Petri net-based scheduling of time constrained single-arm cluster tools with wafer revisiting

Finite capacity PN

Scheduling a cluster tool is to effectively allocate its limited resources to tasks. To do so, this work adopts the resource-oriented PN (ROPN)^39–42 to model the system. It is a finite capacity PN and its basic concept is based on Wu and Zhou ⁴² and Murata. ⁴³ It is denoted as PN = (P, T, I, O, M, K), where P and T are finite sets of places and transitions with P∩T = ∅ and P∪T ≠ ∅; I: P × T → ℕ = {0, 1, 2, …} and O: P × T → ℕ describe the relation from P to T and T to P, respectively; M: P → ℕ is a marking with M(p) being the number of tokens in place p and M₀ being the initial marking; K: P → {1, 2, 3, …} is a capacity function with K(p) representing the maximum number of tokens that can be held by p at a time.

Let ^•t = {p: p ∈ P and I(p, t) > 0} be the pre-set of transition t and t^• = {p: p ∈ P and O(p, t) > 0} be its post-set. Similarly, p’s pre-set ^•p = {t ∈ T: O(p, t) > 0} and post-set p^• = {t ∈ T: I(p, t) > 0}. Then, the transition enabling and firing rules are defined as follows.

Definition 2.1

A transition t ∈ T in a finite capacity PN is said to be enabled if the following conditions are satisfied

M ( p ) ≥ I ( p , t ) , ∀ p ∈ P

(1)

and

K ( p ) ≥ M ( p ) − I ( p , t ) + O ( p , t ) , ∀ p ∈ P

(2)

By Definition 2.1, when Condition (1) is met, t is said to be process-enabled, and while Condition (2) is met, t is said to be resource-enabled. Thus, t is enabled if it is both process- and resource-enabled.

An enabled t ∈ T at marking M can fire. The firing of t transforms the PN from M to M′ according to

M ′ ( p ) = M ( p ) − I ( p , t ) + O ( p , t ) , ∀ p ∈ P

(3)

Modeling activity sequences

In wafer fabrication, ALD is a typical revisiting process and, as done in Wu et al., ²⁹ this work focuses on the ALD process. In an ALD, a wafer visits Step 1 first and then it goes to Steps 2 and 3 sequentially for processing. After being processed by Step 3, it goes back to Steps 2 and 3. This process is repeated for several times. To control the quality, every time the wafer visits Steps 2 and 3, the exactly same processing environment is required. To ensure the processing environment requirement, each step is configured with only one PM. We assume that PM _i is configured for Step i. Thus, as presented in Wu et al., ²⁹ the wafer flow pattern for this process can be denoted as (PM₁, (PM₂, PM₃) ^h , PM₄) with (PM₂, PM₃) ^h being the revisiting process and h ≥ 2. For the simplicity of presentation, this work focuses on the case when h = 2. Note that the obtained results with h = 2 can be extended to cases with h > 2. Then, a single-arm cluster tool with both wafer residency time constraints and wafer revisiting is modeled by an ROPN as follows.

Let ℕ _n  = {1, 2, 3,…, n} and Ω _n  = ℕ _n ∪{0}. As shown in Figure 2, timed place p_n, n ∈ ℕ₄, models the PM for Step n with K(p_n) = 1 representing that there is only one PM. The loadlocks are treated as a processing step with processing time being zero. They are modeled by p₀ with K(p₀) = ∞ representing that the loadlocks can hold all wafers in a tool. Timed place q_n2 with K(q_n2) = 1, n ∈ Ω₄, models the robot waiting before unloading a wafer from p_n. Non-timed places q_n1 and q_n3 with K(q_n1) = K(q_n3) = 1, n ∈ Ω₄, are used to connect two neighboring steps. Place r with K(r) = 1 models the robot (R). When M(r) = 1, the robot is idle and ready for performing a task.

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Figure 2.

PN model for the system with wafer flow pattern (PM₁, (PM₂, PM₃) ^h , PM₄).

With the resources in the system being modeled, transitions are used to model the material flows. Timed transition t_ij, i, j ∈ Ω₄, models the activity for the robot moving from p_i to p_j with a wafer being carried. Timed transition y_ij, i, j ∈ Ω₄, models the activity for the robot moving from p_i to p_j without carrying a wafer. Transitions s_n1 and s_n2 represent the loading and unloading of a wafer into and from p_n, n ∈ Ω₄, respectively. By doing so, the structure of the model is formed.

