Q uantitative E valuation of E mbedded S ystems QUESTION DURING CLASS?

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Q uantitative E valuation of E mbedded S ystems QUESTION DURING CLASS?

Consider a car manufacturing line consisting of... 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

AB C D 10min 20min 10min 20min 30min 20min 5min 3min 4min 15min 1min Disclaimer: no actual car assembly line was studied in order to make this model.

Simulate a few firings assuming sufficient input tokens. Determine the (max,+) matrix. Determine the max. throughput. Determine a periodic schedule for: 1)µ = MCM 2)µ = 2*MCM 3)µ = 3*MCM 4)as a function of µ EXERCISE: Keep your answers for next time!

Q uantitative E valuation of E mbedded S ystems

0 ms y Cycles with a 0 execution time cause livelocks But when logging events, this is mathematically okay...

AB 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.

AB C D 1ms 2ms 4ms u x3x3 y x1x1 x2x2 3ms

AB C D 1ms 2ms 4ms u x3x3 y x1x1 x2x2 3ms...but only when assuming: x 1 (1) = x 2 (1) which is ok for self-timed execution, but not when reasoning about periodic schedules

A B C 1ms 2ms u x3x3 y x1x1 x2x2 3ms