## 3 Scheduling Analysis
### Linear Programs for Solution  
By any one of OBS, OHTS, TBS, and THTS, if an SF\({}^{3}\)-CT operates under the steady state, in a robot task cycle, each step completes four wafers, and four completed wafers are returned to LLs. Also, the time taken to complete four wafers at each step equals the robot cycle time. Thus, we have Propositions 1-4:  
**Proposition 1**. : _For an SF\({}^{3}\)-CT, if the tool is operated by OBS under the steady state, the robot cycle time equals the time for completing four wafers at each step, i.e., Equations (50)-(52) hold:_  
\[\psi_{1}=\theta_{1} \tag{50}\]  
\[\psi_{1}=\theta_{i}\text{, }i\in\mathbf{N}_{n}\backslash\mathbf{N}_{1}\text{ and }n>2 \tag{51}\]  
\[\psi_{1}=\theta_{n} \tag{52}\]  
_where \(\theta_{n}\), \(\theta_{i}\), \(i\in\mathbf{N}_{n}\backslash\mathbf{N}_{1}\) and \(n>2\), and \(\theta_{1}\) are obtained by (3)-(5), respectively._  
**Proposition 2**. : _For an SF\({}^{3}\)-CT, if the tool is operated by OHTS under the steady state, the robot cycle time equals the time for completing four wafers at each step, i.e., Equations (53)-(56) hold:_  
\[\psi_{2}=\theta_{1} \tag{53}\]  
\[\psi_{2}=\theta_{2}\text{, }n\geq 2 \tag{54}\]  
\[\psi_{2}=\theta_{i}\text{, }i\in\{2\text{, }3\text{, }\ldots\text{, }n-1\}\text{ and }n>3 \tag{55}\]  
\[\psi_{2}=\theta_{n} \tag{56}\]  
_where \(\theta_{2}\), \(n>2\); \(\theta_{2}\), \(n=2\); \(\theta_{1}\); \(\theta_{i}\), \(i\in\{2\text{, }3\text{, }\ldots\text{, }n-1\}\) and \(n>3\); and \(\theta_{n}\) are obtained by (14)-(16), (4), and (3), respectively._  
**Proposition 3**. : _For an SF\({}^{3}\)-CT, if the tool is operated by TBS under the steady state, the robot cycle time equals the time for completing four wafers at each step, i.e., Equation (57) holds:_  
\[\psi_{3}=\theta_{1}=\theta_{i} \tag{57}\]  
_where \(\theta_{1}\) and \(\theta_{i}\) are obtained by (23) and (22), respectively._  
**Proposition 4**. : _For an SF\({}^{3}\)-CT, if the tool is operated by THTS under the steady state, the robot cycle time equals the time for completing four wafers at each step, i.e., Equations (58)-(60) hold:_  
\[\psi_{4}=\theta_{1}=\theta_{2}\text{, }n=2 \tag{58}\]  
\[\psi_{4}=\theta_{1}=\theta_{2}\text{, }n>2 \tag{59}\]  
\[\psi_{1}=\theta_{i}\text{, }i\in\mathbf{N}_{n}\backslash\mathbf{N}_{1}\text{ and }n>2 \tag{51}\]0  
_where \(\theta_{i}\), \(i\in\mathbf{N}_{n}\backslash\mathbf{N}_{2}\) and \(n>2\); \(\theta_{2}\), \(n>2\); \(\theta_{2}\), \(n=2\); \(\theta_{1}\), \(n=2\); and \(\theta_{1}\), \(n>2\), are obtained by (34)-(38), respectively._  
Also to meet RTCs, (1) and (2) should be held when any strategy is applied. Note that \(A_{i2}\) and \(A_{i3}\) occur during the processes for unloading wafers from a PM and loading wafers into a PM, respectively. Thus, in terms of the satisfaction of RTCs, it is desired that and \(\omega_{i3}\) (the time taken by \(A_{i2}\) and \(A_{i3}\)) should be as short as possible. Hence, with this observation in mind, according to the above analysis, \(\omega_{i2}\) and \(\omega_{i3}\) are already determined. Due to that, for a steady-state schedule, the cycle time for each processing step is the same, and it should be equal to the robot cycle time; when the robot-waiting time at all the steps is determined, a schedule is obtained. Note that the robot-waiting time is a continuous variable. The problem becomes a continuous optimization problem but not a discrete optimization problem. Hence, given a strategy, to find a schedule is to determine \(\omega_{i1}\) (the time taken by \(A_{i1}\)). Based on Propositions 1-4, we establish four linear programs for different strategies to find optimal and feasible schedules with the objective of minimizing the robot cycle time, i.e., the system cycle time. Based on the property of a schedule, when a schedule is found, the robot's time occupancy is known. **Linear Program 1 (LP1).