With the developed PN structure, by putting a V-token representing a virtual wafer, the initial marking M₀ is set as follows. Set M₀(p_i) = 1, i ∈ ℕ₄\{1}, M₀(p₁) = 0, and M₀(p₀) = n to indicate that there are always wafers in the loadlocks to be processed; M₀(r) = 0; M₀(q_ij) = 0, i ∈ N₄ and j ∈ Ω₃, and M₀(q₀₂) = 1, indicating that the robot is waiting at the loadlocks for unloading a wafer there.

With wafer revisiting process, there are two ways for a token in q₃₃ to go, one is to q₂₁ by firing t₃₂ and the other is to q₄₁ by firing t₃₄, leading to a conflict. To eliminate such a conflict, colors are introduced to the model to form a colored PN as follows.

Task time modeling

In order to schedule the process, it is necessary to model the time taken for performing each activity. From the PN model, both transitions and places represent tasks that take time. As done in Lee and Lee ²⁸ and Wu et al., ²⁹ we assume that the time taken for each activity is deterministic and known. The time taken for the robot to move from Steps i to j, i ≠ j, with or without holding a wafer is assumed to be same ¹⁴ and denoted by µ, that is, it takes µ time units to fire t_ij and y_ij. Note that in the revisiting process, after loading a wafer into p₃/p₂, the robot waits there for the completion of the wafer and it does nothing and takes no time. Similarly, the time taken for the robot to unload (firing s_i2)/load (firing s_i1), a wafer from/into a step is assumed to be the same as well and it is denoted by λ. The time taken for processing a wafer in p_i, i ∈ Ω₄, is α_i with α_i > 0, i ≠ 0, and α₀ = 0.

Let δ_i be the longest time for which a wafer can stay in a PM at Step i without being scrapped after being processed. Then, with wafer residency time constraints, the wafer sojourn time at Step i should be within [α_i, α_i+δ_i], that is, a token should stay in PM _i for at least α_i time units and no more than α_i+δ_i time units. We use τ₁ and τ₄ to denote the wafer sojourn time at Steps 1 and 4, respectively; and τ_2j and τ_3j, j ∈ N₂ to denote the wafer sojourn time for the jth visiting Steps 2 and 3, respectively. Note that no wafer residency time constraint is imposed on Step 0. The symbols and explanation are summarized in Table 1.

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Table 1.

Time durations associated with transitions and places.

Symbol	Transition or place	Actions	Allowed time duration
λ	si 1  ∈ T	Robot loads a wafer into Step i, i ∈ Ω 4	λ
λ	si 2  ∈ T	Robot unloads a wafer from Step i, i ∈ Ω4	λ
µ	tij ∈ T	Robot moves from Steps i to j with a wafer carried, i, j ∈ Ω4	µ
µ	yij ∈ T	Robot moves from Steps i to j without carrying a wafer, i, j ∈ Ω4	µ, i ≠ j; otherwise, 0
τ 1 and τ4	p 1 and p4 ∈ P	A wafer is being processed and waiting in p1 and p4	τ 1  ∈ [α1, α1+δ1] τ4 ∈ [α4, α4+δ4]
τ 2j and τ3j	p 2 and p3 ∈ P	A wafer being processed and waiting in p2 and p3 for the jth time, j ∈ ℕ2	τ 2 j ∈ [α2, α2+δ2] and τ3j ∈ [α3, α3+δ3], j ∈ ℕ2
ωi	qi 2  ∈ P	Robot waits before unloading a wafer from PM i , i ∈ Ω4	≥0
ωi	qi 3  ∈ P	Robot waits after unloading a wafer from PM i , i ∈ Ω4	0
ωi	qi 1  ∈ P	Robot waits before loading a wafer from PM i , i ∈ Ω4	0

From the above analysis, for a single-arm cluster tool with wafer revisiting and residency time constraints, according to Wu et al.^24,29 we have the following schedulability definition.

P-backward scheduling

To make the scheduling analysis, we need to present the p-backward scheduling strategy for a single-arm cluster tool with wafer revisiting. Based on the PN model, we show how a p-backward strategy works. Since a single-arm cluster tool with wafer revisiting is deadlock-prone, to avoid deadlock, the system must start from a proper state. We assume that the system starts from marking M₁ = ({null}, {W₃(1)}, {W₂(2)}, {W₁(1)}) and, at the same time, the robot is idle, or M₁(r) = 1. Note that this marking is consistence with M₀ that is set in the last section and can be reached by applying a p-backward strategy. By starting from M₁ with a p-backward strategy being applied, the system evolves as follows.

