Submitted:
25 July 2024
Posted:
26 July 2024
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Abstract
Keywords:
1. Introduction
1.1. Basic PN Terminology
1.2. Structure of the Article
2. Preliminaries
4. Elements in the support of each minimal T-semiflow can be fired sequentially.
3. Control of ES3PR
- Ni = (PSi ∪ {p0i} ∪ PRi, Ti, Fi, Wi), i ∈ Im
-
P = Ps ∪ P0 ∪ PR is a partition such that
- PS = ∪ i ∈ Im PSi, PSi ≠ ∅ and PSi ∩ PSj = ∅, ∀i ≠ j (i, j ∈ Im)
- PR = ∪ i ∈ Im PRi = {r1, r2, …, rn}, n > 0
- P0 = ∪ i ∈ Im {p0i}
- The elements in P0, PS, and PR are called idle, operation, and resource places, respectively
- The output transitions of an idle place are called source transitions.
- T = ∪ i ∈ Im Ti, Ti ≠ ∅, Ti ∩ Tj = ∅ for all i ≠ j.
- ∀i ∈ Im the subset Ni generated by PSi ∪ {p0i} ∪ Ti is a strongly connected state machine such that every cycle contains p0i
- ∀r ∈ PR, there exists a unique minimal P-semiflow Ir ∈ IN|P| (here IN = {0, 1, 2, …}) such that {r} = ||Ir|| ∩ PR, P0 ∩ ||Ir|| = ∅, PS ∩ ||Ir|| ≠ ∅ and Ir(r) = 1
- PS = ∪ r ∈ PR (||Ir||\{r})
- N is a strongly connected net.
3.1. Siphons in Petri Nets Control
3.2. Procedure of Setting the Supervisor for ES3PR
3.3. Illustrative Example of ES3PR—Siphons and P-Invariants
4. Controllability of Siphons in Generalized Petri Nets
4.1. Controllability of Elementary Siphons
4.2. Controllability of Dependent Siphons
4.3. Illustrative Example of ES3PR—Control
5. Conclusions
Acknowledgments
References
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| No. | Siphons | Traps | Notice |
| 1. |
S1 = {p1, p12} |
Tr1 = {p1, p12} | Eliminate S1, because S1 = Tr1 |
| 2. | S2 = {p2, p11} | Tr2 = {p2, p11} | Eliminate S2 because S2 = Tr2 |
| 3. | S3 = {p4, p13} | Tr3 = {p3, p4, p13} | |
| 4. | S4 = {p5, p7, p16} | Tr4 = {p5, p7, p16} | Eliminate S4 because S4 = Tr4 |
| 5. | S5 = {p4, p8, p15} | Tr5 = {p4, p8, p15} | Eliminate S5 because S5 = Tr5 |
| 6. | S6 = {p3, p9, p14} | Tr6 = {p3, p9, p14} | Eliminate S6 because S6 = Tr6 |
| 7. | S7 = {p5, p8, p15, p16} | Tr7 = {p4, p7, p15, p16} | |
| 8. | S8 = {p4, p9, p14, p15} | Tr8 = {p3, p8, p14, p15} | |
| 9. | S9 = {p7, p8, p9, p10} | Tr9 = {p7, p8, p9, p10} | Eliminate S9 because S9 = Tr9 |
| 10. | S10 = {p5, p9, p14, p15, p16} | Tr10 = {p3, p7, p14, p15, p16} | |
| 11. | S11 = {p1, p2, p3, p4, p5, p6} | Tr11 = {p1, p2, p3, p4, p5, p6} | Eliminate S11 because S11 = Tr11 |
| p1 | p2 | p3 | p4 | p5 | p6 | p7 | p8 | p9 | p10 | p11 | p12 | p13 | p14 | p15 | p16 | |
| I1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| I2 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| I3 | 0 | 0 | 1 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| I4 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
| I5 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
| I6 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
| I7 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| I8 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
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