Problem bank

Every tutorial problem in the unit, with a full walkthrough behind each one.

97 of 97 shown

• 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.

  See encapsulation and the four addresses for the full picture, including what each layer’s package is called.

• T1-Q05

  Tutorial 1, Q5

  recallconceptual

  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.

  See encapsulation and the four addresses for how the logical and port addresses differ in scope.

• T1-Q07

  Tutorial 1, Q7

  analyseconceptual

  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.

  See encapsulation and the four addresses for how the logical address itself works.

• T1-Q09

  Tutorial 1, Q9

  recallconceptual

  What are the responsibilities of the Transport layer?

  • layering
  • port
  • flow-control
  • error-control

  Show walkthrough

  The transport layer has five jobs:

  1. Moving messages — delivers a message from one process to another.
  2. 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.
  3. 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.
  4. Flow control — keeps the sending process from emitting segments faster than the receiving process can handle.
  5. Error control — process-to-process error control, rather than the single-link error control the data link layer provides.

  See encapsulation and the four addresses for the port address in more detail.

• T1-Q10

  Tutorial 1, Q10

  applyconceptual

  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

  1. 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.
  2. 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.
  3. Network layer adds the logical addresses of the two hosts to the segment and hands down a datagram: A B a j Data.
  4. 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:

  $r = \frac{\text{data elements}}{\text{signal elements}}$

  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.

  Full treatment: Signals.

• T2-Q02

  Tutorial 2, Q2

  recallconceptual

  Distinguish between data rate and signal rate.

  • signals
  • bandwidth

  Show walkthrough

  	Data rate	Signal rate
  Definition	Number of data elements (bits) sent in 1 second	Number of signal elements sent in 1 second
  Unit	`bps` (bits per second)	`baud`
  Other names	Bit rate	Pulse rate, modulation rate, baud rate
  Effect of increasing it	Increases the speed of transmission	Increases the bandwidth the link demands

  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.

  Full treatment: Signals.

• T2-Q03

  Tutorial 2, Q3

  recallconceptual

  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.

  Full treatment: Line Coding.

• T2-Q04

  Tutorial 2, Q4

  recallconceptual

  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.

  Full treatment: Signals.

• T2-Q05

  Tutorial 2, Q5

  recallconceptual

  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.

  Full treatment: Signals.

• T2-Q06

  Tutorial 2, Q6

  recallconceptual

  Discuss the steps of PCM.

  • pcm
  • sampling

  Show walkthrough

  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.

  Full treatment: Modulation and PCM.

• T2-Q07

  Tutorial 2, Q7

  recallconceptual

  Which characteristics of an analog signal are changed to represent the digital signal in each of the following digital-to-analog conversions?

  a) ASK b) FSK c) PSK d) QAM

  • modulation
  • ask
  • fsk
  • psk
  • qam

  Show walkthrough

  Each scheme varies a different property of the carrier signal, holding the rest fixed.

  a) Amplitude Shift Keying (ASK). Changes the carrier’s amplitude. Frequency and phase stay constant.

  b) Frequency Shift Keying (FSK). Changes the carrier’s frequency. Amplitude and phase stay constant.

  c) Phase Shift Keying (PSK). Changes the carrier’s phase. Amplitude and frequency stay constant.

  d) Quadrature Amplitude Modulation (QAM). Combines ASK and PSK — it changes two carrier properties at once, amplitude and phase, instead of just one.

  Full treatment: Modulation and PCM.

• T2-Q08

  Tutorial 2, Q8

  analyseconceptual

  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.

  Full treatment: Modulation and PCM.

• T2-Q09

  Tutorial 2, Q9

  applynumeric

  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

  1. 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.

  2. a) ASK with four amplitudes: L = 4.

     $r = \log_2 4 = 2 \text{ bits per baud}$

  3. b) FSK with 8 frequencies: L = 8.

     $r = \log_2 8 = 3 \text{ bits per baud}$

  4. c) PSK with four phases: L = 4.

     $r = \log_2 4 = 2 \text{ bits per baud}$

  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.

  Full treatment: Modulation and PCM.

• T2-Q10

  Tutorial 2, Q10

  applynumeric

  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

  1. 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.

  2. Divide the 12 kHz link into three consecutive 4 kHz slices, starting from the bottom of the band:

     $20 \text{ kHz} + 4 \text{ kHz} = 24 \text{ kHz}, \quad 24 \text{ kHz} + 4 \text{ kHz} = 28 \text{ kHz}, \quad 28 \text{ kHz} + 4 \text{ kHz} = 32 \text{ kHz}$

  Answer20-24 kHz, 24-28 kHz, 28-32 kHz

  Channel	Frequency range
  Channel 1	20-24 kHz
  Channel 2	24-28 kHz
  Channel 3	28-32 kHz

  Full treatment: Multiplexing.

