week02
COMP93341
COMP9334
Capacity Planning for Computer Systems
and Networks
Week 2: Operational Analysis and
Workload Characterisation
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Last lecture
• Modelling of computer systems using Queueing
Networks
• Open networks
• Closed networks
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Open networks
A transaction may visit the CPU and disk multiple times.
An open network is characterised by external transactions.
Example: The server has a CPU and a disk.
Open queueing network
External arrivals
Workload intensity specified by arrival rate
Unbounded number of customers in the system
In equilibrium, flow in = flow out
) throughput = arrival rate
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Closed queuing networks
Closed queueing networks model
• Running batch jobs overnight
• Once a job has completed, a new job starts.
Good performance means high throughput.
#jobs in the system = multi-programming level
Database server for batch jobs
Running batch jobs overnight
E.g. producing managerial reports
Assume once a job has completed, a new job starts
Maintain constant number of customers in the system
A closed queueing network
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This lecture
• The basic performance metrics
• Response time, Throughput, Utilisation etc.
• Operational analysis
• Fundamental Laws relating the basic performance metrics
• Bottleneck and performance analysis
• Workload characterisation
• Poisson process and its properties
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Operational analysis (OA)
• “Operational”
• Collect performance data during day-to-day operation
• Operation laws
• Applications:
• Use the data for building queueing network models
• Perform bottleneck analysis
• Perform modification analysis
• iostat
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Single-queue example (1)
In an observational period of T, server busy for time B
A requests arrived, C jobs 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
server
#requests = A #requests = C
B
Motivating example
• Given
• Observation period = 1 minute
• CPU
• 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 = 1790/60 = 29.83 transactions /s
• Output rate = 1800/60 = 30 transactions/s
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Utilisation law
• The operational quantities are inter-related
• Consider
• Utilisation U = B / T
• Mean service time per completion S = B / C
• Output rate X = C / T
• Utilisation law – Can you relate U, S and X?
• U = S X
• Utilisation law is an example of operational law.
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Application of OA
• Don’t have to measure every operational quantities
• Measure B to deduce U – don’t have to measure U
• Consistency checks
• If U ¹ S X, something is wrong
• Operational laws can be used for performance analysis
• Bottleneck analysis (today)
• Mean value analysis (Later in the course)
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Equilibrium assumption
• OA makes the assumption that
• C = A
• Or at least C » A
• This means that
• The devices and system are in equilibrium
• 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)
• We measure the basic operational quantities for each
device (or other equivalent quantities) over a time of T
• A(j) = Number of arrivals at device j
• B(j) = Busy time for device j
• C(j) = Number of completed jobs for device j
• In addition, we have
• A(0) = Number of arrivals for the system
• C(0) = Number of completions for the system
• 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: If every job arriving at the system will require 3 visits to
the disk (= device j), what is the ratio of C(j) to C(0)?
• We expect C(j)/C(0) = 3.
• V(j) = Visit ratio of device j
• = Number of times a job visits device j
• We have V(j) = C(j) / C(0)
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Forced Flow Law
The forced flow law is
Since
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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
• The quantities “20ms” and “30ms” are the individual service times.
• D(j) = Service demand of a job at device j is the total service
time required by that job
• The service demand for this job = 20ms + 30 ms = 50ms
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Service demand
• Service demand can be expressed in two different
ways
• Ex: A job requires two disk accesses to be completed. One
disk access takes 20ms and the other takes 30ms.
• D(j) = 50ms.
• What are V(j) and S(j)?
• Recall that S(j) = mean service time of device j
• V(j) = 2. S(j) = 25ms.
• Service demand D(j)= V(j) S(j)
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Service demand law (1)
• It is U(j)
Given D(j) = V(j) S(j)
Since
Service demand law
• What is X(j) S(j)?
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Service demand law (2)
• Service demand law D(j) = U(j) / X(0)
• You can determine service demand without knowing the visit ratio
• Over measurement period T, if you find
• B(j) = Busy time of device j
• C(0) = Number of requests completed
• You’ve enough information to find D(j)
• The importance of service demand
• You will see that service demand is a fundamental quantity you
need to determine the performance of a queueing network
• You will use service demand to determine system bottleneck today
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Server example exercise
Measurement time = 1 hr
# I/O/s Utilisation
Disk 1 32 0.30
Disk 2 36 0.41
Disk 3 50 0.54
CPU 0.35
Total # jobs=13680
What is the service time of Disk 2?
What is the service demand of Disk 2?
What is its visit ratio?
