A mixed set drawn from the problem bank, weighted toward the tags your quiz scores are weakest on.
T1-Q01
Tutorial 1, Q1
recallconceptual
List the layers of the OSI model and the TCP/IP protocol suite (the Internet Model).
osi-model
tcp-ip
layering
Show walkthrough
The OSI model has seven layers. The TCP/IP suite has five.
Layer
Job
7 Application
Human-computer interaction layer, where applications reach the network’s services
6 Presentation
Puts data in a usable format; where encryption happens
5 Session
Maintains connections; controls ports and sessions
4 Transport
Transmits data using TCP and UDP
3 Network
Decides which physical path the data takes
2 Data link
Defines the format of data on the network
1 Physical
Transmits the raw bit stream over the medium
TCP/IP, top to bottom: application, transport, network, data link, physical. Q3 below names all five directly when it splits them into network-support and user-support groups.
The TCP/IP model was developed by the U.S. Department of Defense to interconnect different networks and keep data intact in transit, and it predates the OSI model. That is why the two do not line up layer for layer — TCP/IP simply has fewer of them.
See layered models for how the two stacks map onto each other and what each layer’s data unit is called.
T1-Q02
Tutorial 1, Q2
applyconceptual
How do the layers of the OSI correlate to the layers of the TCP/IP protocol suite?
osi-model
tcp-ip
layering
Show walkthrough
OSI’s application, presentation and session layers collapse into a single application layer in TCP/IP. The remaining four layers map one to one.
OSI
TCP/IP
Application, Presentation, Session
Application
Transport
Transport
Network
Network
Data link
Data link
Physical
Physical
TCP/IP is the more simplified model, built to focus on the parts of network communication that matter most, so the three upper OSI layers are combined into one application layer rather than kept separate.
See layered models for the full mapping and what runs at each layer.
T1-Q03
Tutorial 1, Q3
applyconceptual
Which layers of the TCP/IP protocol suite are the network support layers and which layers are the user support layers?
tcp-ip
layering
Show walkthrough
Physical, data link and network are the network support layers. Application is the user support layer. Transport sits between the two and links them.
Group
Layers
Job
Network support
Physical, Data link, Network
Handle the physical side of moving data between devices — electrical specifications, physical connections, physical addressing, transport timing and reliability
User support
Application
The interface with the user; lets unrelated software systems interoperate
Link between the two
Transport
Makes sure what the lower layers transmitted is in a form the upper layer can use
See layered models for what each of the five layers does in full.
T1-Q04
Tutorial 1, Q4
recallconceptual
What is data encapsulation (with respect to Questions 1-3)?
encapsulation
layering
Show walkthrough
Encapsulation is each layer on the sending device adding its own information — a header, and at the data link layer a trailer too — to the message it receives from the layer above, then passing the whole package down to the layer below.
Applied to the layers above, the message grows a piece at a time on the way down:
Application: M
Transport: adds a header → M Ht
Network: adds a header → M Ht Hn
Data link: adds a header and a trailer → M Ht Hn Hl
At the receiving end, each layer strips off the header (and trailer, at the data link layer) meant for it, in the reverse order they were added, until what reaches the top is M again.
What are the responsibilities of the Data Link layer?
layering
mac-address
flow-control
error-control
Show walkthrough
The data link layer has six jobs:
Framing data bits — divides the bit stream handed down from the network layer into manageable frames.
Moving frames — transfers a frame from one hop (node) to the next.
Physical addressing — attaches physical (MAC) addresses so a frame’s sender and receiver can be identified.
Flow control — stops a fast sender overwhelming a slower receiver.
Error control — detects and corrects errors introduced over the physical layer, adding reliability.
Access control — decides which device has control of the link at any given moment, so devices sharing one link do not conflict.
See layered models for where this sits relative to the other six.
T1-Q06
Tutorial 1, Q6
applyconceptual
What is the difference between Network layer and Transport layer delivery?
layering
addressing
ip-address
port
Show walkthrough
Network-layer delivery gets a message to the right host. Transport-layer delivery gets it the rest of the way, to the right process on that host.
Network layer
Transport layer
Provides logical communication between
Hosts
Processes running on different hosts
Address used
Logical address (IP)
Port address
Reaches a specific process?
No — only the destination computer
Yes — the correct process on that computer
A network-layer protocol can only deliver the message to the destination computer. A transport-layer protocol is responsible for delivering it to the specific process the message is for, once it has arrived.
If the Data link layer can detect errors between hops, why do you think we need another mechanism at the Transport layer?
error-control
layering
Show walkthrough
Because the data link layer’s error control only covers one hop at a time. The path between hops — the network layer itself — is unreliable on the internet, and the data link layer never sees far enough to catch errors introduced there.
The data link layer checks each link as a frame crosses it, node to node. It has no visibility past that single link, so an error picked up while a packet travels through an intermediate router, or across the wider path, goes past it unnoticed. Since the network layer offers no reliability guarantee of its own, something has to check the message end to end — and that is the transport layer’s job.
The two also differ in what they check and how carefully:
Data link error control works at the packet or stream level, one hop at a time. Transport error control works at the message level, source to destination.
The data link layer is not concerned with the order packets arrive in. The transport layer is.
T1-Q08
Tutorial 1, Q8
recallconceptual
What are the responsibilities of the Network layer?
layering
routing
ip-address
Show walkthrough
The network layer has three jobs:
Moving packets — delivers individual packets from the source host to the destination host.
Logical addressing — adds a header to each packet carrying the sender’s and receiver’s logical addresses, so source and destination systems can be told apart.
Routing — when independent networks or links are joined into one larger network, the network layer routes packets across that structure so they reach their final destination.
What are the responsibilities of the Transport layer?
layering
port
flow-control
error-control
Show walkthrough
The transport layer has five jobs:
Moving messages — delivers a message from one process to another.
Port addressing — since many processes can be running on the communicating hosts at once, the transport header must carry a service port address in every segment, identifying which process the data is for. Example ports: HTTP 80, SMTP 25.
