## 3 Algorithm
Figure 3 shows a simple cold chain network to illustrate an SCCNMT model, which consists of two suppliers, two logistics companies, three retail stores, and ten arcs. Table 3 shows all the time factors dealt with within the logistics enterprise. Tables 4 and 5 list the data on the capacity and arrival distribution of each arc. \[\left[\sum_{o=1}^{w}\sum_{k=1}^{p}\{f_{o}^{k}\left|q_{o}\in P_{o}^{k}\right.\} \right]\leq M_{t},\text{ for }q=1,\ 2,\...,\ z. \tag{12}\]4 The procedure of the algorithm. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Arc** & **Capacity** & **Probability** \\ \hline  & 0 & 0.7 \\ \(a_{1}\), \(a_{2}\) & 1 & 0.1 \\ \(a_{3}\), \(a_{4}\) & 2 & 0.15 \\  & 3 & 0.05 \\ \hline  & 0 & 0.7 \\ \(a_{5}\), \(a_{6}\) & 1 & 0.2 \\ \(a_{7}\), \(a_{8}\) & 2 & 0.05 \\  & 3 & 0.05 \\ \hline  & 0 & 0.6 \\ \(a_{9}\), \(a_{10}\) & 1 & 0.2 \\  & 2 & 0.05 \\  & 3 & 0.15 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Truck data of the numerical example. Figure 3: An SCCNMT example. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Arc** & **Travel Time** & **Arrival Probability** \\ \hline  & 35 & 0.7 \\ \(a_{1}\), \(a_{2}\) & 37 & 0.1 \\  & 40 & 0.1 \\  & 45 & 0.1 \\ \hline  & 27 & 0.8 \\ \(a_{3}\), \(a_{4}\) & 30 & 0.05 \\  & 32 & 0.05 \\  & 35 & 0.1 \\ \hline  & 35 & 0.9 \\ \(a_{5}\), \(a_{6}\) & 37 & 0.05 \\  & 40 & 0.025 \\  & 45 & 0.025 \\ \hline \hline \end{tabular}
\end{table} 
Table 5: Travel time and arrival probability of each arc. The following algorithm is used to obtain the network reliability of an SCCNMT given \(D=(2,2,2)\), \(I=130\) min, \(M=(3,3,3,3,4,4,4,3,3,3)\). There are 12 delivery paths in this network \(P^{1}_{1}=\{a_{1},\,a_{5}\}\), \(P^{1}_{2}=\{a_{3},\,a_{5}\}\), \(P^{1}_{3}=\{a_{2},\,a_{8}\}\), \(P^{1}_{4}=\{a_{4},\,a_{8}\}\), \(P^{2}_{1}=\{a_{1},\,a_{6}\}\), \(P^{2}_{2}=\{a_{3},\,a_{6}\}\), \(P^{2}_{3}=\{a_{2},\,a_{8}\}\), \(P^{2}_{4}=\{a_{4},\,a_{8}\}\), \(P^{3}_{1}=\{a_{1},\,a_{7}\}\), \(P^{3}_{2}=\{a_{3},\,a_{7}\}\), \(P^{3}_{3}=\{a_{2},\,a_{10}\}\), \(P^{3}_{4}=\{a_{4},\,a_{10}\}\) (Algorithms 4-6). Step 0. Find all feasible flow vectors \(F=\left(f^{1}_{1},f^{1}_{2},f^{1}_{3},f^{1}_{4},f^{2}_{1},f^{2}_{2},f^{2}_{3}, f^{2}_{4},f^{3}_{1},f^{3}_{2},f^{3}_{3},f^{3}_{4}\right)\) satisfying

\[ \begin{split} f^{1}_{1}+f^{1}_{2}+f^{1}_{3}+f^{1}_{4}=2,\\ f^{2}_{1}+f^{2}_{2}+f^{2}_{3}+f^{2}_{4}=2,\\ f^{3}_{1}+f^{3}_{2}+f^{3}_{3}+f^{3}_{4}=2.\end{split}  \tag{22}\]

This step generates 1000 flow vectors, which are shown in the first column of Table 6.

Step 1. Check whether the flow vectors \(F\) exceeds the maximal capacity \(M_{4}\) of each arc or not. (1.1) Check whether the maximal capacity of each arc has been exceeded by the number of vehicles used via

\[ \begin{split}\left[\begin{array}{c}f^{1}_{1}+f^{2}_{1}+f^{3}_{ 1}\\ \end{array}\right]\leq 3,\\ \left[\begin{array}{c}f^{1}_{3}+f^{2}_{3}+f^{3}_{3}\\ \end{array}\right]\leq 3,\\ \left[\begin{array}{c}f^{1}_{2}+f^{2}_{2}+f^{3}_{2}\\ \end{array}\right]\leq 3,\\ \left[\begin{array}{c}f^{1}_{4}+f^{2}_{4}+f^{3}_{4}\\ \end{array}\right]\leq 3,\\ \left[\begin{array}{c}f^{1}_{1}+f^{2}_{1}\\ \end{array}\right]\leq 4,\\ \left[\begin{array}{c}f^{2}_{1}+f^{2}_{2}\\ \end{array}\right]\leq 4,\\ \left[\begin{array}{c}f^{3}_{1}+f^{3}_{2}\\ \end{array}\right]\leq 4,\\ \left[\begin{array}{c}f^{3}_{1}+f^{3}_{2}\\ \end{array}\right]\leq 4,\\ \left[\begin{array}{c}f^{1}_{3}+f^{3}_{4}\\ \end{array}\right]\leq 3,\\ \left[\begin{array}{c}f^{3}_{3}+f^{3}_{4}\\ \end{array}\right]\leq 3.\end{split}  \tag{23}\]

(1.2) Check whether the truck departing from the logistics company can arrive within the given time and use the maximum value for travel time via \begin{table}
\begin{tabular}{c c c} \hline
**Arc** & **Travel Time** & **Arrival Probability** \\ \hline  & 20 & 0.8 \\ \(a_{7},a_{8}\) & 25 & 0.05 \\  & 27 & 0.1 \\  & 30 & 0.05 \\ \hline  & 30 & 0.9 \\ \(a_{9},a_{10}\) & 33 & 0.025 \\  & 35 & 0.05 \\  & 40 & 0.025 \\ \hline \end{tabular}
\end{table}