Analysis of SRPT Scheduling:
Investigating Unfairness
Nikhil Bansal
(Joint work with Mor Harchol-Balter)
Motivation Problem
Client1
Client2
Client3
Aim:
“Good” Scheduling Policy
• Low Response times
• Fair
Server
Time Sharing (PS)
Server shared equally between all the jobs:
 Low response times
 Fair
 Does not require knowledge of sizes
Can we do better ?
Shortest Remaining Proc. Time
Optimal for minimizing mean response times.
Objections:
Knowledge of sizes
Improvements significant ?
Starvation of large jobs
Biggest fear
Questions
 Smalls better
Bigs worse
 How do means compare
 Elephant-mice property and implications
M/G/1 Queue Framework
Arrivals
queue
Server

• Poisson Arrival Process with rate
• Job sizes (S) iid general distribution F
Load(  ) = (arrival rate).E[S]
Queueing Formulas for PS
E[T(x)]: Expected Response time for job of size x
E[T ( x)] PS
x

1 
E[ S ( x)] PS
1

1 
[Kleinrock 71]
Identical for all!
M/G/1 SRPT
x
E[T ( x)]SRPT

 ( t 2 f (t )dt +x 2 (1  F ( x)))
0
2


2(1 ( x))
x
+
dt
0 (1   (t ))
Waiting Time (E[W(x)]) Residence Time (E[R(x)])
x
 ( x)    tf (t )dt
0
 Load up to x
 Variance up to x
 Gains priority after it begins execution
x
dt
 E[ R( x)]SRPT  E[T ( x)] PS 
 1 
0
All-Can-Win under srpt put c
Thm: Every job prefers SRPT, when load <= ½, for
all job size distributions.
Proof: Know that
E[W ( x)]SRPT 
If
E[ R( x)]SRPT  E[T ( x)]PS
1 
( E[T ( x)]PS  E[ R( x)]SRPT )
2
2(1   ( x))
(1   )  2(1   ( x)) 2
Key Observation
E[W ( x)]SRPT  E[T ( x)]PS  E[ R( x)]SRPT
 E[T ( x)]SRPT  E[T ( x)]PS
Holds for all x, if load <= 0.5
What if load > 0.5 ? problem
E[ R( x)]SRPT  E[T ( x)]PS
 ( x )  0 .5
Still holds if
Irrespective of 
The Heavy-Tailed Property: (Elephant -Mice)
1% of the big jobs make up at least 50% of the load.
For a distribution with the HT property,
>99% of jobs better under SRPT
In fact, significantly better,
Under SRPT,
E[S ( x)]
 2 + 2
SRPT
E[S ( x)]PS  1 /(1   )
Bounded by 4
Arbitrarily high
The very largest jobs
 If load <= 0.5, all jobs favor SRPT.
 At any load, > 99% jobs favor SRPT, if HT property.
Moreover significant improvements.
What about the remaining 1% largest jobs?
1. Bounding the damage theorem
Fill in…
2.
E[ S ]SRPT  k (  ) E[ S ]PS
 (1.01   ) 2(1   )

2  2
k() 

(log( 1  ) + 0.01log(
))
2 

2
2 
As   1,
k (  )  0.01
Implication:
Mean slowdown of largest 1% under SRPT: Same as PS
Insert plots here:
1 for BP 1.1 with load 0.9 showing how all
Do better
2 for exp with load 0.9 showing how some do bad.
Other Scheduling Policies
 Non-preemptive:
i. First Come First Serve (FCFS)
ii. Random
iii. Last Come First Serve (LCFS)
iv. Shortest Job First (SJF)

Very bad mean
Performance,
for HT
workloads
Preemptive:
i. Foreground Background (FB)
ii. Preemptive LCFS
Same as PS
Trivially worse
Overload
Add some lines for why good
+ we do work on this in paper
Actual Implementation
Add a plot or couple of lines
Conclusions
• Significant mean performance
improvements.
• Big jobs prefer SRPT under low-moderate
loads.
• Big jobs prefer SRPT even under high loads
for heavy-tailed distributions.
Scratch
Pr { X  x }  x

 1
Under h-t distributions
Job Percentile
SRPT
PS
90%
99%
99.9%
99.99%
100%
1.28
1.62
2.08
2.69
9.54
10
10
10
10
10
Very largest job
Load = 0.9
Heavy-tailed distribution with alpha=1.1
Under light-tailed distributions
Job Percentile
SRPT
PS
90%
3.17
10
95%
4.93
10
99%
11.14
10
99.9%
16.01
10
Load=0.9
Exponential distribution