• T2-Q11

  Tutorial 2, Q11

  applynumeric

  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

  1. Five channels at 100 kHz each need:

     $5 \times 100 \text{ kHz} = 500 \text{ kHz}$

  2. Guard bands sit between adjacent channels, not around the outside, so n channels need n - 1 guard bands. Five channels need 4 guard bands:

     $4 \times 10 \text{ kHz} = 40 \text{ kHz}$

  3. Add the channel bandwidth and the guard-band bandwidth:

     $(100 \times 5) + (4 \times 10) = 500 + 40 = 540 \text{ kHz}$

  Answer540 kHz

  Full treatment: Multiplexing.

• T2-Q12

  Tutorial 2, Q12

  applynumeric

  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

  1. a) Input slot duration. Each input connection runs at 3 kbps, so one bit takes:

     $\frac{1}{3000} \text{ s}$

     The input slot duration equals this bit duration, 1/3000 s.

  2. 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:

     $\frac{1}{3 \times 3000} = \frac{1}{9000} \text{ s}$

  3. c) Frame duration. One frame carries one output slot from each of the three inputs:

     $3 \times \frac{1}{9000} \text{ s} = \frac{1}{3000} \text{ s}$

     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.

  AnswerInput slot: 1/3000 s, output slot: 1/9000 s, frame: 1/3000 s

  Full treatment: Multiplexing.

• T2-Q13

  Tutorial 2, Q13

  applynumeric

  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

  1. a) Input bit duration is the reciprocal of the input bit rate:

     $\frac{1}{1 \text{ Mbps}} = 1 \text{ } \mu\text{s}$

  2. 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:

     $\frac{1 \text{ } \mu\text{s}}{4} = 0.25 \text{ } \mu\text{s}$

  3. c) Output bit rate is the reciprocal of the output bit duration:

     $\frac{1}{0.25 \text{ } \mu\text{s}} = 4 \text{ Mbps}$

     This also equals the sum of the four input bit rates: 1 Mbps × 4 = 4 Mbps.

  4. d) Output frame rate. In synchronous TDM, the frame rate equals the bit rate of any single input channel:

     $\frac{4 \text{ Mbps}}{4} = 1 \text{ Mbps}$

  AnswerInput: 1 µs, output: 0.25 µs, output bit rate: 4 Mbps, frame rate: 1 Mbps

  Full treatment: Multiplexing.

• T2-Q14

  Tutorial 2, Q14

  recallconceptual

  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.

  Full treatment: Multiplexing.

• T2-Q15

  Tutorial 2, Q15

  recallconceptual

  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.

  Full treatment: Multiplexing.

• T2-Q16

  Tutorial 2, Q16

  recallconceptual

  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:

  1. Synchronous Time Division Multiplexing (Sync TDM)
  2. 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.

  Full treatment: Multiplexing.

• T2-Q17

  Tutorial 2, Q17

  analyseconceptual

  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.

  Full treatment: Multiplexing.

• T2-Q18

  Tutorial 2, Q18

  analyseconceptual

  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.

  Full treatment: Multiplexing.

• T2-Q19

  Tutorial 2, Q19

  applynumeric

  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

  1. Ten channels at 4 kHz each:

     $4 \text{ kHz} \times 10 = 40 \text{ kHz}$

  2. Guard bands sit between adjacent channels, so 10 channels need 10 - 1 = 9 guard bands, each 500 Hz = 0.5 kHz:

     $9 \times 0.5 \text{ kHz} = 4.5 \text{ kHz}$

  3. Add the channel bandwidth and the guard-band bandwidth:

     $(4 \times 10) + (9 \times 0.5) = 40 + 4.5 = 44.5 \text{ kHz}$

  Answer44.5 kHz

  Full treatment: Multiplexing.

• T2-Q20

  Tutorial 2, Q20

  applynumeric

  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

  1. Bandwidth available per channel, splitting the 20 kHz passband across 100 channels with no guard bands:

     $\frac{20 \text{ kHz}}{100} = 200 \text{ Hz per channel}$

  2. 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:

     $8 \text{ bit/sample} \times 8 \text{ kHz} = 64 \text{ kbps}$

  3. Divide the bit rate per channel by the bandwidth per channel:

     $\frac{64 \text{ kbps}}{200 \text{ Hz}} = 320 \text{ bits/s/Hz}$

  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.

  Full treatment: Modulation and PCM.

• T2-Q21

  Tutorial 2, Q21

  recallconceptual

  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.