Segmentation and reassembly — splits a message into segments, each given a sequence number, so the destination transport layer can reassemble them in the exact order the sender sent them.
Flow control — keeps the sending process from emitting segments faster than the receiving process can handle.
Error control — process-to-process error control, rather than the single-link error control the data link layer provides.
What is the difference between a port address, a logical address and a physical address?
addressing
mac-address
ip-address
port
Show walkthrough
The three differ in which layer assigns them, how far they reach, and their size.
Physical address
Logical address
Port address
Layer
Data link
Network
Transport
Also known as
Link address, MAC address, hardware address
—
Service point
Scope
The local address of a node; delivers data within one network
Delivers messages across multiple networks
Identifies a process on the station
Size / notation
`6 bytes`, hexadecimal
`32 bits` (IPv4) or `128 bits` (IPv6), dotted-decimal for IPv4
`16 bits` in TCP/IP
Assigned to
A network adapter — no two share one
Every host and router (one IP, one MAC each)
A running process
The physical address is the local address of a node, used by the data link layer to deliver data from one node to another within the same network.
The logical address defines the sender and receiver at the network layer, for delivery across multiple networks. It is needed because a physical address only means something on the one network it belongs to.
The port address (service point) identifies the application process on a station. Since several processes can be running on a host at once, something has to identify which one a piece of data is meant for — that is what the port address does.
Worked example
Tracing one message from process to process
Host A and Host B are connected across the internet. Host A’s logical/physical address pair is A/10; Host B’s is B/20. Host A runs three processes, a, b and c; Host B runs two, j and k. Process a generates data for process j.
Transport layer attaches the port addresses of the two processes (the tutorial’s example values are 1020 and 9566) and hands down a segment: a j Data.
Network layer adds the logical addresses of the two hosts to the segment and hands down a datagram: A B a j Data.
Data link layer adds the physical addresses of the source and the next hop and hands down a frame: 10 20 A B a j Data.
AnswerFrame: 10 20 · A B · a j · Data — physical addresses outermost, then logical, then port, then the data
The tutorial’s own line for the segment step lists the host labels a step early, ahead of where the network layer actually adds them. The reading above follows the pattern its network- and data-link-layer steps establish, which is unambiguous.
T1-Q11
Tutorial 1, Q11
recallconceptual
Name some services provided by the Application layer.
layering
protocol
internet
Show walkthrough
The application layer is the seventh layer of the OSI model, and it gives the user services through logical communication — its interface talks directly to the application and provides the common web application services:
Remote file access and transfer — FTP
Electronic mail — SMTP
Network virtual terminal — TELNET
Web browsing — HTTP
Resolving hostnames to IP addresses — DNS
Managing and monitoring network-connected devices on the internet — SNMP
Each of these is reached through an application-layer port number assigned to that specific service or protocol.
See layered models for where the application layer sits relative to the rest of the stack.
T1-Q12
Tutorial 1, Q12
analyseconceptual
What are the advantages of combining Session, Presentation and Application layers of the OSI model to a single layer in the TCP/IP protocol suite?
layering
osi-model
tcp-ip
Show walkthrough
Compared to the OSI model, TCP/IP drops session and presentation as separate layers and folds their work into a single application layer. Two reasons are given for this:
First, some of what the session layer does is already available in some transport-layer protocols.
Second, applications and protocols at the TCP/IP application layer can handle data formatting, encryption and session management themselves, as needed, without separate layers to do it for them.
That gives five advantages:
Flexibility
Reduced redundancy
Faster processing
Less processing and energy consumption
A simpler model overall
See layered models for the OSI-to-TCP/IP mapping this collapse produces.
T2-Q01
Tutorial 2, Q1
recallconceptual
Distinguish between a Signal Element and a Data Element.
signals
bandwidth
Show walkthrough
Data element
Signal element
What it is
The smallest entity that expresses information — the bit
The shortest unit of a digital signal — one discrete voltage pulse
Role
What you need to send
What you can actually send — the carrier
Analogy
A passenger
The vehicle carrying the passenger
Example
A single binary `1` or `0`
One pulse in the transmitted waveform
A signal element can carry more than one data element, or a single data
element can be spread across more than one signal element. The ratio r
between them says which:
r=signal elementsdata elements
The tutorial gives four cases:
r = 1/1 = 1 — one data element per one signal element
r = 2/1 = 2 — two data elements per one signal element
r = 1/2 = 0.5 — one data element spread across two signal elements
r = 4/3 — four data elements carried across three signal elements
r above 1 means each pulse is doing more than one bit’s worth of work.
r below 1 means a single bit needs more than one pulse to carry it.
The key point the tutorial makes: for a lower baud rate, the same amount of
data can still be transmitted, just spread over less time per pulse by
packing more bits into each signal element. Signal rate is what a channel’s
bandwidth has to keep up with — data rate is just how many bits that signal
rate happens to be carrying.
Define the characteristics of a self-synchronizing signal.
signals
line-coding
Show walkthrough
For the receiver to interpret an incoming signal correctly, its bit
intervals must line up exactly with the sender’s. If the receiver’s clock
runs faster or slower than the sender’s, the intervals drift apart and the
receiver starts misreading bits.
A self-synchronizing signal solves this by carrying its own timing
information inside the data. The receiver extracts the clock directly from
the incoming bit stream, so no separate clock line between sender and
receiver is needed.
In practice this means the encoding must guarantee a transition — a level
change — somewhere in every bit interval, giving the receiver something to
resynchronize against on every single bit. Line codes that do this are
called self-clocking.
What are the differences between parallel and serial transmission?
signals
media
Show walkthrough
Parallel transmission
Serial transmission
How bits move
The bit stream is divided into groups, and one group is sent per clock pulse
One bit is sent per clock pulse
Speed
Faster — several bits transfer at once
Slower — only one bit at a time
Channels needed
`n` communication channels to send `n` bits at a time, raising cost
A single communication channel, which keeps cost down
Converters
None required
Required at both sender and receiver, because the devices themselves work in parallel internally
The trade-off is speed against cost: parallel buys speed by paying for more
physical channels, serial buys a cheap single channel at the cost of sending
one bit at a time.
