Cybernetix – Cost Driven UPPAAL and Insights

Angelika MaderUniversity of Twente

Ametist meeting

December 2002

Dortmund

• recognize order deadlocks a.s.a.p

• reward personalisation, load/unload

Extending the UPPAAL Model

from T. Krilavicius & Y. Usenko

1

3

2

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....

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

   

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

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

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

Alternative Architecture

5 6

1 2

7

3

8

4

5 1 6 2 7 3 8 45 1 6 2 7 3 8 41

5

2

6

3

7

4

8

1

5

2

6

3

7 8

9 9 1

5 6

2

7

3

8

9

5

1

6

2

7

3

8

9

5

1

6

2

7 8

10 10 9

5 6

1

7

2

8

10

5

9

6

1

7

2

8

10

5

9

6

1

7 8

11 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

Cybernetix Architecturesuper single mode

6 -1

5 2

3 4

7

10 -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....

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))

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

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.

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.

2.

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?

Questions

4. More gaps on the belt for alternative architecture:

Probabely advantageous when flip-overs, printers come into the game.

.... extend model

Questions

To what extent can personalisation times vary?

Can super-single mode react better on differentpersonalisation times?

5.

Questions

6. How can model checking contribute?