## 4 Application Example
\begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline  & \multicolumn{3}{c}{**Normally Distributed Activity Duration**} & \multicolumn{3}{c}{**Lognormally Distributed Activity Duration**} \\  & \multicolumn{3}{c}{**Variability**} & \multicolumn{3}{c}{**Variability**} \\ \hline
**Activity** & **Avg. Dur.** & **Duz. Variability** & **(Total) Slack** & **Avg. Dur.** & **Dur. Variability** & **(Total) Slack** \\ \hline \(i\) & \(d_{i}\) & \(CV_{i}\) & \(\sigma_{i}\) & \(s_{i}\) & \(d^{\prime}_{i}\) & \(CV_{i}\) & \(\sigma^{\prime}_{i}\) & \(s_{i}\) \\ \hline
1 & 10 & 0.2 & 2 & 0 & 11.12 & 0.2 & 5.40 & 0 \\
3 & 10 & 0.3 & 3 & 0 & 12.69 & 0.3 & 9.92 & 0 \\
4 & 5 & 0.1 & 0.5 & 5 & 5.13 & 0.1 & 1.20 & 5.98 \\
5 & 10 & 0.2 & 2 & 0 & 11.12 & 0.2 & 5.40 & 0 \\
6 & 8 & 0.1 & 0.8 & 2 & 8.21 & 0.1 & 1.92 & 2.90 \\
7 & 10 & 0.3 & 3 & 0 & 12.69 & 0.3 & 9.92 & 0 \\
8 & 10 & 0.3 & 3 & 0 & 12.69 & 0.3 & 9.92 & 0 \\
9 & 10 & 0.2 & 2 & 0 & 11.12 & 0.2 & 5.40 & 0 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Project network example: deterministic schedule information. Avg.—average; Dur.—duration. A much more accurate estimate of the average project duration with Monte Carlo simulation would be 53.4 days; that is, we miss it by 1.1 days. However, our estimate is substantially closer than the only estimate obtained (\(PD\) = 50 days, which is also the PERT estimate), and it was calculated manually. Now, imagine we want to calculate the project duration standard deviation \(\sigma_{p}\). First, we need to calculate \(\sigma_{I}\) with Equation (13). \[\sigma_{I}=\sqrt{\frac{PD}{\sum d_{j}}\cdot\sum\sigma_{j}^{2}}=\sqrt{\frac{50 }{78}\cdot\sum(3\cdot 2^{2}+4\cdot 3^{2}+0.8^{2})}=5.72\text{ days}.\]  
Then, assuming \(K_{2}=-0.22\) (symmetrical activity duration variability), we use Equation (12) to calculate \(\sigma_{p}\). \[\sigma_{p}=\sqrt{\frac{\sigma_{K}^{2}+\sigma_{I}^{2}}{2}}\left[1+K_{2}\cdot LN \left(S\cdot N_{p}\right)\right]=\sqrt{\frac{5.86^{2}+5.72^{2}}{2}}\cdot[1-0.22 \cdot LN(0.94\cdot 1.56)]=5.24\text{ days}.\]  
The project duration standard deviation obtained by simulation yields 4.88 days. We observe a deviation of just 0.36 days in this case. The only alternative estimate would have been the PERT one, but it does not offer a solution when there are multiple critical paths. If we assume that there is only a single critical path (as all critical paths intentionally had the same standard deviations in our example), then  
\[PERT\ \sigma_{K}=\sqrt{\sum_{one\text{ }path}\sigma_{k}^{2}}=\sqrt{\sum(3\cdot 2 ^{2}+2\cdot 3^{2})}=5.47\text{ days},\]  
which has a deviation of 0.59 days, that is, again worse than our \(\sigma_{p}\) estimate. Finally, the same calculations could have been replicated but assuming activity duration variability modeled with lognormal distributions. In this case, the values of \(K_{1}\) and \(K_{2}\) would change to 1.22 and 0.07, respectively. Additionally, before starting all calculations, the activity \(d_{i}\) and \(\sigma_{i}\) values would need to be calculated with Equations (4) and (5). The results of these variables for all activities are displayed in the rightmost part of Table 1 under the \(d^{\prime}_{i}\) and \(\sigma^{\prime}_{i}\) columns. Then, the deterministic project duration (\(PD\) = 58.7 days) and the activity slacks (\(s_{i}\)) could have been calculated in the same way as normal activity duration variability. Indeed, the rest of the calculation process is exactly analogous to the previous example. The final results would have been as follows:  
\(\mu_{p}\) = 66.8 days (simulation estimate = 68.7 days) better than \(PD\) = 58.7 days. \(\sigma_{p}\) = 18.3 days (simulation estimate = 18.4 days) better than PERT \(\sigma_{K}\) =18.9 days.