1. Smart Scheduling and Dispatching Policies
Lecture 7
Smart Scheduling and Dispatching Policies

2. Single Server Model (M/G/1)
Poisson
arrival
process
w/rate l
Load r = lE[X]<1
X: job size
(service requirement) 1
Bounded Pareto
CPU Lifetimes of UNIX jobs [Harchol-Balter, Downey 96]
Supercomputing job sizes [Schroeder, Harchol-Balter 00]
Web file sizes [Crovella, Bestavros 98, Barford, Crovella 98]
IP Flow durations [Shaikh, Rexford, Shin 99]
Job sizes with huge variance are everywhere in CS:
Huge
Variability
D.F.R.
Top-heavy:
top 1% jobs
make up half
load

3. Outline Smart scheduling
Performance metrics
Policies classification
Examples
Scheduling policies comparison (Fairness, Latency)
Task assignment problem
Supercomputing and web server models
Optimal dispatching/scheduling policies
+ they’ll go out first 
Ladies first!

4. Smart scheduling: Motivation (I)
Why scheduling matters?
Why doesn’t it work?!
Bla, bla, bla…
Grrrrr!
Why doesn’t it work?!
Bla, bla, bla…
Why doesn’t it work?!
!! Delay due to other users who are currently sharing the service !!

5. Smart scheduling: Motivation (II)
The goal of smart scheduling is to reduce mean delay “for free”, i.e., by simply serving jobs in the “right order”, no additional resources
Which is the right order to schedule jobs?
The answer strongly depends on
system load
job size distribution

6. Smart scheduling: Performance metrics (I)
Common metrics to compare scheduling policies
E[T], mean response time
E[N], mean number (of jobs) in system
E[TQ], mean waiting time (= E[T]-E[S], where E[S]=service time)
Slowdown: SD=T/S (response time normalized by the running time)
Meaning: if a job takes twice as long to run due to system load, it suffers from a Slowdown factor of 2, etc.
Job response time should be proportional to its running time. Ideally:
small jobs → small response times
big jobs → big response times

7. Smart scheduling: Performance metrics (II)
Starvation/fairness metrics
A low average Slowdown doesn’t necessarily mean fairness (starvation of large jobs)
Good metric: E[SD(x)] is the expected slowdown of a job of size x, i.e., mean slowdown as a function of x
E[SD]= E[T]/E[S]?
No! First we need to derive:
Then, we get the mean SD:

8. Scheduling policies: classification
Definitions:
Preemptive policy: a job may be stopped and then resumed later from the same point where it was stopped
Size-based policy: it uses the knowledge of job size
Classification
Non-Preemptive, Non-Size-Based Policies
Preemptive, Non-Size-Based Policies
Non-Preemptive, Size-Based Policies
Preemptive, Size-Based Policies
Focus on M/G/1 queue (General job size distribution)
Poisson, λ

9. Non-Preemptive, Non-Size-Based Policies (I)
Non-preemptive policies (each job is run to completion), that don’t assume knowledge of job size, are:
FCFS (First-Come-First-Served) or FIFO
Jobs are served in the order they arrive. Each job is run to completion before next job receives service (e.g., call centers, supercomputing centers)
LCFS (Last-Come-First-Served non-preemptive)
When the server frees up, it always chooses the last arrived job and runs that job to completion (jobs piled onto a stack)
RANDOM
When the server frees up, it chooses a random job to run next (mostly of theoretical interest)

10. Non-Preemptive, Non-Size-Based Policies (II)
Interesting property: All non-preemptive service orders that do not make use of job sizes have the same distribution on the number of jobs in the system (time until completion is equal in distribution for all these policies)
Hence, same E[T], E[N]
What about E[SD]?
For all these policies (in an M/G/1):
Thus,
Since E[S], E[TQ] is the same for each policy → They have the same E[SD]
Proportional to job size variability!
Independently of the job’s size!

