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PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru ­100
Department of Computer Science And Engineering
SCHEME AND SOLUTION FOR IA TEST- 2
Date
: 21-09-2016
Subject & Code : Computer Networks-1 / 10CS55
Name of faculty : Ciji K R
Max Marks : 50
Section
:B&C
Time
: 8:30-10:00AM
Note: Answer all questions
1.a. What is FDM? Briefly explain its multiplexing and demultiplexing phases with diagrams.
6
Ans:
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FDM is an analog technique that can be applied when the b/w of a link (in hertz) is greater than
the combined b/ws of the signals to be transmitted
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In FDM, the signals generated by each sending device modulate different carrier frequencies
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These modulated signals are then combined into a single composite signal that can be transported
by the link
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Carrier frequencies are separated by sufficient b/w to accommodate the modulated signal
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These b/w ranges are the channels through which the various signals travel
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Channels can be separated by strips of unused b/w- guard bands- to prevent signals from
overlapping
Fig: Frequency division multiplexing
Fig: FDM-Multiplexing process
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Each source generates a signal of a similar frequency range
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Inside the multiplexer, these similar signals modulates different carrier frequencies(f1, f2, and f3)
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The resulting modulated signals are then combined into a single composite signal that is sent out
over a media link that has enough b/w to accommodate it
Fig: FDM- Demultiplexing process
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The demultiplexer uses a series of filters to decompose the multiplexed signal into its constituent
component signals
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The individual signals are then passed to a demodulator that separates them from their carriers
and pass them to the output lines
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1.b Four 1-kbps connections are multiplexed together. A unit is 1 bit. Find
(a) the duration of one bit before multiplexing
(b) the transmission rate of the link
(c) the duration of a time slot and
(d) the duration of a frame
Ans:
4
a. The duration of 1 bit before multiplexing is 1 / 1 kbps, or 0.001 s (1 ms).
b. The rate of the link is 4 times the rate of a connection, or 4 kbps.
c. The duration of each time slot is one-fourth of the duration of each bit before multiplexing, or
1/4 ms or 250 μs. Note that we can also calculate this from the data rate of the link, 4 kbps. The bit
duration is the inverse of the data rate, or 1/4 kbps or 250 μs.
d. The duration of a frame is always the same as the duration of a unit before multiplexing, or 1
ms. We can also calculate this in another way. Each frame in this case has four time slots. So the
duration of a frame is 4 times 250 μs, or 1 ms.
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2.a
Differentiate circuit-switched network with packet-switched network.
Ans:
Fig: Circuit swithed network
Fig: Packet switched network
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Circuit- switched network
Packet-switched network
Circuit switching takes place at the physical level
Packet switching takes place at the network level
The actual communication in a circuit-switched
network requires 3 phases: connection setup,
data transfer and connection teardown
In this, only one phase is there called the data
transfer phase
The resources need to be reserved during the
setup phase; the resources remain dedicated for
the entire duration of data transfer until the
teardown phase.
In packet switching, there is no resource
allocation for a packet. Resources are allocated on
demand.
Data transferred between the two stations are not
packetized. The data are a continuous flow sent by
the source station and received by the destination
station
Data are packetized and the packets may travel
different paths to reach their destination
There is no addressing involved during data
transfer. There is end-to-end addressing used
during the setup phase
In datagram networks, each switch (or packet
switch) has a routing table which is based on the
destination address.
The circuit-switched networks are not so efficient
because resources are allocated for the entire
duration of the connection
The efficiency of a datagram network is better
than that of a circuit-switched network; resources
are allocated only when there are packets to be
transferred
The delay in this type of n/w is minimal because
There is no waiting time at each switch. The total
delay is due to the time needed to create the
connection, transfer data, and disconnect the
circuit
There may be greater delay in a datagram n/w
than in a circuit-switched network, because each
packet may experience a wait at a switch before it
is forwarded
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2.b Explain frequency hopping spread spectrum(FHSS) with diagrams
5
Ans:
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The FHSS technique uses M different carrier frequencies that are modulated by the source signal
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At one moment, the signal modulates one carrier frequency; at the next moment, the signal
modulates another carrier frequency
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Although the modulation is done with one carrier frequency at a time, M frequencies are used in
the long run
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The b/w occupied by a source after spreading is
BFHSS >>B
Fig: FHSS
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A pseudorandom code generator called pseudorandom noise(PN), creates a k-bit pattern for
every hopping period Th
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The frequency table uses the pattern to find the frequency to be used for this hopping period and
passes it to the frequency synthesizer
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The frequency synthesizer creates a carrier signal of that frequency, and the source signal
modulates the carrier signal
Fig: Frequency selection in FHSS
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If there are many k-bit patterns and the hopping period is short, a sender and receiver can have
privacy
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If an intruder tries to intercept the transmitted signal, he can only access a small piece of data
because he does not know the spreading sequence to quickly adapt herself to the next hop
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The scheme also has an antijamming effect
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A malicious sender maybe able to send noise to jam the signal for one hopping period(randomly),
but not for the whole period
Fig: FHSS Cycles
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3.a What is TDM? Explain with diagrams. How does statistical TDM overcome the disadvantage of
synchronous TDM?