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Server example solution
Measurement time = 1 hr
# I/O/s Utilisation
Disk 1 32 0.30
Disk 2 36 0.41
Disk 3 50 0.54
CPU 0.35
Total # jobs=13680
Service time = U2/X2 = 0.41/36 = 11.4ms
System throughput = 13680/3600 = 3.8 jobs/s
Service demand = 0.41/3.8 = 108ms
Visit ratio = 36/3.8 = 108 / 11.4 = 9.47
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Little’s law (1)
• Due to J.C. Little in 1961
• A few different forms
• The original form is based on stochastic models
• An important result which is non-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 = Average number of jobs in the device
• When we count the number of jobs 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 jobs
• Navg = Average number of jobs 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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Job index Arrival time Service time Departure time
1 2 2 4
2 6 4 10
3 8 4 14
4 9 3 17
2
3
1 time
2 4 6 10 14 17
4
1
2
3
Consider the single sever queue example from Week 1
Let us use blocks of height 1 to show the time span of the
jobs, i.e. width of each block = response time of the job
2
3
1 time
2 4 6 10 14 17
4
1
2
3
Assuming that in the measurement time interval [0,20]
these 4 jobs arrive arrive and depart from this device, i.e. the
device is in equilibrium.
Total area of the blocks
= Response time of job 1 + Response time of job 2 +
Response time of job 3 + Response time of job 4
= Average response time over the measurement interval *
Number of jobs departing over the measurement interval
This is one interpretation.Let us look at another.
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2
3
1 time
2 4 6 10 14 17
4
1
2
3
2 31 4 time
2 4 6 10 14 17
3 4
4
1
2
3
Let us assume these blocks are “plastic” and let them fall
to the ground. Like this.
There is an interpretation of the height of the graph.
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Job index Arrival time Service time
1 2 2
2 6 4
3 8 4
4 9 3
2 31 4
2 4 6 10 14 17
3 4
4
Interpretation: Height of the graph = number of jobs in the device
E.g. Number of jobs in [9,10] = 3
E.g. Number of jobs in [11,12]= 2 etc.
1
2
3
time
waiting
jobs
Job being
Processed
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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]
= #jobs in [0,1] + #jobs in [1,2] + … + #jobs in [19,20]
= Average number of jobs in [0,20] * 20
2 31 4
2 4 6 10 14 17
3 4
4
1
2
3
time
waiting
jobs
Job being
Processed
Area = Average number of jobs in [0,T] * T
2 31 4
2 4 6 10 14 17
3 4
4
1
2
3
time
waiting
jobs
Job being
Processed
2
3
1 time
2 4 6 10 14 17
4
1
2
3
Area = Average response time over [0,T] *
Number of jobs leaving in [0,T]
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Area = Average response time of all jobs *
Number of jobs leaving in [0,T](Interpretation #1)
= Average number of jobs in [0,T] * T (Interpretation #2)
Since Number of jobs leaving in [0,T] / T
= Device throughput in [0,T]
We have Little’s Law.
Average number of jobs in [0,T]
= Average response time of all jobs * Device throughput in [0,T]
Deriving Little’s Law
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?
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serverqueue
• 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 =
0.8 s
Intuition of Little’s Law
• Little’s Law
• Mean #jobs = Mean response time * Mean throughput
• If # jobs in the device⬆ , then response time ⬆
• And vice versa
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Applicability of Little’s Law
• Little’s Law can be applied at many different levels
• Little’s law can be applied to a device
• Navg(j) = Ravg(j) * X(j)
• A system with K devices
• Navg(j) = #jobs in device j
• Average number of jobs in the system Navg = Navg(1) + …. +
Navg(K)
• Average response time of device j = Ravg(j)
• Average response time of the system = Ravg
• We can also apply it to an entire system
• Navg = Ravg * X(0)
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Using Little’s Law (2)
• The device completes on average 8 requests per second
• On average, there are
• 3.2 requests in the device
• 2.4 requests in the queue
• 0.8 requests in the server
• What is the mean waiting time and mean service time?
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serverqueue
• Hint: You need to draw “boxes” around certain parts of the
device and interpret the meaning of response time for that
box.
Using Little’s Law (2)
• The device completes on average 8 requests per second
• On average, there are
• 3.2 requests in the device
• 2.4 requests in the queue
• 0.8 requests in the server
• What is the mean waiting time and mean service time?
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serverqueue
• Mean throughput X = 8 requests/s
• Mean waiting time = 2.4 / 8 = 0.3 s
• Mean service time = 0.8 / 8 = 0.1 s
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Interactive systems
M users Each user sends a job to
the system
The system sends the
results to the user.
