Quantitative Evaluation of Embedded Systems

QUESTION DURING CLASS?Email : qees3TU@gmail.com

FAIL!

Thank you, Robin Wolffensperger en Ruben Lubben!

Consider a car manufacturing line consisting of...

Exercise:Model a car manufacturing line

• Four assembly robots: A,B,C and D• A production unit that needs 20 minutes to produce a chassis• A production unit that needs 10 minutes to produce a steering installation• A production unit that needs 10 minutes to produce a breaking system• A production unit that needs 20 minutes to produce a body• Three painting units that each need 30 minutes to paint a body• A production unit that needs 15 minutes to produce a radio• Robot A compiles the chassis and the steering installation in 4 min. and sends it to B• Robot B adds the breaking system in 3 min. and sends it to C• Robot C adds a painted body in 5 min. and sends it to D• Robot D adds a radio in 1 min. and sends the car out of the factory• For safety reasons, there can be at most 3 ‘cars’ between A and C, and only 2 between B and D• Every robot can only deal with one of each of the assembled components at a time

Exercise: calculate the first 3 firings of each actor

A B C D

10min

20min 10min 20min

30min20min

5min3min4min

15min

1min

Disclaimer: no actual car assembly line was studied in order to make this model.

Answer: Model a car manufacturing line

Simulate a few firings assuming sufficient input tokens.

Determine the (max,+) matrix.

Determine the max. throughput.Determine a periodic schedule for:1) µ = MCM2) µ = 2*MCM3) µ = 3*MCM4) as a function of µ

EXERCISE:

Keep your answers for next time!

Quantitative Evaluation of Embedded Systems

Recall the characteristic equations…

u(n)Dmax(n)xCy(n)u(n)Bmax(n)xA1)(nx

or for autonomous systems…

(n)xCy(n)(n)xA1)(nx

What about this one?

0 ms

y

(n)x(0)y(n)(n)x(0)1)(nx

0y(n)

Cycles with a 0 execution time cause livelocks

But when logging events, this is mathematically okay...

And this one ?

u(n)Dmax(n)xCy(n)u(n)Bmax(n)xA1)(nx

A B

C D

1ms 2ms

4ms

u y

3ms

Theorem: The number of tokens on any cycle is constant!

Therefore, every cycle must contain at least one token,otherwise a deadlock occurs.

And this one?

A B

C D

1ms 2ms

4ms

u

x3

yx1

x2

3ms u(n)max(n)x2y(n)

u(n)1

1

max(n)x

2

5810

5810

1)(nx

Reducing rows…

A B

C D

1ms 2ms

4ms

u

x3

yx1

x2

3ms

u(n)max(n)x2y(n)

u(n)1

max(n)x2

5101)(nx

...but only when assuming: x1(1) = x2(1)which is ok for self-timed execution,but not when reasoning aboutperiodic schedules

What about reducing columns?

A B

C1ms 2ms

u x3 y

x1

x2

3ms u(n)max(n)x22y(n)

u(n)

1

4max(n)x

1

545

22

1)(nx

And back to (max,+) algebra…

302

01

11

230

121

301

20

ΙAAAA

0

n

n

times