## 3 Scheduling Analysis
### Temporal Properties Under TBS
Thus, we have the following Equations:  
\[\omega_{i2}=max(v-[(2n-1)\mu+(2n-3)\beta+\beta_{0}+\sum_{1}^{i-2}\omega_{i1}+ \sum_{i+1}^{n}\omega_{i2}],0),\ i\in\{3,4,\ldots,n-1\}\ \text{and}\ n>3 \tag{24}\]  
\[\omega_{22}=max(v-[(2n-1)\mu+(2n-3)\beta+\beta_{0}+\sum_{3}^{n}\omega_{i2}],0),n>2 \tag{25}\]\[\omega_{22}=max(v-(3\mu+\beta+\beta_{0}),0),n=2 \tag{26}\]  
When \(i=n\) and \(n>2\), the robot performs task sequence \(\langle M_{n,n-2}\to A_{(n-2)1}\to U_{n-2}\)\(\to M_{n-2,n-1}\to L_{n-1}\to M_{n-1,n-3}\rightarrow\ldots\to M_{2,0}\to U_{0}\to M_{0,1}\to L_{1}\to M_{1,n}\rangle\). Then, \(A_{n2}\) should be performed. And the time taken for \(A_{n2}\) is determined by the longest one between the time taken for a rotation operation of the PM and \(\langle M_{n,n-2}\to A_{(n-2)1}\to U_{n-2}\)\(\to M_{n-2,n-1}\to L_{n-1}\to M_{n-1,n-3}\rightarrow\ldots\to M_{2,0}\to U_{0}\to M_{0,1}\to L_{1}\to M_{1,n}\rangle\). Then, we can obtain \(\omega_{n2}\) as follows:  
\[\omega_{n2}=max(v-[(2n-1)\mu+(2n-3)\beta+\beta_{0}+\sum_{1}^{n-2}\omega_{1}],0 ),n>2 \tag{27}\]  
Similarly, we have  
\[\omega_{12}=max(v-\Big{[}(2n-1)\mu+(2n-2)\beta+\sum_{2}^{n}\omega_{12}\Big{]},0 ). \tag{28}\]  
According to \(\sigma_{14}\), the time taken for task sequence \(\langle M_{i,i-2}\to A_{(i-2)1}\to U_{i-2}\to M_{i-2,i-1}\to L_{i-1}\to M_{i-1,i-3} \rightarrow\ldots\to M_{2,0}\to U_{0}\to M_{0,1}\to L_{1}\to M_{1,n} \to A_{n2}\to U_{n}\to M_{n,n+1}\to L_{n+1}\to M_{n+1,n-1} \rightarrow\ldots\to M_{i+1,i-1}\to A_{(i-12)}\to U_{i-1} \to M_{i-1,i}\to L_{i}\rightarrow\) the internal chamber of the PM rotates \(\rightarrow\) wafer processing \(\rightarrow\) four wafers stay in the PM  
Figure 5: Description of robot movements under the TBS strategy. after processed (i.e., the delay time)) equals the wafer sojourn time of \(W_{i,12}\), \(i\in\mathbf{N}_{n}\backslash\mathbf{N}_{2}\). Furthermore, the time taken for the task sequence \(\langle\)the internal chamber of the PM rotates \(\rightarrow\) wafer processing \(\rightarrow\) four wafers stay in the PM after processed (i.e., the delay time) \(\rightarrow\)\(U_{i}\to M_{i,i+1}\to L_{i+1}\to M_{i+1,i-1}\to A_{(i-1)1} \to U_{i-1}\to M_{i-1,i}\to L_{i}\to M_{i,i-2} \rightarrow\ldots\to M_{2,0}\to U_{0}\to M_{0,1} \to L_{1}\to M_{1,n}\to A_{n2}\to U_{n} \to M_{n,n+1}\to L_{n+1}\to M_{n+1,n-1} \rightarrow\ldots\to M_{i+2,i}\to A_{i2}\rangle\) equals the wafer sojourn time of \(W_{i,34}\), \(i\in\mathbf{N}_{n}\backslash\mathbf{N}_{1}\). Hence, we have the following:  
\[\tau_{i,12}=\alpha_{i}+d_{i}+v+(2n+1)\beta+\beta_{0}+(2n+2)\mu+\sum_{1}^{i-2} \omega_{i1}+\sum_{i-1}^{n}\omega_{i2},\ i\in\mathbf{N}_{n}\backslash\mathbf{N }_{2},\mu>2 \tag{29}\]  
\[\theta_{i}=\Big{[}\alpha_{i}+d_{i}+(2n+5)\beta+\beta_{0}+(2n+5)\mu+v+\sum_{1} ^{i-1}\omega_{i1}+\sum_{i-1}^{n}\omega_{i2}\Big{]}/m_{i},i\in\mathbf{N}_{n} \backslash\mathbf{N}_{1} \tag{22}\]0  
For Step 2, the time taken for task sequence \(\langle M_{2,0}\to U_{0}\to M_{0,1}\to L_{1}\to M_{1,n}\rightarrow\ldots \to M_{i+1,i-1}\to A_{(i-1)2}\to U_{i-1} \to M_{i-1,i}\to L_{i}\to M_{i,i-1}\rightarrow\ldots \to M_{3,1}\to A_{12}\to U_{1} \to M_{1,2}\)\(\to L_{2}\rightarrow\) the internal chamber of the PM rotates \(\rightarrow\) wafer processing \(\rightarrow\) four wafers stay in the PM after processed (i.e., the delay time)\(\rangle\) equals the wafer sojourn time of \(W_{2,12}\) that can be calculated with Equation (31), and the wafer sojourn time of \(W_{2,34}\) is the same as that of \(W_{2,12}\).