List different techniques of serial transmission and explain their differences.
signals
media
Show walkthrough
Serial transmission splits into three techniques.
Asynchronous serial transmission. The bit stream is divided into
8-bit bytes, and each byte is sent independently, whenever it is ready,
regardless of any shared timer. A start bit (0) marks the arrival of a
byte and a stop bit (1) marks its end. The receiver starts a timer at the
start bit and counts bits until it reaches the stop bit — that is the only
synchronization needed, and it only has to hold for one byte at a time. The
extra start and stop bits make this slower than the alternatives, but it is
cheap and effective.
Synchronous serial transmission. Bits are sent one after another with
no start bits, stop bits, or gaps between them. The receiver is responsible
for counting bits and grouping them into bytes, and multiple bytes combine
into frames. Without the extra framing bits, this is faster than
asynchronous transmission — but it cannot support real-time applications
such as television broadcasting, because it has no mechanism to guarantee
an even delay between frames.
Isochronous serial transmission. Real-time audio and video cannot
tolerate uneven delays between frames — television images are broadcast at
30 images per second and must be viewed at that same rate. Synchronization
at the level of individual characters is not enough here; the entire bit
stream has to stay synchronized. Isochronous transmission guarantees that
data arrives at a fixed rate, which is what synchronous transmission alone
cannot promise.
Pulse Code Modulation is the standard technique for turning an analog
signal into digital data, and it runs in three steps.
1. Sampling. The analog signal is measured at equal intervals,
converting it from continuous time to discrete time. This only produces a
series of pulses at the sampled amplitudes — a PAM (Pulse Amplitude
Modulation) signal — not yet a digital signal.
2. Quantization. The sampled PAM pulses can land on non-integer
amplitudes, which cannot be encoded directly. Quantization rounds each
sample’s amplitude to the nearest of a fixed, manageable set of levels.
3. Encoding. Each quantized value is translated into its binary
equivalent: seven bits for the magnitude, plus an eighth bit for the sign.
Sampling turns the signal into discrete points in time. Quantization turns
each point into one of a fixed set of levels. Encoding turns each level
into bits — the three steps run in that order because each one depends on
the last.
Which of the above four are most susceptible to noise? Explain your answer.
modulation
ask
snr
Show walkthrough
Amplitude Shift Keying (ASK) is the most susceptible to noise. It is
the simplest of the four schemes, and that simplicity is exactly what makes
it fragile — its bit detection depends entirely on amplitude, and amplitude
is the property noise disturbs most easily.
Noise here means unintentional voltages introduced onto the line by things
like heat or crosstalk. Those unwanted voltages add directly onto the
signal and shift its amplitude.
ASK reads a bit by measuring amplitude, so any amplitude shift from noise
is read as part of the signal. PSK reads a bit by measuring phase instead,
and noise changes phase far less readily than it changes amplitude.
That difference in what each scheme measures is why ASK is the most
vulnerable to noise corruption of the four.
What is the number of bits per baud for the following techniques?
a) ASK with four different amplitudes
b) FSK with 8 different frequencies
c) PSK with four different phases
modulation
ask
fsk
psk
qam
Show walkthrough
Worked example
Bits per baud for each scheme
Bits per baud is given by r = log2(L), where L is the number of
distinguishable signal elements the scheme uses — not the number of levels
in some other sense, just how many distinct states the receiver has to tell
apart.
a) ASK with four amplitudes:L = 4.
r=log24=2 bits per baud
b) FSK with 8 frequencies:L = 8.
r=log28=3 bits per baud
c) PSK with four phases:L = 4.
r=log24=2 bits per baud
AnswerASK: 2, FSK: 3, PSK: 2 bits/baud
The tutorial solution adds a worked note on how QPSK (four-phase PSK, also
reached via QAM) achieves 2 bits per signal element in practice: two
separate PSK modulators produce an in-phase (I, sine) and a quadrature
(Q, cosine) component with the same frequency and timing. Multiplying the
bit stream’s digital amplitude into each gives I a phase shift of 180°
or 0° and Q a phase shift of 180° or 0°. Adding the two components
back together produces a single signal element with one of four possible
resultant phases — 45°, -45°, 135°, -135° — which is how one signal
element ends up carrying 2 bits.
Assume that a voice channel occupies a bandwidth of 4kHz. We need to
combine three voice channels into a link with a bandwidth of 12kHz, from 20
to 32kHz. Show the configuration, using the frequency domain. Assume there
are no guard bands.
multiplexing
fdm
bandwidth
Show walkthrough
Worked example
Assign each 4 kHz channel a slice of the 12 kHz link
Each voice channel needs 4 kHz of bandwidth. The link provides 12 kHz
total, from 20 kHz to 32 kHz, with no guard bands to reserve space for.
Divide the 12 kHz link into three consecutive 4 kHz slices, starting
from the bottom of the band:
Five Channels, each with a 100kHz bandwidth, are to be multiplexed
together. What is the minimum bandwidth of the link if there is a need for
a guard band of 10kHz between the channels to prevent interference?
multiplexing
fdm
bandwidth
Show walkthrough
Worked example
Total FDM bandwidth with guard bands
Five channels at 100 kHz each need:
5×100 kHz=500 kHz
Guard bands sit between adjacent channels, not around the outside, so n
channels need n - 1 guard bands. Five channels need 4 guard bands:
4×10 kHz=40 kHz
Add the channel bandwidth and the guard-band bandwidth:
From the below figure, the data rate for each input connection is 3kbps.
If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration
of (a) each input slot, (b) each output slot, and (c) each frame?
multiplexing
tdm
Show walkthrough
The figure referenced in the tutorial is not reproduced in the source, but
the solution’s own numbers show a synchronous TDM system with three input
connections, each combined one bit at a time into a single output frame of
three slots.
