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Process Algebra (2IF45) Probabilistic extension

Process Algebra (2IF45) Probabilistic extension. Dr. Suzana Andova. 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

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Process Algebra (2IF45) Probabilistic extension

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  1. Process Algebra (2IF45)Probabilistic extension 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 a s  t  s  t sel_d sel_c 5/6  s  t ½ not_six ½ 1/6 t six 1 h h pass t Process Algebra (2IF45)

  3. Probabilistic LTS sel_d sel_c 1 5/6 2 9 10 ½ not_six ½ 1/6 6 six t 3 5 h 1 1 pass 11 1 4 7  s  t, t  C 8 1 h t For a probabilistic state sP and a set of states C  S we define the total probability of s to reach C as Prob(s,C) =   Process Algebra (2IF45)

  4. Probabilistic LTS sel_d sel_c 1 5/6 2 9 10 ½ not_six ½ 1/6 6 six t 3 5 h 1 1 pass 11 1 4 7 8 1 h Examples: Prob(9, {3,4}) = 1/6 Prob(9, {1}) = 0 Prob(10,{5,6}) = 1 Process Algebra (2IF45)

  5. Probabilistic LTS 1 C1 = {9, 13}, C2 = {10,11,12}, Prob(1, {2}) = 1/3 Prob(1, {3}) = 2/3 Prob(1,C1) = Prob(1,C2) = 0 Prob(5,C2) = 1 Prob(6,C2) = 1 Prob(7,C2) = 1/3 Prob(5,C1) = 0 Prob(6,C1) = 0 Prob(7,C1) = 2/3 Prob(4,C1) = 1 Prob(8,C1) = 1 Prob(4,C2) = 0 Prob(8,C2) = 0 2/3 1/3 2 3 a b a b b 4 5 6 7 8 1 1 2/3 1 1 1/3 9 10 12 11 13 c c c c c c c Process Algebra (2IF45)

  6. Probabilistic LTS 1 E1 = {9,10,11,12,13}, E2={2,3} Prob(1, E2) = 1 Prob(1,E1) = Prob(1,E2) = 0 Prob(5,E1) = 1 Prob(6,E1) = 1 Prob(7,E1) = 1/3 + 2/3 = 1 Prob(4,E1) = 1 Prob(8,E1) = 1 Prob(4,E2) = …=Prob(8,E2)=0 2/3 1/3 2 3 a b a b b 4 5 6 7 8 1 1 2/3 1 1 1/3 9 10 12 11 13 c c c c c c c Process Algebra (2IF45)

  7. Probabilistic LTS 1 E1 = {9,10,11,12,13}, E2={2,3} Prob(1, E2) = 1 Prob(1,E1) = Prob(1,E2) = 0 Prob(5,E1) = 1 Prob(6,E1) = 1 Prob(7,E1) = 1/3 + 2/3 = 1 Prob(4,E1) = 1 Prob(8,E1) = 1 Prob(4,E2) = …=Prob(8,E2)=0 2/3 1/3 2 3 a b a b b 4 5 6 7 8 1 1 2/3 1 1 1/3 9 10 12 11 13 c c c c c c c Process Algebra (2IF45)

  8. Probabilistic LTS 1 E1 = {9,10,11,12,13}, E2={2,3} Prob(1, E2) = 1 Prob(1,E1) = Prob(1,E2) = 0 Prob(5,E1) = 1 Prob(6,E1) = 1 Prob(7,E1) = 1/3 + 2/3 = 1 Prob(4,E1) = 1 Prob(8,E1) = 1 Prob(4,E2) = …=Prob(8,E2)=0 2/3 1/3 2 3 a b a b b 4 5 6 7 8 1 1 2/3 1 1 1/3 9 10 12 11 13 c c c c c c c 2 –a-> 4 2-b->5 3 –a-> 8 3-b->6 3-b->7 Process Algebra (2IF45)

  9. Bisimulation on LTSs • An equivalence relation R on the set of state S = N  P of an PLTS is probabilistic strong bisimulation iff the following conditions hold: • for all states s, tN, whenever (s, t)  R and s –a-> s’ for some a  A and s’ S, then there is a state t’ S such that t –a-> t’ and (s’, t’)  R; • for all states s, tP, Prob(s, C) = Prob(t, C) for all C S/R • Two PLTSs s and t are probabilistic strong bisimilar, s p t, iff there is a probabilistic strong bisimulation R such that (s, t)  R Process Algebra (2IF45)

  10. Probabilistic LTS 1 2/3 1/3 1 2 3 a b a b a b b 4 5 6 7 8 1 1 1 1 2/3 1 1 1/3 9 10 12 11 13 c c c c c c c c Process Algebra (2IF45)

  11. Composing PLTSs 1/2 1 1/2 1 1/2 = b b a a 1/6 1/3 1/3 1/2 1/3 1/2 2/3 1/6 + = d c b a b c d a b d a c

  12. Composing PLTSs: Parallel composition Assume two persons flipping a coin (independently), one coin per person. If they both have the same sides appeared then they can continue (whatever with). If they have different sides thrown they stop. Model the persons and model the system as a composition.

  13. Composing PLTSs 1/6 1/6 1/3 1/3 1/3 1/2 2/3 1/2 || = c c b d a b d a e b d a c c a b c d b d a where a and c communicate in e, and no other communication is defined (in this examples)

  14. Equational theory. Language • Specify processes that can execute certain actions from a given set A • The language of the Probabilistic Basic Process Algebra, namely, the operators in the signature • 0 deadlock constant (inaction) • a._ action prefix for a in A • + non-deterministic choice •  probabilistic choice for   (0,1) Process Algebra (2IF45)

  15. Axioms of PBPA(A) PBPA(A) Signature: 0, a._ , _+_ ,  • (A1) x+ y = y+x • (A2) (x+y) + z = x+ (y + z) • (A3) x + x = x • (A4) x+ 0 = x Process Algebra (2IF45)

  16. Axioms of PBPA(A) PBPA(A) Signature: 0, a._ , _+_ ,  • (PA1) x  y = y 1- x • (PA2) x(y  z) = (xy)  z • where = /( + - ) and  =  + -  • (PA3) x  x = x • (PA4) (x  y) + z = (x + z)  (y + z) Process Algebra (2IF45)

  17. Exercises • Write the process algebraic specification of the PLTSs from the examples. • For the PLTSs on slide 9 derive using the axioms the equality between the two PBPA specifications. Process Algebra (2IF45)

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