Monte Carlo simulation of multiple project tasks – three examples and some general comments

Introduction

In my previous post I  demonstrated the use of a Monte Carlo technique in simulating  a single project task with completion times  described by a triangular distribution.  My aim in that article was to:  a) describe a Monte Carlo simulation procedure in enough detail for someone interested to be able to reproduce the calculations and b) show that it gives sensible results in a situation where the answer is known.  Now it’s time to take things further.  In this post, I present simulations for two tasks chained together in various ways.  We shall see that, even with this small increase in complexity (from one task to two), the results obtained can be surprising.  Specifically, small changes in inter-task dependencies can have a huge effect on the overall (two-task) completion time distribution. Although, this is something that that most project managers have experienced in real life, it is rarely taken in to account by traditional scheduling techniques. As we shall see, Monte Carlo techniques predict such effects as a matter of course.

Background

The three simulations discussed here are built on the example that I used in my previous article, so it’s worth spending a few lines for a brief recap of that example.  The task simulated in the example was assumed to be described by a triangular distribution with  minimum completion time () of 2  hours,   most likely completion time () of 4 hours and   a maximum completion time () of 8 hours.   The resulting triangular probability distribution function (PDF),  –  which gives the probability of completing the task at time t – is shown in Figure 1.

Figure 1 - PDF for triangular distribution (tmin=2, tml=4, tmax=8)

Figure 2 depicts the associated cumulative distribution function (CDF),   which gives the probability that a task will be completed by time t (as opposed to the PDF which specifies the probability of completion at time t). The value of the CDF at t=8 is 1 because the task must finish within 8 hrs.

Figure 2 - PDF for triangular distribution (tmin=2, tml=4, tmax=6)

The equations describing the PDF and CDF are listed in equations 4-7 of my previous article.  I won’t rehash them here as they don’t add anything new  to the discussion – please see the article for all the gory algebraic details and formulas.   Now, with that background, we’re ready to move on to the examples.

Two tasks in series with no inter-task delay

As a first example, let’s look at two tasks that have to be performed sequentially – i.e. the second task starts as soon as the first one is completed. To simplify things, we’ll also assume that they have identical (triangular) distributions as described earlier and shown in Figure 1  (excepting , of course,  that the distribution is shifted to the right for the second task  – since it starts after the first one finishes).  We’ll also  assume that the second task begins right after the first one is completed (no inter-task delay) – yes, this is unrealistic, but bear with me.  The simulation algorithm for the  combined tasks is very similar to the one for a single task (described in detail in my previous post). Here’s the procedure:

1. For each of the two tasks, generate a set of N random numbers. Each random number generated corresponds to the cumulative probability of completion for a single task on that particular run.
2. For each random number generated, find the time corresponding to the cumulative probability by solving equation 6 or 7 in my previous post.
3. Step 2 gives N sets of completion times. Each set has two completion times – one for each tasks.
4. Add up the two numbers in each set to yield the comple. The resulting set corresponds to N simulation runs for the combined task.

I then divided the time interval from t=4 hours  (min possible completion time for both tasks) to t=16 hours (max possible completion time for both tasks) into bins of 0.25 hrs each, and then assigned each combined completion time to the appropriate bin. For example, if the predicted completion time for a particular run was 9.806532 hrs, it was assigned to the bin corresponding to 0.975 hrs.  The resulting histogram is shown in Figure 3 below (click on image to view the full-size graphic).

Figure 3 - Frequency histogram for tasks in series with no inter-task delay

[An aside:  compare the histogram in Figure 3 to the one for a single task (Figure 1):  the distribution for the single task is distinctly asymmetric (the peak is not at the centre of the distribution) whereas the two task histogram is nearly symmetric.  This surprising result is a consequence of the Central Limit Theorem (CLT) – which states that the sum of many identical distributions tends to resemble the Normal (Bell-shaped) distribution, regardless of the shape of the individual distributions.   Note that the CLT holds even though the two task distributions are shifted relative to each other – i.e. the second task begins after the first one is completed.]

The simulation also enables us to compute the cumulative probability of completion for the combined tasks (the CDF). This value of the cumulative probability at a particular bin equals the sum of the number of simulations runs in every bin up to (and including) the bin in question, divided by the total number of simulation runs. In mathematical terms this is:

where   is the cumulative probability at the time corresponding to the  th  bin, , the number of simulation runs in the th  bin and    the total number of simulation runs. Note that this formula is an approximate one because time is treated as a constant within each bin. The approximation can be improved by making the bins smaller (and hence increasing the number of bins).

