1. No category

Answer:
Distributed state abstraction:
The Network OS provides a global view of the network (in a logically centralised way) and presents it as an
(annotated) ‘network graph’ to the network applications.
The Network Operating System functionality is provided by so called ‘Network Controllers’ or ‘Network
Controller Platforms’. The original attempt to implement this was NOX, and many more have been
developed more recently.
The University of Queensland
School of Information Technology and Electrical Engineering
Semester 2, 2014
NOX paper:
Gude, Natasha, et al. "NOX: towards an operating system for networks." ACM SIGCOMM Computer
Communication Review 38.3 (2008): 105-110.
http://yuba.stanford.edu/~casado/nox-ccr-final.pdf
COMS4200/7200 – Tutorial 1, Answers
Configuration abstraction or Specification Abstraction
Simplified, high level abstraction of how to configure network, the specification of ‘what’ the network should
do (goals, policies), not ‘how’ it should be done.
Example: cross-bar model for access control
This configuration or specification abstraction is the northbound interface in the SDN architecture.
There is currently no clear consensus (or standard) on how this should look like.
Q1) What are some of the key motivations behind SDN, i.e. what are the limitations of
current networks and network architectures?
Answer:
 Ossification of current network, slow innovation
 Networks are increasingly large and complex, and are increasingly hard to configure and manage
o Lack of control abstractions, lots of low level, individual device configuration
 Analogy: Like programming without an OS
o This makes it hard to adapt networks to changing demands
 Current network do not support network virtualisation well enough
 Vertically integrated solutions (mainframe analogy), vendor lock-in
Forwarding abstraction:
Should provide an abstraction of the switch functionality, but should not be vendor or implementation
specific. This is essentially the south-bound interface.
OpenFlow provides a forwarding abstraction (not the only one, but currently the pre-dominant one).
Q5) Based on the talk by Scott Shenker, what’s the key value proposition of SDN, in three
words?
Q2) What is the key technical difference of SDN compared to traditional networks?
Answer:
Innovation through abstraction.
Answer:
The key difference of SDN is the separation of the control and data plane, and the logically centralised
control of the network. This is in contrast to the control via distributed protocols (e.g. routing protocols) in
traditional IP networks.
Q6) Given that the Network Operating System provides the abstraction of a global view of
the network as a ‘network graph’, explain how this can be used to implement shortest path
routing.
Answer:
If we have a global view of the network, i.e. the network topology as graph, computing the shortest paths
between any two nodes is relatively easy, e.g. by using Dijkstra’s algorithm. Once routes are computed by
the SDN controller, they need to be installed at the corresponding switches, typically by sending the
corresponding OpenFlow messages.
Q3) What are potential key benefits of SDN?
Answer:





Increased rate of innovation
Greater flexibility and agility through programmability
Simplified network control and management
Increased efficiency and scalability
Lowered cost
Q7) The ONF Whitepaper on SDN provides a brief introduction to OpenFlow, and provides
the definition of a ‘flow’. Compare the granularity of control that OpenFlow provides with
traditional IP routing.
Q4) Based on the talk by Scott Shenker watched during the lecture, what are the 3 key
abstractions in SDN, and how are they implemented?
Answer:
In traditional IP routing, all packets between a source and destination follow the same path through the
network. This is due to the fact that forwarding decisions at each hop are based on the destination IP
address in the packet. In contrast, the matching rules in flow tables of SDN switches are much more finegrained. A flow is defined as a match rule, consisting a large number of packet header fields, including MAC
source and destination address, IP source and destination address, UDP/TCP port numbers, etc. A separate
forwarding action can be defined for each unique flow, i.e. each unique combination of values of all these
header fields.
Talk: http://www.youtube.com/watch?v=YHeyuD89n1Y
1
2
Q3)
a) In an M/M/1 System with arrival rate sand service rate =5 [1/s], to what
percentage is the server utilised on average?
The University of Queensland
School of Information Technology and Electrical Engineering
Answer:
4/5=0.8  80%
Semester 2, 2014
b) When the system is started, what’s the probability that there are no customers arriving
within the next 2 seconds? What’s probability of no customers arriving in the next 2
seconds after the system has been running for 2 minutes?
