# Eight to Late

## The shape of things to come: an essay on probability in project estimation

### Introduction

Project estimates are generally based on assumptions about future events and their outcomes.   As the future is uncertain, the concept of  probability  is  sometimes  invoked in the estimation process.   There’s enough been written about how probabilities can be used in developing estimates; indeed there are a good number of articles on this blog – see this post or this one, for example.  However, most of these writings focus on the practical applications of probability rather than on the concept itself – what it means and how it should be interpreted.  In this article I address the latter point in a way that will (hopefully!) be of interest to those working in project management and related areas.

### Uncertainty is a shape, not a number

Since the future can unfold in a number of different ways one can describe it only in terms of a range of possible outcomes.   A good way to explore the implications of this statement is through a simple estimation-related example:

Assume you’ve been asked to do a particular task relating to your area of expertise.  From experience you know that this task usually takes 4 days to complete. If things go right, however,  it could take as little as 2 days. On the other hand, if things go wrong it could take as long as 8 days.  Therefore, your range of possible finish times (outcomes) is anywhere 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. 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 1.

Figure 1: Likelihood of finishing on day 2, day 4 and day 8.

Figure 1 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 historical data available, and I’ll have more to say about this later…), 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 2. This gives us the well-known triangular distribution.

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

Figure 2: Triangular distribution fitted to points in Figure 1

Of course, this isn’t the only possibility; there are an infinite number of others. Figure 3 is another (admittedly weird) example.

FIgure 3: Another distribution that fits the points in Figure 1

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 (say, if you call in sick for two weeks) or even not be completed at all (say, if you leave for that mythical greener pasture).  Catering for the latter possibility, the shape of the likelihood might resemble Figure 4.

Figure 4: A distribution that allows for a very long (potentially) infinite completion time

From the figures above, we see that uncertainties are shapes rather than single numbers, a notion popularised by Sam Savage in his book, The Flaw of Averages. Moreover, the “shape of things to come” depends on a host of factors, some of which may not even be on the radar when a future event is being estimated.

### Making likelihood precise

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 4.  (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 2 and the weird shape in figure 3) must equal 1.  I won’t go into any further details here, but those interested in the maths for the triangular case may want to take a look at this post  where the details have been worked out.

### The meaning of it all

(Note:  parts of this section borrow from my post on the interpretation of probability in project management)

So now we understand how uncertainty is actually a shape corresponding to a range of possible outcomes, each with their own probability of occurrence.  Moreover, we also know, in principle, how the probability can be calculated for any valid value of time (between 2 and 8 days). Nevertheless, we are still left with the question as to what a numerical probability really means.

As a concrete case from the example above, what do we mean when we say that there is 100% chance (probability=1) of finishing within 8 days?   Some possible interpretations of such a statement include:

1.  If the task is done many times over, it will always finish within 8 days. This is called the frequency interpretation of probability, and is the one most commonly described in maths and physics textbooks.
2. It is believed that the task will definitely finish within 8 days. This is called the belief interpretation.  Note that this interpretation hinges on subjective personal beliefs.
3. Based on a comparison to similar tasks, the  task will finish within 8 days. This is called the support interpretation.

Note that these interpretations are based on a paper by Glen Shafer. Other papers and textbooks frame these differently.

The first thing to note is how different these interpretations are from each other.  For example, the first one offers a seemingly objective interpretation whereas the second one is unabashedly subjective.

So, which is the best – or most correct – one?

A person trained in science or mathematics might  claim that the frequency interpretation wins hands down because it lays out an objective, well -defined procedure for calculating probability: simply perform the same task many times and note the completion times.

Problem is, in real life situations it is impossible to carry out exactly the same task over and over again. Sure, it may be possible to almost the same  task, but even straightforward tasks  such as vacuuming a room or baking a cake can hold hidden surprise (vacuum cleaners do malfunction and a friend may call when one is mixing the batter for a cake). Moreover, tasks that are complex (as is often the case in the project work) tend to be unique and can never be performed in exactly the same way twice.  Consequently, the frequency interpretation is great in theory but not much use in practice.

“That’s OK,”  another estimator might say,” when drawing up an estimate, I compared it to other similar tasks that I have done before.”

This is essentially the support interpretation (interpretation 3 above).  However, although this seems reasonable, there is a problem: tasks that are superficially similar will differ in the details, and these small differences may turn out to be significant when one is actually carrying out the task.  One never knows beforehand which variables are important. For example, my ability to finish a particular task within a stated time depends not only on my skill but also on things such as my workload, stress levels and even my state of mind. There are many external factors that one might not even recognize as being significant. This is a manifestation of the reference class problem.

So where does that leave us? Is probability just a matter of subjective belief?

No, not quite:  in reality, estimators will use some or all of three interpretations to arrive at “best guess” probabilities.  For example, when estimating a project task, a person will likely use one or more of the following pieces of information:

1. Experience with similar tasks.
2. Subjective belief regarding task complexity and potential problems. Also, their “gut feeling” of how long they think it ought to take. These factors often drive excess time or padding that people work into their estimates.
3. Any relevant historical data (if available)

Clearly, depending on the situation at hand, estimators may be forced to rely on one piece of information more than others. However, when called upon to defend their estimates, estimators may use other arguments to justify their conclusions depending on who they are talking to. For example,  in discussions involving managers, they may use hard data presented in a way that supports their estimates, whereas when talking to their peers they may emphasise their gut feeling based on differences between the task at hand and similar ones they have done in the past.  Such contradictory representations tend to obscure the means by which the estimates were actually made.

