Quantitative Evaluation of Embedded Systems

Buffering

Buffering in streaming applications

Image taken from an online tutorial on the VLC media player

Buffering in dataflow graphs

S A CB

26ms

30ms

10ms

• Determine the MCM and choose a period μ ≥ MCM• For each actor a initialize a start-time Ta := 0• Repeat for each arc a—i—b :

Tb := Tb max (Ta + Ea – i μ)until there are no more changes

• Repeat for each actor a:Ta := min{all arcs a-i-b} (Tb - Ea + i μ)

until there are no more changes

• Delayed latency ≤ Toutput + Eoutput + δ·μ - Tinput

Invariants in a periodic schedule

Tb ≥ T

a + Ea – i μ

i ≥ (Ta - T

b + Ea ) / μ

μ = 22ms

St0 = 34ms

A CB

26ms

30ms

t0 = 0ms

t0 = 8ms

10ms i ≥ (Ta – Tc + Ea)/μ

Retaining the invariant when buffering

μ = 22ms

St0 = 34ms

A CB

26ms

30ms

t0 = 0ms

t0 = 8ms

10ms i ≥ (Ta – Tc + Ea)/μi ≥ δ

Retaining the invariant when buffering

A B

Retaining invariant => retaining MCM

By definition of periodic schedule we findfor every arc x-#-y that Ty ≥ Tx + Ex – # μ

Adding these steps over a path B-X-Y----Z-A we find that TA ≥ TB + E(B-X-Y---Z) – #(B-X-Y---Z-A) μ

And so we find that i ≥ (TA - TB + EA)/μ ≥ E(B-X-Y---Z-A) / μ – #(B-X-Y---Z-A)

Which means for the cycle mean on the newly created cycle B-X-Y----Z-A-B:E(B-X-Y---Z-A) / (#(B-X-Y---Z-A) + i) ≤ μ

i

A BReversely, assume that we pick i so that the MCM of the graph does not change.

Then for any cycle B-X-Y----Z-A-B we know:E(B-X-Y---Z-A) / (#(B-X-Y---Z-A) + i) ≤ μ By definition of periodic schedule we havefor every arc x-#-y that Tx ≤ Ty - Ex + # μ

Adding these steps over a path B-X-Y----Z-A we find that TB ≤ TA - E(B-X-Y---Z) + #(B-X-Y---Z-A) μ

And so: (TA - TB + EA)/μ ≤ E(B-X-Y---Z-A) / μ – #(B-X-Y---Z-A) ≤ i

i

Retaining MCM => retaining invariant

How large should a buffer be?

if and only if

the periodic schedule stays unchangedif and only if

the MCM stays unchanged

S A CB

i But be aware of δ…