Note that marking M₈ is equivalent to M₁ in the sense of dynamic behavior of PN. Hence, by transforming M₁ to M₈, a cycle is completed. This is what a p-backward strategy does. With the PN model, it is easy to understand the scheduling strategy.

Cycle time analysis

With the above discussion, we can analyze the cycle time for the system. Let φ_j be the time taken for the robot to perform σ_j, j ∈ ℕ₇. Then, from the time modeling, we have φ₁ = µ+ω₀+λ+µ+λ = 2µ+2λ+ω₀. Similarly, for σ_2–7, we have φ₂ = 2µ+2λ+ω₄; φ₃ = 2µ+2λ+ω₃; φ₄ = 2µ+2λ+ω₂; φ₅ = µ+2λ+α₃; φ₆ = µ+2λ+α₂; and φ₇ =2µ+2λ+ω₁.

Since σ_1–7 form a robot cycle, the robot cycle time is as follows

η = ∑ i = 1 7 φ i = ( 14 λ + 12 μ + α 2 + α 3 ) + ∑ i = 0 4 ω i = η 1 + η 2

(4)

where η₁ = 14λ+12µ+α₂+α₃ is the robot task time in a cycle and η 2 = ∑ i = 0 4 ω i is the robot waiting time in a cycle. Note that η₁ is constant and known in advance and ω_0–4 in η₂ should be determined by a schedule.

Based on the fact that the cycle time of the entire system must be equal to the robot cycle time, ²⁴ with the robot cycle time obtained, we can analyze the wafer sojourn time at each step. It follows from the above analysis that the wafer sojourn time at Step 1 is equal to the robot cycle time minus the time taken for performing σ₁ and the time for firing s₁₂, t₁₂, and s₂₁ in σ₇, that is, we have

τ 1 = η − ( 2 μ + 2 λ + ω 0 ) − ( μ + 2 λ ) = η − 3 μ − 4 λ − ω 0

(5)

Similarly, for Step 4, we have

τ 4 = η − 3 μ − 4 λ − ω 3

(6)

For Step 2, the sojourn time taken for a wafer to visit the step for the first time is different from that for a wafer to visit the step for the second time. We use τ₂₁ and τ₂₂ to denote them, respectively. It follows from the above transition firing analysis that τ₂₁ is equal to the robot cycle time minus the time taken for σ_5–7 and the time taken for firing s₂₂, t₂₃, and s₃₁ in σ₄. Thus, we have

τ 21 = η − ( φ 5 + φ 6 + φ 7 ) − ( λ + μ + λ ) = η − ( μ + 2 λ + α 3 ) − ( μ + 2 λ + α 2 ) − ( 2 μ + 2 λ + ω 1 ) − ( 2 λ + μ ) = η − 5 μ − 8 λ − α 3 − α 2 − ω 1

(7)

When a wafer visits Step 2 for the second time, the robot loads it into PM₂ and waits here for its completion. Thus, we have

τ 22 = α 2

(8)

Similarly, we can calculate the wafer sojourn time for Step 3 and we have

τ 31 = α 3

(9)

τ 32 = η − 5 μ − 8 λ − α 3 − α 2 − ω 2

(10)

With wafer residency time constraints, to make a schedule feasible, the cycle time for each step should be in a permissive range. Thus, we analyze such a range for each step as follows. For Step 1, it follows from the above analysis that to complete a wafer requires three robot moving tasks between steps (t₀₁, t₁₂, and y₂₁), two wafer unloading tasks (s₀₂ and s₁₂), and two wafer loading tasks (s₁₁ and s₂₁). Thus, with the wafer staying time in Step 1, the time taken for this process is as follows

θ 1 = τ 1 + 3 μ + 4 λ + ω 0

(11)

To be feasible, we have τ₁ ∈ [α₁, α₁+δ₁]. Hence, the permissive shortest cycle time at Step 1 is as follows

θ 1 S = α 1 + 3 μ + 4 λ + ω 0

(12)

and the permissive longest cycle time at Step 1 is as follows

θ 1 L = α 1 + δ 1 + 3 μ + 4 λ + ω 0

(13)