11. Preemptive, Non-Size-Based Policies (I)
So far: non-preemptive/non-size-based service
E[T] can be very high when job size variability is high
Intuition: short jobs queue up behind long jobs
Processor-Sharing (PS): when a job arrives, it immediately shares the capacity with all the current jobs (Ex. R.R. CPU scheduling)
+ PS allows short jobs to get out quickly, helps to reduce E[T], E[SD] (compared to FCFS), increases system throughput (different jobs run simultaneously)
- PS is not better than FCFS on every arrival sequence
+ Mean response time for PS is insensitive to job size variability:
E[T]M/G/1/PS= E[S] / (1-ρ)
where ρ is the system utilization (load)

12. Preemptive, Non-Size-Based Policies (II)
Performances of M/G/1/PS system
Response time
Think of Little’s law!
Mean Slowdown
!!! Constant Slowdown (independent of the size x)
!!! In non-preemptive, non-size-based scheduling: E[SD] for small jobs was greater than the one for large jobs
Under PS, all jobs have same Slowdown → FAIR Scheduling

13. Preemptive, Non-Size-Based Policies (III)
Preemptive-LCFS: a new arrival preempts the job in service, when that arrival completes, the preempted job is resumed
E[T(x)], E[SD(x)] as for the PS case
+ wrt PS: less # of preemptions (only 2 per job)
Can we drop E[SD(x)]? → lower SD for smaller jobs
!!! We don’t know the size of the jobs!
FB (Foreground-Background) or LAS - Least Attained Service
Idea: To reduce E[SD] → use knowledge of job’s age (indicator of remaining CPU demand), and serve the job with lowest age

14. Foreground-Background scheduling (cont’d)
Used to control execution of multiple processes on a single processor: two queues (F and B) and one server
Idea of FB: The job with the lowest CPU age gets the CPU to itself.
If several jobs have same lowest CPU age, they share the CPU using PS
Performance depends on how good is the predictor of the age of remaining size (depends on job size distribution)!!
Jobs enter queue F
(PS service)
When a job hits a certain age a, it is moved to queue B
Jobs in B get service only when queue F is empty

15. Non-Preemptive, Size-Based Policies (I)
Size-based policies: special case of Priority Queueing
Often used in computer systems, e.g., database (differentiated levels of service), scheduling of HTTP requests, high/low-priority transactions
Size-based scheduling: it can improve the performance of a system tremendously!
Priority queueing (non-preemptive)
Consider an M/G/1 priority queue with n classes: class 1 has highest priority, n the lowest
Class k job arrival rate is λk= λ pk
Time in queue for jobs of priority k is
Waiting for the job in service
E[TQ(k)]NP-Priority < E[TQ]FCFC
+ k-class job only see load
due to jobs of class ≤ k
Waiting for the jobs in the queue of ≥ priority
Waiting for the jobs of higher priority arriving after k

16. Non-Preemptive, Size-Based Policies (II)
Question: If you want to minimize E[T], who should have higher priority: large or small jobs?
Theorem: Consider an NP-Priority M/G/1 with two classes of jobs: small (S) and large (L). To minimize E[T], class S jobs should have priority over class L jobs (since E[SS]<E[SL])
SJF - Non-preemptive Shortest Job First
Whenever the server is free, it chooses the job with the smallest size (once a job is running, it is never interrupted)
Under heavy-tailed distributions, E[TQ] is smaller than the FCFS one (since most jobs are small)
But, mean delay is proportional to the variance → large delays for very high variance
Small jobs can still get stuck behind a big one (already running) → need of preemption!

17. Preemptive, Size-Based Policies
So far: non-preemptive policies → higher delay under highly variable job size distributions
Preemptive priority queueing
PSJF - Preemptive Shortest Job First
Similar to SJF policy, the job in service is the job with the smallest original size
A preemption occurs if a smaller job arrives
Mean response time far lower than under SJF (PSJF is far less sensitive to variability in job size distr.)
Better compared to non-preemptive, it depends only on the first k priority classes variability!

18. SRPT SRPT - Shortest Remaining Processing Time
Whenever the server is free, the job chosen is the one with shortest remaining processing time
Preemptive policy: a new arrival may preempt the current job in service if it has shorter remaining processing time
Compared to PSJF
SRPT takes into account of remaining service requirement and not just the original job size
Overall mean response time is lower
Compared to FB
In SRPT, a job gains priority as it receives more service
In FB, a job has highest priority when it first enters
In an M/G/1 → E[T(x)]SRPT ≤ E[T(x)]FB

19. Policies comparison: mean response time (I)
M/G/1 queue, job size distribution is Bounded Pareto
FCFS
SJF 24
LAS
LAS
FCFS
SJF 20
LAS
FCFS
SJF 16
LAS
FCFS
SJF 12
Versus ρ
SJF/FCFS delay very high even for low ρ
SRPT/LAS delay slightly increases with ρ
SRPT has the lowest delay
FCFS
SJF
LAS 8
E[T] PS
SRPT
Versus C2
SJF/FCFS delay increases with C2
LAS delay decreases with C2 (DFR needs higher C2)
PS and SRPT are invariant to C2 r
C2 =
(C 2, is the squared coefficient of variation)
Source: Prof. Mor Harchol-Balter,