  Full treatment: Media and Topologies.

• T2-Q22

  Tutorial 2, Q22

  recallconceptual

  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.

  Full treatment: Media and Topologies.

• T2-Q23

  Tutorial 2, Q23

  recallconceptual

  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.

  Full treatment: Media and Topologies.

• T2-Q24

  Tutorial 2, Q24

  analyseconceptual

  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.

  Full treatment: Media and Topologies.

• T2-Q25

  Tutorial 2, Q25

  recallconceptual

  List three traditional switching methods.

  • switching
  • circuit-switching
  • packet-switching

  Show walkthrough

  A switch is a device that creates a temporary connection between two or more devices linked to it. Traditionally, three switching methods matter:

  1. Circuit switching
  2. Packet switching, which itself splits into:
     • Virtual-circuit networks
     • Datagram networks
  3. 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.

  Full treatment: Switching.

• T2-Q26

  Tutorial 2, Q26

  recallconceptual

  What are the two approaches for packet switching?

  • switching
  • packet-switching
  • virtual-circuit

  Show walkthrough

  The two approaches are:

  1. Virtual-circuit networks
  2. Datagram networks

  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.

  Full treatment: Switching.

• T2-Q27

  Tutorial 2, Q27

  analyseconceptual

  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.

  Full treatment: Switching.

• T2-Q28

  Tutorial 2, Q28

  applyconceptual

  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.

  Full treatment: Switching.

• T2-Q29

  Tutorial 2, Q29

  applyconceptual

  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.

  Full treatment: Switching.

• T2-Q30

  Tutorial 2, Q30

  recallconceptual

  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.

  Full treatment: Switching.

• T3-Q01

  Tutorial 3, Q1

  recallconceptual

  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:

  1. 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.
  2. 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

  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

  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.

  See Ethernet Standards, Full Duplex and Switching for how these fit alongside the faster 802.3 standards.

• T3-Q07

  Tutorial 3, Q7

  applyconceptual

  What is switched Ethernet?

  • switched-ethernet
  • switch
  • ethernet

  Show walkthrough

  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.

  MAC Address	Port
  A	1
  B	2
  C	3
  D	4
  E	5
  F	6
  G	7
  H	8

  See Ethernet Standards, Full Duplex and Switching for why splitting a network this way raises effective bandwidth over a shared bus.

• T3-Q08

  Tutorial 3, Q8

  recallconceptual

  What are the advantages of token passing?

  • token-ring
  • fddi

  Show walkthrough

  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.

  See Bridges, Switches and Spanning Tree for the same distinction applied to repeaters, active hubs and switches.

• T3-Q11

  Tutorial 3, Q11

  analyseconceptual

  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.

  See Bridges, Switches and Spanning Tree for how a bridge builds this forwarding table automatically through backward learning.

• T3-Q13

  Tutorial 3, Q13

  applyconceptual

  How does a repeater extend the length of a LAN?

  • repeater
  • hub

  Show walkthrough

  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:

  1. The signal arrives at the repeater, weakened from the distance already travelled.
  2. The repeater detects the signal at the physical layer.
  3. The repeater regenerates the original signal at full strength.
  4. 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.

  See Bridges, Switches and Spanning Tree for how this same regeneration idea distinguishes a repeater from a bridge.

• T3-Q14

  Tutorial 3, Q14

  applyconceptual

  How does a VLAN reduce network traffic?

  • vlan
  • switch

  Show walkthrough

  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 $2 \times 10^8$ m/s.

  • csma-cd
  • ethernet

  Show walkthrough

  Given:

  • Bandwidth = 1 Gbps = $10^9$ bps
  • Frame size = 512 bits
  • Propagation speed = $2 \times 10^8$ m/s
  • Processing delay = 0

  Worked example

  Maximum distance between nodes

  1. Frame transmission time. Tfr = frame size / bandwidth = 512 bits / $10^9$bps = 512 × $10^{-9}$ s.

  2. 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.

  3. Convert Tp to a distance. Distance = propagation speed × Tp = (2 × $10^8$m/s) × (256 × $10^{-9}$ s).

  4. Multiply out. (2 × $10^8$) × (256 × $10^{-9}$) = 512 × $10^{-1}$ = 51.2 m.

  Answer51.2 m

  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:

  1. If the channel is sensed idle for a DIFS, transmit the entire frame. There is no collision detection while transmitting.
  2. 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.
  3. If no ACK is received, a collision is assumed to have happened.
  4. 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.

  See CSMA/CA, RTS/CTS and the Hidden Node Problem for the RTS/CTS handshake that avoids many collisions before they can even happen.