### Summing up

Estimates are invariably made in the face of uncertainty. One way to get a handle on this is by estimating the probabilities associated with possible outcomes.  Probabilities can be reckoned in a number of different ways. Clearly, when using them in estimation, it is crucial to understand how probabilities have been derived and the assumptions underlying these. We have seen three ways in which probabilities are interpreted corresponding to three different ways in which they are arrived at.  In reality, estimators may use a mix of the three approaches so it isn’t always clear how the numerical value should be interpreted. Nevertheless, an awareness of what probability is and its different interpretations may help managers ask the right questions to better understand the estimates made by their teams.

Written by K

April 3, 2012 at 11:40 pm

## On the accuracy of group estimates

### Introduction

The essential idea behind group estimation is that an estimate made by a group is likely to be more accurate than one made by an individual in the group. This notion is the basis for the Delphi method and its variants. In this post, I use arguments involving probabilities to gain some insight into the conditions under which group estimates are more accurate than individual ones.

### An insight from conditional probability

Let’s begin with a simple group estimation scenario.

Assume we have two individuals of similar skill who have been asked to provide independent estimates of some quantity, say  a project task duration. Further, let us assume that each individual has a probability $p$ of making a correct estimate.

Based on the above, the probability that they both make a correct estimate, $P(\textnormal{both correct})$,  is:

$P(\textnormal{both correct}) = p*p = p^2$,

This is a consequence of our assumption that the individual estimates are independent of each other.

Similarly,  the probability that they both get it wrong, $P(\textnormal{both wrong})$, is:

$P(\textnormal{both wrong}) = (1-p)*(1-p) = (1-p)^2$,

Now we can ask the following question:

What is the probability that both individuals make the correct estimate if we know that they have both made the same estimate?

This can be figured out using Bayes’ Theorem, which in the context of the question can be stated as follows:

$P(\textnormal{both correct\textbar same estimate})= \displaystyle{\frac{ P(\textnormal{same estimate\textbar both correct})*P(\textnormal{both correct})}{ P(\textnormal{same estimate})}}$

In the above equation, $P(\textnormal{both correct\textbar same estimate})$ is the probability that both individuals get it right given that they have made the same estimate (which is  what we want to figure out). This is an example of a conditional probability – i.e.  the probability that an event occurs given that another, possibly related event has already occurred.  See this post for a detailed discussion of conditional probabilities.

Similarly, $P(\textnormal{same estimate\textbar both correct})$ is the conditional probability that both estimators make the same estimate given that they are both correct. This probability is 1.

Question: Why?

Answer: If both estimators are correct then they must have made the same estimate (i.e. they must both within be an acceptable range of the right answer).

Finally, $P(\textnormal{same estimate})$ is the probability that both make the same estimate. This is simply the sum of the probabilities that both get it right and both get it wrong. Expressed in terms of $p$ this is, $p^2+(1-p)^2$.

Now lets apply Bayes’ theorem to the following two cases:

1. Both individuals are good estimators – i.e. they have a high probability of making a correct estimate. We’ll assume they both have a 90% chance of getting it right ($p=0.9$).
2. Both individuals are poor estimators – i.e. they have a low probability of making a correct estimate. We’ll assume they both have a 30% chance of getting it right ($p=0.3$)

Consider the first case. The probability that both estimators get it right given that they make the same estimate is:

$P(\textnormal{both correct\textbar same estimate})= \displaystyle\frac{1*0.9*0.9}{0.9*0.9+0.1*0.1}= \displaystyle \frac{0.81}{0.82}= 0.9878$

Thus we see that the group estimate has a significantly better chance of being right than the individual ones:  a probability of 0.9878 as opposed to 0.9.

In the second case, the probability that both get it right is:

$P(\textnormal{both correct\textbar same estimate})= \displaystyle \frac{1*0.3*0.3}{0.3*0.3+0.7*0.7}= \displaystyle \frac{0.09}{0.58}= 0.155$

The situation is completely reversed: the group estimate has a much smaller chance of being right than an  individual estimate!

In summary:  estimates provided by a group consisting of individuals of similar ability working independently are more likely to be right (compared to individual estimates) if the group consists of  competent estimators and more likely to be wrong (compared to individual estimates) if the group consists of  poor estimators.

### Assumptions and complications

I have made a number of simplifying assumptions in the above argument. I discuss these below with some commentary.