6
Ans:
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TDM is a digital process that allows several connections to share the high b/w of a link
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Instead of sharing a portion of the b/w as in FDM, time is shared; each connection occupies a
portion of the time in the link
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The TDM is, in principle, a digital multiplexing technique; digital data from different sources are
combined into one time shared link
Fig: TDM
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In synchronous TDM, the data flow of each input connection is divided into units, where each
input occupies one input time slot
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A unit can be 1 bit, one character or one block of data
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Each input unit becomes one output unit and occupies one output time slot
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However, the duration of an output time slot is n times shorter than the duration of an input time
slot
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If an input time slot is T s, the output time slot is T/n s, where n is the number of connections
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In other words, a unit in the output connection has a shorter duration; it travels faster
Fig: Synchronous TDM
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In synchronous TDM, each input has a reserved slot in the output frame; this can be inefficient if
some i/p lines have no data to send
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In statistical TDM, slots are dynamically allocated to improve b/w efficiency
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Only when an i/p line has a slot’s worth of data to send, it is given a slot in the o/p frame
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In statistical multiplexing, the no of slots in each frame is less than the no of i/p lines
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The multiplexer checks each i/p line in a round-robin fashion; it allocates a slot for an i/p line if the
line has a data to send; otherwise, it skips the line and checks the next line
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In statistical TDM, a slot needs to carry data as well as the address of the destination
Fig: TDM slot comparison
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3.b What is a virtual circuit network? List five characteristics of the same.
4
Ans:
A virtual-circuit network is a cross between a circuit-switched network and a datagram network. It has
some characteristics of both.
Characteristics of VCNs:
1. As in a circuit-switched network, there are setup and teardown phases in addition to the data
transfer phase
2. Resources can be allocated during the setup phase, as in a circuit-switched n/w , or on demand as
in a datagram n/w
3. As in a datagram n/w, data are packetized, and each packet carries an address in the header,
which has the local jurisdiction(it defines what should be the next switch and the channel on which
the packet is being carried)
4. As in a circuit-switched n/w, all packets follow the same path established during the connection
5. A VCN is normally implemented in the data-link layer, while a circuit switched n/w is implemented
in the physical layer and a datagram n/w in the network layer
Fig: Virtual circuit network
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4.
Explain the working of CSMA/CD with diagrams.
10
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CSMA/CD is Carrier Sense Multiple Access with Collision Detection
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The CSMA method does not specify the procedure following a collision; CSMA/CD augments the
algorithm to handle the collision
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In this method, a station monitors the medium after it sends a frame to see if the transmission
was successful. If so, the station is finished. If there is a collision, the frame is sent again
•
Figure shows the collision of the first bit in CSMA/CD
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At time t1, station A has executed its persistence procedure and starts sending the bits of its frame
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At time t2, station C has not yet sensed the first bit sent by A. Station C executes its persistence
procedure and starts sending the bits in its frame, which propagate both to the left and to the
right
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The collision occurs some time after time t2.