• T3-Q17

  Tutorial 3, Q17

  applynumeric

  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.

• T4-Q01

  Tutorial 4, Q1

  recallconceptual

  Define the random access mechanism and give examples.

  • multiple-access
  • random-access
  • aloha
  • csma

  Show walkthrough

  Random access. The method which allows the stations to access the transmission medium (channel) randomly at any time is known as the random access method.

  Random-access mechanisms

  • No superiority. No station is superior to another station, and none is assigned control over another.
  • Random access. There is no scheduled time for a station to transmit. Transmission is random among the stations.
  • Contention methods. No rules specify which station should send next. Stations compete with one another to access the medium.
  • No controlling. Each station has the right to the medium without being controlled by any other station.
  • Collision. If more than one station tries to send, there is an access conflict — a collision.

  Examples of random-access methods

  • ALOHA
  • Carrier Sense Multiple Access (CSMA)
  • Carrier Sense Multiple Access with Collision Detection (CSMA/CD)
  • Carrier Sense Multiple Access with Collision Avoidance (CSMA/CA)

• T4-Q02

  Tutorial 4, Q2

  recallconceptual

  Define controlled access mechanism and give examples.

  • multiple-access
  • controlled-access
  • polling
  • reservation
  • token-passing

  Show walkthrough

  Controlled access. In controlled access, the stations consult one another to find which station has the right to send.

  Controlled-access mechanisms

  • Consulting. The stations consult each other to find which station has the right to send.
  • Permission. Controlled-access protocols grant permission to send to only one node at a time, to avoid collision of messages on the shared medium.
  • Authorization. A station cannot send data unless it is authorized by the other stations.

  Examples of controlled-access methods

  There are three controlled-access methods:

  1. Reservation
  2. Polling
  3. Token passing

• T4-Q03

  Tutorial 4, Q3

  analyseconceptual

  Compare and contrast a random access protocol with a controlled access protocol. Define why collision is an issue in a random access protocol but not in a controlled access protocol.

  • multiple-access
  • random-access
  • controlled-access
  • collision

  Show walkthrough

  Aspect	Random access protocol	Controlled access protocol
  Control	There is no control; access is based on contention.	Either a central authority (in polling) or other stations (in reservation and token passing) control the access.
  Superiority	No station is superior to another station.	The authorized station has the superiority.
  Collision	Collision is common.	Collision-free.
  Examples	ALOHA, CSMA, CSMA/CD, and CSMA/CA.	Reservation, Polling and Token Passing.

  Why collision occurs in random access but not in controlled access:

  • In a random-access protocol, a device assumes it can transmit if it detects no signal on the medium. If multiple independent devices decide to transmit “at the same time” their messages will interfere with each other, or collide.
  • In a controlled-access protocol, a device waits until it has explicit permission to transmit. Without authorization, no station is allowed to transmit data.
  • Since multiple devices cannot independently decide to transmit in a controlled-access network, there can never be a collision.

• T4-Q04

  Tutorial 4, Q4

  applynumeric

  In a CSMA/CD network with a data rate of 10 Mbps, the minimum frame size is found to be 512 bits for the correct operation of the collision detection process. What should be the minimum frame size if we increase the data rate to 100 Mbps?

  • csma-cd
  • frame-size
  • data-rate
  • propagation-delay

  Show walkthrough

  Given:

  • Data rate, $R_1 = 10$ Mbps $= 10 \times 10^6$ bps
  • Minimum frame size, $L_1 = 512$ bits

  Worked example

  Minimum frame size at 100 Mbps

  1. Frame transmission time. Tframe = L / R = 512 / (10 × $10^6$) = 51.2 × $10^{-6}$ s = 51.2 μs.

  2. Apply the CSMA/CD rule. In a CSMA/CD network, the frame transmission time must be at least two times the maximum propagation time to detect the collision: Tframe = 2 × Tprop.

  3. Solve for the maximum propagation time. Tprop = Tframe / 2 = (L1 / R1) / 2 = (512 / (10 × $10^6$)) / 2 = 25.6 × $10^{-6}$ s = 25.6 μs.

  4. Apply the new data rate, $R_2 = 100$ Mbps $= 100 \times 10^6$ bps. The minimum frame size to detect the collision is L2 = 2 × R2 × Tprop = 2 × 100 × $10^6$× 25.6 × $10^{-6}$ = 5120 bits.

  Answer5120 bits

  Answer: minimum frame size = 5120 bits.

• T4-Q05

  Tutorial 4, Q5

  applynumeric

  In a CSMA/CD network with a data rate of 10 Mbps, the maximum distance between any station pair is found to be 2500 m for the correct operation of the collision detection process. What should be the maximum distance if we increase the data rate to 100 Mbps?