Similarly, the cycle time for completing a wafer at Step 4 is as follows

θ 4 = τ 4 + 3 μ + 4 λ + ω 3

(14)

The permissive shortest cycle at Step 4 is as follows

θ 4 S = α 4 + 3 μ + 4 λ + ω 3

(15)

and the permissive longest cycle time at Step 4 is as follows

θ 4 L = α 4 + δ 4 + 3 μ + 4 λ + ω 3

(16)

For Step 2 in the revisiting process, we analyze the wafer cycle as follows. By firing s₂₁ (σ₈ = 〈s₂₁ (λ)〉), a wafer is loaded into Step 2 for processing for the first time, which is followed by transition firing sequence σ₉ = 〈y₂₀ → s₀₂ → t₀₁ → s₁₁ → y₁₄ → s₄₂ → t₄₀ → s₀₁ → y₀₃ → s₃₂ → t₃₄ → s₄₁ → y₄₂〉. During the time for performing σ₉, the wafer just loaded into Step 2 is being processed. Hence, the time taken for performing σ₉ is the wafer sojourn time for the first visiting Step 2, that is, τ₂₁. Then, transition firing sequence σ₁₀ = 〈t₂₃ (µ) → s₃₁ (λ) → y₃₃ (0) → waiting for the completion at Step 3 (α₃) → s₃₂ (λ) → t₃₂ (µ) → s₂₁ (λ) (loading the wafer into Step 2 to be processed for the second time) → y₂₂ (0) → waiting for the completion of the wafer at Step 2 (τ₂₂) → s₂₂ (λ) → t₂₃ (µ) → s₃₁ (λ) → y₃₁ (µ) → waiting in q₁₂ (ω₁) → s₁₂ (λ) → t₁₂ (µ)〉 is performed. By performing σ_8–10, a cycle is completed at Step 2. Thus, the time taken for completing a wafer at Step 2 is as follows

θ 2 = τ 21 + τ 22 + α 3 + 5 μ + 8 λ + ω 1

(17)

With τ₂₁ ∈ [α₂, α₂+δ₂] and τ₂₂ = α₂, the permissive shortest cycle at Step 2 is as follows

θ 2 S = α 2 + α 2 + α 3 + 5 μ + 8 λ + ω 1 = 2 α 2 + α 3 + 5 μ + 8 λ + ω 1

(18)

and the permissive longest cycle time at Step 2 is as follows

θ 2 L = α 2 + δ 2 + α 2 + α 3 + 5 μ + 8 λ + ω 1 = 2 α 2 + δ 2 + α 3 + 5 μ + 8 λ + ω 1

(19)

For Step 3, similar to Step 2, we analyze the process as follows. We have the following transition firing sequence σ₁₁ = 〈s₃₁ (a wafer is loaded into Step 3 to be processed for the first time, λ) → y₃₃ (0) → waiting for the completion of the wafer at Step 3 (τ₃₁) → s₃₂ (λ) → t₃₂ (µ) → s₂₁ (λ) → y₂₂ (0) → waiting for the completion of the wafer at Step 2 (α₂) → s₂₂ (λ) → t₂₃ (µ) → s₃₁ (λ)〉. Then, transition firing sequence σ₁₂ = 〈y₃₁ → s₁₂ → t₁₂ → s₂₁ → y₂₀ → s₀₂ → t₀₁ → s₁₁ → y₁₄ → s₄₂ → t₄₀ → s₀₁ → y₀₃〉 is performed. During the time for performing σ₁₂, a wafer for its second visiting of Step 3 is completed. Thus, the time taken for doing so is τ₃₂. Finally, to complete a cycle, transition firing sequence σ₁₃ = 〈s₃₂ (λ) → t₃₄ (µ) → s₄₁ (λ) → y₄₂ (µ) → waiting in q₂₂ (ω₂) → s₂₂ (λ) → t₂₃ (µ)〉 is performed. Thus, the cycle time for Step 3 is as follows

θ 3 = τ 31 + τ 32 + α 2 + 5 μ + 8 λ + ω 2

(20)

With τ₃₁ = α₃ and τ₃₂ ∈ [α₃, α₃+δ₃], the permissive shortest cycle at Step 3 is as follows

θ 3 S = 2 α 3 + α 2 + 5 μ + 8 λ + ω 2

(21)

and the permissive longest cycle time at Step 3 is as follows

θ 3 L = 2 α 3 + δ 3 + α 2 + 5 μ + 8 λ + ω 2

(22)

With the above timeliness analysis, we discuss the scheduling problem next. Note that the above-derived expressions are functions of robot waiting time. Hence, the scheduling problem is to decide the robot waiting time ω_i’s to obtain a schedule if it exists.