The user after a thinking
time, sends another job to
the system.
– Thinking time = time
spent by the user
An interactive system is
an example of closed
system.
results jobs
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Interactive systems (Time line)
results jobs
User 1
User 1 sends a
job to the
computer
system
The time the
job spends in
the computer
system
Results are
returned to
the user
Thinking time
time
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Interactive system (1)
• M interactive clients
• Z = mean thinking time
• R = mean response time
of the computer system
• X0 = throughput
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Interactive system (2)
• Mavg = mean #
interactive clients
• Z = mean thinking time
• X0 = throughput
• Apply Little’s Law to
the interactive part, we
have Mavg = Z * X0
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Interactive system (3)
• Navg = average # clients
in the computer system
• R = mean response time
at the computer system
• X0 = throughput
• Apply Little’s Law to the
computer system, we
have Navg = R * X0
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Interactive system (4)
• Mavg = X0 * Z
• Navg = X0 * R
• The system is closed, the
total number of users M is
a constant, we have
• M = Mavg + Navg
• Therefore,
• M = X0 * (Z+R)
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The operational laws
• These are the operational laws
• Utilisation law U(j) = X(j) S(j)
• Forced flow law X(j) = V(j) X(0)
• Service demand law D(j) = V(j) S(j) = U(j) / X(0)
• Little’s law N = X R
• Interactive response time M = X(0) (R+Z)
• Applications
• Mean value analysis (later in the course)
• Bottleneck analysis
• Modification analysis
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Bottleneck analysis – motivation
D(j) Utilisation
Disk 1 79ms 0.30
Disk 2 108ms 0.41
Disk 3 142ms 0.54
CPU 92ms 0.35
Service demand law: D(j) = U(j) / X(0)
==> U(j) = D(j) X(0)
Utilisation increases with increasing
throughput and service demand
Utilisation vs. throughput plot U(j) = D(j) X(0)
Observation:For all system throughput:
Utilisation of Disk 3 > Utilisation of Disk 2 >
Utilisation of CPU > Utilisation of Disk 1
Disk 3
Disk 2
Disk 1
CPU
What
determines
this order?
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Bottleneck analysis
• Recall that utilisation is the busy time of a device divided by
measurement time
• What is the maximum value of utilisation?
• Based on the example on the previous slide, which device
will reach the maximum utilisation first?
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Bottleneck (1)
• Disk 3 has the highest service demand
• It is the bottleneck of the whole system
Operational law:
Utilisation limit:
}
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Bottleneck (2)
Should hold for all K devices in the system
Bottleneck throughput is
limited by the maximum
service demand
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Bottleneck exercise
D(j) Utilisation
Disk 1 79ms 0.30
Disk 2 108ms 0.41
Disk 3 142ms 0.54
CPU 92ms 0.35
The maximum system throughput is 1 / 0.142 = 7.04 jobs/s.
What if we upgrade Disk 3 by a new disk that is 2 times faster,
which device will be the bottleneck after the upgrade? You
can assume that service time is inversely proportional to disk
speed.
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Another throughput bound
• Little’s law
Previously, we have
Therefore:
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Throughput bounds
Throughput
N
Bound 1
Bound 2. Slope =
Actual throughput
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Bottleneck analysis
• Simple to use
• Needs only utilisation of various components
• Assumes service demand is load independent
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Modification analysis (1)
• (Reference: Lazowska Section 5.3.1)
• A company currently has a system (3790) and is considering switching
to a new system (8130). The service demands for these two systems
are given below:
System
Service demand (seconds)
CPU Disk
3790 4.6 4.0
8130 5.1 1.9
• The company uses the system for interactive application with a think
time of 60s.
• Given the same workload, should the company switch to the new
system?
• Exercise: Answer this question by using bottleneck analysis. For each
system, plot the upper bound of throughput as a function of the number
of interactive users.
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Modification analysis (2)
Slope = 1/67
Slope = 1/68.6 1/4.6
1/5.1
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Operational analysis
• These are the operational laws
• Utilisation law U(j) = X(j) S
• Forced flow law X(j) = V(j) X(0)
• Service demand law D(j) = V(j) S(j) = U(j) / X(0)
• Little’s law N = X R
• Interactive response time M = X(0) (R+Z)
• Operational analysis allows you to bound the system
performance but it does NOT allow you to find the
throughput and response time of a system
• To order to find the throughput and response time, we
need to use queueing analysis
• To order to use queueing analysis, we need to specify the
workload
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Workload analysis
• Performance depends on workload
• When we look at performance bound earlier, the bounds depend
on number of users and service demand
• Queue response time depends on the job arrival rate and job
service time
• One way of specifying workload is to use probability
distribution.
