[LITMUS^RT] Question from experimenting with Litmus-RT Performance
Yu-An(Victor) Chen
chen116 at usc.edu
Tue Jan 12 04:18:12 CET 2016
Hi,
I am doing some experiment with rt-xen and litmus-rt. What I am trying to
do is to see the schedubility of real time tasks of 1vm while the other vm
is fully utilized. Both guest VMs use litmus-rt.
The setup is the following:
Using Xen 4.5.0.
1. 2vm sharing core 0-7 ( both vm can access core0-7) , with RTDS
scheduler, both has period of 4000us and budget of 2000(us)
2. Dom0 using one core from CPU 8-15, with RTDS scheduler, period of
10000us and budget of 10000us
3. both guest vm have ubuntu 12.04 and are using "litmus-rt-2014.2.patch"
and with Geoffrey's patch for IPI interrupt (https://github.com/LITMUS-RT
/liblitmus/pull/1/files)
the taskset is generated as followed:
a taskset is composed of a collection of real-time tasks, and each
real-time task is a sequence of jobs that are released periodically... All
jobs are periodic, where each job Ti is defined by a period (and deadline) p
i and a worse-case execution time ei, with pi ≥ ei ≥ 0 and pi, ei ∈ integers.
Each job is comprised of a number of iterations of floating point
operations during each job. This is based on the base task.c provided with
the LITMUSRT userspace library.
The period for a task is from uniform distribution (10ms,100ms)
and the utilization rate of a task is also from a uniform distribution
(0.1,0.4)
In my experiment:
Step0: disable networking and other unused services.
Step1: I loaded VM#2 with constant running task with total utilization of 4
cores.
Step2: In VM#1 I many run iterations of tasksets from total utilization
rate 0.2 cores all the way to 4.6 and record their schedulbility using
st_trace.
In my results, I do see that schedulbility do drop to zeros at either total
util rate of 4.2 or 4.4. (we use worst-case execution time for benchmarking
the base amount of computation that is why it takes more than total util
rate of 4 for schedubility to drop to 0)
What puzzle me is why are there two groups of results as shown in the
attached graph?(one group that has schedubility of 0 at total util rate of
4.2 and another group that has schedubility of 0 at total util rate of 4.4)
( I used " * " in the legend to indicate the groups)
Shouldn't each run should be somewhat close to each other or showing
randomness instead of seeing the two groups of performance curves?
I wonder if my base computation is wrong but that still does not explain
why they are two types of performance curves.
Any advice or suggestion on how I can go about this will be helpful!
Thank you!
Victor
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