CSP-case-study-1

# Tm train t l m 2 reftm refusedtm train t l 4 train 2

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booked.t.m TRAIN ( t, l, m ) 2 ref.t.m refused.t.m TRAIN ( t, l ))) 4

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& \$ % Train (2) TRAIN ( t, l, l 0 ) = reverse.t rel.t.l 0 TRAIN ( t, ¯ l ) 2 ( goto.t.β arrive.t.l leave.t.l rel.t.l TRAIN ( t, l 0 ) ) Note: Communication with driver has one of the forms ”reverse”, ”book”, ”booked”, ”refused”, goto . All other events are communications with a crossing point. 5
& \$ % Crossing Points (1) Crossing points also have unique names (typicaly α , β ). they have two roles. Firstly they act as intermediaries between TRAINs and LINEs. Secondly they only control the timing of transitions from one line to another—specially they only allow one train to be using them at any time (to avoid crashes). 6

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& \$ % Crossing Points (2) CP ( α ) = CP 1( α ) k CP 2( α ), where CP 1( α ) = 2 S ( l )= α,t T req.t.l get. ( N ( l )) ( con. ( N ( l )) con.t.l CP 1( α ) 2 dis. ( N ( l )) ref.t.l CP 1( α ) ) 2 ( 2 S ( l )= α,t T rel.t.l rel.N ( l ) CP 1( α ) ) CP 2( α ) = 2 E ( l )= α,t T arrive.t.l ( 2 S ( m )= α enter.t.m leave.t.l CP 2( α ) ) 7
& \$ % The Whole System The solution to the problem is thus TRAINS || CPS || LINKS where TRAINS , CPS and LINKS are respectively the parallel composition of all TRAINs, CPs and LINEs.

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