ᎏᎏ X CA/U ⫻ X CO/EXCA/E/XCA/U
4.10 Dealing with Uncertainty
4.10.2 Program Evaluation and Review Technique and Extensions
Two common analytical approaches are used to assess uncertainty in projects. Both are based on the cen- tral limit theorem, which states that the distribution of the sum of independent random variables is approximately normal when the number of terms in the sum is sufficiently large.
The first approach yields a rough estimate and assumes that the duration of each project activity is an independent random variable. Given probabilistic durations of activities along specific paths, it follows that elapsed times for achieving events along those paths are also probabilistic. Now, suppose that there are n activities in the project, k of which are critical. Denote the durations of the critical activities by the random variables di with mean d苶iand variance s2i, i1,…, k. Then the total project length is the random variable
Xd1d2…dk
It follows that the mean project length, E[X], and the variance of the project length, V [X], are given by E[X ]d苶1d苶2…d苶k
V [X ]s12s22…s2k
These formulas are based on elementary probability theory, which tells us that the expected value of the sum of any set of random variables is the sum of their expected values, and the variance of the sum of independent random variables is the sum of the variances.
Now, invoking the central limit theorem, we can use normal distribution theory to find the probabil- ity of completing the project in less than or equal to some given time τas follows:
P(X τ)P 冢 冣P冢Z 冣 (4.10)
where Z is the standard normal deviate with mean 0 and variance 1. The desired probability in Equation (4.10) can be looked up in any statistics book.
τE[X ] V [X ]1/2 τE[X ]
V [X ]1/2 XE[X ]
V [X ]1/2
Project length (weeks)
Frequency
17 18 19 20 21 22 23 24 25 26 27 28 29 1
2 3 4 5 6 7 8 9
0
FIGURE 4.33 Frequency distribution of project length for simulation runs.
Continuing with the example project, if (based on the simulation) the mean time of the critical path is 22.5 weeks and the variance is (2.9)2, then the probability of completing the project within 25 weeks is found by first calculating
z 0.86
and then looking up 0.86 in a standard normal distribution table. Doing so, we find that P(Z 0.86) 0.805, so the probability of finishing the project in 25 weeks or less is 80.5%. This solution is depicted in Figure 4.34.
If, however, the mean project length, E[X], and the variance of the project length, V[X], are calculated using the assumption that the critical activities are only those that have a zero slack in the deterministic CPM analysis (A–C–F–G), we get
E[X ]584522
V [X ]1\2兹1苶2苶苶0.3苶3苶2苶苶0.6苶6苶2苶苶0.3苶3苶21.285
On the basis of this assumption, the probability of completing the project within 25 weeks is P冢Z 冣P (Z 2.33)0.99
This probability is higher than 0.805, which was computed using data from the simulation in which both sequences A–C–F–G and A–D–F–G were critical.
The procedure above is, in essence, PERT. Summarizing for an AON network:
1. For each activity i, assess its probability distribution or assume a beta distribution and obtain esti- mates of ai, biand mi. These values should be supplied by the project manager or experts who work in the field.
2. If a beta distribution is assumed for activity i, then use the estimates a, b, and m to compute the variance sˆ2iand mean dˆifrom Equations (4.1) and (4.2). These values then are used in place of the true but unknown values of s2iand d苶i, respectively, in the above formulas for V [X] and E[X].
3. Use CPM to determine the critical path given dˆi, i1,…, n.
4. Once the critical activities are identified, sum their means and variances to find the mean and vari- ance of the project length.
5. Use Equation (4.10) with the statistics computed in step 4 to evaluate the probability that the proj- ect finishes within some desired time.
1.2852522 2522.5 2.9
13 14 15 16 17 18 19 20 21 22 23 24 0.00
0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18
Project duration (weeks)
Probability density function
0.805
25 26 27 28 29 30 31 32
FIGURE 4.34 Example of probabilistic analysis with PERT.
Using PERT, it is possible to estimate completion time for a desired completion probability. For example, for a 95% probability the corresponding z value is z.951.64. Solving for the time τ,for which the prob- ability to complete the project is 95%, we get
z0.95 1.64
or
τ(1.64)(2.5)22.527.256 weeks
A shortcoming of the standard PERT calculations is that they ignore all activities that are not on the crit- ical path. A more accurate analytical approach is to identify each sequence of activities that lead from the start node of the project to the finish event, and then to calculate separately the probability that the activ- ities that compose each sequence will be completed by a given date. This step can be done as above by assuming that the central limit theorem holds for each sequence and then applying normal distribution theory to calculate the individual path probabilities. It is necessary, though, to make the additional assumption that the sequences themselves are statistically independent to proceed. This means that the time to traverse each path in the network is independent of what happens on the other paths. Although it is easy to see that this is rarely true because some activities are sure to be on more than one path, empir- ical evidence suggests that good results can be obtained if there is not too much overlap.
Once these calculations are performed, assuming that the various sequences are independent of each other, the probability of completing the project by a given date is set equal to the product of the individ- ual probabilities that each sequence is finished by that date. That is, given n sequences with completion times X1, X2,…, Xn, the probability that X is τis found from
P(X1 τ)P(X1 τ)P(X2 τ)…P(Xn τ) (4.11) where now the random variable Xmax{X1, X2,…, Xn}.