Schedulability and scheduling algorithm

Let θ be the production cycle time of the system. Then, the production cycle time for all the steps must be θ. Also, the robot cycle time should be equal to the production cycle time too, that is, we have

θ = θ 1 = θ 2 = θ 3 = θ 4 = η

(23)

Based on the above analysis, to make equation (23) hold, we need to decide ω_i’s in η₂. Also, to schedule a cluster tool, the production rate should be maximized, which requires to minimize η₂. Furthermore, with wafer residency time constraints, schedule feasibility is essential. To be feasible, a schedule should ensure that θ_i is in an acceptable interval. Let π_iL and π_iU denote the lower and upper bounds of θ_i. It follows from the above analysis that when ω₀ = ω₁ = ω₂ = ω₃ = ω₄ = 0, θ_iS and θ_iL are the lower and upper bounds of θ_i, respectively. Thus, by removing the robot waiting time from equations (12), (13), (15), (16), (18), (19), (21), and (22), we have the following lower and upper bounds of θ_i

π 1 L = α 1 + 3 μ + 4 λ

(24)

π 1 U = α 1 + δ 1 + 3 μ + 4 λ

(25)

π 2 L = 2 α 2 + α 3 + 5 μ + 8 λ

(26)

π 2 U = 2 α 2 + δ 2 + α 3 + 5 μ + 8 λ

(27)

π 3 L = 2 α 3 + α 2 + 5 μ + 8 λ

(28)

π 3 U = 2 α 3 + δ 3 + α 2 + 5 μ + 8 λ

(29)

π 4 L = α 4 + 3 μ + 4 λ

(30)

π 4 U = α 4 + δ 4 + 3 μ + 4 λ

(31)

Let π_Lmax = max{π_iL, i ∈ ℕ₄} and π_Umin = min{π_iU, i ∈ ℕ₄}, we have the following lemma.

Proof

It follows from the above discussion that θ = η ≥ η₁. When η = η₁, we have ω₀ = ω₁ = ω₂ = ω₃ = ω₄ = 0. In this case, if π_Umin = π_kU, k ∈ {1, 4}, by equations (5) and (6), we have τ_k = η₁ − 3µ − 4λ > π_kU − 3µ − 4λ = α_k+δ_k+3µ+4λ − 3µ − 4λ = α_k+δ_k. This implies that the wafer residency time constraints are violated for Step k, k ∈ {1, 4}. If π_Umin = π_2U, by equation (7), we have τ₂₁ = η₁ − 5µ − 8λ − α₃ − α₂ > π_2U − 5µ − 8λ − α₃ − α₂ = 2α₂+δ₂+α₃+5µ+8λ − 5µ − 8λ − α₃ − α₂ = α₂+δ₂. Hence, the wafer residency time constraints are violated for Step 2. If π_Umin = π_3U, by equation (7), we have τ₃₂ = η₁ − 5µ − 8λ − α₃ − α₂ > π_3U − 5µ − 8λ − α₃ − α₂ = 2α₃+δ₃+α₂+5µ+8λ − 5µ − 8λ − α₃ − α₂ = α₃+δ₃. The wafer residency time constraints are violated for Step 3.

Then, we discuss the case when η > η₁. In this case, we have η 2 = η − η 1 = ∑ i = 0 4 ω i and ω_i ≤ η − η₁, ∀i ∈ Ω₄. If π_Umin = π_kU, k ∈ {1, 4}, by equations (5) and (6), we have τ_k = η − 3µ − 4λ − ω_k − 1 ≥ η − 3µ − 4λ − (η − η₁) = η₁ − 3µ − 4λ > π_kU − 3µ − 4λ = α_k+δ_k+3µ+4λ − 3µ − 4λ = α_k+δ_k. The residency time constraints are violated for Step k, k ∈ {1, 4}. If π_Umin = π_2U, by equation (7), we have τ₂₁ =η − 5µ − 8λ − α₃ − α₂ − ω₁ ≥ η − 5µ − 8λ − α₃ − α₂ − (η − η₁) = η₁ − 5µ − 8λ − α₃ − α₂ > π_2U − 5µ − 8λ − α₃ − α₂ = 2α₂+δ₂+α₃+5µ+8λ − 5µ − 8λ − α₃ − α₂ = α₂+δ₂. The residency time constraints are violated for Step 2. If π_Umin = π_3U, by equation (10), we have τ₃₂ = η − 5µ − 8λ − α₃ − α₂ − ω₂ ≥ η − 5µ − 8λ − α₃ − α₂ − (η − η₁) = η₁ − 5µ − 8λ − α₃ − α₂ > π_3U − 5µ − 8λ − α₃ − α₂ = 2α₃+δ₃+α₂+5µ+8λ − 5µ − 8λ − α₃ − α₂ = α₃+δ₃. The residency time constraints are violated for Step 3. Thus, in any case, no matter how ω_i’s are set, a feasible schedule cannot be found.

