An introduction to Monte Carlo simulation of project tasks

Introduction

In an essay on the uncertainty of project task estimates, I described how a task estimate corresponds to a probability distribution. Put simply, a task estimate is actually a range of possible completion times, each with a probability of occurrence specified by a distribution. If one knows the distribution, it is possible to answer questions such as: “What is the probability that the task will be completed within x days?”

The reliability of such predictions depends on how faithfully the distribution captures the actual spread of task durations – and therein lie at least a couple of problems. First, the probability distributions for task durations are generally hard to characterise because of the lack of reliable data (estimators are not very good at estimating, and historical data is usually not available). Second, many realistic distributions have complicated mathematical forms which can be hard to characterise and manipulate.

These problems are compounded by the fact that projects consist of several tasks, each one with its own duration estimate and (possibly complicated) distribution. The first issue is usually addressed by fitting distributions to point estimates (such as optimistic, pessimistic and most likely times as in PERT) and then refining these estimates and distributions as one gains experience. The second issue can be tackled by Monte Carlo techniques, which involve simulating the task a number of times (using an appropriate distribution) and then calculating expected completion times based on the results. My aim in this post is to present an example-based introduction to Monte Carlo simulation of project task durations.

Although my aim is to keep things reasonably simple (not too much beyond high-school maths and a basic understanding of probability), I’ll be covering a fair bit of ground. Given this, I’d best to start with a brief description of my approach so that my readers know what coming.

Monte Carlo simulation is an umbrella term that covers a range of approaches that use random sampling to simulate events that are described by known probability distributions. The first task then, is to specify the probability distribution. However, as mentioned earlier, this is generally unknown for task durations. For simplicity, I’ll assume that task duration uncertainty can be described accurately using a triangular probability distribution – a distribution that is particularly easy to handle from the mathematical point of view. The advantage of using the triangular distribution is that simulation results can be validated easily.

Using the triangular distribution isn’t a limitation because the method I describe can be applied to arbitrarily shaped distributions. More important, the technique can be used to simulate what happens when multiple tasks are strung together as in a project schedule (I’ll cover this in a future post). Finally, I’ll demonstrate a Monte Carlo simulation method as applied to a single task described by a triangular distribution. Although a simulation is overkill in this case (because questions regarding durations can be answered exactly without using a simulation), the example serves to illustrate the steps involved in simulating more complex cases – such as those comprising of more than one task and/or involving more complicated distributions.

So, without further ado, let me begin the journey by describing the triangular distribution.

The triangular distribution

Let’s assume that there’s a project task that needs doing, and the person who is going to do it reckons it will take between 2 and 8 hours to complete it, with a most likely completion time of 4 hours. How the estimator comes up with these numbers isn’t important at this stage – maybe there’s some guesswork, maybe some padding or maybe it is really based on experience (as it should be). What’s important is that we have three numbers corresponding to a minimum, most 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 1 below.

Figure 1: Triangular Distribution

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 $t_{max}$ 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 2

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 $t_{max}$ .

Another quantity of interest is the cumulative distribution function (or CDF) which is the probability, $P$ , that the task is completed by a 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 3 and 4.

Figure 3 - Triangular PDF (tmin=2, tml=4, tmax=8)

Figure 4 – Triangular CDF (tmin=2, tml=4, tmax=8)

Monte Carlo Simulation of a Single Task

OK, so now we get to the business end of this essay – the simulation. I’ll first outline the simulation procedure and then discuss results for the case of the task described in the previous section (triangular distribution with $t_{min}=2$ , $t_{ml}=4$ and $t_{max}=8$ ). Note that I used TK Solver – a mathematical package created by Universal Technical Systems – to do the simulations. TK Solver has built-in backsolving capability which is extremely helpful for solving some of the equations that come up in the simulation calculations. One could use Excel too, but my spreadsheet skills are not up to it :-(.

So, here’s my simulation procedure:

Generate a random number between 0 and 1. Treat this number as the cumulative probability, $P(t)$ for the simulation run. [Technical Note: I used the random number generator that comes with the TK Solver package (the algorithm used by the generator is described here). Excel’s random number generator is even better.]
Find the time, $t$ , corresponding to $P(t)$ by solving equations (6) or (7) for $t$ . The resulting value of $t$ is the time taken to complete the task. [Technical Note: Solving equation (6) or (7) for $t$ isn’t straightforward because $t$ appears in several places in the equations. One has two options to solve for $t$ a) Use numerical techniques such as the bisection or Newton-Raphson method or b) use the backsolve (goal seek) functionality in Excel or other mathematical packages. I used the backsolving capability of TK Solver to obtain t for each random value of P generated. TK Solver backsolves equations automatically – no fiddling around with numerical methods – which makes it an attractive option for these kinds of calculations.]
Repeat steps (1) and (2) N times, where N is a “sufficiently large” number – say 10,000.

I did the calculations for $N=10000$ using the triangular distribution with parameters $t_{min}=2$ , $t_{ml}=4$ and $t_{max}=8$ . This gave me 10,000 values of $P(t)$ and $t$ .

As an example of a simulation run proceeds, here’s the data from my first simulation run: the random number generator returned 0.490474700804856 (call it 0.4905). This is the value of $P(t)$ . The time corresponding to this cumulative probability is obtained by solving equation (7) numerically for $t$ . This gave $t = 4.503057452476027$ (call it 4.503) as shown in Figure 5. This is the completion time for the first run.