1. The main assumption is that individuals work independently. This assumption is not valid for many situations. For example, project estimates are often made  by a group of people working together.  Although one can’t work out what will happen in such situations using the arguments of the previous section, it is reasonable to assume that given the right conditions, estimators will use their collective knowledge to work collaboratively.   Other things being equal,  such collaboration would lead a group of skilled estimators to reinforce each others’ estimates (which are likely to be quite similar) whereas less skilled ones may spend time arguing over their (possibly different and incorrect) guesses.  Based on this, it seems reasonable to conjecture that groups consisting of good estimators will tend to make even better estimates than they would individually whereas those consisting of poor estimators have a significant chance of making worse ones.
2. Another assumption is that an estimate is either good or bad. In reality there is a range that is neither good nor bad, but may be acceptable.
3. Yet another assumption is that an estimator’s ability can be accurately quantified using a single numerical probability.  This is fine providing the number actually represents the person’s estimation ability for the situation at hand. However, typically such probabilities are evaluated on the basis of  past estimates. The problem is, every situation is unique and history may not be a good guide to the situation at hand. The best way to address this is to involve people with diverse experience in the estimation exercise.  This will almost often lead to  a significant spread of estimates which may then have to be refined by debate and negotiation.

Real-life estimation situations have a number of other complications.  To begin with, the influence that specific individuals have on the estimation process may vary – a manager who is  a poor estimator may, by virtue of his position, have a greater influence than others in a group. This will skew the group estimate by a factor that cannot be estimated.  Moreover, strategic behaviour may influence estimates in a myriad other ways. Then there is the groupthink factor  as well.

…and I’m sure there are many others.

Finally I should mention that group estimates can depend on the details of the estimation process. For example, research suggests that under certain conditions competition can lead to better estimates than cooperation.

### Conclusion

In this post I have attempted to make some general inferences regarding the validity of group estimates based on arguments involving conditional probabilities. The arguments suggest that, all other things being equal, a collective estimate from a bunch of skilled estimators will generally be better than their individual estimates whereas an estimate from a group of less skilled estimators will tend to be worse than their individual estimates. Of course, in real life, there are a host of other factors  that can come into play:  power, politics and biases being just a few. Though these are often hidden, they can  influence group estimates in inestimable ways.

### Acknowledgement

Thanks go out to George Gkotsis and Craig Brown for their comments which inspired this post.

Written by K

December 1, 2011 at 5:16 am

## The drunkard’s dartboard revisited: yet another Excel-based example of Monte Carlo simulation

(Note: An Excel sheet showing sample calculations and plots discussed in this post can be downloaded here.)

### Introduction

Some months ago, I wrote a post explaining the basics of Monte Carlo simulation using the example of a drunkard throwing darts at a board. In that post I assumed that the darts could land anywhere on the dartboard with equal probability. In other words, the hit locations were assumed to be uniformly distributed. In a comment on the piece, George Gkotsis challenged this assumption, arguing that that regardless of the level of inebriation of the thrower, a dart would be more likely to land near the centre of the board than away from it (providing the player is at least moderately skilled). He also suggested using the Normal Distribution to model the spread of hits, with the variance of the distribution serving as a rough measure of the inaccuracy (or drunkenness!) of the drunkard. In George’s words:

I would propose to introduce a ‘skill’ factor, which represents the circle/square ratio (maybe a normal-Gaussian distribution). Of course, this skill factor would be very low (high variance) for a drunken player, but would still take into account the fact that throwing darts into a square is not purely random.

In this post I revisit the drunkard’s dartboard, taking into account George’s suggestions.

### Setting the stage

To keep things simple, I’ll make the following assumptions:

Figure 1: The dartboard

1. The dartboard is a circle of radius 0.5 units centred at the origin (see Figure 1)
2. The chance of a hit is greatest at the centre of the dartboard and falls off as one moves away from it.
3. The distribution of hits is a function of distance from the centre but does not depend on direction. In mathematical terms, for a given distance $r$ from the centre of the dartboard, the dart can land at any angle $\theta$ with equal probability, $\theta$ being the angle between the line joining the centre of the board to the dart and the x axis. See Figure 2 for graphical representations of a hit location in terms of $r$ and $\theta$. Note that that the $x$ and $y$ coordinates can be obtained using the formulas $x = r\cos\theta$ and $y= r\sin\theta$ as s shown in Figure 2.
4. Hits are distributed according to the Normal distribution with maximum at the centre of the dartboard.
5. The variance of the Normal distribution is a measure of inaccuracy/drunkenness of the drunkard: the more drunk the drunk, the greater the variation in his aim.

Figure 2: The coordinates of a hit location

These assumptions are consistent with George’s suggestions.

### The simulation

[Note to the reader: you may want to download the demo before continuing.]

The steps of a simulation run are as follows:

1. Generate a number that is normally distributed with a zero mean and a specified standard deviation. This gives the distance, $r$, of a randomly thrown dart from the centre of the board for a player with a “inaccuracy factor” represented by the standard deviation. Column A in the demo contains normally distributed random numbers with zero mean and a standard deviation of 0.2 . Note that I selected the latter number for no other reason than the results show up clearly on a fixed-axis plot shown in Figure 2.
2. Generate a uniformly distributed random number lying between 0 and $2\pi$. This represents the angle $\theta$. This is the content of column B of the demo.
3. The numbers obtained from steps 1 and 2 for completely specify the location of a hit. The location’s $x$ and $y$ coordinates can be worked out using the formulas $x = r\cos\theta$ and $y= r\sin\theta$. These are listed in columns C and D in the Excel demo.
4. Re-run steps 1 through 4 as many times as needed. Note that the demo is set up for 5000 runs. You can change this manually or, better yet, automate it. The latter is left as an exercise for you.