This lemma shows that if the robot cycle time is longer than the upper bound of cycle time for any step, then the system is not schedulable. Next, we discuss the schedulability and scheduling problem when η₁ ≤ π_Umin. We have the following two cases

① [ π 1 L , π 1 U ] ∩ [ π 2 L , π 2 U ] ∩ [ π 3 L , π 3 U ] ∩ [ π 4 L , π 4 U ] ≠ ∅

② [ π 1 L , π 1 U ] ∩ [ π 2 L , π 2 U ] ∩ [ π 3 L , π 3 U ] ∩ [ π 4 L , π 4 U ] = ∅

For Case 1, we have the following lemmas.

Proof

In this case, the cycle time θ = η = η₁. Then, for Step i, i ∈ {1, 4}, we have τ_i = η₁ − 3µ − 4λ ≥ π_iL − 3µ − 4λ = α_i and τ_i = η₁ − 3µ − 4λ ≤ π_iU − 3µ − 4λ = α_i+δ_i. For Step 2, we have τ₂₁ = η₁ − 5µ − 8λ − α₃ − α₂ ≥ π_2L − 5µ − 8λ − α₃ − α₂ = α₂ and τ₂₁ = η₁ − 5µ − 8λ − α₃ − α₂ ≤ π_2U − 5µ − 8λ − α₃ − α₂ = α₂+δ₂, and τ₂₂ = α₂. For Step 3, we have τ₃₁ = α₃, and τ₃₂ = η₁ − 5µ − 8λ − α₃ − α₂ ≥ π_3L − 5µ − 8λ − α₃ − α₂ = α₃ and τ₃₂ = η₁ − 5µ − 8λ − α₃ − α₂ ≤ π_3U − 5µ − 8λ − α₃ − α₂ = α₃+δ₃. Thus, for all the steps, the wafer residency time constraints are satisfied and the obtained schedule is feasible.

For the case that there is an overlap for the permissive cycle time ranges of all the steps, Lemma 3 shows that if the robot cycle time is in the overlap range, the system is schedulable and the robot do not need to wait anywhere.

For Case 2, there must exist at least one i ∈ N₄ such that π_iU < π_Lmax. Let E = {i|π_iU < π_Lmax, i ∈ N₄} and F = N₄\E. Then, for Step i, i ∈ N₄, we set ω_i − 1 as follows.

ω i − 1 = { 0 , i ∈ F π Lmax − α i − δ i − 4 λ − 3 μ , i ∈ E ∩ { 1 , 4 } π Lmax − 2 α 2 − δ 2 − α 3 − 5 μ − 8 λ , i ∈ E ∩ { 2 } π Lmax − 2 α 3 − δ 3 − α 2 − 5 μ − 8 λ , i ∈ E ∩ { 3 }

(32)

Then, we have the following lemma.