Worked example
Input slot, output slot, and frame duration
a) Input slot duration. Each input connection runs at 3 kbps, so one
bit takes:
30001 s
The input slot duration equals this bit duration, 1/3000 s.
b) Output slot duration. Three input streams are interleaved onto one
output line, so each output slot is one-third the length of an input slot:
3×30001=90001 s
c) Frame duration. One frame carries one output slot from each of the
three inputs:
3×90001 s=30001 s
The frame duration comes out equal to the input slot duration — one full
frame is sent in exactly the time it takes one input to produce one bit,
which is what interleaving three streams onto one line has to achieve.
The figure below shows synchronous TDM with a data stream for each input
and one data stream for the output. The unit of data is 1 bit. Find (a) the
input bit duration, (b) the output bit duration, (c) the output bit rate,
and (d) the output frame rate.
multiplexing
tdm
Show walkthrough
The figure referenced in the tutorial is not reproduced in the source. The
solution’s own numbers show four input streams, each running at 1 Mbps,
combined one bit at a time into a single output stream.
Worked example
Input/output bit duration, bit rate, and frame rate
a) Input bit duration is the reciprocal of the input bit rate:
1 Mbps1=1μs
b) Output bit duration. One bit from each of the four input streams is
combined into a single output bit, so the output bit duration is one
quarter of the input bit duration:
41μs=0.25μs
c) Output bit rate is the reciprocal of the output bit duration:
0.25μs1=4 Mbps
This also equals the sum of the four input bit rates: 1 Mbps × 4 = 4 Mbps.
d) Output frame rate. In synchronous TDM, the frame rate equals the
bit rate of any single input channel:
List the main multiplexing techniques discussed in the lecture.
multiplexing
fdm
tdm
Show walkthrough
The lecture covers three main multiplexing techniques.
1. Frequency Division Multiplexing (FDM). Signals are combined in the
frequency domain. It only works when the link’s bandwidth exceeds the
combined bandwidth of the signals being transmitted, and channels need
guard bands between them to stop signals overlapping.
2. Synchronous Time Division Multiplexing (Sync TDM). A high-capacity
link’s usage is divided into fixed time slots, one per device. It only
works with a digital signal. Fixed slots make it simple to implement, but
they can waste bandwidth when a device has nothing to send, and every
device has to stay strictly synchronized.
3. Statistical (Asynchronous) Time Division Multiplexing (STDM).
Frames are built from a statistical analysis of how likely each input line
is to be transmitting at a given moment, and the multiplexer keeps scanning
input lines until a frame fills up before sending it. Slots are variable
rather than fixed, which is more complex to implement but guarantees the
full capacity of the link is used efficiently.
Which of them (Q14) are used for combining analog signals?
multiplexing
fdm
Show walkthrough
Frequency Division Multiplexing (FDM) is the one used for combining
analog signals.
FDM is an analog technique. It applies whenever a link’s bandwidth, in
hertz, is greater than the combined bandwidths of the signals it needs to
carry. It works by assigning each signal its own frequency band, which lets
several analog signals travel over the same medium at once without
interfering — as long as guard bands separate the bands.
Which of them (Q14) are used for combining digital signals?
multiplexing
tdm
Show walkthrough
Time Division Multiplexing (TDM) is the digital multiplexing
technique — it combines several low-rate channels into one high-rate
channel. The two variants named in Q14 both do this:
Synchronous Time Division Multiplexing (Sync TDM)
Asynchronous Time Division Multiplexing (Async TDM)
Both work in the time domain, dividing transmission time into slots
assigned to different channels, rather than dividing the frequency band the
way FDM does.
Which of the three multiplexing techniques is common for fiber optic
links? Explain why.
multiplexing
fibre
Show walkthrough
Wavelength Division Multiplexing (WDM) is the technique the tutorial
gives as common for fiber optics. The tutorial notes that WDM is used for
analog signals — it is not one of the three techniques named in Q14 (FDM,
Sync TDM, and Async TDM), so treat this question as introducing a fourth
technique specific to fiber, not as picking one of the earlier three.
WDM carries multiple data channels over a single fiber simultaneously, each
on a different wavelength of light, which multiplies the fiber’s overall
data capacity without needing extra physical cable.
The advantages the tutorial lists for WDM on fiber:
High bandwidth
Efficient use of the medium
Flexible
Long-distance transmission
Using different wavelengths, effectively different colors of light, lets a
fiber system reach data rates far beyond what a single wavelength could
carry — which is why WDM is the preferred technique for modern optical
communication systems.
Distinguish between synchronous TDM vs. statistical TDM.
multiplexing
tdm
Show walkthrough
Aspect
Synchronous TDM
Statistical TDM
Time slot allocation
Fixed, predetermined slot within the TDM frame
Allocated dynamically, based on statistical analysis
Allocation logic
Every input channel gets a fixed slot in the frame, whether or not it has data to send
A slot is only allocated when data is actually present
Bandwidth efficiency
Potential wastage when a channel has no data
Minimal waste — more efficient than synchronous TDM
Timing predictability
Predictable — the receiver knows exactly when to expect each channel's data
Not fixed — a given channel's slot timing varies
Addressing
Not needed
The receiver needs addressing to identify which channel's data is in each slot
Suited to
Continuous, consistent transmission, such as voice communication
Bursty, unpredictable transmission, such as computer networks
The trade-off is the same one that shows up across multiplexing generally:
synchronous TDM buys predictable timing and a simple receiver at the cost of
wasted slots, while statistical TDM buys efficiency at the cost of needing
per-slot addressing.
Assume that a voice channel occupies a bandwidth of 4 kHz. We need to
multiplex 10 voice channels with guard bands of 500 Hz using FDM.
