Outline. Time Petri Net State Space Reduction Using Dynamic Programming and Time Paths. Petri Net. Petri Net. Time Petri Net.

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1 Outline Using and Time Paths Humboldt-Universität zu Berlin Institut of Computer Science Unter den Linden 6, Berlin, Germany IFORS 2005, Hawaii July 11-15, 2005 Definition () The structure N =(P, T, F, V, m 0 )isa (PN), iff P, T and F are finite sets, } P set of places set of vertices(nodes) T set of transitions P T =, P T, F set of edges (arcs) F (P T ) (T P) anddom(f ) cod(f )=P T V : F N + (weights of edges) m 0 : P N (initial marking) m 0 =(0, 1, 1) m 0 =(0, 1, 1) h 0 =(, 0, 0, 0) p-marking t-marking Definition (Time Petri net) The structure Z =(P, T, F, V, m o, I ) is called a Time Petri net (TPN) iff: S(Z) :=(P, T, F, V, m o ) is a PN (skeleton of Z) I : T Q + 0 (Q+ 0 { }) and I 1 (t) I 2 (t) for each t T,whereI(t) =(I 1 (t), I 2 (t)).

2 state Definition (state) Let Z =(P, T, F, V, m o, I )beatpnandh : T R + 0 {#}. z =(m, h) is called a state in Z iff: m is a p-marking in Z. h is a t-marking in Z. Definition (state changing) Let Z =(P, T, F, V, m o, I )beatpn, z =(m, h), z =(m, h )betwostates. Then z =(m, h) changes into z =(m, h )by firing a transition / time elapsing (m 0, 0 0 (m 1, (m 1, z (m 2, z (m 2, t4 (m 3, 0.0 z t4 (m 3, 0.0 (m 4, 4.3

3 z t4 (m 4, 4.3 t1 (m 5, 0.0 z t4 t1 (m 5, 0.0 t2 (m 6, 0.0 Transition sequences, Runs Reachable state, Reachable marking, State space Definition transition sequence: σ =(t 1,, t n ) run: σ(τ) =(τ 0, t 1,τ 1,,τ n 1, t n,τ n ) τ feasible run: z 0 0 z t 1 τ 0 1 z1 z t 1 n τ n zn z n feasible transition sequence : σ is feasible if there ex. a feasible run σ(τ) Definition z is reachable state in Z if there ex. a feasible run σ(τ) and σ(τ) z 0 z m is reachable marking in Z if there ex. a reachable state z in Z with z =(m, h) The set of all reachable states in Z is the state space of Z ( denoted: StSp(Z) ). Qualitative Properties Quantitative Properties static properties: being/having homogenous ordinary free choice extended simple conservative deadlocks, etc. decidable without knowledge of the state space! dynamic properties: being/having bounded live reachable marking/state place- or transitions invariants, etc. decidable, if at all (TPN is equiv. to TM!), with implicit/explicit knowledge of the state space each time proposition as having/computing (min-/max) time length of path path between two states/markings with min-/max time length set of all reachable markings within a period looking for efts andlfts leading to certain qualitative/quantitative properties etc. decidable, if at all, with implicit/explicit knowledge of the state space

4 Parametric Description of the State Space Let Z =[P, T, F, V, m 0, I ]beatpnandσ =(t 1,, t n )bea transition sequence in Z. δ(σ) =[m σ, Σ σ, B σ ] is the parametric description of σ, if σ m 0 m σ Σ σ (t) is a parametrical t marking B σ is a set of conditions (a system of inequalities) Obviously σ (m σ, Σ σ )=:z σ, z 0 StSp(Z) = z σ. σ σ =(e) = δ(σ) =C e = {((0, 1, 1), (x }{{} 1,,,x 1 )) 0 x }{{} 1 3 }{{} m σ Σ σ B σ } Theorem (1) Let Z be a TPN and σ =(t 1,, t n ) be a feasible transition sequence in Z, with a run σ(τ) as an execution of σ, i.e. τ z 0 t 0 0 τ n t n zn =(m n, h n ), and all τ i R + 0. Then, there exists a further feasible run σ(τ ) of σ with such that τ0 z 0 t0 τn tn z n =(m n, h n). Theorem (1 continuation) τ z 0 t 0 0 τ n t n zn =(m n, h n ),τ i R + 0. τ0 z 0 t0 τn tn zn =(mn, hn), τi N. 1. Foreachi, 0 i n the time τi is a natural number. 2. For each enabled transition t at marking m n (= mn) it holds: 2.1 h n (t) = h n (t). 2.2 n = n τ i i=1 τ i i=1 3. For each transition t T holds: tisreadytofireinz n iff t is ready to fire in z n,too. Theorem (2 similar to 1) Let Z be a TPN and σ =(t 1,, t n ) be a feasible transition sequence in Z, with a run σ(τ) as an execution of σ, i.e. τ z 0 t 0 0 τ n t n zn =(m n, h n ), and all τ i R + 0. Then, there exists a further feasible run σ(τ ) of σ with such that τ0 z 0 t0 τn tn z n =(m n, h n). Theorem (2 continuation) 1. Foreachi, 0 i n the time τi is a natural number. 2. For each enabled transition t at marking m n (= mn) it holds: 2.1 h n (t) = h n (t). 2.2 n = n τ i i=1 τ i i=1 3. For each transition t T holds: tisreadytofireinz n i t is ready to fire in z n,too.