It is instructive to visualize the resulting hits using a scatter plot. Among other things this can tell you, at a glance, if the results make sense. For example, we would expect hits to be symmetrically distributed about the origin because the drunkard’s throws are not biased in any particular direction around the centre). A non-symmetrical distribution is thus an indication that there is an error in the calculations.

Now, any finite collection of hits is unlikely to be perfectly symmetrical because of outliers. Nevertheless, the distributions should be symmetrical on average. To test this, run the demo a few times (hit F9 with the demo open). Notice how the position of outliers and the overall shape of the distribution of points changes randomly from simulation to simulation. In all cases, however, there is a clear maximum at the centre of the dartboard with the probability of a hit falling with distance from the centre.

Figure 3: Scatter plot for standard deviation=0.2

Figure 3 shows the results of simulations for a standard deviation of 0.2. Figures 4 and 5 show the results of simulations for standard deviations of 0.1 and 0.4.

Figure 4: Scatter plot for standard deviation=0.1

Note that the plot has fixed axes- i.e. the area depicted is the 1×1 square that encloses the dartboard, regardless of the standard deviation. Consequently, for larger standard deviations (such as 0.4) many hits will be out of range and will not show up on the plot.

Figure 5: Scatter plot for standard deviation=0.4

### Closing remarks

As I have stressed in my previous posts on Monte Carlo simulation, the usefulness of a simulation depends on the choice of an appropriate distribution. If the selected distribution does not reflect reality, neither will the simulation. This is true regardless of whether one is simulating a drunkard’s wayward aim or the duration of project task. You may have noted that the assumption of normally-distributed hits has no justification whatsoever; it is just as arbitrary as my original assumption of uniformity. In fact, the hit locations of drunken dart throws is highly unlikely to be either uniform or Normal. Nevertheless, I hope that some of my readers will find the above example to be of pedagogical value.

### Acknowledgement

Thanks to George Gkotsis for his comment which got me thinking about this post.

Written by K

November 3, 2011 at 4:59 am

## Uncertainty about uncertainty

### Introduction

More often than not, managerial decisions are made on the basis of uncertain information. To lend some rigour to the process of decision making, it is sometimes assumed that uncertainties of interest can be quantified accurately using probabilities. As it turns out, this assumption can be incorrect in many situations because the probabilities themselves can be uncertain.   In this post I discuss a couple of ways in which such uncertainty about uncertainty can manifest itself.

### The problem of vagueness

In a paper entitled, “Is Probability the Only Coherent Approach to Uncertainty?”,  Mark Colyvan made a distinction between two types of uncertainty:

1. Uncertainty about some underlying fact. For example, we might be uncertain about the cost of a project – that there will be a cost is a fact, but we are uncertain about what exactly it will be.
2. Uncertainty about situations where there is no underlying fact.  For example, we might be uncertain about whether customers will be satisfied with the outcome of a project. The problem here is the definition of customer satisfaction. How do we measure it? What about customers who are neither satisfied nor dissatisfied?  There is no clear-cut definition of what customer satisfaction actually is.

The first type of uncertainty refers to the lack of knowledge about something that we know exists. This is sometimes referred to as epistemic uncertainty – i.e. uncertainty pertaining to knowledge. Such uncertainty arises from imprecise measurements, changes in the object of interest etc.  The key point is that we know for certain that the item of  interest has well-defined properties, but we don’t know what they are and hence the uncertainty. Such uncertainty can be quantified accurately using probability.

Vagueness, on the other hand, arises from an imprecise use of language.  Specifically, the term refers to the use of criteria that cannot distinguish between borderline cases.  Let’s clarify this using the example discussed earlier.  A popular way to measure customer satisfaction is through surveys. Such surveys may be able to tell us that customer A is more satisfied than customer B. However, they cannot distinguish between borderline cases because any boundary between satisfied and not satisfied customers is arbitrary.  This problem becomes apparent when considering an indifferent customer. How should such a customer be classified – satisfied or not satisfied? Further, what about customers who choose not to respond? It is therefore clear that any numerical probability computed from such data cannot be considered accurate.  In other words, the probability itself is uncertain.

### Ambiguity in classification

Although the distinction made by Colyvan is important, there is a deeper issue that can afflict uncertainties that appear to be quantifiable at first sight. To understand how this happens, we’ll first need to take a brief look at how probabilities are usually computed.

An operational definition of probability is that it is the ratio of the number of times the event of interest occurs to the total number of events observed. For example, if my manager notes my arrival times at work over 100 days and finds that I arrive before 8:00 am on 62 days then he could infer that the probability my arriving before 8:00 am is 0.62.   Since the probability is assumed to equal the frequency of occurrence of the event of interest, this is sometimes called the frequentist interpretation of probability.

The above seems straightforward enough, so you might be asking: where’s the problem?