COMS4200/7200 – Tutorial 2, Answers
Answer:
We have a Poisson process here, which has no memory, i.e. the probability of k customers arriving in an
interval of t seconds does not depend on the position of this interval on the time axis.
Q1) Classify the following Queuing Systems using Kendall’s Notation:
P (t , k ) 
a) Data packets arrive at a router with exponential inter-arrival time. The packet length is
independent from the arrival time and has also an exponential distribution. The time it
takes to forward a packet is proportional to its length. We assume that the router can
buffer an unlimited number of packets.
e  t (  t ) k
k!
In this case we have: t= 2s, k=0, =4
P(2,0)=e-t
Answer:
M/M/1
(M/M/1/FIFO/∞/∞)
= e-
c) What’s the probability that there are more than two customers in the system?
b) Consider the express checkout line at a supermarket with 5 checkout operators.
Customers arrive according to a Poisson process and the service time has an unknown
distribution. We assume that the queue can fit a maximum of 100 customers (including
customers being served).
Answer:
General M/M/1 results:
P0 = 1-0.2
Pk = (1-)k
P1= (1-)1 = 0.16
P2 = (1-)2 = 0.128
Answer:
M/G/5/100
(M/G/5/FIFO/100/∞)
Probability that there are up to 2 customers in the system: P0 + P1 + P2 = 0.488
Probability that there are more than 2 customers in the system: 1-(P0 + P1 + P2) = 0.512
c) Cars arrive at a car wash station with an average rate of 5 per hour. It takes exactly 5
minutes for each car to be washed by the single station. There is unlimited space for cars
to queue.
Q4) Suppose we have an M/M/2/2 system where a maximum of two customers are
allowed in the system. Since we have two servers, this means we have no queuing space.
Answer:
G/D/1
(G/D/1/FIFO/∞/∞)
a) Draw a state transition diagram for this scenario (Markov chain)
Q2) Skiers arrive at a Ski lift with an average rate of 10 per minute. (The inter-arrival times
follow a normal distribution.) The lift can take 1 Skier every 4 seconds. Observation of the
queue shows that on average, there are 100 people queuing. How long is the average wait
in the queue?
Answer:

0
Answer:
We can use Little’s Law, which is independent of the distribution of arrival and service times.
Little: N=λT
Since we can apply Little’s Law to any sub-system, we apply it to the queue.
NQ=λTQ
TQ= NQ/λ = 100/10=10 minutes
1

1

2

2
b) Find the probabilities for each of the states of the system?
The University of Queensland
School of Information Technology and Electrical Engineering
Answer:
Equilibrium conditions:
State 0: P0*P1*P1=P0(/)
Semester 2, 2014
State 1: P1*P1*P0*P2*2
P2=(P1(P0*P0(/)()P0*P0()

Normalisation: P0+P1+P2 = 1
COMS4200/7200 – Tutorial 3, Answers
P0+P0() + P0() = 1

Q1) Classify the following Queuing Systems using Kendall’s Notation:
P0 =1/ (1+)
P1=(1+)
P2=0.5* (1+)
A call centre has 10 phone lines and employs 4 operators during daytime. The time
between calls is assumed to have an exponential distribution. The distribution of the call
durations is not known.
Answer:
When all operators are busy, a caller is put on hold. If all 10 phone lines are busy, a caller is rejected, i.e.
gets the busy tone. We can assume a FIFO queue.
Q5) A communication line capable of transmitting at a rate of 50Kbit/s will be used to
accommodate 10 sessions, each generating Poisson traffic at a rate of 150 packets/min.
Packet lengths are exponentially distributed with a mean of 1000 bits.
M/G/4/10
a) In a first scenario, we split the channel into 10 TDM sub-channels of equal capacity,
with 5 Kbit/s each. Each sub-channel is allocated to a session. What’s the average delay T
that packets experience in this scenario?
Q2) What’s the ‘offered load’ in Erlang for the following system?
Answer:
We can model this scenario as 10 separate M/M/1 queues, with min = 2.5/s. The transmission rate of
each sub-channel is 5Kbit/s, with an average packet size of 1000 bits, this results in a service rate of s

From our analysis of M/M/1 queues we know T=1/()
a) Calls arrive with an average rate of 10 per minute. We assume the average call duration
is 30 seconds.