5 The theorem 1 solves the following problem : Input: a TPN, a transition sequence σ =(t 1,...,t n )and a sequence of (n +1)realnumbers, ( ˆβ(x 0 ), ˆβ(x 1 ),, ˆβ(x n )) subject to a certain finite set VC of conditions (inequalities). The solving of the output is the problem P : Problem P : Compute a sequence of (n +1)integers, (β (x 0 ),β (x 1 ),,β (x n )) subject to VC 1. a sequence of (n +1)integers, (β (x 0 ),β (x 1 ),,β (x n )) subject to VC. The solution strategy for the problem P is a typical dynamic programming s one. 1 VC is a certain finite superset of the set VC σ =(t 1 t 3 t 4 t 2 t 3 ) σ =(t 1 t 3 t 4 t 2 t 3 ) 0.7 σ(τ) :=z 0 t1 0.0 t3 0.4 t4 1.2 t2 0.5 t3 1.4 z ( continuation ) σ =(t 1 t 3 t 4 t 2 t 3 ) m σ =(1, 2, 2, 1, 1) x 4 + x 5 x 5 Σ σ = x 5 x 5 x 0 + x 1 + x 2 + x 3 + x 4 + x 5 and

6 ( continuation ) ( continuation ) The run σ(τ) with σ(τ) = B σ = { 0 x, 0 x 2, 0 x + x + x x 1, x 2 2, x 2 + x x 2, x 3 2, x 0 + x 1 + x 2 + x x 3, x 4 2, x 0 + x 1 + x 2 + x 3 + x 4 5 }. 0 x 4, x 5 2, x 0 + x 1 + x 2 + x 3 + x 4 + x x 5, x 0 + x 1 5 x 4 + x z 0 t1 0.0 t3 0.4 t4 1.2 t2 0.5 t3 1.4 (m, ) 4.2 is feasible. ( continuation ) (m, ) (m, ) }{{}}{{} z 0 σ(?) z z 0 σ( ˆβ) z (m, ) 5.0 }{{} z 0 σ(?) z ( continuation ) I x 0 x 1 x 2 x 3 x 4 x 5 Σ σ (t 1 )Σ σ (t 2 )Σ σ (t 5 ) ˆβ = β β β β β β β ( continuation ) II x 0 x 1 x 2 x 3 x 4 x 5 Σ σ (t 1 )Σ σ (t 2 )Σ σ (t 5 ) ˆβ = β β β β β β β ( continuation ) Hence, the runs σ(τ1 1 ):=z 0 t1 0 t3 1 t4 1 t2 0 t3 1 z σ(τ) = z t1 0.0 t3 0.4 t4 1.2 t2 0.5 t3 1.4 z σ(τ2 ):=z 0 1 t1 0 t3 0 t4 2 t2 0 t3 2 z are feasible in Z, too.

