## 3 Algorithm

Table 5: _Cont._\[\max\{f_{1}^{1},f_{2}^{1}\}\leq \left\lceil\frac{15}{130-45-(20+15)-45}\right\rceil, \tag{24}\] \[\max\{f_{1}^{2},f_{2}^{2}\}\leq \left\lceil\frac{15}{130-45-(20+15)-45}\right\rceil,\] \[\max\{f_{1}^{3},f_{2}^{3}\}\leq \left\lceil\frac{15}{130-45-(20+15)-45}\right\rceil,\] \[\max\{f_{3}^{1},f_{4}^{1}\}\leq \left\lceil\frac{15}{130-45-(20+15)-45}\right\rceil,\] \[\max\{f_{3}^{2},f_{4}^{2}\}\leq \left\lceil\frac{15}{130-45-(20+15)-45}\right\rceil,\] \[\max\{f_{3}^{3},f_{4}^{3}\}\leq \left\lceil\frac{15}{130-45-(20+15)-45}\right\rceil.\]

Steps 1.1 and 1.2 generate 780 flow vectors, which are shown in the second and the third column of Table 6.

Step 2. Generate the capacity vectors \(X=(x_{1},x_{2},\,\ldots,\,x_{10})\). (2.1) Transform each \(F\) into the capacity vector \(X=(x_{1},x_{2},x_{3},x_{4})\) before the logistic company via

\[ \begin{split} x_{1}=&\left\lceil f_{1}^{1}+f_{2}^{ 1}+f_{3}^{3}\right\rceil,\\ x_{2}=&\left\lceil f_{3}^{1}+f_{3}^{2}+f_{3}^{3} \right\rceil,\\ x_{3}=&\left\lceil f_{2}^{1}+f_{2}^{2}+f_{2}^{ 2}\right\rceil,\\ x_{4}=&\left\lceil f_{4}^{1}+f_{4}^{2}+f_{4}^{ 3}\right\rceil.\end{split}  \tag{25}\]

(2.2) Transform each \(F\) into the capacity vector \(X=(x_{5},x_{6},x_{7},x_{8},x_{9},x_{10})\) after the logistic company and use the maximal value for travel time via

\[ \begin{split} x_{5}=&\left\lceil\frac{15^{*}\max\{ f_{1}^{1},f_{2}^{1}\}}{130-(45+(20+15)+45)}\right\rceil,\\ x_{6}=&\left\lceil\frac{15^{*}\max\{f_{1}^{1},f_{2}^{ 2}\}}{130-(45+(20+15)+45)}\right\rceil,\\ x_{7}=&\left\lceil\frac{15^{*}\max\{f_{1}^{3},f_{2}^{ 3}\}}{130-(45+(20+15)+45)}\right\rceil,\\ x_{8}=&\left\lceil\frac{15^{*}\max\{f_{3}^{1},f_{4}^{ 1}\}}{130-(45+(20+15)+45)}\right\rceil,\\ x_{9}=&\left\lceil\frac{15^{*}\max\{f_{3}^{2},f_{4}^{ 2}\}}{130-(45+(20+15)+45)}\right\rceil,\\ x_{10}=&\left\lceil\frac{15^{*}\max\{f_{3}^{3},f_{4}^{ 3}\}}{130-(45+(20+15)+45)}\right\rceil.\end{split}  \tag{26}\]

This step transforms 712 flow vectors into capacity vectors \(X\) in \(\boldsymbol{\Omega}\). (2.3) Use the method in Table 1 to eliminate other non-MCVs in \(\boldsymbol{\Omega}\). This step obtains 47 capacity vectors in \(\boldsymbol{\Omega}_{\min}\), which are shown in the fourth column of Table 6.

Step 3. There are 47 MCVs that are shown in Table 6. The demand reliability \(R_{2,\,2,\,2}\) in an SCCNMT according to Table 4 using the RSDP can be obtained via

\[R_{(2,\,2,\,2)}=\Pr\left\{\left\lfloor\sum_{i=1}^{\sigma}\left[X\middle|X \geq X_{i}\right\rceil\right\}=0.6321\right. \tag{27}\]Step 1. Calculate the currently allocatable travel time \(I^{\prime}\) via Equation (28), by referring to Table 1 for the unloading and loading time. \[I=130-(20+2\times 15)=130-(20+30)=80. \tag{28}\]

Step 2. Find the travel time upper bound vectors. (2.1) Find the feasible travel time vectors \(V\) for each route using Equations (29) and (30),

\[ \begin{split} 35\leq v_{1}\leq 45,\\ 35\leq v_{2}\leq 45,\\ 27\leq v_{3}\leq 35,\\ 35\leq v_{5}\leq 45,\\ 20\leq v_{7}\leq 30,\\ 20\leq v_{8}\leq 30,\\ 30\leq v_{9}\leq 40,\\ 30\leq v_{10}\leq 40.\end{split}  \tag{29}\]

\begin{table}
\begin{tabular}{l l l l l} \hline Step 1 & Step 1.1 & Step 1.2 & Step 2 & Step 2.3 \\ Find Feasible Flow & Delete the Vectors & Delete the Vectors & Transform \(F\) to Delivery & Check if \(X\) is the MCV \\ Vectors \(F\) & Exceed \(M_{i}\) & Conserv



\[v_{1}+v_{5} \leq 80, \tag{30}\] \[v_{3}+v_{5} \leq 80,\] \[v_{2}+v_{8} \leq 80,\] \[v_{4}+v_{8} \leq 80,\] \[v_{1}+v_{6} \leq 80,\] \[v_{3}+v_{6} \leq 80,\] \[v_{2}+v_{9} \leq 80,\] \[v_{4}+v_{9} \leq 80,\] \[v_{1}+v_{7} \leq 80,\] \[v_{3}+v_{7} \leq 80,\] \[v_{3}+v_{5} \leq 80,\]0 \[v_{3}+v_{5} \leq 80,\]1

This step generates 8 travel time vectors. (2.2) Obtain travel time vectors \(V\) using step 2.1 and using the method in Table 2 to eliminate other non-travel time upper bound vectors. In this case, the number of travel time upper bound vectors is the same as the number of travel time vectors, i.e., 7. Step 3.