The problem is that events can generally be classified in several different ways and the computed probability of an event occurring can depend on the way that it is classified. This is called the reference class problem.   In a paper entitled, “The Reference Class Problem is Your Problem Too”, Alan Hajek described the reference class problem as follows:

“The reference class problem arises when we want to assign a probability to a proposition (or sentence, or event) X, which may be classified in various ways, yet its probability can change depending on how it is classified.”

Consider the situation I mentioned earlier. My manager’s approach seems reasonable, but there is a problem with it: all days are not the same as far as my arrival times are concerned. For example, it is quite possible that my arrival time is affected by the weather: I may arrive later on rainy days than on sunny ones.  So, to get a better estimate my manager should also factor in the weather. He would then end up with two probabilities, one for fine weather and the other for foul. However, that is not all: there are a number of other criteria that could affect my arrival times – for example, my state of health (I may call in sick and not come in to work at all), whether I worked late the previous day etc.

What seemed like a straightforward problem is no longer so because of the uncertainty regarding which reference class is the right one to use.

Before closing this section, I should mention that the reference class problem has implications for many professional disciplines. I have discussed its relevance to project management in my post entitled, “The reference class problem and its implications for project management”.

### To conclude

In this post we have looked at a couple of forms of uncertainty about uncertainty that have practical implications for decision makers. In particular, we have seen that probabilities used in managerial decision making can be uncertain because of  vague definitions of events and/or ambiguities in their classification.  The bottom line for those who use probabilities to support decision-making is to ensure that the criteria used to determine events of interest refer to unambiguous facts that are appropriate to the situation at hand.  To sum up: decisions made on the basis of probabilities are only as good as the assumptions that go into them, and the assumptions themselves may be prone to uncertainties such as the ones described in this article.

Written by K

September 29, 2011 at 10:34 pm

## The drunkard’s dartboard: an intuitive explanation of Monte Carlo methods

(Note to the reader: An Excel sheet showing sample calculations and plots discussed in this post  can be downloaded here.)

Monte Carlo simulation techniques have been applied to areas ranging from physics to project management. In earlier posts, I discussed how these methods can be used to simulate project task durations (see this post and this one for example). In those articles, I described simulation procedures in enough detail for readers to be able to reproduce the calculations for themselves. However, as my friend Paul Culmsee mentioned, the mathematics tends to obscure the rationale behind the technique. Indeed, at first sight it seems somewhat paradoxical that one can get accurate answers via random numbers. In this post, I illustrate the basic idea behind Monte Carlo methods through an example that involves nothing more complicated than squares and circles. To begin with, however, I’ll start with something even simpler – a drunken darts player.

Consider a sozzled soul who is throwing darts at a board situated some distance from him.  To keep things simple, we’ll assume the following:

1. The board is modeled by the circle shown in Figure 1, and our souse scores a point if the dart falls within the circle.
2. The dart board is inscribed in a square with sides 1 unit long as shown in the figure, and we’ll assume for simplicity that the  dart always  falls somewhere  within the square (our protagonist is not that smashed).
3. Given his state, our hero’s aim is not quite what it should be –  his darts fall anywhere within the square with equal probability. (Note added on 01 March 2011: See the comment by George Gkotsis below for a critique of this assumption)

Figure 1: "Dartboard" and enclosing square

We can simulate the results of our protagonist’s unsteady attempts by generating two sets of  uniformly distributed random numbers lying between 0 and 1 (This is easily done in Excel using the rand() function).  The pairs of random numbers thus generated – one from each set –  can be treated as the (x,y) coordinates of  the dart for a particular throw. The result of 1000 pairs of random numbers thus generated (representing the drunkard’s dart throwing attempts) is shown in Figure 2 (For those interested in seeing the details, an Excel sheet showing the calculations for 100o trials can be downloaded here).

Figure 2: Result for 1000 throws

A trial results in a “hit” if it lies within the circle.  That is, if it satisfies the following equation:

$(x-0.5)^2 + (y-0.5)^2 < 0.25\ldots \ldots (1)$

(Note:  if we replace “<”  by “=”  in the above expression, we get the equation for a circle of radius 0.5 units, centered at x=0.5 and y=0.5.)

Now, according to the frequency interpretation of probability, the probability of the plastered player scoring a point is approximated by the ratio of the number of hits in the circle to the total number of attempts. In this case, I get an average of 790/1000 which is 0.79 (generated from 10 sets of 1000 trials each). Your result will be different from because you will generate different sets of random numbers from the ones I did. However, it should be reasonably close to my result.

Further, the frequency interpretation of probability tells us that the approximation becomes more accurate as the number of trials increases. To see why this is so, let’s increase the number of trials and plot the results. I carried out simulations for 2000, 4000, 8000 and 16000 trials. The results of these simulations are shown in Figures 3 through 6.

Figure 3: Result for 2000 throws

Figure 4: Result for 4000 throws

Figure 5: Result for 8000 throws

Figure 6: Result for 16000 throws

Since a dart is equally likely to end up anywhere within the square, the exact probability of a hit is simply the area of the dartboard (i.e. the circle)  divided by the entire area over which the dart can land. In this case, since the area of the enclosure (where the dart must fall) is 1 square unit, the area of the dartboard is actually equal to the probability.  This is easily seen by calculating the area of the circle using the standard formula $\pi r^2$ where $r$ is the radius of the circle (0.5 units in this case). This yields 0.785398 sq units, which is reasonably close to the number that we got for the 1000 trial case.  In the 16000 trial case, I  get a number that’s closer to the exact result: an average of 0.7860 from 10 sets of 16000 trials.