20. Policies comparison: mean response time (II)
Weibull distribution, ρ=0.7
Fast increase with C2
Invariant to job size variability
Requires higher C2 to perform well
Source: Prof. Mor Harchol-Balter,

21. Exercise
M/G/1 queue
Job size distribution: Bounded Pareto
The load is ρ = 0.9
The (very) biggest job in the job size distribution has size x = 1010
Question: E[T(x)] is lower under SRPT scheduling or under PS scheduling?
Source: Prof. Mor Harchol-Balter,

22. Exercise: Solution Small jobs should favor SRPT
Large jobs have the lowest priority under SRPT, but they get treated equally under PS (equal time-sharing)
Thus, it seems much better for “Mr. Max” to go to the PS queue
E[T(x)]PS should be far lower than E[T(x)]SRPT
Source: Prof. Mor Harchol-Balter,

23. Exercise: Solution (cont’d)
The largest job prefers SRPT to PS, but almost all jobs ( %) prefer SRPT to PS by more than a factor of 2
99% of jobs prefer SRPT to PS by more than a factor of 5
But how can this be? Can every job really do better in expectation under SRPT than under PS?
All-Can-Win Theorem! (for BP distribution holds for ρ < 0.96)
< 5 times
Source: Prof. Mor Harchol-Balter,

24. SRPT: Fairness SRPT is optimal wrt mean response time
In practice, not used for scheduling job
Job size is not always known
PS is preferred in web servers, unless serving static requests
What about Fairness?
A policy is fair if each job has the same expected SD, regardless of its size
SRPT vs PS? SRPT worse with large jobs?
All-Can-Win Theorem: in an M/G/1, if ρ<0.5, → E[T(x)]SRPT ≤ E[T(x)]PS (for all distribution, for all x)
Intuition: Once a large job starts to get the service, it gains priority; under light load even a job of large size x could do worse under PS than under SRPT (because of higher residence time)

25. Summary on scheduling single server (M/G/1): E[T]
Poisson
arrival
process
Load r <1
Smart scheduling greatly improves mean response time (e.g., SRPT)
Variability of job size distribution is key
Let’s order the policies based on E[T]:
LOW E[T] HIGH E[T]
SRPT < LAS < PS < SJF < FCFS
Requires D.F.R.
(Decreasing
Failure Rate)
~E[S2]
(shorts caught behind longs)
Insensitive
to E[S2]
Surprisingly bad:
(E[S2] term)
OPT for all
arrival
sequences
No “Starvation!” Even the biggest jobs prefer SRPT to PS
Source: Prof. Mor Harchol-Balter,

26. (Sometimes scheduling policy is fixed – legacy system)
Multiserver Model
Server farms:
+ Cheap
+ Scalable capacity
Sched. policy
Routing
(assignment)
policy
Incoming
jobs:
Poisson
Process
Sched. policy
Router
Sched. policy
2 Policy Decisions
(Sometimes scheduling policy is fixed – legacy system)

27. Outline & Review of scheduling in single-server Supercomputing
FCFS
Router
Supercomputing
Web server farm model PS
Router
IV. Towards Optimality …
SRPT
Router
&
Metric:
Mean
Response
Time,
E[T]

28. Supercomputing Model Jobs are not preemptible.
FCFS
Router
Routing
(assignment)
policy
Poisson
Process
Jobs are not preemptible.
Jobs processed in FCFS order.
Assume hosts are identical.
Jobs i.i.d. ~ G: highly variable size distribution.
Size may or may not be known. Initially assume known.

29. Q: Compare Routing Policies for E[T]?
FCFS
Router
Routing
policy
Poisson
Process
Jobs i.i.d. ~ G: highly variable
Supercomputing
Round-Robin
2. Join-Shortest-Queue
Go to host w/ fewest # jobs.
Least-Work-Left
Go to host with least total work.
5. Central-Queue-Shortest-Job (M/G/k/SJF)
Host grabs shortest job when free.
6. Size-Interval Splitting
Jobs are split up by size among hosts.