Calculate required bandwidth.
multiplexing
fdm
bandwidth
Show walkthrough
Worked example
Total FDM bandwidth for 10 channels
Ten channels at 4 kHz each:
4 kHz×10=40 kHz
Guard bands sit between adjacent channels, so 10 channels need 10 - 1 = 9
guard bands, each 500 Hz = 0.5 kHz:
9×0.5 kHz=4.5 kHz
Add the channel bandwidth and the guard-band bandwidth:
We need to transmit 100 digitized voice channels using a pass-band channel
of 20 kHz. What should be the ratio of bits/Hz if we use no guard band?
pcm
sampling
bandwidth
Show walkthrough
Worked example
Bits per second per hertz for 100 digitized voice channels
Bandwidth available per channel, splitting the 20 kHz passband across
100 channels with no guard bands:
10020 kHz=200 Hz per channel
Each digitized voice channel’s bit rate. The solution assumes the standard
telephone PCM figures — an 8 kHz sampling rate and 8 bits per sample:
8 bit/sample×8 kHz=64 kbps
Divide the bit rate per channel by the bandwidth per channel:
200 Hz64 kbps=320 bits/s/Hz
Answer320 bits/s/Hz
The 8 kHz sampling rate and 8 bits/sample are not given in the question
itself — they come from the standard PCM telephony figures the tutorial
solution assumes, the same 64 kbps used throughout this unit for a
digitized voice channel.
What is the significance of the twisting in twisted pair cable?
media
twisted-pair
Show walkthrough
Twisting minimizes interference — noise and crosstalk — between the two
adjacent wires that make up a pair.
Because both wires are twisted together, external influences such as noise
and crosstalk affect them equally, rather than affecting the wire closest
to the interference source more strongly, as would happen with a pair of
untwisted parallel wires. The receiver reads the difference between the two
wires, and since the interference on each wire is nearly identical, that
difference cancels the unwanted signal out.
By reducing crosstalk and electromagnetic interference this way, twisting
prevents signal degradation and lets the cable maintain high data
transmission rates over longer distances.
What is the purpose of cladding in optical fiber cable?
media
fibre
Show walkthrough
Cladding exists to propagate the signal effectively down the fiber’s core.
It has a lower density than the core, and that density difference is what
does the work.
The core has a higher refractive index than the cladding. At the boundary
between them, light undergoes total internal reflection instead of escaping
— so the light signal stays inside the core and keeps traveling forward
rather than leaking out.
That mechanism is what lets light propagate through the fiber with minimal
loss, even where the fiber bends.
Name the advantages of optical fiber over twisted pair and coaxial cable.
media
fibre
coaxial
twisted-pair
Show walkthrough
Fiber-optic cable is used widely in telephone and cable television
networks, and the tutorial gives six advantages over copper media.
Immunity to noise. Fiber carries light rather than electricity, so it
is resistant to electromagnetic interference.
Greater transmission distance. Lower signal attenuation means fiber
covers longer distances than other guided media before the signal degrades
too far to use.
Enhanced security. Fiber is difficult to tap into.
Compact and durable. It is smaller and lighter than copper, and immune
to corrosion.
High bandwidth. Fiber provides the highest bandwidth of any
transmission system covered.
Fewer repeaters needed. Fewer repeaters are needed to cover a given
distance, which makes fiber more efficient for extensive networks.
The tutorial ties the distance advantage to a specific number: fiber’s low
signal attenuation lets it transmit up to 50 km without needing
regeneration, while coaxial and twisted-pair cable typically need
regeneration roughly every 5 km.
How does sky propagation differ from line-of-sight propagation?
media
propagation
Show walkthrough
Sky (ionospheric) propagation
Line-of-sight propagation
Mechanism
Higher-frequency radio waves travel up to the ionosphere and are bounced back down towards Earth
Very high frequency signals travel in a straight line
Antenna requirement
None beyond transmitting upward — no direct antenna alignment needed
Both antennas must be in view of each other
Frequency range
2-30 MHz
Not stated as a range in the tutorial
Stability
Changes with time of day and weather, because it depends on the ionosphere
Not weather-dependent in the same way
Other names
Sky wave propagation
Space waves, or tropospheric propagation
Range limit
Not distance-limited by line of sight
Limited by the curvature of the Earth for ground-based stations — about 50 km
The distinction comes down to what each method uses to get the signal from
transmitter to receiver: sky propagation bounces the signal off a layer of
the atmosphere, while line-of-sight propagation relies on the signal
traveling in a direct, unobstructed straight line.
A switch is a device that creates a temporary connection between two or
more devices linked to it. Traditionally, three switching methods matter:
Circuit switching
Packet switching, which itself splits into:
Virtual-circuit networks
Datagram networks
Message switching
Of the three, circuit switching and packet switching are the two in common
use today. Packet switching is the one that further divides into the two
subcategories above.
They differ in how packets get routed and whether any connection state
persists across the transfer. A virtual-circuit network establishes a
connection path before any data is sent. A datagram network routes each
packet independently, with no connection set up in advance.
Compare and contrast a circuit switched network and a packet switched
network.
switching
circuit-switching
packet-switching
Show walkthrough
Aspect
Circuit switching
Packet switching
Connection establishment
A physical circuit is established before transmission begins
No physical circuit is established beforehand — no setup or teardown phase
Data format
Continuous bits
Packets
Phases
Three: setup, data transfer, teardown
No formal phases
Resource utilization
Less efficient
More efficient
Layer of operation
Physical layer
Data link layer (virtual-circuit approach) or network layer (datagram approach)
Application
Telephone networks — bi-directional, fast, real-time transfer
The Internet
The dividing line running through all six rows is the same one: circuit
switching pays an up-front setup cost for a dedicated, predictable path,
while packet switching skips that cost and routes each unit of data as it
goes, at the price of less predictable resource use.
What is the role of the address field in a packet travelling through a
datagram network?
switching
packet-switching
Show walkthrough
A datagram network is connectionless — packet switching done at the
network layer. No virtual connection exists between source and destination,
so there is no setup or teardown phase to go through first.
When a source creates a datagram, it writes its own address into the
source address field and the ultimate destination’s address into the
destination address field. Switches hold routing tables that map
destination addresses to the output port each one should be forwarded
through.
When a switch receives the packet, it reads the destination address and
looks it up in the routing table to find the matching output port.
This address stays the same for the packet’s entire journey — unlike a
virtual-circuit network, where the address used for switching changes at
each hop.
