## 3 Scheduling Analysis
### Temporal Properties Under OBS  
Let \(\theta_{i}\) be the time for processing four wafers at Step \(i\). Under the steady state, to complete four wafers by a PM at Step \(n\), by OBS, with the robot-waiting activities being considered, task sequence \(\sigma_{9}=\langle U_{n}\to M_{n,n+1}\to L_{n+1}\to M_{n+1,n}\to A_{n2}\to U_{n} \to M_{n,n+1}\to L_{i+1}\to M_{n+1,n-1}\to A_{(n-1)1}\to U_{n-1}\to L_{n}\to M_{n -1,n}\to L_{n}\to M_{n,n-1}\to A_{(n-1)2}\to U_{n-1}\to M_{n-1,i}\to A_{n3} \to L_{n}\to L_{n}\to\) the internal chamber of the PM rotates \(\rightarrow\) wafer processing \(\rightarrow\) four wafers stay in the PM after processed (i.e., the delay time)\(\rangle\) should be performed. From Figure 3, we can see the task sequence of the robot between PM\({}_{1}\) and PM\({}_{2}\) and between PM\({}_{1}\) and LL as well. In the figure, the red arrows present the wafer movement direction; blue arrows indicate the next step in the process; blank circles represent empty LLs; purple/blue circles represent wafers. In the steps with a yellow circular background, the internal chamber of the PM rotates just before starting to process wafers. Note that different colored wafers are used to clarify the robot task sequence and do not represent different wafers. With \(m_{n}\) parallel PMs serving for Step \(n\), according to \(\sigma_{9}\), the time for processing four wafers at the step can be calculated using Equation (3):  
\[\theta_{n}=(\alpha_{n}+d_{n}+8\beta+7\mu+v+\omega_{n2}+\omega_{n3}+\omega_{(n- 1)1}+\omega_{(n-1)2})/m_{n} \tag{3}\]  
Furthermore, to complete four wafers by a PM of Step \(i\), \(i\in\{3,\ldots,n-1\}\) and \(n>2\), by OBS, with the robot-waiting activities being considered, task sequence \(\sigma_{10}=\langle U_{i}\to M_{i,i+1}\to L_{i+1}\to M_{i+1,i}\to A_{i2}\to U_{i} \to M_{i,i+1}\to A_{(i+1)3}\to L_{i+1}\to M_{i+1,i-1}\to A_{(i-1)1} \to U_{i-1}\to M_{i-1,i}\to L_{i}\to M_{i,i-1}\to M_{i-1,i}\to M_{i-1,i} \to M_{i-1,i}\to M_{i-1,i}\to A_{i3}\to L_{i}\to\) wafer processing \(\rightarrow\) four wafers stay in the PM after processed (i.e., the delay time)\(\rangle\) should be performed. With \(m_{i}\) parallel PMs serving for Step \(i\), according to \(\sigma_{10}\), the time for processing four wafers at the step can be calculated using Equation (4):  
\[\theta_{i}=(\alpha_{i}+d_{i}+8\beta+7\mu+v+\omega_{(i+1)3}+\omega_{i2}+\omega _{i3}+\omega_{(i-1)1}+\omega_{(i-1)2})/m_{i},\,i\in\{3,\ldots,n-1\}\text{ and }n>2 \tag{4}\]  
Similarly, to complete four wafers by a PM of Step 1, by OBS, with the robot-waiting activities being considered, task sequence \(\sigma_{11}=\langle U_{1}\to M_{1,2}\to L_{2}\to M_{2,1}\to A_{12}\to U_{1} \to M_{1,2}\to A_{23}\to L_{2}\to M_{2,0}\to U_{0} \to M_{0,1}\to L_{1}\to M_{1,0}\to U_{0} \to M_{0,1}\to A_{13}\to L_{1}\rightarrow\) the internal chamber of the PM rotates \(\rightarrow\) wafer processing \(\rightarrow\) four wafers stay in the PM after processed (i.e., the delay time)\(\rangle\) should be performed. With \(m_{1}\) parallel PMs serving for Step 1, according to \(\sigma_{11}\), the time for processing four wafers at the step can be calculated with Equation (5):  
\[\theta_{1}=(\alpha_{1}+d_{1}+2\beta_{0}+6\beta+7\mu+v+\omega_{23}+\omega_{12} +\omega_{13})/m_{1} \tag{5}\]In \(\sigma_{9}\), \(\sigma_{10}\), and \(\sigma_{11}\), \(A_{i2}\), \(i\in\mathbf{N}_{n}\backslash\{0\}\) is caused by the rotation of the internal chamber of a PM for Step \(i\). Notice that after the robot performs an unloading task at the PM in \(\sigma_{9}\), \(\sigma_{10}\), or \(\sigma_{11}\) for the first time, the internal chamber of the PM starts to rotate. At the same time, the robot performs task sequence \(\langle M_{i,j+1}\to L_{i+1}\to M_{i+1,j}\rangle\). Then, \(A_{i2}\) should be performed, and the time taken for \(A_{i2}\) is determined by the longest one between the time taken for a rotation operation of the PM and \(\langle M_{i,j+1}\to L_{i+1}\to M_{i+1,j}\rangle\).