7 Dynamic programming Dynamic programming Input: Where is the here? Let us consider the tableau I again! The TPN Z 2, the transition sequence σ =(t 1, t 3, t 4, t 2, t 3 ) the six (6 =n +1,i.e. n =5) elapses of time ˆβ(x 0 )=0.7, ˆβ(x 1 )=0.0, ˆβ(x 2 )=0.4, ˆβ(x 3 )=1.2, ˆβ(x 4 )=0.5, ˆβ(x 5 )=1.4, which are real numbers and the run σ( ˆβ) =(0.7, t 1, 0.0, t 3, 0.4, t 4, 1.2, t 2, 0.5, t 3, 1.4) is a feasible one in Z 2. Dynamic programming Six elapses of time β (x 0 ),β (x 1 ),,β (x 5 ) which are integers, σ(β ) is a feasible run in Z 2. The set of transitions which are ready to fire after σ( ˆβ) is the same as the set of transitions which are ready to fire after σ(β ). I x 0 x 1 x 2 x 3 x 4 x 5 Σ σ (t 1 )Σ σ (t 2 )Σ σ (t 5 ) ˆβ = β β β β =P I x 0 x 1 x 2 x 3 x 4 x 5 Σ σ (t 1 )Σ σ (t 2 )Σ σ (t 5 ) ˆβ = β β β β I x 0 x 1 x 2 x 3 x 4 x 5 Σ σ (t 1 )Σ σ (t 2 )Σ σ (t 5 ) ˆβ = β β β β β β β = β Σ σ (t 1 )=x 4 + x 5, Σ σ (t 2 )=Σ σ (t 3 )=Σ σ (t 4 )=x 5 Σ σ (t 5 )=x 1 + x 2 + x 3 + x 4 + x 5

8 The set of its critical states is the singleton S o = {5}. The set of its terminal states is the singleton S t = {0}. The set of non-terminal states is S =S\ S t = {1, 2,...,5}. The T-linker L T has the form L T (z(s o )) = z o =z(s o ). The transition function t is defined as t(s) := s - 1, s S. The linker L is clearly given by z(s) = L(s, {(s,z(s )) s t(s)}), s S = L(s, z(t(s))) = L(s,z(s-1)) := β s The time length of the run σ( ˆβ) is l σ(β ) = ˆβ(x 0 )+ ˆβ(x 1 )+ ˆβ(x 2 )+ ˆβ(x 3 )+ ˆβ(x 4 )+ ˆβ(x 5 )=4.2 In tableau I: The time length of the run σ(β )isl σ(β ) = 4 In tableau II: The time length of the run σ(β )isl σ(β ) = 5 i.e. l σ(β ) = =l σ(β ) =4.2 5 = l σ(β ) Corollary Each feasible t-sequence σ in Z can be realized with an integer run. Each reachable marking in Z can be found using integer runs only. If z is reachable in Z, then z and z are reachable in Z, too. The length of the shortest and longest time path between two arbitrary p-markings are natural numbers. Definition A state z =(m, h) inatpnisaninteger one iff for all enabled transitions t at m holds: h(t) N. ( ) Theorem ( 3 ) Let Z be a FTPN. The set of all reachable integer states in Z is finite if and only if the set of all reachable markings in Z is finite. Remark: Theorem 3 can be generalized for all TPNs (applying a further reduction).

9 (The FTPN Z 2 and its reachability graph(s) ) Definition Basis) 1z 0 RG(Z) Step) Let z be in RG(Z) already. 1. for i=1 to ndo t if z i z possible in Z then z RG(Z) end 1 2. if z z possible in Z then z RG(Z) = The reachability graph is a weighted directed graph. (The infinite TPN Z 3 and its reachability graph RG(Z 3 )) Let Z =(P, T, F, V, I, m o ) be a bounded TPN. The following problems can be decided/computed with the knowledge of its RG, amongst others: Result: Input: Solution: z and z - two states (in Z). Is there a path between z and z in RG(Z)? If yes, compute the path with the shortest time length. By means of prevalent methods of the graph theory, e.g. Bellman-Ford algorithm (the running time is O( V E ) andrg(z) =(V, E) ) Result: Input: Solution: m and m - two markings (in Z). Is there a path between m and m in RG(Z)? If yes, compute the path with the shortest time length. By means of prevalent methods of the graph theory, for computing all-pairs shortest paths. The running time is polynomial, too. Definition The longest path between two states (vertices in RG(Z)) z and z is lp(z, z )with lp(z, z ):= max τ i σ(τ), if a cycle is reachable starting on z before reaching z, if z σ(τ) z

10 Result: Result: Input: z and z - two states (in Z). Input: m and m - two states (in Z). Is there a path between z and z in RG(Z)? If yes, compute the path with the longest time length. Is there a path between z and z in RG(Z)? If yes, compute the path with the longest time length. Solution: By means of prevalent methods of the graph theory, e.g. Bellman-Ford algorithm (polyn. running time). or by computing all strongly connected components of RG(Z). (linear running time) Solution: By means of prevalent methods of the graph theory, e.g. Bellman-Ford algorithm. or by computing all strongly connected components of RG(Z). The of a TPN is a nonoptimization truncated decision problem The minimal and the maximal time length of a path between two markings in a TPN is a natural number (if finite) = it can be computed in polynomial/linear time (with res. to the RG) Thank you!