Proof

We first show that when ω₄ ≥ 0, the obtained schedule is feasible. It follows from equation (32) and ω 4 = π Lmax − η 1 − ∑ i = 0 3 ω i that we have θ = η = π_Lmax. Then, for Step i, i ∈ {1, 4}, if i ∈ F, τ_i = π_Lmax − 3µ − 4λ ≥ π_iL − 3µ − 4λ = α_i and τ_i = π_Lmax − 3µ − 4λ ≤ π_iU − 3µ − 4λ = α_i+δ_i. If i ∈ E, τ_i = π_Lmax − 3µ − 4λ − (π_Lmax − α_i − δ_i − 4λ − 3µ) = α_i+δ_i. For Step 2, we have τ₂₂ = α₂. If 2 ∈ F, τ₂₁ = π_Lmax − 5µ − 8λ − α₃ − α₂ ≥ π_2L − 5µ − 8λ − α₃− α₂ = α₂ and τ₂₁ = π_Lmax − 5µ − 8λ − α₃ − α₂ ≤ π_2U − 5µ − 8λ − α₃ − α₂ = α₂+δ₂. If 2 ∈ E, τ₂₁ = π_Lmax − 5µ − 8λ − α₃ − α₂ − (π_Lmax − α₂ − δ₂ − α₂ −α₃ − 4λ − 3µ) = α₂+δ₂. For Step 3, we have τ₃₁ = α₃. If 3 ∈ F, τ₃₂ = π_Lmax − 5µ − 8λ − α₃ − α₂ ≥ π_3L − 5µ − 8λ − α₃ − α₂ = α₃ and τ₃₂ = π_Lmax − 5µ − 8λ − α₃ − α₂ ≤ π_3U − 5µ − 8λ − α₃ − α₂ = α₃+δ₃. If 3 ∈ E, τ₃₂ = π_Lmax − 5µ − 8λ − α₃ − α₂ − (π_Lmax − α₃ − δ₃ − α₂ − α₃ − 4λ − 3µ) = α₃+δ₃. Thus, the wafer residency time constraints are satisfied for all the steps and the obtained schedule are feasible.

Now we show that the system is not schedulable if ω₄ < 0. When ω₄ < 0, the obtained schedule is meaningless, or we cannot find a feasible schedule such that θ = η = π_Lmax. Then, there are two choices: (1) η < π_Lmax and (2) η > π_Lmax.

If η < π_Lmax, we assume that π_iL = π_Lmax, i ∈ N₄. If i ∈ {1, 4}, for Step i, we have τ_i = η − 3µ − 4λ − ω_i − 1 < π_Lmax − 3µ − 4λ − ω_i − 1 = π_iL − 3µ − 4λ − ω_i − 1 ≤ α_i. If π_2L = π_Lmax, we have τ₂₁ = η − 5µ − 8λ − α₃ − α₂ − ω₁ < π_Lmax − 5µ − 8λ − α₃ − α₂ − ω₁ = π_2L − 3µ − 4λ − ω₁ ≤ α₂. If π_3L = π_Lmax, we have τ₃₂ = η − 5µ − 8λ − α₃ − α₂ − ω₂ < π_Lmax − 5µ − 8λ − α₃ − α₂ − ω₂ = π_3L − 3µ − 4λ − ω₂ ≤ α₃. This implies that the wafer residency time constraints are violated for at least one step. Thus, a feasible schedule cannot be found; it is not schedulable.

If η > π_Lmax, we assume that η = π_Lmax+Δη and φ_i − 1 = ω_i − 1+Δω_i − 1, where ω_i − 1 is obtained via equation (32), i ∈ E. Then, according to equation (4), we have

∑ k = 1 4 ( ω k − 1 + Δ ω k − 1 ) ≤ η − η 1 = ( π Lmax − η 1 ) + Δ η

(33)

If π_iL = π_Lmax, i ∈ E, and i ∈ {1, 4}, we have π_Lmax+Δη = α_i+δ_i+4λ+3µ+ω_i −1+Δω_i − 1 = π_Lmax+Δω_i −1. If π_iL = π_Lmax, i ∈ E, and i ∈ {2, 3}, we have π_Lmax+Δω = α_i+δ_i+8λ+5µ+α₂+α₃+ω_i −1+Δω_i −1 = π_Lmax+Δω_i −1. Thus, we have Δη = Δω_i −1, π_iL = π_Lmax and i ∈ E. Since ω₄ < 0, ∑ k = 1 4 ω k − 1 > π Lmax − η 1 must hold, leading to the following result

∑ k = 1 4 ( ω k − 1 + Δ ω k − 1 ) ≤ η − η 1 > ( π Lma x − η 1 ) + Δ η

(34)

This contradicts to equation (33), that is, a feasible schedule cannot be found.

For the case that there is not an overlap for the permissive cycle time ranges of all the steps, Lemma 4 shows that if the robot has enough idle time (ω₄ ≥ 0), the system is still schedulable, otherwise it is not schedulable.