30. Supercomputing model (II)
High
E[T]
Low
Round-Robin
Jobs assigned to hosts (servers) in a cyclical fashion
2. Join-Shortest-Queue
Go to host with fewest # jobs
3. Least-Work-Left (equalize the total work)
Go to host with least total work (sum of sizes of jobs there)
4. Central-Queue-Shortest-Job (M/G/k/SJF)
Host grabs shortest job when free
5. Size-Interval Splitting
Jobs are split up by size among hosts. Each host is assigned to a size interval (e.g., Short/Medium jobs go to the first host, Long jobs go to the second host)
Hp: Job size is known!

31. What if job size is not known?
The TAGS algorithm “Task Assignment by Guessing Size” s
Host 1 m
Host 2
Outside
Arrivals
Host 3
Answer:
When job reaches size limit for host, then it
is killed and restarted from scratch at next host.
Explain where used – Microsoft example, UNIX example, supercomputing centers actually do this if you can’t predict the runtime of your job, although they do it preemptively.
[Harchol-Balter, JACM 02]

32. Results of Analysis Bounded Pareto Jobs Random Least-Work-Left TAGS
2 hosts only, system load = JSQ is in between LWL and Random. Mean job size = 3000.
High
variability
Lower
variability

33. Supercomputing model (III)
Summary
This model is stuck with FCFS at servers. It is important to find a routing/dispatching policy that helps smalls not be stuck behind bigs → Size-Interval Splitting
By isolating small jobs, can achieve effects of smart single-server policies
Greedy routing policies (JSQ, LWL) are poor (don’t provide isolation for smalls, not good under high variability workloads)
Don’t need to know size (TAGS = Task Assignment by Guessing Size)

34. Web server farm model (I)
Examples: Cisco Local Director, IBM Network Dispatcher, Microsoft SharePoint, etc.
Router
Routing
policy
Poisson
Process PS
HTTP requests are immediately dispatched to server
Requests are fully preemptible
Processor-Sharing (HTTP request receives “constant” service)
Jobs i.i.d. with distribution G (heavy tailed job size distr. for Web sites)

35. Web server farm model (II)
Random
2. Join-Shortest-Queue
Go to host with fewest # jobs
3. Least-Work-Left
Go to host with least total work
4. Size-Interval Splitting
Jobs are split up by size among hosts
E[T]
JSQ
LWL
RAND
SIZE
8 servers, r = .9, C2=50
Shortest-Queue is better (high variance distr.)
Same for E[T], but not great
Source: Prof. Mor Harchol-Balter,

36. Optimal dispatching/scheduling scheme (I)
What is the optimal dispatching + scheduling pair?
Central-queue-SRPT looks very good
Is Central-queue-SRPT always optimal for server farm?
No!! It does not minimize E[T] on every arrival sequence!
Practical issue: jobs must be immediately dispatched (cannot be held in a central queue)!!
Assumptions:
Jobs are fully preemptible within queue
Jobs size is known
SRPT

37. Optimal dispatching/scheduling scheme (II)
Claim:
The optimal dispatching/scheduling pair, given immediate dispatch, uses SRPT at the hosts
Router
SRPT
Immediately
Dispatch Jobs
Incoming
jobs
Intuition: SRPT is very effective at getting short jobs out → it reduces E[N], thus the mean response time E[T] (Little’s law)
→ narrow search to policies with SRPT at hosts!

38. Optimal dispatching/scheduling scheme (III)
Optimal immediate dispatching policy is not obvious!
RANDOM task assignment performs well: each queue looks like an M/G/1/SRPT queue with arrival rate λ/k
Idea: short jobs spread out over SRPT servers → IMD algorithm (Immediate Dispatching)
Divide jobs into size classes (e.g., small, medium, large) and assign jobs to the server with fewest # of jobs of that size class
Each server should have some small, some medium and some large jobs (so that SRPT can be maximally effective)
IMD performance is as good as Central-Queue-SRPT
Almost no stochastic analysis (analysis available for worst-cases)!