As we see from Figure 6, in the 16000 trial case, the entire square is peppered with closely-spaced “dart marks” – so much so, that it looks as though the square is a uniform blue.  Hence, it seems intuitively clear that as we increase, the number of throws, we should get a better approximation of the area and, hence, the probability.

There are a couple of points worth mentioning here. First, in principle this technique can be used to calculate areas of figures of any shape. However, the more irregular the figure, the worse the approximation – simply because it becomes less likely that the entire figure will be sampled correctly by “dart throws.” Second,  the reader may have noted that although the 16000 trial case gives a good enough result for the area of the circle, it isn’t particularly accurate considering the large number of trials. Indeed, it is known that the “dart approximation” is not a very good way of calculating areas – see this note for more on this point.

Finally, let’s look at connection between the general approach used in Monte Carlo techniques and the example discussed above  (I use the steps described in the Wikipedia article on Monte Carlo methods as representative of the general approach):

1. Define a domain of possible inputs – in our case the domain of inputs is defined by the enclosing square of side 1 unit.
2. Generate inputs randomly from the domain using a certain specified probability distribution – in our example the probability distribution is a pair of independent, uniformly distributed random numbers lying between 0 and 1.
3. Perform a computation using the inputs – this is the calculation that determines whether or not a particular trial is a hit or not (i.e.  if the x,y coordinates obey inequality  (1) it is a hit, else  it’s a miss)
4. Aggregate the results of the individual computations into the final result – This corresponds to the calculation of the probability (or equivalently, the area of the circle) by aggregating the number of hits for each set of trials.

To summarise: Monte Carlo algorithms generate random variables (such as probability) according to pre-specified distributions.  In most practical applications  one will use more efficient techniques to sample the distribution (rather than the naïve method I’ve used here.)  However, the basic idea is as simple as playing drunkard’s darts.

Acknowledgements

Thanks go out to Vlado Bokan for helpful conversations while this post was being written and to Paul Culmsee for getting me thinking about a simple way to explain Monte Carlo methods.

Written by K

February 25, 2011 at 4:51 am

## Monte Carlo simulation of risk and uncertainty in project tasks

### Introduction

When developing duration estimates for a project task, it is useful to make a distinction between the  uncertainty inherent in the task and uncertainty due to known risks.  The former is uncertainty due to factors that are not known whereas the latter corresponds uncertainty due to events that are known, but may or may not occur. In this post, I illustrate how the two types of uncertainty can be combined via Monte Carlo simulation.  Readers may find it helpful to keep my introduction to Monte Carlo simulations of project tasks handy, as I refer to it extensively in the present piece.

### Setting the stage

Let’s assume that there’s a 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 here – maybe there’s some guesswork, maybe some padding or maybe it is really based on experience (as it should be).  For simplicity we’ll assume the probability distribution for the task duration is triangular. It is not hard to show that, given the above mentioned estimates, the probability, $p(t)$,  that the task will finish at time $t$ is given by the equations below (see my introductory post for a detailed derivation):

$p(t)=\frac{(t-2)}{6}\dots\ldots(1)$ for  2 hours $\leq t \leq$ 4 hours

And,

$p(t)=\frac{(8-t)}{12}\dots\ldots(2)$ for  4 hours $\leq t \leq$ 8 hours

These two expressions are sometimes referred to as the probability distribution function (PDF).  The PDF described by equations (1) and (2) is illustrated in Figure 1. (Note: Please click on the Figures to view full-size images)

Figure 1: Probability distribution for task

Now, a PDF tells us the probability that the task will finish at a particular time $t$. However, we are more interested in knowing whether or not the task will be completed by time $t$. – i.e. at or before time $t$. This quantity, which we’ll denote by $P(t)$ (capital P), is sometimes known as the cumulative distribution function (CDF). The CDF  is obtained by summing up the probabilities from $t=2$ hrs to time $t$.  It is not hard to show that the CDF for the task at hand is given by the following equations:

$P(t)=\frac{(t-2)^2}{12}\ldots\ldots(3)$ for  2 hours $\leq t \leq$ 4 hours

and,

$P(t)=1- \frac{(8-t)^2}{24}\ldots\ldots(4)$ for  4 hours $\leq t \leq$ 8 hours

For a detailed derivation, please see my introductory post. The CDF for the distribution is shown in Figure 2.

Figure 2: CDF for task

Now for the complicating factor: let us assume there is a risk that has a bearing on this task.  The risk could be any known factor that has a negative impact on task duration. For example, it could be that a required resource is delayed or that the deliverable will fails a quality check and needs rework. The consequence of the risk – should it eventuate – is that the task takes longer.  How much longer the task takes depends on specifics of the risk. For the purpose of this example we’ll assume that the additional time taken is also described by a triangular distribution with a minimum, most likely and maximum time of 1, 2 and 3 hrs respectively.  The PDF $p_{r}(t)$ for the additional time taken due to the risk is:

$p_{r}(t)=(t-1)\dots\ldots(5)$ for  1 hour $\leq t \leq$ 2 hours

And

$p_{r}(t)=(3-t)\dots\ldots(6)$ for  2 hrs $\leq t \leq$ 3 hours

The figure for this distribution is shown in Figure 3.