Answer: Offered load = average number of circuits (servers) that would be used if no calls were rejected.
T=1/(5-2.5) [s] = 0.4s
A=λ/µ=10/(1/0.5) = 5 Erlang
b) In this scenario, the 10 sessions share the 50Kbit/s channel via statistical multiplexing,
i.e. there is one single queue where packets from all the 10 sessions are queued. (e.g. in a
router) The single server (transmission line) is fed from this queue. What’s the average
delay T that packets experience in this scenario?
Q3) We have a call centre with the following properties:
•
•
•
•
Answer:
We can model Scenario b) as a single M/M/1 queue with parameters min = 25/s and s.
T=1/() = 1/(50-25)[s] = 0.04s
(We use the fact that merging Poisson processes result in a new Poisson process)

This shows that decreased delay is one of the general benefits of packet switching over circuit switching.
The same model and argument can be applied in any situation where we evaluate the use of one fast server
versus multiple slower servers (and individual queues) with equal total capacity.
Two operators
Two phone lines
Call arrivals are Poisson, with an average rate of 20 per hour
Service time (for each operator) has exponential distribution with Ts = 3 min
a) What type of queuing system do we have?
Answer: M/M/2/2
3
1
b) If we ring the call centre, what’s the probability of an operator answering the phone,
instead of getting a busy tone?
M / G /1 :
Answer: We need to find the “Blocking Probability” PB. The Erlang B formula gives us this result.
Utilisation factor: ρ =
m=2, λ=20/hour, µ=1/3min = 20/hour
λ/µ=ρ=20/20=1
PB =
Pollaczek-Khinchin formula :
(λ / µ ) m / m!
λ Var (TS ) +
m
∑ (λ / µ )
n
/ n!
TQ =
n =0
PB =
λ
µ
1 / 2 ⋅1
0.5
=
= 0.2 = 20%
1
2
1 / 1 + 1 / 1 + 1 / 2 2.5
λ
µ2
2(1 − ρ )
2
0
NQ =
The probability of getting an operator to answer our call is 1-PB =1 – 0.2 = 0.8.
λ2 Var (TS ) + ρ 2
2(1 - ρ )
b) Through observation over a few days, you find that the Variance of Alice’s service time
is 0.01 [h2]. Bob’s service time has a variance of 0.04 [h2]. How much longer is the queue
in front of Bob’s terminal on average?
c) What’s the probability that there are two customers in the system, i.e. P2=?
Answer:
We use the Pollaczek-Khinchin formula.
λ=9 per hour
µ=10 per hour
Answer: This is exactly the blocking probability! The probability that a customer is rejected is the probability
that both servers are busy, which is P2.
We also could have used the results from the previous Tutorial, Question 4. There we found for a M/M/2/2
system:
Alice:
ρ=λ/µ
NQ =
P0 =1/ (1+ρ+0.5 ρ2)
P1=ρ/(1+ρ+0.5 ρ2)
P2=0.5* ρ2/(1+ρ+0.5 ρ2)
Bob:
For ρ=λ/µ = 20/20 =1, we get P2 = 0.5/(1+1+0.5) = 0.2
NQ =
Q4) Alice and Bob are 2 checkout operators at Coles. On average they are both able to
handle 10 customers per hour with customers arriving with exponentially distributed interarrival times. (With the same average arrival rate of 9 customers per hour at both
terminals.) The store management notices that the queue at Bob’s checkout terminal is on
average significantly longer than the one at Alice’s terminal.
λ2 Var (TS ) + ρ 2 81⋅ 0.01 + 0.9 2
=
= 8.1
2(1 - ρ )
0.2
λ2 Var (TS ) + ρ 2 81 ⋅ 0.04 + 0.9 2
=
= 20.25
2(1 - ρ )
0.2
Difference: 20.25 – 8.1 = 12.15
Bob’s Queue has on average 12.15 customers more waiting than Alice’s queue.
You can verify the results with the Excel spread sheet available on the course web site.
a) How can you explain this?
Answer: We have 2 individual M/G/1 systems with the same λ and µ.
The Pollaczek-Khinchin formula tells us that for M/G/1 systems, a high variance of service times increases
the average Queuing Delay and the average Queue Length. We can therefore say that Alice works more
consistently whereas Bob has bursts of intense activity and periods where he takes it easy.
