Week 2 math new readings
5. Project management and the importance of assessing risk
project management: systematic, phased approach to defining, organizing, planning, monitoring
and controlling projects, there are 2 different approaches:


Program evaluation and review technique (PERT)
Critical path method (CPM)
Now they are combined into one single approach (PERT/CPM) involving 5 steps
1) Defining the work breakdown structure, breakdown into activities
2) Diagramming the network, establish the precedence relationships and draw the implied
network
3) Developing the schedule, estimate durations incorporating uncertainty
 We can identify the EST/EFT and LST/LFT, the implied slacks and the critical path
 Now if each activity that makes up the project had to be done in sequence, working on only
one activity at a time, the completion time would equal the sum of the durations of all the
activities, i.e. 4+1+3+6+2+3+2 = 21 weeks. However, our network visualizes that some
activities can be undertaken simultaneously; we can achieve less than 21 weeks
 Each sequence of activities between the project’s start and finish is called a path. The network
describing our Case Software Design contains three such path:
(1) 2-3-5-6-7 => Completion time: 1+3+2+3+2 = 11 weeks
(2) 1-3-5-6-7 => Completion time: 4+3+2+3+2 = 14 weeks
(3) 1-4-6-7 => Completion time: 4+6+3+2 = 15 weeks
The critical path is the path that takes the longest to complete => path (3)
 Consequently, the durations of the activities along this critical path determine the minimum
completion time of the project. Of course, if any of these critical activities is delayed, the entire
project will be delayed beyond 15 weeks.
 As we have seen, the maximum amount of time that an activity can be delayed without
delaying the whole project is called the slack of that activity. By definition, critical activities
have zero slack.
 Non-critical activities have some degree of allowed schedule slippage without delaying the
whole project
4) Analyzing cost-time trade-offs: addresses crashing (=shorten project duration in the most
economical way)
5) Assessing risks: estimates usually include high degree of uncertainty, especially if durations of
projects have not been experience several times (no past data available)
+ Some activities can also be inherently uncertain (ex: activities that depend on the weather)
 activity durations 𝐷𝑖 become random variables to be modelled with probability
distributions (This in turn implies that all the earliest and latest start and finish times, the
slacks, and the project completion time, will become random variables as well.)
Now imagine that management insists that the project be completed within 16 weeks. Obviously,
we would then like to determine what the probability is of actually completing the project within
this deadline! => Quantify the risk associated with the project’s timing
 We have to determine the probability distribution of the project completion time, starting from
the probability distribution of the underlying activity durations (we have 2 approaches).
6. Statistical analysis
6.i) Modelling the probability distribution of the activity durations
start by specifying the probability distributions use Beta distribution by obtaining 3 estimates
for each activity:
1)optimistic duration (o) the shortest possible time,
2) the most likely duration (m) the mode of the activity’s probability distribution: can fall
anywhere between the optimistic and pessimistic estimates o and p
3) pessimistic duration (p) the longest time required to complete the activity
=> the calculate the mean and the standard deviation for each activity
Beta distribution can have various shapes, does not have to be symmetric, it is usually skewed to
the right due to the potential delays.
The upper endpoint p is much further away from the mode
m than the lower endpoint o.
If the endpoints are equidistant to the mode, the beta
distribution becomes symmetric, and very similar to the
well-known normal distribution.
Note: The standard deviation (or its square, i.e. the variance) of an activity’s duration depends on
the range between the optimistic and pessimistic times: the wider o and p are apart, the more
uncertain a project manager must be about the actual duration of that activity, and this translates
into a larger standard deviation.
6.ii) Determining the probability distribution of the minimum completion time
If we want to determine the probability distribution of the project completion time, the starting
point is to have a look at the probability distribution of the sum of the durations of our critical
activities
 we will try to determine the shape, the mean and the variance of the probability distribution
of the project completion time on the basis of this probability distribution.
Strong Assumptions
1) Generalizing: if noncritical activities happen to have longer durations than expected, they may
become critical. Conversely, if critical activities happen to have shorter durations than
expected, they may become noncritical.
 even though all activities durations will become random and could fluctuate, we assume that
our project completion time is always determined by the sum of the durations of the critical
path activities in the baseline scenario
 In other words: we always see the baseline critical path as THE critical path
2) Regarding the shape, we assume that the number of critical activities is sufficiently large to
make the project completion time approximately normally distribution (due to the central limit
theorem when we have enough activities)
Regarding the mean (or expected value): irrespective of the nature of the distributions involved,
and without further assumptions, the mean of a sum of random variables is simply equal to the
sum of the means of those random variables.