Figure 3: Probability distribution of additional time due to risk

The CDF for the additional time taken if the risk eventuates (which we’ll denote by $P_{r}(t)$) is given by:

$P_{r}(t)=\frac{(t-1)^2}{2}\ldots\ldots(7)$ for  1 hour $\leq t \leq$ 2 hours

and,

$P_{r}(t)=1- \frac{(3-t)^2}{2}\ldots\ldots(8)$ for  2 hours $\leq t \leq$ 3 hours

The CDF for the risk consequence is shown in Figure 4.

Figure 4: CDF of additional time due to risk

Before proceeding with the simulation it is worth clarifying what all this means, and what we want to do with it.

Firstly, equations 1-4 describe the inherent uncertainty associated with the task while equations 5 through 8 describe the consequences of the risk, if it eventuates.

Secondly, we have described the task and the risk separately. In reality, we need a unified description of the two – a combined distribution function for the uncertainty associated with the task and the risk taken together.  This is what the simulation will give us.

Finally, one thing I have not yet specified is the probabilty that the risk will actually occur. Clearly, the higher the probability, the greater the potential delay. Below I carry out simulations for risk probabilities of varying from 0.1 to 0.5.

That completes the specification of the  problem – let’s move on to the simulation.

### The simulation

The simulation procedure I used  for the zero-risk case  (i.e. the task described by equations 1 and 2 ) is as follows :

1. Generate a random number between 0 and 1.  Treat this number as the cumulative probability, $P(t)$ for the simulation run. [You can generate random numbers in Excel using the  rand() function]
2. Find the time, $t$,  corresponding to $P(t)$ by solving equations (3) or (4) for $t$. The resulting value of $t$ is the time taken to complete the task.
3. Repeat steps (1) and (2)  for a sufficiently large number of trials.

The frequency distribution of completion times for the task, based on 30,000 trials is shown in Figure 5.

Figure 5: Simulation histogram for zero-risk case

As we might expect,  Figure 5 can be translated to the probability distribution shown in Figure 1 by a straightforward normalization – i.e. by dividing each bar by the total number of trials.

What remains to be done is to  incorporate the risk (as modeled in equations 5-6) into the simulation. To simulate the task with the risk, we simply do the following for each trial:

1. Simulate the task without the risk as described earlier.
2. Generate another random number between 0 and 1.
3. If the random number is less than the probability of the risk, then simulate the  risk. Note that since the risk is described by a triangular function, the procedure to simulate it is the same as that for the task (albeit with different parameters).
4. If the random number is greater than the probability of the risk, do nothing.
5. Add the results of 1 and 4. This is the outcome of the trial.
6. Repeat steps 1-5 for as many trials as required.

I performed simulations for the task with risk probabilities of 10%, 30% and 50%. The frequency distributions of completion times for these are displayed in Figures 6-8 (in increasing order of probability). As one would expect, the spread of times increases with increasing probability. Further, the distribution takes on a distinct second peak as the probability increases: the first peak is at $t=4$, corresponding to the most likely completion time of the risk-free task and the second at $t=6$ corresponding to the most likely additional time of 2 hrs if the risk eventuates.

Figure 6: Simulation histogram (10% probability of risk)

Figure 7: Simulation histogram (30% probability of risk)

Figure 8: Frequency histogram (50% probability of risk)

It is also instructive to compare average completion times for the four cases (zero-risk and 10%, 30% and 50%). The average can computed from the simulation by simply adding up the simulated completion times (for all trials) and dividing by the total number of simulation trials (30,000 in our case). On doing this, I get the following:

Average completion time for zero-risk case = 4.66 hr

Average completion time with 10% probability of risk =  4.89 hrs

Average completion time with 30% probability of risk =  5.36 hrs

Average completion time with 50% probability of risk=  5.83 hrs

No surprises here.

One point to note is that the result obtained from the simulation for the zero-risk case compares well with the exact formula for a triangular distribution (see the Wikipedia article for the triangular distribution):

$t_{av} = \frac{t_{worst}+t_{best}+t_{most likely}}{3}=\frac{8+2+4}{3}=4.67$ hrs

This serves as a sanity check on the simulation procedure.

It is also interesting to compare the cumulative probabilities of completion in the zero-risk and high risk (50% probability) case. The CDFs for the two are shown in Figure 9. The co-plotted CDFs allow for a quick comparison of completion time predictions. For example, in the zero-risk case, there is about a  90% chance that the task will be completed in a little over 6 hrs whereas when the probability of the risk is 50%, the 90% completion time increases to 8 hrs (see Figure 9).

Figure 9: CDFs for zero risk and 50% probability of risk cases

### Next steps and wrap up

For those who want to learn more about simulating project uncertainty and risk, I can recommend the UK MOD paper – Three Point Estimates And Quantitative Risk Analysis A Process Guide For Risk Practitioners.  The paper provides useful advice on how three point estimates for project variables should be constructed. It also has a good discussion of risk and how it should be combined with the inherent uncertainty associated with a variable. Indeed, the example I have described above  was inspired by the paper’s discussion of uncertainty and risk.