The resulting cumulative probability function is shown in Figure 4. This allows us to answer questions such as:  “What is the probability that the tasks will be completed within 10 days?”. Answer:  .698, or approximately 70%. (Note:  I obtained this number by interpolating between values obtained from equation (1), but this level of precision is uncalled for, particularly because the formula is an approximate one)

Figure 4 - CDF for tasks in series with no inter-task delay

Many project scheduling techniques compute average completion times for component tasks and then add them up to get the expected completion time for the combined task. In the present case the average works out to 9.33 hrs (twice the average for a single task). However, we see from the CDF that there is a significant probability (.43) that we will not finish by this time – and this in a “best-case ” situation where the second task starts immediately after the first one finishes!

[An aside: If one applies the  well-known PERT formula   to each of the tasks, one gets an expected completion time  of 8.66 hrs for the combined task.  From the CDF one can show that there is a  probability of non-completion of 57%  by t=8.66 hours (see Figure 4) – i.e. there’s a greater than even chance of not finishing by this time!]

As interesting as this case is, it is somewhat unrealistic because successor tasks seldom start the instant the predecessor is completed. More often than not, there is a cut-off time before which the successor cannot start – even if there are no explicit dependencies between the two tasks.  This observation is a perfect segue into my next example, which is…

Two tasks in series with a fixed earliest start for the successor

Now we’ll introduce a slight complication: let’s assume, as before, that the two tasks are done one after the other but that the earliest the second task can start is at hours  (as measured from the start of the first task). So, if the first task finishes within 6 hours, there will be a delay between its completion and the start of the second task. However, if the first task takes longer than 6 hours to finish, the second task will start soon after the first one finishes.  The simulation procedure is the same as described in the previous section excepting for the last step – the completion time for the combined task is given by:

, for t   6 hrs and , for t < 6 hrs

I divided the time interval from t=4hrs to t=20 hrs into bins of 0.25 hr duration (much as I did before) and then assigned each combined completion time to the appropriate bin. The resulting histogram is shown in Figure 5.

Figure 5 - Frequency histogram for tasks in series with inter-task delay

Comparing Figure 5 to Figure 3, we see that the earliest possible finish time now increases from 4 hrs to 8 hrs. This is no surprise, as we built this in to our assumption.  Further, as one might expect, the distribution is distinctly asymmetric – with a minimum completion time of 8 hrs, a most likely time between 10 and 11 hrs and a maximum completion time of about 15 hrs.

Figure 6 shows the cumulative probability of completion for this case.

Figure 6 - CDF for tasks in series with inter-task delay

Because of the delay condition, it is impossible to calculate the average completion from the formulas for the triangular distribution – we have to obtain it from the simulation.  The average can be calculated from the simulation adding up all completion times and dividing by the total number of simulations, . In mathematical terms this is:

where is the average time,    the completion time for the th simulation run and    the total number of simulation runs.

This procedure gave me a value of about 10.8  hrs for the average.  From the CDF in Figure 6 one sees that the probability that the combined task will finish by this time is only 0.60 – i.e. there’s only a 60% chance that the task will finish by this time.  Any naïve estimation of time would do just as badly unless, of course, one is overly pessimistic and assumes a completion time of 15 – 16 hrs.

From the above it should be evident that the simulation allows one to associate an uncertainty (or probability) with every estimate. If management imposes a time limit of 10 hours,  the project manager can refer to the CDF in Figure 6 and figure out the probability of completing the task by that time (there’s a 40 % chance of completion by 10 hrs).  Of course, the reliability of the numbers depend on how good the distribution is. But  the assumptions that have gone into the model are known –  the optimistic, most likely and pessimistic times and the form of the distribution – and these can be refined as one gains experience.

Two tasks in parallel

My final example is the case of two identical tasks performed in parallel. As above, I’ll assume the uncertainty in each task is characterized by a triangular distribution with ,   and  of 2, 4 and 8 hours respectively. The simulation procedure for this case is the same as in the first example, excepting the last step. Assuming the simulated completion times for the individual tasks are and , the completion time for the combined tasks is given by the greater of the two – i.e. the combined completion time is given by .

To plot the histogram shown in Figure 7 , I divided the interval from t=2 hrs to t=8 hrs into bins of 0.25 hr duration each (Warning: Note the difference in the time axis scale  from Figures 3 and 5!).

Figure 7 - Frequency histogram for tasks in parallel

It is interesting to compare the above  histogram with that for an individual task with the same parameters (i.e. the example that was used in my previous post). Figure 8 shows the histograms for the two examples on the same plot (the combined task in red and the single task in blue). As one might expect, the histogram for the combined task is shifted to the right, a consequence of the nonlinear condition on the completion time.

