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Probabilistic Verification Tong Wang tw2436 Yihan Zou yz2575 Hang Yin hy2368 Miaoqiong Wang mw2908 of Go-Back-N
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Assumption Components of states Flow chart of algorithm Analysis Agenda
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Assumptions RTT >> Tx: multiple messages on channel FC and BC are FIFO queues: packets arrive in order Sender/ Receiver can only process one packet at a time Sender always has packet to send Timeout occur Stopping conditions Initial state: ( 0 0 (0,1,2,3) ( ) )
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Source
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Receiver
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Timeout Backward channel is empty Both forward and backward channel is empty
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Stopping points 3 lost packages on the channel ACK0 is successfully received
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Global state (Tx, Rx, FC, BC) Four stacks: stack(i) contains states going through i low probability transitions
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Initial (0 0 (0,1,2,3) ()) Get M0, Get M1, Get M2, Get M30 step 1Pop (0 0 (0,1,2,3) ()) (0 1 (1,2,3) (0)) Get M0, Get M1, Get M2, Get M3, Acpt M00 (0 0 (1,2,3) ()) Get M0, Get M1, Get M2, Get M31 step 2Pop (0 1 (1,2,3) (0)),(1 1 (1,2,3,4) ()) is accepted (0 2 (2,3) (0,1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M10 (0 1 (1,2,3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M01 (0 1 (2,3) (0)) Get M0, Get M1, Get M2, Get M3, Acpt M01 (0 0 (1,2,3) ()) Get M0, Get M1, Get M2, Get M31 step 3Pop (0 2 (2,3) (0,1)),(1 2 (2,3,4) (1)) is accepted (0 3 (3)(0,1,2)) Get M0, Get M1, Get M2, Get M3, Acpt M0, AcptM1, AcptM2 0 (0 2 (2,3) (1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M11 (0 2 (3) (0,1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M11 (0 1 (1,2,3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M01 (0 1 (2,3) (0)) Get M0, Get M1, Get M2, Get M3, Acpt M01 (0 0 (1,2,3) ()) Get M0, Get M1, Get M2, Get M31
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Stopping point analysis
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step 49Pop (0 3 () (1,2)),(2 3 (4,5) (2)) is accepted (0 4 () (2,3)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M2, Acpt M32 (0 3 (3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 3 () (2)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 2 (3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 2 () (1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 1 (3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 1 () (0)) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 0 (3) ()) Get M0, Get M1, Get M2, Get M33 (0 0 () ()) Get M0, Get M1, Get M2, Get M33 (0 0 (0,1,2,3) ()) Get M0, Get M1, Get M2, Get M33 (0 4 () (1,2,3)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M2, Acpt M33 (0 3 (3) (1,2)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 3 () (0,1,2)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 2 (2,3) (1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 2 (3) (0,1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 1 (1,2,3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 1 (2,3) (0)) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 0 (1,2,3) ()) Get M0, Get M1, Get M2, Get M33
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step 50Pop (0 4 () (2,3)),(3 4 (4,5,6) (3)) is accepted (0 4 () (3)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M2, Acpt M3 3 (0 3 (3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 3 () (2)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 2 (3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 2 () (1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 1 (3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 1 () (0)) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 0 (3) ()) Get M0, Get M1, Get M2, Get M33 (0 0 () ()) Get M0, Get M1, Get M2, Get M33 (0 0 (0,1,2,3) ()) Get M0, Get M1, Get M2, Get M33 (0 4 () (1,2,3)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M2, Acpt M33 (0 3 (3) (1,2)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 3 () (0,1,2)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M1, Acpt M23 (0 2 (2,3) (1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 2 (3) (0,1)) Get M0, Get M1, Get M2, Get M3, Acpt M0, Acpt M13 (0 1 (1,2,3) ()) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 1 (2,3) (0)) Get M0, Get M1, Get M2, Get M3, Acpt M03 (0 0 (1,2,3) ()) Get M0, Get M1, Get M2, Get M33
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Table1(157 steps) Table2(50 steps) VS
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Self Critique 1. Additional assumptions about channel RTT>> Tx, push 4 msg into channel, omit the possibility of Rx prompt reply (concurrency) RTT < timeout interval, omit the possibility of pre- mature timeout (fixed already)
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Self Critique (cont’d) 2. No mapping of msg # from Real Number to finite set of number {0,1,2,3} Hard to justify stopping point, successful reception of ACK(100) does not guarantee the successful reception of ACK(101)
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Self Critique (cont’d) 3. Insufficient justification of stopping We stop verification when first message is successfully received, and take it for granted that the following messages can be received (need to verify!)
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Self Critique (cont’d) Our valuables: Stack processing technique Under our assumptions, reasonable results are obtained Succinct implementation of program
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Thank you!
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