Process Algebra (2IF45) Analysing Probabilistic systems Dr. Suzana Andova

1 Probabilistic LTS Process Algebra (2IF45) Basic ingredients of a PLTS: states non-detereministic states set N probabilistic states set P transitions action transitions labelled with actions and t  P probabilistic transitions labelled with probabilities and t  N For a probabilistic state s,   = 1  s  ts  t  s  ts  t a s  ts  t

2 Process Algebra (2IF45) Composing PLTSs 1/2 a b + = 1/3 c d 2/3 1/3 ab 1/6 a 1/3 d c c d b

3 Strong Probabilistic bisimulation on PLTSs Process Algebra (2IF45) 1 2/3 b /3 a 9 c 1 10 c c b /3 12 c c a c b 1 c 2/3 4 1 b 1 a 1 c 1

4 Process Algebra (2IF45) 1.A chatting philosopher is a person dedicated to two activities: thinking and chatting. A philosopher uses his phone for chatting. He can decide to pick up the phone with probability pi, or stay thinking with probability 1-pi. Once he starts chatting, he end the call with probability ro, or keep chatting with probability 1-ro. 2.There is a switch which allocates connection to a philosopher, and also deallocating a connection. Our switcher is capable of handling only one connection at time. 3.We consider a system of two philosophers and one switcher 4.First, we compute Phil 1 || Phil 2, where Phil i = Think i Chatting Philosophers example (partially)

5 Analysing PLTSs – main ingredients Process Algebra (2IF45) The set of all paths in x starting in p?! What can we measure on x? Do we need schedulers for it? n p ks 0 x

6 Example 1 (cont.) Process Algebra (2IF45) Property1: A path has a trace c*a Property2: A path has a trace c*b Property3: A path has a trace (cc)*a Property4: A path has a trace (cc)*b Property5: A path reaches a deadlock state

7 Example 1 (cont.) Process Algebra (2IF45) Property1: A path has a trace c*a n p ks 0 x p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/

8 Example 1 (cont.) Process Algebra (2IF45) Property1: A path has a trace c*a n p ks 0 x p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/ Prob(SetPaths1) = ?

9 Example 1 (cont.) Process Algebra (2IF45) Property1: A path has a trace c*a n p ks 0 x p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/ Prob(SetPaths1) = 1/3 + 1/6x1/3 + (1/6)^2x1/3 + …. =  k  0 1/3x(1/6)^k = (1/3)/ (1-1/6) = 2/5

10 Example 1 (cont.) Process Algebra (2IF45) Property2: A path has a trace c*b n p ks 0 x p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/

11 Example 1 (cont.) Process Algebra (2IF45) Property2: A path has a trace c*b n p ks 0 x p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/ Prob(SetPaths2) = ?

12 Example 1 (cont.) Process Algebra (2IF45) Property3: A path has a trace (cc)*a or (cc)*b n p ks 0 x p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/ kss 0 p 1/2 1/6 a b c

13 Example 1 (cont.) Process Algebra (2IF45) Property3: A path has a trace (cc)*a or (cc)*b n p ks 0 x p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/ kss 0 p 1/2 1/6 a b c Prob(SetPaths3) = ?

14 Example 2 Process Algebra (2IF45) The set of all paths in y starting in p? What can we measure on y? Do we need schedulers for it? n p ks 0 y k p kss 0 p 1/3 1/2 1/6 a b c kss 0 p 1/2 1/6 a b c 1/ b b

15 Example 2 (cont.) Process Algebra (2IF45) Property1: A path has a trace c*a? First select a scheduler, then compute this set, and its probability 1.1. Scheduler  1.2. Scheduler  Scheduler  2

16 Example 2 (cont.) – scheduler  Process Algebra (2IF45) Computation tree CT y (p,  ) n p ks 0 y k p ksn 0 p 1/3 1/2 1/6 a c ksn 0 p 1/2 1/6 b c 1/

17 Example 2 (cont.) – scheduler  Process Algebra (2IF45) Prob  (p, trace = c*a) = 12/35 How? n p ks 0 y k p ksn 0 p 1/3 1/2 1/6 a c ksn 0 p 1/2 1/6 b c 1/3.... Property1: A path has a trace c*a

18 Example 2 (cont.) – scheduler  1 Process Algebra (2IF45) n p ks 0 y k p kss 0 p 1/3 1/2 1/6 a c kss 0 p 1/2 1/6 a b c 1/ b Computation tree CT y (p,  1 )

19 Example 2 (cont.) – scheduler  1 Process Algebra (2IF45) n p ks 0 y k p kss 0 p 1/3 1/2 1/6 a c kss 0 p 1/2 1/6 a b c 1/ b Property1: A path has a trace c*a Prob  1 (p, trace = c*a) = 2/5 How? Can we do better?

Process Algebra (2IF45) Abstraction on PLTSs

21 Towards probabilistic branching bisimulation Process Algebra (2IF45) We consider again hiding of internal behaviour Again in the style of branching bisimulation, which is: -congruence -easy to axiomatize -rather intuitive -preserve properties

22 Towards probabilistic branching bisimulation Process Algebra (2IF45) Recall Branching bisimulation on LTss s t s’  ts t’s’ a t’’ a 

23 Towards probabilistic branching bisimulation Process Algebra (2IF45) Recall Branching bisimulation on LTss s t s’  ts t’s’ a t’’ a  Recall Strong Probabilistic bisimulation on PLTss s t C1 (eq. class ) s s’ a t’ a t C2 (eq. class ) 11 11 22 22

24 Towards probabilistic branching bisimulation Process Algebra (2IF45) Recall Branching bisimulation on LTss s t s’  ts t’s’ a t’’ a  Recall Strong Probabilistic bisimulation on PLTss s t C1 (eq. class ) s s’ a t’ a t C2 (eq. class ) 11 11 22 22 Combining them into Probabilistic Branching Bisimulation

25 Missing ingredients: Process Algebra (2IF45) 1.We need a notion of  for action transitions, just like in BB on LTSs 2.We need to compute probability to go to next eq. class from a probabilistic state, just like in PSB on PLTSs. 3. And something more… 

26 Missing ingredients Process Algebra (2IF45) s 0 a u 1 k 0 a r 1 n p 1  m q 1  Relate probabilistic and non-deter. states!

27 Missing ingredients: Process Algebra (2IF45) 1.We need a notion of  for action transitions, just like in BB on LTSs OK! Our unobservable paths are now: p0  n1  p1  n2…  pk or p0  n1  p1  n2…  nk 2.We need to compute probability to go to next eq. class from a probabilistic state, just like in PSB on PLTSs. But also for non-dterministic states. 1 if n  C Prob(n, C) = 0 if n  C 

28 Probabilistic Branching Bisimulation Process Algebra (2IF45) Definition : An equivalence relation R ⊆ S × S is a probabilistic branching bisimulation iff for every (s, t) ∈ R the following two conditions hold: (i)if s –-> s′ for a ∈ A or a= , then there exist states t0,..., tn, t′ such that t = t > t > … tn –-> t’ and (s, ti) ∈ R for all 0 ≤ i ≤ n, and (s′, t′) ∈ R, (ii) for all equivalence classes of states M ∈ S/R, Prob(s,M) = Prob(t,M). States s and t are branching bisimilar, denoted by s ∼ pbb t, if (s, t) ∈ R for some branching bisimulation relation R. a  or  a

29 Examples: Probabilistic Branching Bisimulation Process Algebra (2IF45) Distributed page (part 2)