{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lec09_fixpriority2 - Priority-Driven Scheduling of Periodic...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Priority-Driven Scheduling of Periodic Tasks (2) - Chapter 6 - Schedulable utilization bound Simpler method for the schedulabiity check
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 Utilization A periodic task’s utilization U i of an active resource is the ratio between its execution time and period: U i = C i /p i Given a set of periodic tasks on an active resource, e.g. the CPU, the CPU’s utilization is equal to the sum of periodic tasks’ utilization: = i i i p C U Can we find a bound called “schedulable utilization bound” under which a task set is guaranteed to be schedulable? e schedulabl is set task , if bound i i i U p C U = Processor utilization factor For a given algorithm A, we are interested in finding its schedulability bound (e.g., the schedulability bound of EDF is 1)
Background image of page 2
3 Processor utilization factor Entire Set of (e 1 ,…e n , p 1 , … p n ) Entire set of (e 1 ,…e n ) with fixed (p 1 , … p n) Fully utilized Find the minimum utilization factor among all fully utilized dots (barelly schedulable task set) Which pattern of e and p values? Now, we can consider only (e and p) combinations that make the system barely schedulable. How to find a (e and p) combination that has the minimal utilization factor? Always start with examples intuition generic theorem 0 0 3 6 9 7 U = 2/3 + 2/7 0 0 3 6 9 7 U = (2- )/3 + (2+2 )/7 3 / 7 / ) 3 / 7 ( /7 7/3 : increase /3 : decrease = <
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4 Which pattern of p values? For e values: “No-overflow theorem” What about p values? 0 0 3 6 9 7 U = 1/3 + 4/7 0 0 3 6 9 7 U = 1/(3*2) + (4+1)/7 Decrease: 1/3 – 1/(2*3)=1/(2*3) Increase: 1/7 Since 7 > 2*3, U decreases Transform (Ratio 3) to 2 2 2 1 2 2 1 1 2 2 1 2 1 2 1 1 1 2 1 2 1 1 2 1 2 1 1 2 1 1 2 2 1 10 ; 3, . . 2 4 10 4 10 , 2 2 4 2 10 2 2 2 * 4 2 4 2 4 2 ( ) ( ) ( ) ( ) 0 2 4 10 2 * 4 10 ( ) 0, 2 2 p p e e p where e g p p p p e e e p if p p e e e e e p p p p e e if p p p p p Since p + = = + = + + = = + + = + + + + + + > = > > 2 2 2 1 1 1 3, 3 2. ,1.5 1 and 2 2 2 p p p we have Thus p p p = > > > > = ; p 2 > 2p 1 p 1 is doubled, so we obtain ratio 2 among the periods
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}