Livelock example. p 1. Assume this as initial marking. t 1 R. t 2. t 4. t 3. t 6. t 5. t 7. t 8. (taken from Fundamentals of SE by C.
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1 p 1 Livelock example (taken from Fundamentals of SE by C. Ghezzi) p 1 Assume this as initial marking. t 1 R t 3 t 4 t 5 t 6 t 7 t 8 Slide No.106
2 Livelock example analysis The tokens in R cannot be divided between transitions t 5 and t 6 any more - either one of the transitions will have both tokens at a given moment and the PN will evolve. Some possible firing sequences: <t 1, t 3, t 5, t 7, t 1, t 3,...> <, t 4, t 6, t 8,, t 4,...> <t 1,, t 3, t 4, t 5, t 7,, t 4,...> Slide No.107
3 Starvation example (taken from Fundamentals of SE by C. Ghezzi) t 1 p i t 3 t 4 Assume this as initial marking. Slide No.108
4 Starvation example analysis Starvation of the right side of the previous starvation example. The initial firing sequence could be one of the following: <t 1,, t 3, t 4 > or <, t 1, t 3, t 4 > or etc. At this point the part of the system will go into starvation and the repeated firing sequence will be: <t 1, t 3 > (once the token from p i is used up it can never be replenished) Slide No.109
5 PN extensions Assigning values to tokens Associating predicates and functions with transitions Prioritising transitions Timed PNs Slide No.110
6 Extending PNs (example) (1) Consider the following system described using an FSM: B_done wait digit 1 digit 0 state A A_done state B This situation cannot be adequately modelled using basic PN notation (see next slide). Slide No.111
7 Extending PNs (example) (2) p 1 t 3 p 2 t 1 p 3 t 4 System states: <p 1 >: wait <p 2 >: state A <p 3 >: state B IT IS POSSIBLE TO ASSUME THAT Firing sequence leading to state A: <t 1 > Firing sequence leading to state B: < >...BUT THIS IS NOT EVIDENT FROM THE ABOVE PN. Slide No.112
8 Assigning values to PN tokens Giving definite values to PN tokens could result in the following diagram: p t 3 t 1 t 4 p 2 p 3 It is now clear that the system accepts the digits 0 and 1. Which transitions t 1 or these fire is, however, still unclear. Slide No.113
9 Predicate association (2) However, if we extend the concept of PN transitions to include predicate association and associate predicate p 1 =1 with t 1 and predicate p 1 =0 with the situation becomes clear. p t 3 t 1 t 4 p p 1 =1 p 1 =0 2 p3 Slide No.114
10 Function association However, sometimes the tokens themselves get changed while passing through a transition in this way the tokens do not only serve as control markers but also as data carriers. In the example if we associate the function p 2 =p 1 and p 3 =p 1 then the tokens are left unchanged as follows. Firing sequence was: <t 1, > or <, t 1 > t 3 t 1 p 1 p p 1 =1 p 1 =0 2 p 3 t Slide No.115
11 Textbook example (taken from Fundamentals of SE by C. Ghezzi) Transition association table: trans. predicates functions t 1 p 2 >p 1 p 4 :=p 2 +p 1 p 1 p 2 p p 3 =p 2 p 4 :=p 3 -p 2 p 5 :=p 2 +p 3 t 1 Firing sequence and outcome: Starting <t 1 > p 4 :=7 or p 4 :=10 if p 4 :=7 then < > is disabled. if p 4 :=10 and < > then p 4 :=0 and p 5 :=8 Starting < > p 4 :=0 and p 5 :=8 after <t 1 > p 4 :=10 <t 1 >: non-determinate <, t 1 >: p 4 :=10 and p 5 :=8 p 4 p 5 Slide No.116
12 Prioritising PN transitions Re-definition of firing rule: If at any one moment more than one transition is enabled, then only the one with the highest priority can fire. Priorities are generally assigned in ascending integer order (1-least, 2-higher, etc.) Priorities can be defined either statically or dynamically as functions of transitions Slide No.117
13 PN prioritisation example p 1 p 2 p 3 p t 1 t 3 Assuming the following static priorities: t 1 : priority = 2; : priority = 1; t 3 : priority = 3 ONLY the following firing sequence is possible: <t 3, t 1, > Slide No.118
14 Introducing timing in PNs The idea is to associate a minimum and maximum response time (t min, t max ) with each transition. Any transition t i can fire if t i min< t < t i max Any timed PN regular PN if for each of its transitions t min = 0 and t max = Slide No.119
15 Timed PN example p 1 p 2 p 3 p Arrival of tokens 0 t 1 t min = 3 t max = 6 priority = 2 t min = 0 t max = 8 priority = 1 t 3 t min = 1 t max = 4 priority = 3 p 2 &p 4 < > <t 3 > <t 3 > <t 3 > < t 1 > < t 1 > < > < > none p 4 none <t 3 > <t 3 > <t 3 > none none none none none p 2 < > < > < > < t 1 > < t 1 > < t 1 > < > < > none none none none none none none none none none time Slide No.120
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