• We will look at a well-known arrival process called Poisson
process today.
• We will first begin by looking at exponential distribution.
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Exponential distribution (1)
• A continuous random variable is exponentially distributed
with rate l if it has probability density function
Probability that x £ X £ x + dx is
f(x) dx = l exp(- lx) dx
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Exponential distribution – cumulative distribution
• The cumulative distribution function F(x)= Prob(X £ x) is:
What is Prob(X ³ x)?
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Arrival process
• Each vertical arrow in the time line below depicts the arrival
of a customer
• An arrival can mean
• A telephone call arriving at a call centre
• A transaction arriving at a computer system
• A customer arriving at a checkout counter
• An HTTP request arriving at a web server
• The inter-arrival time distribution will impact on the response time.
• We will study an inter-arrival distribution that results from a large number
of independent customers.
time
t1 t2 t3 t4 t5
Inter-arrival time
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Many independent arrivals (1)
• Assume there is a large pool of N customers
• Within a time period of d (d is a small time period), there is a probability
of pd that a customer will make a request (which gives rise to an
arrival)
• Assuming the probability that each customer makes a request is
independent, the probability that a customer arrives in time period d is
Npd
• If a customer arrives at time 0, what is the probability that the next
customer does not arrive before time t
time
0 t
No arrival!
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Many independent arrivals (2)
• Divide the time t into intervals of width d
time
0 t
d
• No arrival in [0,t] means no arrival in each interval d
• Probability of no arrival in d = 1 – Npd
• There are t / d intervals
• Probability of no arrival in [0,t] is
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Exponential inter-arrival time
• We have showed that the probability that there is no arrival
in [0,t] is exp(- N p t)
• Since we assume that there is an arrival at time 0, this
means
Probability(inter-arrival time > t) = exp(- N p t)
• This means
Probability(inter-arrival time £ t) = 1 – exp(- N p t)
• What this shows is the inter-arrival time distribution for
independent arrival is exponentially distributed
• Define: l = Np
• l is the mean arrival rate of customers
Two different methods to describe arrivals
Method 1: Continuous probability distribution of inter-arrival
time
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time
Inter-arrival time
Two different methods to describe arrivals
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Method 2: Use a fixed time interval (say t), and count the
number of arrivals within t.
time
5 arrivals in t 8 arrivals in t 6 arrivals in t
• The number of arrivals in t is random
• The number of arrivals must be an non-negative integer
• We need a discrete probability distribution:
• Prob[#arrivals in t = 0]
• Prob[#arrivals in t = 1]
• etc.
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Poisson process (1)
• Definition: An arrival process is Poisson with parameter l if
the probability that n customer arrive in any time interval t
is
Example:
Example:
l= 5 and t = 1
Note: Poisson is a
discrete probability
distribution.
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Poisson process (2)
• Theorem: An exponential inter-arrival time distribution with
parameter l gives rise to a Poisson arrival process with
parameter l
• How can you prove this theorem?
• A possible method is to divide an interval t into small time intervals
of width d. A finite d will give a binomial distribution and with d è 0,
we get a Poisson distribution.
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Customer arriving rate
• Given a Poisson process with parameter l, we know that
the probability of n customers arriving in a time interval of t
is given by:
• What is the mean number of customers arriving in a time
interval of t?
• That’s why l is called the arrival rate.
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Customer inter-arrival time
• You can also show that if the inter-arrival time distribution
is exponential with parameter l, then the mean inter-arrival
time is 1/l
• Quite nicely, we have
Mean arrival rate = 1 / mean inter-arrival time
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Application of Poisson process
• Poisson process has been used to model the arrival of
telephone calls to a telephone exchange successfully
• Queueing networks with Poisson arrival is tractable
• We will see that in the next few weeks.
• Beware that not all arrival processes are Poisson! Many
arrival processes we see in the Internet today are not
Poisson. We will see that later.
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References
• Operational analysis
• Lazowska et al, Quantitative System Performance, Prentice Hall, 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. Note that Menasce doesn’t cover certain aspects of performance bounds.
So, you will also need to read Sections 5.1-5.3 of Lazowska.
• Alternative 2: You can read Harcol-Balter, Chapters 6 and 7. The treatment is
more rigorous. You can gross over the discussion mentioning ergodicity.
• Little’s Law (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
• Tutorial exercises based on this week’s lecture are available from course
web site
• We will discuss the questions in next week’s tutorial time
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