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Station C detects a collision at time t3 when it receives the first bit of A’s frame. It immediately
aborts the transmission
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Station A detects collision at time t4 when it receives the first bit of C’s frame. It also immediately
aborts transmission
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Station A transmits for the duration t4-t1; C transmits for the duration t3-t2
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For the protocol to work, the length of any frame divided by the bit rate must be more than either
of these durations
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At time t4, the transmission of A’s frame , though incomplete, is aborted; At time t3, the
transmission of C’s frame, though incomplete, is aborted;
Fig: Collision of the first bit in CSMA/CD
Persistence methods:
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The p-persistent method is used if the channel has time slots with a slot duration equal to or
grater than the maximum propagation time
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The p-persistent method combines the advantages of the other two strategies
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It reduces the chance of collision and improves efficiency. In this method, after the station finds
the line idle, it follows these steps:
1. With probability p, the station sends its frame
2. With probability q=1-p, the station waits for the beginning of the next time slot and checks
the line again
a. If the line is idle, it goes to step1
b. If the line is busy, it acts as though a collision has occurred and uses the back-off
procedure
Fig: Flow diagram for CSMA/CD
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Initialize K=0, which is the number of attempts
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Apply any of the persistence methods to transmit a frame
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In CSMA/CD transmission and collision detection is a continuous process; we do not send the
entire frame and then look for a collision; the station transmits and receives continuously and
simultaneously(using two different ports); a loop is used to show that transmission is a continuous
process
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We constantly monitor to detect one of two conditions: either transmission is finished or a
collision is detected. Either event stops transmission
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When we come out of the loop, if a collision has not been detected, then the transmission is
complete; the entire frame is transmitted. Otherwise, a collision has occurred
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In case of collision, send a short jamming signal that enforces the collision in case other stations
have not yet sensed the collision
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Increment K by one. If it’s less than Kmax, then a random number between 0 and 2K-1 is randomly
chosen and multiplied by Tp (max propagation time) or Tfr(average time required to send out a
frame) to find TB which is the back-off time
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Repeat the procedure form the step which applies the persistence method to send the frame
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After a maximum number of retransmission attempts Kmax, a station must give up and try later
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The value of Kmax is usually chosen as 15
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5.a A slotted ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the
throughput if the system(all stations together) produces
a) 1000 frames per second
b) 500 frames per second
c) 250 frames per second
4
Ans:
Solution
The frame transmission time is 200/200 kbps or 1 ms.
a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case
S = G× e−G or S = 0.368 (36.8 percent). This means that the throughput is 1000 × 0.0368 = 368 frames.
Only 386 frames out of 1000 will probably survive.
If the system creates 500 frames per second, this is (1/2) frame per millisecond. The load is (1/2). In this
case S = G × e−G or S = 0.303 (30.3 percent). This means that the throughput is 500 × 0.0303 = 151.
Only 151 frames out of 500 will probably survive.
c. If the system creates 250 frames per second, this is (1/4) frame per millisecond. The load is (1/4). In
this case S = G × e −G or S = 0.195 (19.5 percent). This means that the throughput is 250 × 0.195 = 49. Only
49 frames out of 250 will probably survive.
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5.b. Explain any two controlled access methods with neat diagrams.
6
Ans:
In controlled access, the stations consult one another to find which station has the right to send. A station
cannot send unless it has been authorized by other stations.
Two controlled access methods are Reservation and Polling.
Reservation
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In this method, a station needs to make a reservation before sending data
Time is divided into intervals. In each interval, a reservation frame precedes the data frames sent
in that interval
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If there are N stations in the system, there are exactly N reservation minislots in the reservation
frame
Each minislot belongs to a station. When a station needs to send a data frame, it makes a
reservation in its own minislot
The stations that have made reservations can send their data frames after the reservation frame
Figure 12.18 shows a situation with 5 stations and a 5 minislot reservation frame
In the 1st interval, only stations 1,3 and 4 have made reservations. In the second interval, only
station 1 has made a reservation
Fig: Reservation access method
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Polling
Poling works with topologies in which one device is designated as a primary station and the other
devices are secondary stations
All data exchanges must be made through the primary device even when the ultimate destination
is a secondary device
The primary device controls the link; the secondary devices follow its instructions
It is up to the primary device to determine which device is allowed to use the channel at a given
time. The primary device, therefore, is always the initiator of a session
If the primary wants to receive data, it asks the secondaries if they have anything to send; this is
called poll function
If the primary wants to send data, it tells the seconday to to get ready to receive; this is called the
select function
The select function is used whenever the primary device has something to send
Since the primary controls the link, it knows when the link is available; but doesn’t know whether
the target device is ready to receive
So the primary must alert the secondary to the upcoming transmission and wait for an ack of the
secondary’s ready status
Before sending the data, the primary creates and transmits a select(SEL) frame, one field of which
includes the address of the intended secondary
The poll function is used by the primary device to solicit transmissions from the secondary devices
When the primary is ready to receive data, it must ask (poll) each device in turn if it has anything
to send
When the first secondary is approached, it responds either with an NAK frame if it has nothing to
send, or with data(in the form of a data frame) if it has.
If the response is negative(NAK), then the primary polls the next secondary in the same manner
until it finds one with data to send
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When the response is positive(a data frame), the primary reads the frame and returns an ack(ACK
frame), verifying its receipt
Fig: Select and poll functions in polling access method
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B.E Semester V