Figure 5

After completing 10,000 simulation runs, I sorted these into bins corresponding to time intervals of .25 hrs, starting from $t$ =2hrs through to $t$ =8 hrs. The resulting histogram is shown in Figure 6. Each bar corresponds to the number of simulation runs that fall within that time interval.

Figure 6: Distribution of simulation runs

As one might expect, this looks like the triangular distribution shown in Figure 4. There is a difference though: Figure 4 plots probability as a continuous function of time whereas Figure 6 plots the number of simulation runs as a step function of time. To convince ourselves that the two are really the same, lets look at the cumulative probability at $t_{ml}$ – i.e the probability that the task will be completed within 4 hrs. From equation 6 we get $P(t_{ml})=0.3333$ . The corresponding number from the simulation is simply the number of simulation runs that had a completion time less than or equal to 4 hrs, divided by the total number of simulation runs. For my simulation this comes out to be 0.3383. The agreement’s not perfect, but is convincing enough. Just to be sure, I performed the simulation a number of times – generating several sets of random numbers – and took the average of the predicted $P(t_{ml})$ . The agreement between theory and simulation improved, as expected.

Wrap up

A limitation of the triangular distribution is that it imposes an upper cut-off at $t_{max}$ . Long-tailed distributions may therefore be more realistic. In the end, though, the form distribution is neither here nor there because the technique can be applied to any distribution. The real question is: how do we obtain reliable distributions for our estimates? There’s no easy answer to this one, but one can start with three point estimates (as in PERT) and then fit these to a triangular (or more complicated) distribution. Although it is best if one has historical data, in the absence this one can always start with reasonable guesses. The point is to refine these through experience.

Another point worth mentioning is that simulations can be done at a level higher than that of an indivdual task. In their brilliant book – Waltzing With Bears: Managing Risk on Software Projects – De Marco and Lister demonstrate the use of Monte Carlo methods to simulate various aspects of project – 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 (although it is a lot more work) – 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.

I would be remiss if I didn’t mention some of the various Monte Carlo packages available in the market. I’ve never used any of these, but by all accounts they’re pretty good: see this commercial package or this one, for example. Both products use random number generators and sampling techniques that are far more sophisticated than the simple ones I’ve used in my example.

Finally, I have to admit that the example described in this post is a very complicated way of demonstrating the obvious – I started out with the triangular distribution and then got back the triangular distribution via simulation. My point, however, was to illustrate the method and show that it yields expected results in a situation where the answer is known. In a future post I’ll apply the method to more complex situations- for example, multiple tasks in series and parallel, with some dependency rules thrown in for good measure. Although, I’ll use the triangular distribution for individual tasks, the results will be far from obvious: simulation methods really start to shine as complexity increases. But all that will have to wait for later. For now, I hope my example has helped illustrate how Monte Carlo methods can be used to simulate project tasks.

Note added on 21 Sept 2009:

Follow-up to this article published here.

Note added on 14 Dec 2009:

See this post for a Monte Carlo simulation of correlated project tasks.

Written by K

September 11, 2009 at 11:05 pm

Posted in Estimation, Monte Carlo Simulation, portfolio management, Probability, Project Management, Statistics

15 Responses

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links for 2009-09-11 « Blarney Fellow

September 12, 2009 at 12:35 pm

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Hi Kailash,
Brilliant post and an great example to help us understand. But the only concern is the 3 point estimates (PERT). In an earlier post you had provided us links to a RAND corporation link about how PERT was a commercial makeover by the US DoD. PERT has several problems like “Merge Bias”, focussing only on critical path, neglecting correlation coefficients etc.. How do you still think 3 point estimates are a good start for Monte Carlo analysis? Will the final results from the simulation not have these as well?

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Prakash

September 14, 2009 at 3:10 am

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Prakash,

That’s an excellent point – and one that I should’ve explained in the post.

Anyway, here goes:

First, PERT – as it is commonly used- calculates expected task durations (typically via 3 point estimates) and then treats these as deterministic – i.e. the expected (or average) duration is treated as the task duration. This is OK for single tasks, as long as one understands that one is looking at the average completion time (over several trials). However, when one chains several tasks together, as in a project , it is no longer true that expected values add up (except in special cases). This point will be made clear in my next post on the topic, where I will deal with the simulation of multiple tasks (actually, only 2 tasks – but that will be enough to illustrate my point).

Second, the method described above is actually more realistic than a simple PERT treatment because the output is a range of possible completion times – each with a probability attached to it. Consequently, one can answer questions such as: “What is the probability of completion by time t?”. Now, of course, this question can be answered straight off (without simulation) for a single task described by a triangular distribution, but is well nigh impossible to answer straight-off for multiple tasks. Simulation is the only way for the latter.

Third, when simulating multiple tasks one can add in realistic inter-task events – such as delays etc. The merge path problem you mention is a consequence of ignoring slack path contributions to schedule risk. Simulations can take these into consideration as a matter of course.

Finally, as I’ve mentioned in the post, I’ve used the triangular distribution for illustrative purposes – the method described can be applied to any distribution.

Hope this addresses your question – do let me know if it doesn’t. Thanks again for raising an excellent point!

Regards,

Kailash.

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K

September 14, 2009 at 5:57 am

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