the successful ones.  Notwithstanding the popular quote about there being many roads to success, in reality there are a great many more roads to failure.

Of course, that’s precisely why organisations are never “left to their own devices.” Indeed, a strategic plan specifies actions that are intended to make a successful state more likely than an unsuccessful one. However, no plan can guarantee success; it can, at best, make it more likely. As in the marble game, success is ultimately a matter of chance, even when we take actions to make it more likely.

If we accept this, the key question becomes: how can one design a strategy that improves the odds of success?  The marble analogy suggests a way to do this is to:

1. Define success in terms of an end state that is a natural extension of your current state.
2. Devise a plan to (approximately) achieve that end state. Such a plan will necessarily build on the current state rather than change it wholesale. Successful change is an evolutionary process rather than a revolutionary one.

My contention is that these points are often ignored by management strategists. More often than not, they will define an end state based on a textbook idealisation, consulting model or (horror!) best practice. The marble analogy shows why copying others is unlikely to succeed.

Figure 4 shows a variant of the marble game in which we have two sets of marbles (or organisations!), one blue, as before, and the other red.

Figure 4: Two distinct configurations of marbles (or organisations)

Now, it is considerably harder to align an additional marble with both sets of marbles than the blue one alone. Here’s why…

To align with both sets, the new marble has to end up close to the point that lies at the intersection of the blue and red lines in Figure 5. In contrast, to align with the blue set alone, all that’s needed is for it to get close to any point on the blue line.

QED!

Figure 5: Why copying others is not a good idea (see text for explanation)

Finally, on a broader note, it should be clear that the arguments made above go beyond organisational strategies. They apply to pretty much any planned action, whether at work or in one’s personal life.

So, to sum up: when developing an organisational (or personal) strategy, the first step is to understand where you are and then identify the minimal actions you need to take in order to get to an “improved” state that is consistent with  your current one. Yes, this is akin to the incremental and evolutionary approach that Agilistas and Leaners have been banging on about for years. However, their prescriptions focus on specific areas: software development and process improvement.  My point is that the basic principles are way broader because they are a direct consequence of a fundamental fact regarding the relative likelihood of order and disorder in a toddler’s room, an organisation, or even the universe at large.

Written by K

April 4, 2017 at 9:16 pm