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Transcript of Cybernetix – Cost Driven UPPAAL and Insights Angelika Mader University of Twente Ametist meeting...
![Page 1: Cybernetix – Cost Driven UPPAAL and Insights Angelika Mader University of Twente Ametist meeting December 2002 Dortmund.](https://reader036.fdocuments.us/reader036/viewer/2022083009/5697bfd41a28abf838cac90f/html5/thumbnails/1.jpg)
Cybernetix – Cost Driven UPPAAL and Insights
Angelika MaderUniversity of Twente
Ametist meeting
December 2002
Dortmund
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• recognize order deadlocks a.s.a.p
• reward personalisation, load/unload
Extending the UPPAAL Model
from T. Krilavicius & Y. Usenko
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3 32 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....
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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
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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
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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.
GOOGOODD NEWNEWSS
The super-single mode meets the theoretical lower bound, if the personalisation time is not too low.
here we abstract away from the
belt
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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 holdscondition 1 holds, if the personalisation time is long enough
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Alternative Architecture
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• 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
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Cybernetix Architecturesuper single mode
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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
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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....
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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))
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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
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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 + 1 throughput: 1/3( 1 – 3/(p+4))
k = (p-2) / 3 + 2 throughput: 1/3( 1 – 3/(p+7))
Result: more stations give more throughput, even if the schedule is not
theoretically optimal any more.
2.
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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 + 1 throughput: 1/4( 1 + 1/(p+5))
k = (p+2) / 4 + 2 throughput: 1/4( 1 + 1/(p+9))
Result: more stations do not give more throughput.
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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?
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Questions
4. More gaps on the belt for alternative architecture:
Probabely advantageous when flip-overs, printers come into the game.
.... extend model
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Questions
To what extent can personalisation times vary?
Can super-single mode react better on differentpersonalisation times?
5.
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Questions
6. How can model checking contribute?