I.e. Alice’s service rate has a lower variance than Bob’s
2
3
Q5) Decode the following ‘object’ encoded in ASN.1 BER Transfer Syntax.
The University of Queensland
School of Information Technology and Electrical Engineering
00000110 00000111 00101011 00000110 00000001 00000010 00000001 00000001 00000011
Semester 2, 2014
The following partial view of the OID tree is provided:
1 – ISO
1.3 – ISO Member Body
1.3.6 – US DoD
1.3.6.1 – Internet
1.3.6.1.2 – Management
1.3.6.1.2.1 – MIB-II
1.3.6.1.2.1.1 – System
1.3.6.1.2.1.1.1 – SysDescription
1.3.6.1.2.1.1.3 – SysUptime
1.3.6.1.2.1.2 – Interface
1.3.6.1.2.1.2.2.1.2.1 – IfDescription1
1.3.6.1.2.1.2.2.1.2.2 – IfDescription2
1.3.6.1.2.1.2.2.1.4.1 – IfMTU1
1.3.6.1.2.1.2.2.1.4.2 – IfMTU2
1.3.6.1.2.1.2.2.1.5.1 – IfSpeed1
1.3.6.1.2.1.2.2.1.5.2 – IfSpeed2
1.3.6.1.2.1.2.2.1.6.1 – IfIpAddress1
COMS4200/7200 – Tutorial 4, Answers
Q1) What are the 5 conceptual areas of network management as defined by ISO?
Answer:
Performance Management
Fault Management
Configuration Management
Accounting Management
Security Management
Q2) What are the components (entities) of a general network management architecture?
Answer:
T: Universal, primitive type, Object Identifier
L:7
V : {1 3 6 1 2 1 1 3}
Answer:
Management station
Managed entity
Agent
Management Protocol
Management Data (Management Information)
 System Uptime object in MIB-2
00 0 00110 00000111 00101011 00000110 00000001 00000010 00000001 00000001 00000011
T=OID
L= 7
1,3
6
1
2
1
1
3
Q3) What is the following OID: {1 3 6 1 2 1 1}?
Q6) Decode the following ‘object’ encoded in ASN.1 BER Transfer Syntax.
Answer:
System in MIB-II
00000011 00000010 00000111
10000001
Q4) Encode the following in ASN.1 BER Transfer Syntax.
Type
Counter32
Answer:
00000011
00000010
00000111
Bit String length=2
last 7 bits unused
Value
1000 (decimal)
10000001
1
Answer:
Bit String = ‘1’
T
L
V
01 0 00001
00000100
00000000 00000000 00000011 11101000
Application Specific
Primitive
Code 1: Counter32
Length 4 bytes (32 bits)
Value 1000 in binary representation
1
2
Q7) What are the main improvements of SNMPv2c over SNMPv1?
The University of Queensland
School of Information Technology and Electrical Engineering
Answer:
Adds GetBulk message
Allows downloading multiple objects (e.g. a table) with one request
Semester 2, 2014
Adds Inform message
For sending trap messages from manager to manager. Acknowledged.
Adds RMON functionality
Remote Monitoring probes (“super SNMP agent”) can be deployed to monitor entire subnets continuously
and independently of manager
COMS4200/7200 – Tutorial 5, Answers
Q1) Describe two major differences between the ECN method and the RED method of
congestion avoidance.
Answer:
Q8) What is the key difference between SNMP and OSI Network Management (CMIP) in
terms of data model and transport protocol?
First, the ECN method explicitly sends a congestion notification to the source by setting a bit, whereas RED
implicitly notifies the source by simply dropping one of its packets. Second, the ECN method drops a packet
only when there is no buffer space left, whereas RED drops packets before all the buffer are exhausted.
Answer:
SNMP:
Static database
No support for object oriented data model. (no methods, no inheritance)
Cannot express relationships between objects.
Q2) A token bucket scheme is used for traffic shaping. A new token is put into the bucket
every 5 µsec. Each token is good for one short packet, which contains 48 bytes of
data. What is the maximum sustainable data rate?
Connection-less, unreliable transport protocol (UDP).