3) When it comes to the variance of the probability distribution of project completion time, we
assume independence: the duration of one activity is unrelated to that of another activity
SUMMARIZING: under the three stated assumptions (or approximations), our project completion
time will ...
-obey a normal distribution;
- with a mean equal to the sum of the mean durations of the (assumedly) critical activities 1, 4, 6
and 7
- with a variance equal to the sum of the variances of the (assumedly) critical activities 1, 4, 6 and
7
IF-function, column N identifies the critical activities with a “1”, critical
activity might show an extremely small nonzero slack, something like
–63
1.32*10 we defined cell N4 as “=IF(M4<0.0001,1,0)”
SUMPRODUCT calculate mean completion time of our project in cell
N12 and its variance in cell N13, by adding the means and variances of
the critical activities.
6iii) Analyzing managerial questions
how likely are we to complete the project within 16 weeks? (16 = Completion Time)
𝑃(𝐶𝑇 ≤ 16) = 𝑃 (
𝐶𝑇 − 𝑀𝑒𝑎𝑛 16 − 𝑀𝑒𝑎𝑛
16 − 𝑀𝑒𝑎𝑛
≤
) = 𝑃(𝑧 ≤
)
𝑆𝑡𝑑𝑒𝑣
𝑆𝑡𝑑𝑒𝑣
𝑆𝑡𝑑𝑒𝑣
NORMDIST functionin Excel it takes as input z value and returns the corresponding cumulative
standard normal probability
Column M shows the corresponding z-transformation: be sure to note the absolute references
(the “$”) w.r.t. cells N12 and N14!
Column N then converts this z-value into the desired probability, using the NORMSDIST function.
As you can see, the probability to complete the
project within its expected completion time of
15 weeks (=the mean) is exactly 50%; this is an
obvious consequence of the symmetry of the
normal distribution. More interesting, the
probability to complete the project within the
16 week deadline stated by management is
estimated to be 65.4%. Alternatively, there is a
34.6% probability that the deadline will be
exceeded. Of course, as the deadline increases,
the probability of timely project completion
rises; the probability to complete the project
within 23 weeks is estimated at virtually 100%.
what is the expected loss caused by the possible delay in the project? => Column O and P
If each week of delay costs €100,000 of revenue loss
Column O gives us the probability that the project will be completed during a specific week. if
there is a 0.654 probability to finish the project within 16 weeks and a 0.787 probability to do so
within 17 weeks, then the probability that the project will be completed during week 17 is
obviously 0.787 – 0.654 = 0.133 (remember the time conventions defined at the start of p. 4!). T
his 1-week delay carries an expected loss of 0.133 * €100,000, as programmed in column P.
The probability to complete the project during week 18 is 0.883 – 0.787 = 0.096; but since this
involves a 2-week delay, the implied expected loss is not simply 0.096 * €100,000, but 0.096 * 2 *
€100,000: hence the factor (L20-16) in the formula for cell P20, total expected loss is simply the
sum of the expected losses associated with increasing delays
6.iv) Drawbacks of statistical analysis we base everything on 3 main assumptions
1) Critical path remains the same because of our assumption even though activities durations
fluctuate we assume that the fluctuations will have a minor impact, but we cannot be sure
about it
2) have assumed that, while the durations of the individual activities obey a beta distribution, the
CLT allows us to approximate the project completion time with a normal distribution.most
of the times the durations of individual activities are very skewed
3) Independence assumptionsometimes interdependencies exist between activities
=> Using a simulation allows us to do away with assumptions 1 and 2, but we stick to
independence assumption.
7. Simulation
Using simulation software, the duration for each activity is randomly chosen from its probability
distribution + the critical path of the network is determined and the completion time of the
project computed.
 The procedure is then repeated (“replicated”) many times, which results in a simulated
probability distribution for the project’s completion time
we will use UMS Tools in Excel: From now on, we will explicitly regard the durations Di as
uncertain quantities, i.e. we will treat them as random variables. This means that we will assign
random values to these durations, drawn from their respective beta distributions, as defined by
the estimates o, m and p THEN we perform a “one-shot” simulation/experiment/trial, run this for
10000 replications each time with different random values
7.i) Step 1: Set up the simulation worksheet

Simulate the duration: DRAWBETA(Alpha,Beta,A,B)
𝑚𝑒𝑎𝑛−𝑜
o Alpha= (
𝑝−𝑜
(𝑝−𝑚𝑒𝑎𝑛)
) ∗ ((𝑚𝑒𝑎𝑛 − 𝑜) ∗ 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒−1
𝑝−𝑚𝑒𝑎𝑛)
o Beta= 𝐴𝑙𝑝ℎ𝑎 ∗ ( 𝑚𝑒𝑎𝑛−𝑜 )
o A=o
o B=p
FOR EXAMPLE for activity 1
In column G, the Alpha-argument has been programmed. Be sure to check how eq. (13.1) has been
implemented, by referring to columns B, D, E and F.