39. + Summary Supercomputing Web server farm model FCFS PS
Router PS
Router
Need Size-interval splitting
to combat job size variability
and enable good performance
Job size variability is not an issue
LWL, JSQ, performs well
Optimal dispatching/scheduling pair
SRPT
Router +
Both have similar worst-case E[T]
Almost exclusively worst-case analysis, so hard to compare with above results
Need stochastic research
Source: Prof. Mor Harchol-Balter,

40. Exercises Ex. 1 – Slowdown
Jobs arrive at a server which services them in FCFS order. The average arrival rate is λ = 1/2 job/sec. The job sizes (service times) are independently and identically distributed according to random variable S where: S=1 with prob. ¾, S=2 o.w.
Suppose: E[T] = 29/12. Compute the mean slowdown, E[SD], where the slowdown of job j is defined as Slow(j) = T(j)/S(j), where T(j) is the response time of job j and S(j) is the size of job j.
Solution:
Recall the definition of response time for a FCFS queue: T = TQ + S. Here, TQ is the waiting or queueing time. Thus,

41. Exercises
Since the server is FCFS, a particular job’s waiting time is independent of its service time. This fact allows us to break up the expectation, giving us:
The distribution of S is given, so we calculate E[S] and E[1/S] using the definition of expectation:
E[S] =5/4 and E[1/S]=7/8
Then, we get E[SD] =1+[(29/12 – 5/4)7/8]=97/48.
If the service order had been SJF, would the same technique have worked for computing mean slowdown?
In the SJF case, S and TQ are not independent, so we can’t split the expectation as we did above. The reason why they are not independent is because job size affects the queueing order: short jobs get to jump to the front of the queue under SJF, and hence their TQ is shorter.

42. Exercises Ex. 2 – FCFS-SJF-RR CPU scheduling
Compute the average waiting time for processes with the following next CPU burst times (ms) and ready queue order:
P1: 20
P2: 12
P3: 8
P4: 16
P5: 4

43. Exercises + Very simple algorithm - Long waiting time! Solution FCFS:
Average waiting time: 148/5=29.6 P1 P2 P3 P4 P5 20 32 40 56 60
+ Very simple algorithm
- Long waiting time!

44. Exercises + Shorter average waiting time - Requires future knowledge
Solution SJF:
Waiting time:
T1=40
T2=12
T3=4
T4=24
T5=0
Average waiting time: 16 P5 P2 P3 P4 P1 4 12 40 24 60
+ Shorter average waiting time
- Requires future knowledge

45. Exercises
Solution RR scheduling: Give each process a unit of time (time slice, quantum) of execution on CPU. Then move to next process in the queue. Continue until all processes completed.
Hp: Time quantum of 4. P1 P2 P3 P4 P5 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60
P5 completes
P3 completes
P2 completes
P4 completes
P1 completes

46. Exercises Same exercise with other scheduling disciplines!
Waiting time:
T1: 60-20=40
T2: 44-12=32
T3: 32-8=24
T4: 56-16=40
T5: 20-4=16
Average waiting time: 30.4 P1 P2 P3 P4 P5 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60
P5 completes
P3 completes
P2 completes
P4 completes
P1 completes
Ave. waiting time high
+ Good ave. response time (Important for interactive/time-sharing systems)
Use of smaller quantum (overhead increase)
Same exercise with other scheduling disciplines!

47. Exercises
Ex. 3 - LCFS
Derive the mean queueing time E[TQ]LCFS. Derive this by conditioning on whether an arrival finds the system busy or idle.
Solution:
With probability 1 − ρ, the arrival finds the system idle. In that case E[TQ] = 0.
With probability ρ, the arrival finds the system busy and has to wait for the whole busy period started by the job in service.
The job in service has remaining size Se. Thus the arrival has to wait for the expected duration of a busy period started by Se, which we denote by E[B(Se)] = E[Se]/(1−ρ).

48. Exercises
You can derive this fact by first deriving the mean length of a busy period started by a job of size x, namely E[B(x)] = x /(1−ρ) , and then deriving E[B(Se)] by conditioning on the probability that Se equals x.
Putting these two pieces together, we have
As expected, this is exactly the mean queueing time under FCFS.

49. Exercises Ex. 4 – Server Farm
Suppose you have a distributed server system consisting of two hosts. Each host is a time-sharing host. Host 1 is twice as fast as Host 2.
Jobs arrive to the system according to a Poisson process with rate λ = 1/9.
The job service requirements come from some general distribution D and have mean 3 seconds if run on Host 1.
When a job enters the system, with probability p = 3/4 it is sent to Host 1, and with probability 1 − p = 1/4 is sent to Host 2.
Question: What is the mean response time for jobs?

50. Exercises Solution: The mean response time is simply: PS
E[T] = ¾ (Mean response time at server 1)+ ¼ (Mean response time at server 2)
But server 1 (2) is just an M/G/1/PS server, which has the same mean response time as an M/M/1/FCFS server, namely just
Thus, PS
Poisson (1/9)
3/4
1/4
3 sec.
6 sec.