[SOLVED] 程序代写 COMP9334 2

Capacity Planning of Computer Systems and Networks
Week 1B: Queuing networks. Operational analysis

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• Solve capacity planning by solving a number of performance analysis problems
• Performancemetrics
• Response time, waiting time • Throughput
• SingleserverFIFOqueue • A server = A processing unit
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This lecture
• Queueing networks • Operational analysis
• Fundamentallawsrelatingthebasicperformancemetrics
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Modelling computer systems
• Single server queue considers only a component within a computer system
• AcomponentcanbeaCPU,adisk,atransmissionchannel
• A request may require multiple resources
• E.g.CPU,disk,networktransmission
• We model a computer systems with multiple resources
by a Queueing Networks (QNs)
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Pictorial representation of single server queues
Waiting line
Arriving customers
Arriving requests
Requests waiting to be processed
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Pictorial representation of queues
Systems with m servers
Waiting line
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A simple database server
The server has a CPU and a disk.
pen queueing network
xternal arrivals
orkload intensity specified by arrival rate
A transaction may visit the CPU and disk multiple times. nbouT1n,2d02e2dnumCbOeMPr93o34fcustomersinthesystem 7

Database servers for batch jobs
• Example:Batchprocessingsystem
• E.g.Forsummarizationdatafromdatabases
• Noon-linetransactions
atabase server for batch jobs
unning batch jobs overnight
E.g. producing managerial reports
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Open vs. closed queueing networks (1)
pen queueing network
Open queueing network
• External arrivals
• Workload intensity
specified by arrival rate
xternal arrivals
abase server for batch jobs
orkload intensity specified by arrival rate nboundednumberostomersinthesystem
equilibrium, flow in = flow out throughput = arrival a
ning batch jobs overnight
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Closed queueing network • No external arrivals
• Workload intensity
specified by customer
population
E.g. producing managerial reports

Open vs. closed queueing networks (2)
en queueing network
Open queueing network • Possibly unbounded #customers
• For stable equilibrium
Throughput = arrival rate
ernal arrivals
abase server for batch jobs
rkload intensity specified by arrival rate boundednumberofomersinthesystem
quilibrium, flow in = flow out throughput = arrival ra
ning batch jobs overnight
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Closed queueing network • Known #customers
• Throughput depends on
#customers etc.

E.g. producing managerial reports

Open vs. closed queueing networks – Terminology
en queueing network
ernal arrivals
abase server for batch jobs
rkload intensity specified by arrival rate boundednumberofomersinthesystem
quilibrium, flow in = flow out throughput = arrival ra
ning batch jobs overnight
Work in a closed queueing network is called jobs
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Work in an open queueing network is called transaction

E.g. producing managerial reports

DB server – mixed model
• Theserverhasboth
• Externaltransactions
• Batchjobs
Mixed queueing network
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Different techniques are needed to analyse open and closed TqruaensuaecitniognnetwoMrkasximum Average Minimum
Service Level Agreements
Group Response Time (sec) Throughput

owed into the database. If the MPL is too
suffer, since not all DBMS resources he other hand, if the MPL is too high,
up in an external queue. The application can then control the order in which transactions are executed by scheduling
DB server – Multi-programming level
ontrol on scheduling. The question of L to achieve both goals simultaneously
incoming transactions
• Somedatabaseserver
ot just for databases but in system de-
the external queue.
management systems (DBMS)
in we study this problem in the context
set an upper limit on the number
loads, both via extensive experimenta- oretic analysis.
MPL=4 DBMS
of active transactions within the
external queue
o most critical factors in adjusting the
of resources that the workload utilizes
• Thisupperlimitiscalledmulti- the transactions’ service demands. We
programming level (MPL)
ased controller, augmented by queue- for automatically adjusting the MPL. methods to the specific problem of ex- of transactions. We find that external nearly as effective as internal prioriti- egative consequences, when the MPL
Figure 1. Simplified view of the mechanism used in external scheduling. A fixed limited number of trans- actions (MPL=4) are allowed into the DBMS simul- taneously. The remaining transactions are held back in an external queue. Response time is the time from when a transaction arrives until it completes, includ- ing time spent queueing externally to the DBMS.
• AhelppagefromSAPexplainingMPL
Examples of recent work on external scheduling come
• http://dcx.sap.com/1200/en/dbadmin_en12/running-s-3713576.html
• PicturefromSchroderetal.“Howtodetermineagoodmulti-
eb applications are largely dependent
se, whperoe gthreamamjoirnitgy loefvtehel froeqruexstternal scheduling”
from many areas including storage servers, web servers, and
database servers. For example, Jin et al. [9] develop an ex-
rants CCR-0133077, CCR-0311383, 0313148,
ternal scheduling front-end to provide proportional sharing
among the requests at a storage service utility. Blanquer et
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will On t ent c MP em, n Here work
g the e tw ber
ty of ck b
h Digital Greenhouse Grant.
MPL is already met, all remaining transactions are queued
al. [4] study external scheduling for quality of service pro-