Answer:
With a token every 5 μsec, 200,000 packets/sec can be sent. Each packet holds 48 data bytes or 384 bits.
The data rate is then 76.8 Mbps.
OSI NM:
Dynamic database
Support of OO features, e.g. inheritance
Can express relationships between objects.
Q3) A computer on a 6-Mbps network is regulated by a token bucket. The token bucket is
filled at a rate of 1 Mbps. It is initially filled to capacity with 8 megabits. How long can
the computer transmit at the full 6 Mbps?
Uses connection-oriented, reliable transport protocol.
Answer:
The naive answer says that at 6 Mbps it takes 4/3 sec to drain an 8 megabit bucket. However, this answer is
wrong, because during that interval, more tokens are generated. The correct answer S = B/(M − R), where B
– bucket size, M – network data rate, R – flow average rate. Substituting, we get S = 8/(6 − 1) or 1.6 sec.
Q4) Consider the user of differentiated services with expedited forwarding. Is there a
guarantee that expedited packets experience a shorter delay than regular packets?
Why or why not?
Answer:
There is no guarantee. If too many packets are expedited, their channel may have even worse performance
than the regular channel. DiffServ does not guarantee quality of service. It only reserves resources in routers
(bandwidth, buffers, CPU) for particular traffic class but if offered load exceeds the reserved resources the
quality of service will be poor.
3
1
Q5) In the figure below suppose a new flow E is added that takes a path from R1 to R2 to
R6. How does the max-min bandwidth allocation change for the five flows?
The University of Queensland
School of Information Technology and Electrical Engineering
Semester 2, 2014
COMS4200/7200 – Tutorial 6, Answers
Q1) For the following Chord ring with m=6, write down the finger tables for Node 20 and for
Node 41.
6
60
Answer:
10
Allocation for flow A will be 1/2 on links R1R2 and R2R3. Allocation for flow E will be 1/2 on links R1R2 and
R2R6. All other allocations remain the same.
50
Q6) Discuss the advantages and disadvantages of credits versus sliding windows
protocols.
14
44
20
41
25
Answer:
34
The sliding window is simpler, having only one set of parameters (the window edges) to manage. Windows
are already used for flow control so congestion control windows can be easily added. However, the credit
scheme is more flexible, allowing a dynamic management of the buffering, separate from the
acknowledgements.
30
Answer:
Finger table entries for Node n:
Q7) Explain why congestion avoidance algorithms (congestion control laws) decide on
how much bandwidth is given to each flow. Also discuss why efficiency, fairness, and
convergence are important in these algorithms.
Finger table entry i: successor (i=n+2^i)
For i=0..m-1
All calculations are done modulo 2^m=64
Answer:
Node 20:
i=0: successor(20 + 1) = 25
i=1: successor(20 + 2) = 25
i=2: successor(20 + 4) = 25
i=3: successor(20 + 8) = 30
i=4: successor(20 + 16) = 41
i=5: successor(20 + 32) = 60
The algorithms decide how much to crease or increase the current bandwidth so they decide about the
bandwidth allocation. The allocation for all flows (aggregate bandwidth) should be efficient (close to the link
capacity) but it also needs to be fair for the flows (or at least do not starve some flows). The algorithms need
to converge to the operating point in order to avoid oscillation in bandwidth allocation.
Q8) Some policies for fairness in congestion control are Additive Increase Additive
Decrease (AIAD), Multiplicative Increase Additive Decrease (MIAD), and Multiplicative
Increase Multiplicative Decrease (MIMD). Discuss these three policies in terms of
convergence and stability.
Node 41:
i=0: successor(41 + 1) = 44
i=1: successor(41 + 2) = 44
i=2: successor(41 + 4) = 50
i=3: successor(41 + 8) = 50
i=4: successor(41 + 16) = 60
i=5: successor(41 + 32) = 10
Answer:
In AIAD and MIMD, the users will oscillate along the efficiency line, but will not converge. MIAD will converge
just like AIMD. None of these policies are stable. Decrease policy in AIAD and MIAD is not aggressive, and
increase policy in MIAD and MIMD is not gentle.
2
1
Q2) For a wireless link with the following parameters, what is the maximum distance between
sender and receiver (Free Space Loss Model)?
Q3) Under what condition would it be possible to use CSMA/CD for IEEE 802.11 LANs?
