Process Algebra (2IF45) Analysing Probabilistic systems

Process Algebra (2IF45)

Analysing Probabilistic systems

Dr. Suzana Andova

2

Probabilistic LTS


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 t

s t

as t

3 Process Algebra (2IF45)

Composing PLTSs

1/2

a b

1/2+ =

1/3

c d

2/3 1/3

a b

1/61/6

a

1/3

dc cdb

1/3

a b

2/3 1/2

c d

1/2|| =

1/3

c b

1/31/6

a

1/6

ac

db de

1

a a dd bc

11

b c

1 11 1

4

Strong Probabilistic bisimulation on PLTSs


1 2/3

b32

5

1/3

a

9

c

1

10

c c

b

76

11

1/3

12

c c

a

8

1

13

c

b

1

c

2/3

4 1

b

1

a

1

c

1

5 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 Phil1 || Phil2, where Phili = Thinki

Chatting Philosophers example (partially)

6

Analysing PLTSs – main ingredients


The set of all paths in x starting in p?!What can we measure on x? Do we need schedulers for it?

n

p

k s

0

x

7

Example 1 (cont.)


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

8

Example 1 (cont.)



n

p

k s

0

x p

k s s

0p

1/31/2 1/6

a b c

k s s

0p

1/2 1/6

a b c

1/3

.

.

.

9

Example 1 (cont.)



n

p

k s

0

x p

k s s

0p

1/31/2 1/6

a b c

k s s

0p

1/2 1/6

a b c

1/3

.

.

.Prob(SetPaths1) = ?

10

Example 1 (cont.)



n

p

k s

0

x p

k s s

0p

1/31/2 1/6

a b c

k s s

0p

1/2 1/6

a b c

1/3

.

.

.Prob(SetPaths1) = 1/3 + 1/6x1/3 + (1/6)^2x1/3 + ….

= k0 1/3x(1/6)^k = (1/3)/ (1-1/6) = 2/5

11

Example 1 (cont.)



n

p

k s

0

x p

k s s

0p

1/31/2 1/6

a b c

k s s

0p

1/2 1/6

a b c

1/3

.

.

.

12

Example 1 (cont.)



n

p

k s

0

x p

k s s

0p

1/31/2

1/6

a b c

k s s

0p

1/2 1/6

a b c

1/3

.

.

.Prob(SetPaths2) = ?

13

Example 1 (cont.)


Property3: A path has a trace (cc)*a or (cc)*b

n

p

k s

0

x

p

k s s

0p

1/31/2 1/6

a b c

k s s

0p

1/2 1/6

a b c

1/3

.

.

.

k s s

0p

1/31/2 1/6

a b c

14

Example 1 (cont.)


Property3: A path has a trace (cc)*a or (cc)*b

n

p

k s

0

x

p

k s s

0p

1/31/2 1/6

a b c

k s s

0p

1/2 1/6

a b c

1/3

.

.

.

k s s

0p

1/31/2 1/6

a b c

Prob(SetPaths3) = ?

15

Example 2


The set of all paths in y starting in p?What can we measure on y? Do we need schedulers for it?

n

p

k s

0

y

k

p

k s s

0p

1/31/2 1/6

a

b

c

k s s

0p

1/2 1/6

ab c

1/3

.

.

.

b

b

16

Example 2 (cont.)


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 1

1.3. Scheduler 2

17

Example 2 (cont.) – scheduler


Computation tree CTy(p, )

n

p

k s

0

y

k

p

k s n

0p

1/31/2 1/6

a c

k s n

0p

1/2 1/6

b

c

1/3

.

.

.

18

Example 2 (cont.) – scheduler


Prob(p, trace = c*a) =

12/35 How?

n

p

k s

0

y

k

p

k s n

0p

1/31/2 1/6

a c

k s n

0p

1/2 1/6

b

c

1/3

.

.


19

Example 2 (cont.) – scheduler 1


n

p

k s

0

y

k

p

k s s

0p

1/31/2 1/6

a c

k s s

0p

1/2 1/6

ab c

1/3

.

.

.

b

Computation tree CTy(p, 1)

20

Example 2 (cont.) – scheduler 1


n

p

k s

0

y

k

p

k s s

0p

1/31/2 1/6

a c

k s s

0p

1/2 1/6

ab c

1/3

.

.

.

b


Prob1(p, trace = c*a) = 2/5

How? Can we do better?


Abstraction on PLTSs

22

Towards probabilistic branching bisimulation


We consider again hiding of internal behaviourAgain in the style of branching bisimulation, which is:

-congruence-easy to axiomatize-rather intuitive-preserve properties

23



Recall Branching bisimulation on LTss

st

s’

t s

t’ s’

a

t’’

a

24




st

s’

t s

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 )

11

22

25




st

s’

t s

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 )

11

22

Combining theminto

Probabilistic Branching Bisimulation

26

Missing ingredients:


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…

27

Missing ingredients


s

0

a

u

1

k

0

a

r1

n

p

1

m

q

1

Relate probabilistic and non-deter.states!

28

Missing ingredients:


1. We need a notion of for action transitions, just like in BB on LTSsOK! Our unobservable paths are now: p0 n1 p1 n2… pkor 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

29

Probabilistic Branching Bisimulation


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 = t0 -------> t1 ------> … 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 or

30

Examples: Probabilistic Branching Bisimulation


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Process Algebra (2IF45) Analysing Probabilistic systems

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