Example 5 Consider the simple project in Figure 4.35. If no uncertainty exists in activity durations, then the critical path is A-B and exactly 17 weeks are required to finish the project. Now if we assume that the durations of all four activities are normally distributed (the corresponding means and standard devia- tions are listed under the arrows in Figure 4.35), then the durations of the two sequences are also nor- mally distributed (i.e., N(µ,σ)), with the following parameters:
length(AB)X1⬃N(17, 3.61) length(CD)X2⬃N(16, 3.35)
τ22.5 2.9
1 4
3 2
A B
C D
[8,2] [9,3]
[6,1.5]
[10,3]
FIGURE 4.35 Stochastic network.
The accompanying probability density functions are plotted in Figure 4.36. It should be clear that the project can end in 17 weeks only if both A–B and C–D are completed within that time. The probability that A–B finishes within 17 weeks is
P(X1 17)P冢Z 冣P(Z 0)0.5
and similarly for C–D,
P(X2 17)P冢Z 冣P(Z 0.299)0.62
Using Equation (4.11), we can now determine the probability that both sequences finish within 17 weeks:
P(X 17)P (X1 17)P(X2 17)(0.5)(0.62)0.31
Thus, the probability that the project will finish by week 17 is ~31%. A similar analysis for 20 weeks yields P(X 20) 0.7 70%.
The approach that is based on calculating the probability of each sequence completing by a given due date is accurate only if the sequences are independent. This is not the case when one or more activities are members of two or more sequences. Consider, for example, the project in Figure 4.37. Here, activity E is a member of the two sequences that connect the start of the project (event 1) to its termination node (event 5). The expected lengths and standard deviations of these sequences are
Sequence Expected Length Standard Deviation
A–B–E 8 9 3 20 兹2苶2苶苶32苶苶苶42苶5.39 C–D–E 10 6 3 19 兹3苶2苶苶1.5苶2苶苶苶42苶5.22 The probability that the sequence A–B–E will be completed in 17 days is calculated as follows:
z 0.5565 implying that P0.29
which is obtained from a standard normal distribution table by noting that P(Z z)1P(Z z) 1720
5.39
1716 3.35
1717 3.61
0.12 0.1 0.08 0.06 0.04 0.02 0
Probability density
Duration of C - D Duration of A - B
−2 8 18 28 38
FIGURE 4.36 Performance time distribution for the two sequences.
Similarly, the probability that the sequence C–D–E will be completed in 17 days is calculated by deter- mining z(17 19)/5.22 0.383 and then using Table 4.9 (C-1) a standard normal distribution table to find P0.35.
Thus, the simple PERT estimate (based on the critical sequence A–B–E) indicates that the probability of completing the project in 17 days is 29%. If both sequences A–B–E and C–D–E are taken into account, then the probability of completing the project in 17 days is estimated as
P(XABE 17)P(XCDE 17)(0.29)(0.35)0.1 or 10%
assuming that the two sequences are independent. However, because activity E is common to both sequences, the true probability of completing the project in 17 days is somewhere between 10 and 29%.
The next question that naturally arises is what to do if only the parameters of the distribution are known but not its form (e.g., beta, normal), and the number of activities is too small to rely on the cen- tral limit theorem to give accurate results. In this case, Chebyshev’s inequality can be used to calculate project duration probabilities (see Montgomery and Runger, 2003). The underlying theorem states that if X is a random variable with mean µ and variance σ2, then for any k0,
P(円Xµ円kσ) An alternative form is
P(円Xµ円kσ)1
Based on the second inequality, the probability of a random variable being within ±3 standard deviations of its mean is at least 8/9, or 89%. Although this might not be a tight bound in all cases, it is surprising that such a bound can be found to hold for all possible discrete and continuous distributions.
To illustrate the effect of uncertainty, consider the example project. Four sequences connect the start node to the finish node. The mean length and the standard deviation of each sequence are summarized in Table 4.10.
The probability of completing each sequence in 22 weeks is computed next and summarized in Table 4.11.
Based on the simple PERT analysis, the probability of completing the project in 22 weeks is 0.5. If both sequences A–C–F–G and A–D–F–G are considered and assumed to be independent, the probability is reduced to (0.5)(0.73) 0.365.
Because three activities (A, F, G) are common to both sequences, the actual probability of completing the project in 22 weeks is closer to 0.5 than to 0.365. Based on the data in Figure 4.33, we see that in 24 of 50 simulation runs, the project duration was 22 weeks or less. This implies that the probability of com- pleting the project in 22 weeks is 24/50 0.48, or 48%.
Continuing with this example, if the Chebyshev’s inequality is used for the critical path (µ 22, σ 1.285), then the probability of completing the project in, say, 22 (2)(1.285) 24.57 weeks is approximately
1冢 冣12 234 0.75
k12 k12
1 4
3 2
A B
C D
[8,2] [9,3]
[6,1.5]
[10,3]
5 [3,4]
FIGURE 4.37 Stochastic network with dependent sequences.
By way of comparison, using the normal distribution assumption, the corresponding probability is
P冢Z 冣P(Z 2)0.97
Of the two, the Chebyshev estimate is likely to be more reliable given that there are only a few activities on the critical path.
Because uncertainty is bound to be present in most activities, it is possible that after determining the critical path with CPM, a noncritical activity may become critical as certain tasks are completed. From a practical point of view, this suggests the basic advantage of ES schedules. Starting each activity as soon as possible reduces the chances of a noncritical activity becoming critical and delaying the project.