Figure 8 - Histograms for tasks in parallel (red) and single task (blue)

What about the average? I calculated the average as before, by using equation (2) from the previous section. This gives an average of 5.38 hrs (compared to 4.67 hrs for either task, taken individually).   Note that the method to calculate the average is the same regardless of the form of the distribution. On the other hand,  computing the average from the equations would be a complicated affair, involving a stiff dose of algebra with an optional  sprinkling of  calculus.  Even worse – the calculations would vary from distribution to distribution. There’s no question that simulations are much easier.

The CDF for the overall completion time is also computed easily using equation (1). The resulting plot is shown in Figure 9  (Note the difference in the time axis scale  from Figures  4 and 6!). There are no surprises here – excepting how easy it is to calculate once the simulation is done.

Figure 9 - CDF for tasks in parallel

Let’s see what time corresponds to a 90% certainty of completion. A rough estimate for this number can be obtained from Figure 9 – just find the value of t (on the x axis) corresponding to a cumulative probability of 0.9 (on the y axis).  This is the graphical equivalent of solving the CDF for time, given the cumulative probability is 0.9. From Figure 9, we get a time of approximately 6.7 hrs. [Note: we could get a more accurate number by fitting the points obtained from equation (1) to a curve and then calculating the time corresponding to ]. The interesting thing is that the 90% certain completion time is not too different from that of a single task (as calculated from equation 7 of my previous post) – which works out to 6.45 hrs.

Comparing the two histograms in Figure 8, we expect the biggest differences in cumulative probability to occur at about the t=4 hour mark, because by that time the probability for the individual task has peaked whereas that for the combined task is yet to peak. Let’s see if this is so: from Figure 8, the cumulative probability for t=4  hrs is about .15 and from the CDF for the triangular distribution (equation 6 from my previous post), the cumulative probability at t=4 hours  (which is the most likely time) is .333 – double that of the combined task.  This, again, isn’t too surprising (once one has Figure 8 on hand). The really nice thing is that we are able to attach uncertainties to all our estimates.

Conclusion

Although the examples discussed above are simple – two identical tasks with uncertainties described by a triangular distribution – they serve to illustrate some of the non-intuitive outcomes when tasks have dependencies.   It is also worth noting that although the distribution for the individual tasks is known, the only straightforward way to obtain the distributions for the combined tasks (figures 3, 5 and 7) is through simulations. So, even these simple examples are a good demonstration of the utility of Monte Carlo techniques. Of course, real projects are way more complicated, with diverse tasks distributed in many different ways.   To simplify simulations in such cases,  one could  perform  coarse-grained simulations on a small number of high-level tasks,  each consisting of a number of  low-level, atomic tasks. The high-level tasks could be constructed in such a way as to focus attention on areas of greatest complexity, and hence greatest uncertainty.

As I have mentioned several times in this article and the previous one: simulation results are only as good as the distributions on which they are based. This begs the question: how do we know what’s an appropriate distribution for a given situation? There’s no one-size-fits-all answer to this question. However, for project tasks there are some general considerations that apply. These are:

1. There is a minimum time () before which a task cannot cannot be completed.
2. The probability will increase from 0 at to a maximum at a “most likely” completion time, . This holds true for most atomic tasks – but may be not for composite tasks which consist of many smaller tasks.
3. The probability decreases as time increases beyond ,  falling to 0 at a time much larger than .   This is simply another way of saying that the distribution has a long (but not necessarily infinite!) tail.

Asymmetric triangular distributions (such as the one used in my examples) are the simplest distributions that satisfy these conditions. Furthermore, a three point estimate is enought to specify a triangular distribution completely – i.e. given a three point estimate there is only one triangular distribution that can be fitted to it. That said, there are several other distributions that can be used; of particular relevance are certain long-tailed distributions.

Finally, I should mention that I make no claims about the efficiency or accuracy of the method presented here:  it should be seen  as  a demonstration rather than a definitive technique.  The many commercial Monte Carlo tools available in the market probably offer far more comprehensive, sophisticated and reliable algorithms (Note:  I ‘ve never used any of them, so I can’t make any recommendations!).  That said, it is always helpful to know the principles behind such tools,  if for no other reason than to understand how they work and, more important,  how to use them correctly.  The material discussed in this and the previous article came out of my efforts to develop an understanding Monte Carlo techniques and how they can be applied to various aspects of project management (they can also be applied to cost estimation, for example).  Over the last few weeks  I’ve spent many enjoyable evenings developing and running these simulations, and learning from them.  I’ll  leave it here with the hope that you find my articles helpful in your own explorations of the wonderful world of Monte Carlo simulations.