Up to now, we present the conditions under which a feasible schedule can be found for a single-cluster tool with wafer revisiting and residency time constraints. Note that, by above analysis, for each case, we present a schedule, and then check its feasibility. Thus, if a feasible schedule exists, a feasible schedule can be found by simply setting the robot waiting time. In summary, an algorithm is presented as follows to find a schedule if it exists.

Example 1

The wafer processing time at PM_1–4 is α₁ = 120 s, α₂ = 40 s, α₃ = 45 s, and α₄ = 125 s, respectively. After a wafer is completed, it can stay in PM_1–4 for at most δ₁ = 30 s, δ₂ = 20 s, δ₃ = 20 s, and δ₄ = 30 s, respectively. The robot task time is λ = µ = 3 s.

For this example, we have π_1L = 141 s, π_1U = 171 s, π_2L = 164 s, π_2U = 184 s, π_3L = 169 s, π_3U = 189 s, π_4L = 146 s, π_4U = 176 s, π_Lmax = 169 s, and η₁ = 163 s. We can check that [π_1L, π_1U]∩[π_2L, π_2U]∩[π_3L, π_3U]∩[π_4L, π_4U] ≠ ∅ and η₁ ≤ π_Lmax. Then, according to Lemma 2, we set ω₀ = ω₁ = ω₂ = ω₃ = 0 and ω₄ = π_Lmax − η₁ = 6 s and a feasible schedule is obtained with cycle time θ = η = π_Lmax = 169 s. The Gantt chart for the obtained schedule is shown in Figure 3.

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Figure 3.

Gantt Chart for the schedule of Example 1.

Example 2

The wafer processing time at PM_1–4 is α₁ = 95 s, α₂ = 35 s, α₃ = 30 s, and α₄ = 100 s, respectively. After a wafer is completed, it can stay in PM_1–4 for at most δ₁ = 30 s, δ₂ = 20 s, δ₃ = 20 s, and δ₄ = 30 s, respectively. The robot task time is λ = µ = 3 s.

We have π_1L = 116 s, π_1U = 146 s, π_2L = 139 s, π_2U = 159 s, π_3L = 134 s, π_3U = 154 s, π_4L = 121 s, π_4U = 151 s, π_Lmax = 139 s, and η₁ = 143 s. Thus, [π_1L, π_1U]∩[π_2L, π_2U]∩[π_3L, π_3U]∩[π_4L, π_4U] ≠ ∅ and π_Lmax ≤ η₁ ≤ π_Umin hold, and it is schedulable. Then, according to Lemma 3, we set ω₀ = ω₁ = ω₂ = ω₃ = ω₄ = 0 s and a feasible schedule is obtained with cycle time θ = η = η₁ = 143 s. The Gantt chart for the obtained schedule is shown in Figure 4.

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Figure 4.

Gantt Chart for the schedule of Example 2.

Example 3

The wafer processing time at PM_1–4 is α₁ = 115 s, α₂ = 40 s, α₃ = 45 s, and α₄ = 125 s, respectively. After a wafer is completed, it can stay at PM_1–4 for at most δ₁ = 30 s, δ₂ = 20 s, δ₃ = 20 s, and δ₄ = 30 s, respectively. The robot task time is λ = µ = 3 s.

For this example, we have π_1L = 136 s, π_1U = 166 s, π_2L = 164 s, π_2U = 184 s, π_3L = 169 s, π_3U = 189 s, π_4L = 146 s, π_4U = 176 s, π_Lmax = 169 s, and η₁ = 163 s. It can be checked that [π_1L, π_1U]∩[π_2L, π_2U]∩[π_3L, π_3U]∩[π_4L, π_4U] = ∅ and E = {1}. According to Lemma 4, we set ω₀ = 3, ω₁ = ω₂ = ω₃ = 0, and ω₄ = 3. Since ω₄ ≥ 0, the system is schedulable and the obtained schedule is feasible with cycle time θ = η = π_Lmax = 169 s. Based on the PN model, we show how this schedule is implemented as follows. In order to clearly illustrate the state of the system, let γ_i, i ∈ Ω₇, represent the time point i. According to the system modeling, we have M₀ = ({null}, {V₃(1)}, {V₂(2)}, {V₁(1)}) and we assume the start time is γ₀ = 0. Then, the system evolves as follows.

Through the above sequence, a cycle of the system is formed, which demonstrates that the cycle time is 169 s. The Gantt chart for the obtained schedule is shown in Figure 5.

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Figure 5.

Gantt Chart for the schedule of Example 3.