- Similarly, in column H, the Beta-argument has been programmed on the basis of eq. (13.2). - The
DRAWBETA-function is called in column I: note how e.g. cell I4 refers to the arguments Alpha, Beta,
A and B for Activity 1, as defined by cells G4, H4, B4 and D4.
- Columns J to Q are exactly like columns G to N in the page 13 spreadsheet. The only difference is
the fact that in columns L and N, the EFT- and LST-definitions now refer to the randomly drawn
durations in column I instead of the expected durations in column E.
Press the function key [Fn]-[F9]UMS tool generates new random durations for each of the
activities as in the following excel output
Note that, in this particular trial, the random durations are such that path (2) happens to be
critical, rather than our “baseline” path (3). The project completion time is 13.620 weeks, as
reported in cell A13. Pressing the keys [Fn]-[F9] a few times will simulate additional trials, each
with different activity durations and therefore a different project completion time; most of them
probably with critical path (3) rather than (2).
7.ii) Step 2: replicate the model
 Select the outcome cell A13 (the so-called reference cell) with your mouse, go to the Add-Ins
tab and select the UM Simulation Tool. This will open the UM Simulation Tool window. In its last
line, change the number of trials per run from the default value of 100 into 10,000 and push the
Simulate button, the section of graphs allows you to see the overall distribution
7.iii) Step 3: Analyzing the results do the assumptions hold?
our simulation suggests that the mean project completion time is 15.586 weeks, i.e. clearly more
than 15=> Due violation of our first assumption.
Under this assumption, the project completion time is always determined by path (3),(= the
“baseline” critical path. But whenever path (2) consisting of activities 1-3-5-6-7 is longer than path
(3), it will become the critical path; the actual project completion time will then be determined by
the length of path (2) instead of the shorter path (3), and will therefore be longer than our
statistical analysis assumes.
 Regarding assumption 1: the more often the near-critical path (2) becomes critical, the more
our statistical analysis will underestimate the mean completion time
Additionally, as already noted, the probability distribution of project completion times is visibly
skewed to the right, rather than symmetric (as would be implied by the normal distribution).
 violation of our second assumption: with project completion time being the sum of a very small
number of activity durations, some of which are strongly skewed to the right, the central limit
theorem does not have enough “bite” to fully work itself out.
We can also answer managerial questions like “determine the probability to complete the project
within the 16 week deadline imposed by management” by adding a cell with the formula
=IF(A13<16,1,0) and then use that cell as reference to run the simulation in UMS tool
We can also finally test “how often will paths (3), (2) or even (1) be critical?”
(1) activities 2-3-5-6-7
(2) activities 1-3-5-6-7
(3) activities 1-4-6-7
1) Q4 to Q10 are selected as reference cells since they indicate whether an activity is critical or
not
2) UMS Tool window can only display those statistics for one of the activitiesTo choose which
one, you can use the “arrow buttons” to identify an Active Row anywhere between 1 (i.e. the
first reference cell, Q4) up to 7 (i.e. the seventh and last reference cell, Q10), before running
the simulation. run our simulations seven times, each time focusing on the descriptive
statistics for one reference cell i.e. activity!
3) it is possible to display the most important descriptive statistics directly in cells of your Excel
worksheet (rather than in the UMS Tool window) for all seven reference cells simultaneously.
As you can see, in cell R4 we ask for the mean of Active Row 1 i.e. cell Q4 as “=mean(1)”;
similarly up to cell R10, which gives us the mean for Active Row 7 i.e. cell Q10 as “=mean(7)”
From our network, it should be clear that Activities 6 and 7 will always be critical, since they are
part of all three paths => we observe a mean value of 1 in cells R9 and R10 above.
- Cell R5, reporting a mean value of 0, indicates that Activity 2 will never be critical. Since this
activity is ONLY in path (1), we conclude that this path will never be the longest= the critical.
With path (1) removed from the rolls, Activity 1 must always be critical, since it belongs to both
the remaining paths (2) and (3).
 And indeed, as already noticed, we observe a mean value of 1 in cell R4 above.
- This brings us to the most important question: how often is path (3) critical, and how often path
(2)? Well, our network implies that Activity 4 is ONLY in path (3).
=> Consequently, the mean value of 0.6426 reported in cell R7 implies that the probability that the
“baseline” path (3) indeed turns out critical can be estimated at around 64%.
=> Consistent with this, the probability that our “near-critical” path (2) turns out to be critical is
equal to the remaining 36%. See cells R6 and R8!