## 3 Scheduling Analysis
### Temporal Properties Under THTS
When \(n=2\), for Step 1, to complete four wafers by a PM, the task sequence \(\sigma_{20}=\langle U_{1}\rightarrow\) the robot rotates \(\to L_{1}\to M_{1,2}\to L_{2}\rightarrow\) the internal chamber of the PM rotates \(\to L_{2}\to M_{2,0}\to U_{0}\rightarrow\) the robot rotates \(\to L_{0}\to M_{0,1}\to A_{12}\to U_{1}\)\(\rightarrow\) the robot rotates \(\to L_{1}\rightarrow\) 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 \(\sigma_{19}\) and \(\sigma_{20}\), the time for processing four wafers at Step 1 can be calculated with Equations (37) and (38):  
\[\theta_{1}=\left[\alpha_{1}+d_{1}+(2n+3)\beta+\beta_{0}+(2n+3)\mu+\upsilon+ \sum_{1}^{n}\omega_{i2}]/m_{1},n>2\right. \tag{37}\]  
\[\theta_{1}=(\alpha_{1}+d_{1}+7\beta+\beta_{0}+6\mu+2\upsilon+\omega_{12})/m_{ 1},n=2 \tag{38}\]  
By observing \(\sigma_{16}\), the time taken for the robot-waiting activity \(A_{i2}\), \(i\in\{3,\ldots,n-1\}\) and \(n>3\), is determined by the longest one between the time taken for a rotation operation of the PM and \(\langle M_{i,i-2}\to A_{(i-2)1}\to U_{i-2}\to M_{i-2,i-1}\to L_{i-1}\to M_{i-1,i-3} \rightarrow\ldots\to M_{3,0}\to U_{0}\to M_{0,1}\to A_{11} \to U_{1}\)\(\rightarrow\) the robot rotates \(\to L_{1}\to M_{1,2}\to L_{2}\to M_{2,n}\to A_{n2}\to U_{n} \to M_{n,n+1}\to L_{n+1,n-1}\rightarrow\ldots\to M_{i+2,i}\rangle\). When \(n>2\), \(A_{i2}\) is determined by the longest one between the time taken for a rotation operation of the PM and \(\langle M_{n,n-2}\to A_{(n-2)1}\to U_{n-2}\to M_{n-2,n-1}\to L_{n-1}\to M_{n-1,n-3} \rightarrow\ldots\to M_{3,0}\to U_{0}\to M_{0,1}\to A_{11} \to U_{1}\)\(\rightarrow\) the robot rotates \(\to L_{1}\to M_{1,2}\to L_{2}\to M_{2,n}\)). Thus, \(\omega_{i2}\), \(i\in\{3,\ldots,n-1\}\) and \(n>3\), and \(\omega_{n2}\), \(n>2\) can be obtained with Equations (39) and (40):  
\[\omega_{i2}=max(\upsilon-[(2n-1)\mu+(2n-3)\beta+\beta_{0}+\sum_{1}^{i-2}\omega_{i1 }+\sum_{i+1}^{n}\omega_{i2}],0),i\in\{3,4,\ldots,n-1\}\text{ and }n>3 \tag{39}\]  
\[\omega_{n2}=max(\upsilon-[(2n-1)\mu+(2n-3)\beta+\beta_{0}+\sum_{1}^{n-2}\omega_{i1 }],0),n>2 \tag{40}\]Observing \(\sigma_{17}\) and \(\sigma_{18}\), after the robot performs a loading task at Step 2 for the first time so as to complete a robot rotation operation, the internal chamber of the PM starts to rotate. Hence, the time taken for the robot-waiting activity \(A_{22}\), \(n>2\), is determined by the longest one between the time taken for a rotation operation of the PM and \(\langle M_{2,n}\to A_{n2}\)\(\rightarrow\)\(U_{n}\)\(\rightarrow\)\(M_{n,n+1}\)\(\rightarrow\)\(L_{n+1}\)\(\rightarrow\)\(M_{n+1,n-1}\)\(\rightarrow\)\(\ldots\)\(\rightarrow\)\(M_{5,3}\)\(\rightarrow\)\(A_{32}\)\(\rightarrow\)\(U_{3}\)\(\rightarrow\)\(L_{4}\)\(\rightarrow\)\(M_{4,2}\)). When \(n=2\), the robot just needs to wait for the completion of the internal rotation of the PM. Therefore, \(\omega_{22}\) should be calculated using Equations (41) and (42):  
\[\omega_{22}=\max(v-[(2n-3)\mu+2(n-2)\beta+\sum_{3}^{n}\omega_{2}],0),n>2 \tag{41}\]  
\[\omega_{22}=v,\,n=2 \tag{42}\]  
Observing \(\sigma_{19}\) and \(\sigma_{20}\), after the robot performs a loading task at Step 1 for the first time so as to complete a robot rotation operation, the internal chamber of the PM starts to rotate.