What is the role of the address field in a packet travelling through a
virtual-circuit network?
switching
packet-switching
virtual-circuit
Show walkthrough
A virtual-circuit network mixes characteristics of circuit switching and
datagram networks, and does its switching at the data link layer. It
requires a virtual connection to be set up between the communicating nodes
before any data can move.
It uses two kinds of addressing:
Global addressing. Source and destination need unique addresses —
their IP addresses.
Local addressing. A small address carried in a frame between two
switches, called the virtual-circuit identifier (VCI).
There are three phases: setup request, setup acknowledgement, and data
transfer.
Setup request: global (IP) addresses identify the source and
destination.
Setup acknowledgement: the path is established and a VCI is assigned.
Data transfer: local addresses (VCIs) are used for switching between
adjacent switches, because they are cheaper to process than global
addresses.
The VCI changes at every hop, while the global addresses stay constant
— the opposite of a datagram network, where the address is fixed for the
whole journey.
List four major components of a packet switch and their functions.
switching
packet-switching
Show walkthrough
A packet-switch’s structure differs from a circuit-switch’s. The tutorial
gives four major components.
1. Input port. Handles the physical and data link functions of the
packet switch. It holds buffers (queues) that hold a packet before it moves
into the switching fabric.
2. Output port. Performs the same functions as the input port, in
reverse order: outgoing packets are queued first, then encapsulated into a
frame, then transmitted.
3. Routing/switching processor. Performs the data link layer’s
functions and searches the routing (forwarding) table to find the next-hop
address for a packet.
4. Switching fabric. Moves a packet from its input queue to the
correct output queue. Different switching techniques do this move — the
tutorial names the Crossbar Switch and the Banyan Switch.
Briefly describe the services provided by the data link layer.
data-link-layer
framing
flow-control
error-control
Show walkthrough
The physical layer only pushes a raw stream of bits onto the wire. The data
link layer sits directly on top of it and adds six services:
Framing. Divides the stream of bits handed down from the network layer
into manageable frames.
Moving frames. Transfers each frame from one hop to the next.
Physical addressing. Provides physical (MAC) addresses that identify
the sender and the receiver.
Flow control. Stops the sender from overwhelming the receiver with too
much data too quickly.
Error control. Detects, and adds reliability against, errors picked up
on the physical link.
Access control. Decides which device has control over the link at any
given moment, so devices sharing one link do not conflict.
Six services because each one solves a separate problem — addressing and
error control both live inside the frame, but neither can substitute for the
other, and none of the six is optional.
See Framing for how framing and addressing actually
work.
T3-Q02
Tutorial 3, Q2
recallconceptual
Define framing and the reason for its need.
framing
data-link-layer
Show walkthrough
Framing is the process, in the data link layer, of dividing the stream of
bits received from the network layer into frames. Every frame carries four
parts:
Frame header — the source and destination address.
Payload field — the message being delivered.
Trailer — the error detection and correction bits.
Flag — marks the beginning and end of the frame.
Framing is needed for two reasons the tutorial gives directly:
It adds the physical sender and destination addresses. The destination
address is where the frame should go; the sender address is what an
acknowledgement (ACK) gets sent back to.
It divides the message into appropriately sized frames, which is what
makes flow control and error control efficient. Smaller frames mean a
corrupted or lost unit costs less to detect and resend.
See Framing for the flag pattern and bit stuffing
that make the boundary between frames unambiguous.
T3-Q03
Tutorial 3, Q3
analyseconceptual
Compare and contrast byte-oriented and bit-oriented protocols. Which category
has been popular in the past (explain the reason)? Which category is popular
now (explain the reason)?
framing
data-link-layer
Show walkthrough
Byte-oriented (character-oriented)
Bit-oriented
Unit carried
8-bit characters from a coding system, e.g. ASCII
A stream of individual bits, with no fixed character boundary
Popular when
In the past
Now
Why
Early data exchange was text-based, and 8-bit ASCII characters were the natural unit to send
Modern data includes text, graphics, audio and video, none of which map naturally onto ASCII characters. A bit pattern represents all of them, and is more versatile and efficient
Character-oriented protocols were popular when the data link layer only ever
carried text. Bit-oriented protocols are the more common choice now, because
they handle any data type as a plain sequence of bits rather than forcing it
into character-sized units.
The lecture slides do not develop this byte-oriented vs bit-oriented
distinction — the Framing topic page covers it as
supplementary material for exactly this reason. The comparison and reasoning
above follow the tutorial’s own solution, which is the authority for this
question.
T3-Q04
Tutorial 3, Q4
analyseconceptual
Compare and contrast the three most common error detection methods.
error-detection
parity
vrc
lrc
crc
Show walkthrough
VRC (parity)
LRC
CRC
What it adds
1 extra bit per byte
A redundant row of bits across the whole block
n − k redundancy bits from polynomial division
How it works
Even parity makes the total number of 1s even; odd parity makes it odd
The receiver recomputes the LRC and compares it to the one appended at the sender. A match means no error is detected
The message is treated as a polynomial and divided by a fixed generator; the remainder becomes the check bits
Reliability
About 50% — one of the oldest and simplest methods
Better than VRC, particularly at catching burst errors
Detects more error types than either VRC or LRC — the strongest of the three
Simplest and least reliable, to most powerful, left to right — this is the tutorial's own ordering.
CRC in more detail, since it is the most involved of the three. With
message length k and codeword length n:
Encode. Multiply P(x) by x^(n−k). Divide [M(x) × x^(n−k)] by the
generator G(x) and take the remainder R(x). The codeword is
T(x) = [M(x) × x^(n−k)] + R(x).
Decode. Divide the received T(x) by G(x). A zero remainder means no
error was detected; anything else means an error was found.
See Error Detection for a full worked CRC
division and the CRC burst-error guarantee.