Answer:
Wireless transceivers would need to be full-duplex, i.e. able to send and listen at the same time, in order to
detect collisions.
The hidden node problem, which prevents sending nodes from detecting collisions that happen at remote
receivers, would need to be solved somehow. This however is a fundamental problem of the wireless medium.
Frequency: 900 MHz
Transmitter Antenna Gain (TxG): 15dBi
Transmitter losses (TxL): 2 dB
Transmitter output power (TxP): 100 milliwatts
Receiver Losses (RxL): 1 dB
Receiver Antenna Gain (RxG): 15 dBi
Miscellaneous loses (ML): 4 dB
Receiver Sensitivity (RxS): -75 dBm
Q4) Explain why IEEE 802.11 uses acknowledgements?
Link Margin (LM): 10 dB
Answer:
The wireless medium is much more unreliable than a wired Ethernet for example. Reasons for this are
interference, multi-path propagation, signal attenuation and collisions (802.11 has no way of knowing if there
was a collision or not, except via ACKs.)
The cost of leaving the responsibility of implementing reliability entirely to higher layers (i.e. TCP), would very
costly. Lost frames on the wireless link would incur end-to-end retransmissions at the transport layer.
Answer:
RxP = RxS + Link Margin = -75 dBm + 10 dB = -65 dBm
RxP = TxP + TxG - TxL - FSL - ML + RxG - RxL
FSL = RxG
FSL = 15 dBi
- RxP + TxP +
+ 65 dBm + 20 dBm +
TxG 15 dBi -
TxL 2 dB -
ML 4 dB -
Q5) At the moment, station A is transmitting a frame using IEEE 802.11 (DCF mode). Stations
B and C are within range of A and of each other. Both want to send a frame and are waiting
for the channel to become free.
The size of the contention window is currently 31 slots. What is the probability that B and C
will collide twice in a row after the channel becomes free.
RxL
1dB = 108 dB
Convert to non-dB ratio:
X dB = 10 log10(P1/P2)
(P1/P2) = 10^(X/10)
Answer:
The probability of a first collision PC1 is the probability that B and C pick the same random number in the range
1..31. PC1 = 1/31 ≈ 0.0323.
108 dB:
(P1/P2) = 10^(108/10) = 10^10.8
After the collision, the contention window size is increased to 63 slots.
n
(The size of the contention window is 2 -1, with n being increased by one each time a collision happens, up to a
maximum of 1023: 31, 63, 127, 255, 511, 1023 )
The probability of a second collision is the probability that B and C pick the same random number in the range
1..63.
PC2 = 1/63 ≈ 0.0159
Free Space Loss Model:
L = (4πdf/c)^2 = 10^10.8
d = L^0.5 * c/4πf
Speed of Light (approximately):
Since the two events are independent, the probability of two collisions in a row is PC12= PC1 * PC2 ≈ 0.0005
c = 3 * 10^8 m/s
d = 10^5.4 * 3*10^8/ (4* 3.1416 * 900*10^6) = 6.66 km
Q6) The Hidden Node Problem was briefly discussed during the Lecture. Due to limited
transmission range, nodes believe the channel is free when in effect another node is already
transmitting to the intended destination node. The end result is a collision at the receiver
which is not detected by either of the two senders.
a) Can you think of a scenario where the opposite is happening, i.e. a node believes that the
channel is not free, when in effect it could send to the destination and no collision would occur
(at the destination). Draw a figure that shows such a scenario.
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Answer:
This is called the “Exposed Terminal Problem”. In the scenario shown below, A is transmitting to B and C wants
to transmit to D. When C listens to the channel, it detects that it is busy (due to A transmitting), and therefore will
defer transmission. However, in this case A could transmit to B at the same time as C is transmitting to D and no
collision would occur at the receivers, since C is B is out of range of C and D is out of range of A. This problem is
not destructive in the sense that it does not cause any collisions. However it causes underutilisation of the
medium.
Transmission
Range of A
Transmission
Range of C
A
C
B
D
b) Explain how RTS/CTS can be used to address this problem?
Answer:
When a node hears an RTS from a neighboring node, but not the corresponding CTS, that node can deduce that
it is an exposed node (out of range) and is permitted to transmit to other neighboring nodes.
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