Operational analysis (OA)
• “Operational”
• Collectperformancedataduringday-to-dayoperation
• Operation laws
• Applications:
• Usethedataforbuildingqueueingnetworkmodels • Performbottleneckanalysis
• Performmodificationanalysis
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Single-queue example (1)
#requests = A
#requests = C
In an observational period of T, server busy for time B A requests arrived, C requests completed
A, B and C are basic measurements
Deductions: Arrival rate l = A/T Output rate X = C/T
Utilisation U = B/T
Mean service time per completed request = B/C
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Motivating example
• Observationperiod=1minute
• Busy for 36s.
• 1790 requests arrived
• 1800 requests completed • Find
• Mean service time per completion = 36/1800 = 0.02s
• Utilisation = 36/60 = 60%
• Arrival rate =
• Output rate =
1790/60 = 29.83 requests /s
1800/60 = 30 requests/s
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Utilisation law
• The operational quantities are inter-related
• Consider
• UtilisationU=B/T
• MeanservicetimepercompletionS=B/C • OutputrateX=C/T
• Utilisation law – Can you relate U, S and X? • U=SX
• Utilisation law is an example of operational law.
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Application of OA
• Don’thavetomeasureeveryoperationalquantities • MeasureBtodeduceU-don’thavetomeasureU
• Consistencychecks
• IfU1SX,somethingiswrong
• Operationallawscanbeusedforperformanceanalysis • Bottleneckanalysis(Lecture2A)
• Meanvalueanalysis(Laterinthecourse)
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Equilibrium assumption
• OA makes the assumption that • C=A
• OratleastC»A
• This means that
• Thedevicesandsystemareinequilibrium
Arrival rate of requests to a device = Output rate of requests for that device = Throughput of the device
The above statement also applies to the system, i.e. replace the word “device” by “system”
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OA for Queueing Networks (QNs)
The computer system has K devices, labelled as 1,…,K.
The convention is to add an additional device 0 to represent the outside world.
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OA for QNs (cont’d)
• Wemeasurethebasicoperationalquantitiesforeach device (or other equivalent quantities) over a time of T
• A(j)=Numberofrequestarrivingatdevicej
• B(j)=Busytimefordevicej
• C(j)=Numberofcompletedrequestsfordevicej
• Inaddition,wehave
• A(0)=Numberofarrivalsatthesystem
• C(0)=Numberofcompletionsforthesystem
• Question: What is the relationship between A(0) and C(0) for a closed QNs?
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Visit ratios
• A job arriving at the system may require multiple visits to a device in the system
• Example:Ifeveryjob(ortransaction)arrivingatthesystemwill require 3 visits to the disk (= device j), what is the ratio of C(j) to C(0)?
• WeexpectC(j)/C(0)=3. • V(j)=Visitratioofdevicej
= Number of times a job (transaction) visits device j • We have V(j) = C(j) / C(0)
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Forced Flow Law
The forced flow law is
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Service time versus service demand
• Ex: A job requires two disk accesses to be completed. One disk access takes 20ms and the other takes 30ms.
• Service time = the amount of processing time required per visit to the device
• Thequantities“20ms”and“30ms”aretheindividualservicetimes.
• D(j) = Service demand of a job at device j is the total service time required by that job
• Theservicedemandforthisjob=20ms+30ms=50ms
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Service demand
• Service demand can be expressed in two different ways
• Ex:Ajobrequiresthreediskaccessestobecompleted.One disk access takes 20ms and the others take 30ms and 28ms.
• What is D(j)? 78ms.
• What are V(j) and S(j)?
• Recall that S(j) = mean service time of device j
• V(j) = 3. S(j) = 26ms.
• Service demand D(j) = V(j) S(j)
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Service demand law (1)
Given D(j) = V(j) S(j) Since
• What is X(j) S(j)?
• It is U(j)
Service demand law
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Service demand law (2)
• Service demand law D(j) = U(j) / X(0)
• Youcandetermineservicedemandwithoutknowingthevisitratio
• OvermeasurementperiodT,ifyoufind
• B(j) = Busy time of device j
• C(0) = Number of requests completed
• You’veenoughinformationtofindD(j) • The importance of service demand
• Youwillseethatservicedemandisafundamentalquantityyou need to determine the performance of a queueing network
• Youwilluseservicedemandtodeterminesystembottleneckin Lecture 2A
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Server example exercise
# I/O per second
Utilisation
Total # jobs=13680
What is the service time of Disk 2? What is the service demand of Disk 2? What is its visit ratio?
Measurement time = 1 hr
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Server example solution
Service time System throughput Service demand Visit ratio
= U2/X2 = 0.41/36 = 11.4ms = 13680/3600 = 3.8 jobs/s
= 0.41/3.8 = 108ms
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Measurement time = 1 hr
# I/O per second
Utilisation
Total # jobs=13680
= 36/3.8 = 108 / 11.4 = 9.47