Of course, as with any quantitative predictions of project variables, the numbers are only as reliable as the assumptions that go into them, the main assumption here being the three point estimates that were used to derive the distributions for the task uncertainty and risk (equations 1-2 and 5-6).  Typically these are obtained from historical data. However, there are well known problems associated with history-based estimates. For one, as we can never be sure that the historical tasks are similar to the one at hand in ways that matter (this is the reference class problem).  As Shim Marom warns us in this post, all our predictions depend on the credibility of our estimates.  Quoting from his post:

Can you get credible three point estimates? Do you have access to credible historical data to support that? Do you have access to Subject Matter Experts (SMEs) who can assist in getting these credible estimates?

If not, don’t bother using Monte Carlo.

In closing, I hope my readers will find this simple example useful in understanding how uncertainty and risk can be accounted for using Monte Carlo simulations. In the end, though, one should always keep in mind that the use of sophisticated techniques does not magically render one immune to the GIGO principle.

Written by K

February 17, 2011 at 9:55 pm

## Six ways in which project estimates go wrong

Despite the increasing focus on  project estimation, the activity still remains  more guesswork than art or science.  In his book on the fallacies of software engineering,  Robert Glass has this to say about it:

Estimation, as you might imagine, is the process by which we determine how long a project will take and how much it will cost. We do estimation very badly in the software field. Most of our estimates are more like wishes than realistic targets. To make matters worse, we seem to have no idea how to improve on those very bad practices. And the result is, as everyone tries to meet an impossible estimation target, shortcuts are taken, good practices are skipped, and the inevitable schedule runaway becomes a technology runaway as well…

Moreover, he suggests that poor estimation is one of the top two causes of project failure.

Now, there are a number of reasons why project estimates go wrong,  but in my experience there are half-dozen standouts. Here they are, in no particular order:

1.  False analogies: Project estimates based on historical data are generally considered to be more reliable than those developed using other methods such as expert judgement (see this article, from the MS Project support site for example). This is fine and good as long as one uses data from historical projects that are identical to the one at hand in relevant ways. Problem is, one rarely knows what is relevant and what isn’t.  It is all too easy too select a project that is superficially similar to the one at hand, but actually differs in critical ways.  See my posts on false analogies and the reference class problem for more on this point.

2.  False precision: Project estimates are often quoted as single numbers rather than ranges. Such estimates are incorrect because they ignore the fact that uncertain quantities should be quantified by a range of numbers (or more accurately, a distribution) rather than  point values. As Dr. Sam Savage emphasises in his book, The Flaw of Averages: an uncertain quantity is a shape, not a number (see my review of the book for more on this point). In short, an estimate quoted as a single number is almost guaranteed to be incorrect.

3.  Estimation by decree: It should be obvious that estimation must be done by those who will do the work. Unfortunately this principle is one of the first to be sacrificed on Death March Projects. In such projects, schedules  are shoe-horned into  predetermined timelines, with estimates cooked up by those who have little or no idea of the actual effort involved in doing the work.

4.   Subjectivity: This is where estimates are plucked out of thin air and “justified” based on gut-feel and other subjective notions. Such estimates are prone to overconfidence and a range of other cognitive biases. See my post on cognitive biases as project meta-risks for a detailed discussion of how these biases manifest themselves in project estimates.

5.  Coordination neglect: Projects consist of diverse tasks that need to be coordinated and integrated carefully. Unfortunately, the time and effort needed for coordination and integration  is often underestimated (or even totally overlooked)  by project decision makers. This is referred to as coordination neglect. Coordination neglect is a problem in projects of all sizes, but  is generally more significant for projects involving large teams (see this paper for an empirical study of the effect of team size on coordination neglect).  As one might imagine, coordination neglect also becomes a significant problem in projects that consist of a large number of dependent tasks  or have a large number of external dependencies.

6.  Too coarse grained: Large tasks are made up of smaller tasks strung together in specific ways. Consequently, estimates for large tasks should be built up from estimates for the smaller sub-tasks. . Teams often short-circuit the process by attempting to estimate the large task directly. Such estimates usually turn out to be incorrect because sub-tasks  are overlooked. Another problem is coordination neglect  between sub-tasks, as discussed in the earlier point. It is true – – the devil is always in the details.

I should emphasise that the above list based on personal experience, not on any systematic study.

I’ll conclude this piece with another fragment from Glass, who  is not very optimistic about improvements in the area of project estimation. As he states in his book:

The bottom line is that, here in the first decade of the twenty-first century, we don’t know what constitutes a good estimation approach, one that can yield decent estimates with good confidence that they will really predict when a project will be completed and how much it will cost. That is a discouraging bottom line. Amidst all the clamor to avoid crunch mode and end death marches, it suggests that so long as faulty schedule and cost estimates are the chief management control factors on software projects, we will not see much improvement.

True enough,  but being aware of the ways in which estimates can go wrong is the first step towards improving them.

Written by K

October 1, 2010 at 4:55 am