## 4 Application Example  
In this section, we develop a manual example with a small project schedule. The purpose is to illustrate how \(\mu_{p}\) and \(\sigma_{p}\) could be easily calculated by practitioners. All necessary auxiliary calculations and intermediate steps are also included. The results of the \(\mu_{p}\) and \(\sigma_{p}\) are eventually compared with the PERT estimates and the (more accurate) estimates obtained by Monte Carlo simulation. Let us refer to the nine-activity schedule network represented in Figure 1. This project consists of activities with varied duration variabilities (represented by different gray tones). Initially, we assume the activity duration variability is normally distributed. This way, the deterministic activity durations are \(d_{i}=\mu_{i}\), and the activity duration standard deviations are \(\sigma_{i}=\mu_{i}\cdot CV_{i}\). Both sets of values are represented for all activities in the leftmost part of Table 1. From them, calculating the activity slacks is straightforward (indeed, all are equal to zero, apart from the slacks of activities 4 and 6, which are \(s_{4}=d_{5}-d_{4}\) and \(s_{6}=d_{6}-d_{5}\), respectively). Figure 1: Example project schedule. In this case, the deterministic project duration \(PD\) coincides with the sum of \(d_{1}+d_{2},d_{3}+d_{5}+d_{7}\), or \(d_{8}+d_{9}=50\) days (or time units). From the initial set of all nine activities (\(i\)), we know that the critical activities (\(k\)) are those with \(s_{i}=0\), that is, \(K=\{1,2,3,5,7,8,9\}\) (all \(i\) other than activities 4 and 6). The second step involves calculating the approximated critical path duration standard deviation \(\sigma_{K}\) with Equation (7). \[\sigma_{K}=\sqrt{\frac{PD}{\sum d_{k}}\cdot\sum o_{k}^{2}}=\sqrt{\frac{50}{7 \cdot 10}\cdot\sum(3\cdot 2^{2}+4\cdot 3^{2})}=5.86\text{ days}.\]  
The longest chain of activities is five activities long; thus, \(m=5\) and, since \(n_{i}=9\) activities (all initial activities), the value of the serial-parallel indicator with Equation (8) is  
\[SP=\frac{m-1}{n_{i}-1}=\frac{5-1}{9-1}=0.5\]  
As \(\sigma_{K}\) and \(SP\) are known, we can calculate the slack admissibility threshold \(\tau\) with Equation (6). By assuming \(K_{0}=1.5\), then  
\[\tau=K_{0}(1-SP)\sigma_{K}=1.5(1-0.5)5.86=4.39\text{ days}.\]  
The only non-critical activity whose total slack is lower than 4.39 is activity 6 (\(s_{6}=2\) days). Activity 4 with a slack \(s_{4}=5>\tau=4.39\) is not considered any further. Hence, the set of critical + subcritical activities is \(J=\{1,2,3,5,6,7,8,9\}\), that is, all but activity 4. The cardinality of \(J\) (the number of elements in the set) is then \(n_{j}=8\). Before calculating \(\mu_{p}\), it is necessary to calculate the value of a few more auxiliary variables. \(n_{j}=8\) act; \(\Sigma d_{j}=7\times 10+8=78\) days; \(\Sigma\sigma_{j}=3\times 2+4\times 3+0.8=18.8\) days; \(\Sigma s_{j}=7\times 0+2=2\) days. \(K_{1}=1.12\), as we are assuming symmetrical (normal) activity duration variability. \[N_{p}=\frac{\sum d_{j}}{PD}=\frac{78}{50}=1.56\text{ {paths}, with Equation \eqref{eq:Np}.}\]  
\[S=1-\frac{\sum s_{j}}{\tau\cdot n_{j}}=1-\frac{2}{4.39\cdot 8}=0.94,\text{ with Equation \eqref{eq:Np}.}\]  
With all this information, we can now easily calculate \(\mu_{p}\) with Equation (9). \[\mu_{p}=PD\cdot\left[1+K_{1}\sqrt{\frac{N_{p}}{n_{j}}}.\frac{\sum\sigma_{j}}{ \sum d_{j}}.LN\left(S\cdot N_{p}\right)\right]=50\cdot\left[1+1.12\sqrt{\frac {1.56}{8}}.\frac{18.8}{78}.LN\left(0.94\cdot 1.56\right)\right]=52.3\text{ days}.\]