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
Dont have to measure every operational quantities
Measure B to deduce U dont have to measure U
Consistency checks
If U 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 (contd)
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
Youve 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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Littles 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 Littles
Laws 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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Littles Law (2)
X = Throughput of the device
Ravg = Average response time of the jobs
Navg = Average number of jobs in the device
Littles Law (for OA) says that
Navg = X * Ravg
We will argue the validity of Littles 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 Littles Law.
Average number of jobs in [0,T]
= Average response time of all jobs * Device throughput in [0,T]
Deriving Littles Law
Using Littles 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 Littles Law, average response time = Navg/X = 3.2 / 8 =
0.8 s
Intuition of Littles Law
Littles 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 Littles Law
Littles Law can be applied at many different levels
Littles 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 Littles 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 Littles 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 Littles 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 Littles 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)
Littles 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
Littles 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)
Littles 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 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 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?
Thats 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 doesnt 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.
Littles 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 weeks lecture are available from course
web site
We will discuss the questions in next weeks tutorial time
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