## 3 Network Model
### Modulation Format Conversion  
As described in Section 2, the use of MFC to divide the routing path of a traffic request into multiple MFSs can break through the limitation of the longest length segment on spectrum efficiency in traditional optical networks and further achieve efficient utilization of network resources. We assume that MFC is implemented by connecting different transceivers in a back-to-back configuration [54], and the nodes capable of performing MFC are referred to as CNs. Different selections of CNs and MFs lead to different results of network resource occupancy, which is illustrated by an example shown in Figure 3. The complexity of deploying transceivers is not considered here, which will be put into a later study. Each traffic request is defined by (source node, destination node, and traffic rate). In Figure 3, suppose the distances between nodes 1, 2, 3, and 4 are 240 km, 1500 km, and 460 km, respectively. The source and destination nodes of traffic request A (1, 4, 6) are node 1 and node 4, respectively. The traffic rate is 6 Tb/s, and the routing path is 1\(\rightarrow\)2\(\rightarrow\)3\(\rightarrow\)4. Four options to choose CNs on the routing path are discussed. * Option 1: no CN is set between node 1 and node 4, i.e., no MFC is performed on the entire routing path. The MF is selected based on the distance from node 1 to node 4, which is 2200 km in Figure 3. From Table 2, BPSK (supporting 100 Gb/s) should be selected as the MF, so \(60\times 3\) SBs are required to carry A on all links, implying that \(2\times 3\) SLs are occupied;
* Option 2: node 2 is set as the CN, and MFC is performed at node 2. The distance from node 1 to node 2 is 240 km, so DP-16 QAM (supporting 800 Gb/s) is used as  
\begin{table}
\begin{tabular}{c c c} \hline \hline
**Modulation Format** & **Supported Traffic Rate** & **Achievable Transmission** \\  & **(Gb/s)** & **Distance (km)** \\ \hline BPSK & 100 & 4000 \\ QPSK & 200 & 2000 \\ DP-QPSK & 400 & 1000 \\ DP-8 QAM & 600 & 500 \\ DP-16 QAM & 800 & 250 \\ DP-32 QAM & 1000 & 125 \\ \hline
**2 Tb/s** & **240 km** & **960 km** \\
**Case 1** & **20.5 Bs** & **20.5 Bs** \\
**Case 2** & **10.5 Bs** & **10  
the MF, occupying 1 SL (8 SBs). QPSK (supporting 200 Gb/s) is selected because the distance from node 2 to node 4 is 1960 km, and 1 SL (30 SBs) is occupied. Therefore, \(1+1\times 2\) SLs (\(8+30\times 2\) SBs) are required to carry A on the routing path;
* Option 3: node 3 is set as the CN, and MFC is performed at node 3. The distance from node 1 to node 3 is 1740 km, which determines the selection of QPSK (supporting 200 Gb/s) as MF, so 1 SL (30 SBs) is occupied for carrying A. For the distance of 460 km from node 3 to node 4, the DP-8 QAM (supporting 600 Gb/s) is selected and 1 SL (10 SBs) is required on the link. As a result, \(1\times 2+1\) SLs (\(30\times 2+10\) SBs) on the routing path are occupied in Option 3;
* Option 4: both node 2 and node 3 are set as CNs, and MFC is available at both nodes. The distance from node 1 to node 2 is 240 km, so DP-16 QAM (supporting 800 Gb/s) is selected and 1 SL (8 SBs) is occupied. 1 SL (30 SBs) is occupied when QPSK (supporting 200 Gb/s) is selected according to the distance of 1500 km from node 2 to node 3. Moreover, 1 SL (30 SBs) is occupied on the 460 km link between node 3 and node 4 with the adoption of DP-8 QAM (supporting 600 Gb/s). Therefore, there are \(1+1+1\) SLs (\(8+30+10\) SBs) occupied on the routing path when both node 2 and node 3 are selected as CNs. This example illustrates all possible options for setting CNs on the routing path of A. No CN is set in option 1, one of the intermediate nodes on the routing path is set as a CN in option 2 and option 3, and both intermediate nodes on the routing path are set as CNs in option 4. It is worth noting that CNs are set in the last three options to achieve MFC, and 6, 3, 3, and 3 SLs (180, 68, 70, and 48 SBs) are occupied corresponding to these four different options to carry traffic request A.