Little’s law (1)
• Due to J.C. Little in 1961
• Afewdifferentforms
• The original form is based on stochastic models
• Animportantresultwhichisnon-trivial
• All the other operational laws are easy to derive, but Little’s
Law’s derivation is more elaborate. • Consider a single-server device
• Navg=Averagenumberofrequestsinthedevice
• When we count the number of requests in a device, we include
the one being served and those in the queue waiting for service
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Little’s Law (2)
• X = Throughput of the device
• Ravg = Average response time of the requests
• Navg = Average number of requests in the device • Little’s Law (for OA) says that
Navg = X * Ravg
We will argue the validity of Little’s Law using a simple example.
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Consider the single sever queue example from Week 1
Request index
Arrival time
Service time
Departure time
Let us use blocks of height 1 to show the time span of the requests, i.e. width of each block = response time of the request
246 10 1417
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2 4 6 10 14 17
Assuming that in the measurement time interval [0,20]
these 4 requests arrive and depart from this device, i.e. the device is in equilibrium.
Total area of the blocks
= Response time of request 1 + Response time of request 2 +
Response time of request 3 + Response time of request 4 = Average response time over the measurement interval *
Number of requests completed over the measurement interval
This is one interpretation. Let us look at another.
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Let us assume these blocks are “plasticine” and let them fall to the ground. Like this.
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There is an interpretation of the height of the graph.
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Request index
Arrival time
Service time
Requests waiting
to be processed
246 10 1417 time
Request being processed
Interpretation: Height of the graph = #requests in the device E.g. Number of requests in [9,10] = 3
E.g. Number of requests in [11,12] = 2 etc.
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246 10 1417 time
waiting requests
Request being processed
Again, consider the measurement time interval of [0,20].
Area under the graph in [0,20]
= Height of the graph in [0,1] + Height of the graph in [1,2] + …
Height of the graph in [19,20]
= #reqs in [0,1] + #reqs in [1,2] + … + #reqs in [19,20]
= Average number of requests in [0,20] in the device * 20
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Area = Average response time over [0,T] * Number of requests completed in [0,T]
waiting requests
246 10 1417 time
Area = Average number of requests in [0,T] * T
being processed
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Deriving Little’s Law
Area = Average response time of all jobs *
Number of requests completed in [0,T] (Interpretation #1)
= Average #requests in [0,T] * T (Interpretation #2) Since Number of requests completed in [0,T] / T
= Device throughput in [0,T] We have Little’s Law.
Average number of requests in [0,T]
= Average response time of all reqs * Device throughput in [0,T]
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Using Little’s Law (1)
• A device consists of a server and a queue
• The device completes on average 8 requests per second • On average, there are 3.2 requests in the device
• What is the response time of the device?
• Mean throughput X = 8 requests/s
• Mean number of requests Navg = 3.2 requests
• By Little’s Law, average response time = Navg/X = 3.2 / 8 = 0.4 s
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Intuition of Little’s Law
• Little’s Law
• Mean#requests=Meanresponsetime*Meanthroughput
• If #requests in the device ⬆ , then response time ⬆ • Andviceversa
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Applicability of Little’s Law
• Little’sLawcanbeappliedatmanydifferentlevels
• Little’slawcanbeappliedtoadevice • Navg(j)=Ravg(j)*X(j)
• A“box”withKdevices
• Box3inthenextslidehas4devices
• Navg(j)=#requestsindevicej
• AveragenumberofrequestsintheboxNavg=Navg(1)+….+ Navg(K)
• Averageresponsetimeofthebox=Ravg
• Response time of a box = Departure time from the box – Arrival time
to the box
• Wecanalsoapplyittoanentiresystem
• Navg=Ravg*X(0)
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Using Little’s Law (2)
queue server
• The device completes on average 8 requests per second
• On average, there are
• 3.2requestsinthedevice • 2.4requestsinthequeue • 0.8requestsintheserver
• What is the mean waiting time and mean service time?
• Hint: You need to draw “boxes” around certain parts of the device and interpret the meaning of response time for that box.
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Using Little’s Law (2)
queue server
• The device completes on average 8 requests per second • On average, there are
• 3.2requestsinthedevice • 2.4requestsinthequeue • 0.8requestsintheserver
• What is the mean waiting time and mean service time?
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• MeanthroughputX=8requests/s • Meanwaitingtime=2.4/8=0.3s • Meanservicetime=0.8/8=0.1s

References
• Operationalanalysis
• Lazowska et al, Quantitative System Performance,, 1984. (Classic text on performance analysis. Now out of print but can be download from http://www.cs.washington.edu/homes/lazowska/qsp/
• Chapters 3 and 5 (For Chapter 5, up to Section 5.3 only)
• Alternative 1: You can read Menasce et al, “Performance by design”, Chapter
3. From beginning of Chapter 3 to Section 3.2.4.
• Alternative 2: You can read Harcol-Balter, Chapter 6. The treatment is more rigorous. You can gross over the discussion mentioning ergodicity.
• Little’sLaw(Optional)
• I presented an intuitive “proof”. A more formal proof of this well known Law is
in Bertsekas and Gallager, “Data Networks”, Section 3.2
• Revisionquestionsbasedonthisweek’slectureareavailablefromcourse
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