1. Cybernetix – Cost Driven UPPAAL and Insights Angelika Mader University of Twente Ametist meeting December 2002 Dortmund

2. recognize order deadlocks a.s.a.p reward personalisation, load/unload Extending the UPPAAL Model from T. Krilavicius & Y. Usenko 1 3 2 332  restrictions in the model  still long model checking times  guide model checking in the direction of the super-single mode,  in this way make use of branch-and-bound  finds ssm-like schedules pretty fast, for checking all still too long  could not compete with SPIN...  beginning of thinking....

3. On Optimal Cybernetix Schedules Schedules in the Cybernetix case study have 3 phases:  start-up phase: the machine is filled with the initial batches  cyclical phase: the schedule has a periodicity  end-phase: the last cards are removed from the machine 1. Should be proved start-up phase and end-phase happen only once, cyclical phase determines the long-term efficiency restrict further considerations to the cyclical phase

4. experiments: optimal schedules of a fixed (rather small) number of cards  too much weight to the start-up and end-phase. searching for optimal cycles (i.e. cheapest cycles) : start to count time from the moment when the initial load of cards is in the machine search for repetition of a state (modulo batch number) Consequences for model checking

5. Theoretical lower bound 2. (parallelism argument) p: personalisation time k: number of personalisation stations 1 personalisation station can personalise 1 card and get a new one in p+1 time. k personalisation stations can personalise k cards and get new ones in p+1 time. GOO D NEW S The super-single mode meets the theoretical lower bound, if the personalisation time is not too low. here we abstract away from the belt

6. Observation A smart-card-personalisation schedule is theoretically optimal iff  as soon as a smart card is personalised it leaves the personalisation station  a personalisation station is empty only as long as it (minimally) takes to get a new card 3. super-single mode: condition 2 above always holds condition 1 holds, if the personalisation time is long enough

7. Alternative Architecture 56 12 7 3 8 4 51627384516273841 5 2 6 3 7 4 8 1 5 2 6 3 78 991 56 2 7 3 8 9 5 1 6 2 7 3 8 9 5 1 6 2 7 8 10 9 5 6 1 7 2 8 5 9 6 1 7 2 8 5 9 6 1 7 8 11 10 5 6 9 7 1 8 11 5 10 6 9 7 1 8 11 5 10 6 9 7 8 12 1 move 1 time unit k parallel unload/load k*2 time units (k-1)*2 moves(k-1)*2 time units 4k-1 time units cycle length: max{ 4k-1, p+1 } optimal, if p+1  4k-1  p  4k-2

8. Cybernetix Architecture super single mode 6 5 2 3 4 710-5 9 6 7 8 11 … k+1 load/unloads (k-1 of them parallel)(k+1) * 2 time units k+1 moves (after each load/unload)(k+1) * 1 time units ----------------------- 3k+3 time units cycle length: max{ p+1, 3k+3 } optimal schedule for p+1  3k+3  p  3k+2 cycle begin cycle end see the schedule in the handouts....

9. First Results Cybernetix architecture / super-single mode theoretically optimal schedule with personalisation time p number of stations k  p-2 / 3 throughput (for greatest k) k/p+1 = 1/3( 1 – 3/(p+1)) alternative architecture / schedule theoretically optimal schedule with personalisation time p number of stations k  p+2 / 4 throughput (for greatest k) k/p+1 = 1/4( 1 + 1/(p+1))

10. Questions 1.Which architecture/schedule is better? Cybernetix better than alternative: (p-2)/3 > (p+2)/4p > 14 alternative better than Cybernetix: (p-2)/3 < (p+2)/4p < 14

11. Questions So far: relations for personalisation time and number of stations to get a theoretically optimal schedule. But: can’t more stations give more throughput, even if the schedule is not theoretically optimal any more? Cybernetix: k = (p-2) / 3 throughput: 1/3( 1 – 3/(p+1)) k = (p-2) / 3 + 1throughput: 1/3( 1 – 3/(p+4)) k = (p-2) / 3 + 2throughput: 1/3( 1 – 3/(p+7)) Result: more stations give more throughput, even if the schedule is not theoretically optimal any more. 2.

12. Questions So far: relations for personalisation time and number of stations to get a theoretically optimal schedule. But: can’t more stations give more throughput, even if the schedule is not theoretically optimal any more? alternative: k = (p+2) / 4 throughput: 1/4( 1 + 1/(p+1)) k = (p+2) / 4 + 1throughput: 1/4( 1 + 1/(p+5)) k = (p+2) / 4 + 2throughput: 1/4( 1 + 1/(p+9)) Result: more stations do not give more throughput. 2.

13. Questions 3.Consider faulty cards: How fast can we get back to the basic schedule? What are the conditions for chronological order when faulty cards appear?

14. Questions 4. More gaps on the belt for alternative architecture: Probabely advantageous when flip-overs, printers come into the game..... extend model

15. Questions To what extent can personalisation times vary? Can super-single mode react better on different personalisation times? 5.

16. Questions 6. How can model checking contribute?