T3-Q05
Tutorial 3, Q5
analyseconceptual
Compare and contrast flow control and error control. How does error control
complement flow control?
flow-control
error-control
Show walkthrough
Flow control
Error control
What it manages
How much data the sender can push before waiting for an acknowledgement
Detecting and correcting errors in the data received
Why it is needed
A receiver processes data more slowly than it arrives, and has only a limited buffer to hold the backlog
Bits get corrupted in transit, independently of how fast they arrive
What breaks without it
The buffer overflows and frames are discarded
Corrupted data reaches the layer above as though it were correct
Tutorial 3 asks for this comparison directly, plus how the two complement each other.
They complement each other rather than overlapping. Flow control decides when
frames are allowed to keep arriving, which stops the buffer from discarding
them before anything gets a chance to check them. Error control then decides
whether what did arrive is trustworthy. Put simply:
Flow control handles the speed of transmission.
Error control handles the integrity of transmission.
Together they are what makes reliable delivery possible — flow control alone
would still let corrupted frames through, and error control alone would still
let a buffer overflow.
See ARQ and Flow Control for how ACK, NAK
and timeout-triggered retransmission carry out this error control in
practice.
T3-Q06
Tutorial 3, Q6
recallconceptual
Standard Ethernet defines several Physical layer implementations. Discuss the
most common implementation categories?
ethernet
10base5
10base-t
Show walkthrough
Standard Ethernet defines several physical-layer implementations. Four are
in common use:
Name
Media
Max length
Line encoding
10Base5
Thick coaxial cable
500 m
Manchester
10Base2
Thin coaxial cable
185 m
Manchester
10Base-T
2 pairs of UTP
100 m
Manchester
10Base-F
2 fibre strands
2000 m
Manchester
10Base5 (Thick Ethernet, Thicknet). Bus topology, with an external
transceiver connected to the coax handling transmit, receive and collision
detection.
10Base2 (Thin Ethernet, Cheapernet). Also bus topology, but with a
thinner, more flexible cable that can bend close to each station. The
transceiver sits on the NIC inside the station rather than externally.
10Base-T (twisted-pair Ethernet). Star topology: every station connects
to a hub over two pairs of twisted cable, giving separate send and receive
paths.
10Base-F (fibre Ethernet). The most common 10 Mbps fibre variant, also
star topology, using a pair of fibre-optic cables per station and a
NIC-mounted transceiver.
Switched Ethernet connects every device on the same network so they can all
reach one another, but instead of broadcasting every frame to every station,
the switch reads the destination address in each frame and sends it only out
the port that address lives on.
Operation. When a frame arrives on one port, the switch checks the
destination address and sends the frame out the corresponding port, not every
port.
Addressing and forwarding. A MAC address table (forwarding table)
records which port each address lives on, and is what the switch consults on
every frame.
Bandwidth sharing. Bandwidth is shared only between a station and the
switch — a dedicated link — not across every device on the network the way an
unswitched shared bus would be.
Worked example. Given the table below, when Host E sends data to Host D,
the switch looks up D’s address, finds it on port 4, and forwards the frame
directly there rather than to every other port.
Token passing. A token circulates regularly from one node to the next. A
node holding the token that has data to send releases it onto the network. A
node with nothing to send just passes the token on to the next node.
Advantages:
Fewer collisions. A node only releases data once it actually holds the
token, so two nodes never transmit into the same slot by accident.
Ring outperforms bus under heavy traffic. Because access is ordered by
the token rather than contested, token passing makes ring topology handle
heavy load better than bus topology, where every node still competes for
the same shared medium.
No server required. Nothing central has to coordinate which node
transmits when — the token itself is the coordination.
Full bandwidth use under heavy demand. Because there is no time lost to
collisions or contention, the channel’s bandwidth can be used fully when
demand is high.
T3-Q09
Tutorial 3, Q9
applyconceptual
In the event of failure, how does the FDDI network automatically reconfigure?
fddi
token-ring
Show walkthrough
The problem with a plain token ring. If the link between two adjacent
stations fails, the whole ring fails — a single break anywhere stops data
from completing the loop.
The fix: a dual ring. The Fiber Distributed Data Interface (FDDI)
uses two rings instead of one. Data on the two rings flows in opposite
directions, which is why FDDI networks are called counter-rotating.
Automatic reconfiguration. Normally the primary ring carries all the
data, and the secondary ring sits in reserve. If a fault occurs on the
primary ring, FDDI automatically reconfigures to route traffic over the
secondary ring instead. This is called self-healing, and it is why FDDI
is described as a self-healing network.
How it works. When a break occurs on the primary ring, the nodes
adjacent to that break loop the signal back onto the secondary ring, so
traffic still completes a full path around the network and connectivity is
maintained.
T3-Q10
Tutorial 3, Q10
analyseconceptual
Compare and contrast between a hub (repeater) and a bridge.
hub
bridge
repeater
Show walkthrough
Hub (repeater). Operates at the physical layer. It receives a
signal and, before it becomes too weak or corrupted, regenerates the
original bit pattern — extending the physical length of the network. A
repeater does not understand a complete frame, and cannot filter packets. A
hub is a multiport repeater: it broadcasts whatever one station sends to
every other station attached to it, which is why it is the usual way to
connect several machines to the same LAN.
Bridge. Operates at the data link layer. It can read the destination
address of a frame and decide whether to forward or discard it. It
understands the complete frame, which is what makes packet filtering
possible, and it is mostly used inside LANs.
Aspect
Repeater/hub
Bridge
Layer of operation
Physical layer
Data link layer
Frame understanding
No
Yes
Filtering capability
No
Yes
Addressing
Not aware
MAC-address aware
Function
Signal regeneration
Intelligent forwarding
The difference underneath all of these rows is the same one: a hub only
knows about voltage levels on a wire, while a bridge reads the frame that
those voltage levels encode.
Compare and contrast between a bridge and a switch (Layer 2).
bridge
switch
Show walkthrough
Bridge
Switch (Layer 2)
Speed
Relatively slow
Faster, and built to handle packets with more sophistication
Ports
A limited number
Can have many
Switching method
Store and forward
Store and forward, cut-through, or fragment-free
Collision
Collision domain becomes smaller
Probability of collision reduced tremendously
Similarities. A switch is, in general, a multi-port bridge. Both operate
at the physical and data link layers. Both make filtering decisions based on
the physical (MAC) address of the frame they receive, and both keep a
forwarding table of MAC address to port.
Key difference. Switches operate at higher speeds and with more
sophisticated switching methods — able to use cut-through or fragment-free
switching as well as store-and-forward — which is what makes them more
efficient in modern networks than a bridge covering the same job.
See Bridges, Switches and Spanning Tree
for cut-through vs store-and-forward in more detail, and how a switch’s
forwarding table is built through backward learning.
T3-Q12
Tutorial 3, Q12
applyconceptual
What do we mean when we say that a bridge can filter traffic? Why is
filtering important?
bridge
switch
Show walkthrough
What filtering means. A bridge has access to station physical
addresses, so instead of repeating every frame everywhere, it can forward a
frame only to the segment where the destination actually sits. That
selective forwarding is what “filtering” refers to.
Why filtering matters:
Controlling congestion. Filtering keeps a frame from reaching network
segments where it is not needed, so segments are not carrying traffic
they have nothing to do with.
Improving security. Restricting which ports receive a given piece of
traffic also restricts who can see it.
How filtering works. When a bridge receives a frame, it reads the
destination MAC address and checks its forwarding table:
Source and destination on the same segment — the frame is not
forwarded (blocked). It is already where it needs to be.
Destination on a different segment — the frame is forwarded to the
correct port.
Destination unknown — the frame is flooded to every port except the
one it arrived on.
The problem. A signal can only travel a fixed distance before
attenuation threatens the integrity of the data it carries.
How a repeater works. A repeater receives that signal and, before it
becomes too weak or corrupted, regenerates the original bit pattern from
scratch, then sends the refreshed signal onward.
The process, step by step:
The signal arrives at the repeater, weakened from the distance already
travelled.
The repeater detects the signal at the physical layer.
The repeater regenerates the original signal at full strength.
The repeater transmits the regenerated signal to the next segment.
Result. Because each segment starts fresh at the repeater rather than
continuing to degrade, a LAN can span a distance greater than any single
cable segment could support on its own — the network is divided into
repeater-extended segments instead.
VLAN operation. A VLAN breaks up a large network into smaller,
independent broadcast domains, called workgroups.
Traffic reduction mechanism. A broadcast message from one workgroup
member only needs to reach the other members of that workgroup. That
eliminates the need to broadcast the message, and all of the overhead
traffic that comes with it, to every other station on the network.
Result. Confining broadcasts this way reduces overall traffic on the
network.
Example. Instead of a broadcast reaching all 100 connected devices, a
VLAN can limit it to the 10 devices actually in that workgroup — a 90%
reduction in that message’s network load. This 90% figure describes that one
example, not a fixed rule for every VLAN; the actual reduction depends on how
large the workgroup is relative to the whole network.
See VLANs and the 802.1Q Tag for how stations are
actually grouped into a VLAN, and how a trunk link carries several VLANs over
one physical connection.
T3-Q15
Tutorial 3, Q15
applynumeric
For a CSMA/CD network with 1 Gbps bandwidth, if a frame size of 512 bits
is used, what is the maximum possible distance between nodes? Assume zero
processing delay and EM propagation delay is 2×108 m/s.
Apply the CSMA/CD worst-case rule. A sender must still be transmitting
when word of a collision at the far end of the network gets back to it, so
the frame time must be at least twice the one-way propagation time:
Tfr = 2 × Tp. That gives Tp = Tfr / 2 = 256 × 10−9 s.
Convert Tp to a distance.Distance = propagation speed × Tp = (2 × 108m/s) × (256 ×10−9 s).
The 0.5 factor in a distance-from-frame-size calculation like this one is
always this same worst-case rule in disguise: Tfr ≥ 2 × Tp, so Tp is
half the frame’s transmission time. See
CSMA/CD and the Ethernet Frame for that
rule derived the other way around — from a given Tp to the minimum frame
size it forces.
T3-Q16
Tutorial 3, Q16
applyconceptual
How collision can be detected in CSMA/CA networks?
csma-ca
csma-cd
Show walkthrough
CSMA/CA has no way to sense a collision directly the way a wired network
does. A collision is detected using an ACK (acknowledgement) instead —
its absence is what a collision looks like.
Sender procedure:
If the channel is sensed idle for a DIFS, transmit the entire frame.
There is no collision detection while transmitting.
If the channel is sensed busy: start a random backoff timer, which counts
down only while the channel is idle, and transmit once it expires.
If no ACK is received, a collision is assumed to have happened.
Increase the random backoff interval and repeat the procedure.
Receiver procedure. If the frame is received correctly, return an ACK
after a SIFS. The ACK is required specifically because of the hidden
terminal problem — a sender may have no way of hearing a collision itself,
so a missing ACK is the only signal it has.
Key terms:
DIFS — Distributed Inter Frame Space, the idle time checked before a
new transmission can start.
SIFS — Short Inter Frame Space, the gap between receiving a frame and
sending its ACK.
Collision is inferred from the absence of an ACK, not detected while
transmission is in progress — the opposite of CSMA/CD, where the sender
detects the collision itself, on the wire, as it happens.
Consider 3 network configurations below. Assuming the cables/lines provide a
capacity of 100 Mbps. What will be the throughput for each configuration?
ethernet
switched-ethernet
Show walkthrough
The source solution is incomplete for this question. What follows is only what the tutorial actually gives.
The tutorial solution for this question is cut off before it gives the three
network configurations themselves, and before it finishes the calculation for
any of them. What survives in the source material is three partial numbers,
with no working attached and no diagram of what each configuration actually
looks like:
Configuration 1: 100 Mbps / 12, with the rest of the calculation missing.
Configuration 2: 25 Mbps.
Configuration 3: 100 Mbps.
That is the entire extent of what the source gives. The network layouts these
three numbers are supposed to describe do not survive in the material at all,
so there is nothing here to reconstruct them from, and no safe way to fill in
the missing working without guessing at configurations that were never seen.
Check the tutorial recording, or ask in class, for the full configurations
and the complete calculation.