  • csma-cd
  • maximum-distance
  • data-rate
  • propagation-delay

  Show walkthrough

  Given:

  • Data rate, $R_1 = 10$ Mbps $= 10 \times 10^6$ bps
  • Maximum distance, $d_1 = 2500$ m

  Analysis. d = v × Tprop (distance = propagation speed × propagation time). From Q4, Tprop = (L / R) / 2, so d = (v / 2) × (L / R).

  Key insight: if the data rate increases by a factor of 10 while the minimum frame size stays constant, the maximum distance decreases by a factor of 10. For a given network with constant minimum frame size and propagation speed, maximum distance is inversely proportional to the data rate.

  Worked example

  Maximum distance at 100 Mbps

  1. At 10 Mbps, the maximum distance for correct operation of the collision detection process is 2500 m.

  2. Increasing the data rate to 100 Mbps is a 10 times increase.

  3. The maximum distance decreases by 10 times: d2 = d1 / 10 = 2500 / 10 = 250 m.

  Answer250 m

  Answer: maximum distance = 250 m.

• T4-Q06

  Tutorial 4, Q6

  applynumeric

  The topology figure this question refers to did not extract from the source material — it is in the tutorial PDF. What survives is the text of the question itself:

  In the figure, the data rate is 10 Mbps, the distance between station A and C is 2000 m, and the propagation speed is $2 \times 10^8$ m/s. Station A starts sending a long frame at time $t_1 = 0$; station C starts sending a long frame at time $t_2 = 3\ \mu s$. The size of the frame is long enough to guarantee the detection of collision by both stations. Find:

  a) The time when station C hears the collision ($t_3$) b) The time when station A hears the collision ($t_4$) c) The number of bits station A has sent before detecting the collision d) The number of bits station C has sent before detecting the collision

  • csma-cd
  • collision-detection
  • propagation-delay

  Show walkthrough

  Given:

  • Data rate, $R = 10$ Mbps $= 10 \times 10^6$ bps
  • Distance between station A and C, $d = 2000$ m
  • Propagation speed, $v = 2 \times 10^8$ m/s
  • A starts sending a long frame at $t_1 = 0$
  • C starts sending a long frame at $t_2 = 3\ \mu s$
  • Propagation time, Tprop = d / v = 2000 / (2 × $10^8$) = 10 μs

  a) The time when station C hears the collision ($t_3$):

  When the first bit of A reaches C, station C hears the collision. Time taken for the first bit of A to reach C: Tprop = d / v = 2000 / (2 × $10^8$) = 10 μs.

  The time when station C hears the collision: t3 = t1 + Tprop = 0 + 10 = 10 μs.

  b) The time when station A hears the collision ($t_4$):

  When the first bit of C reaches A, station A hears the collision. Time taken for the first bit of C to reach A: Tprop = d / v = 2000 / (2 × $10^8$) = 10 μs.

  The time when station A hears the collision: t4 = t2 + Tprop = 3 + 10 = 13 μs.

  c) The number of bits station A has sent before detecting the collision:

  Number of bits transmitted = R × Tframe.

  (t4 - t1) × R = (13 - 0) × $10^{-6}$× 10 × $10^6$ = 130 bits.

  The number of bits station A has sent before detecting the collision = 130 bits.

  d) The number of bits station C has sent before detecting the collision:

  (t3 - t2) × R = (10 - 3) × $10^{-6}$× 10 × $10^6$ = 70 bits.

  The number of bits station C has sent before detecting the collision = 70 bits.

• T4-Q07

  Tutorial 4, Q7

  recallconceptual

  How is the preamble field of the IEEE 802.3 MAC frame different from the SFD field?

  • ethernet
  • frame-format
  • preamble
  • sfd

  Show walkthrough

  Aspect	Preamble	Start Frame Delimiter (SFD)
  Length	7 bytes	1 byte
  Purpose	Used for synchronization	Used to determine the beginning of the frame
  Pattern	Contains 7 bytes of alternating 0s and 1s (`10101010...`)	1 byte (`10101011`)
  Function	Acts as a synchronization pulse	Last two bits set to `11` to indicate the next field is the destination address
  Timing	—	Last chance for the receiver to synchronize its input timing

  Detailed explanation. In the frame, the first 7 bytes — that is, 56 bits of alternating 0s and 1s — represent the Preamble field, mainly used for synchronization.

  After counting 56 bits, the receiver reads the next 1 byte (8 bits, 10101011), which represents the SFD field. This signals the beginning of the frame.

  The SFD warns the station or stations that this is the last chance for synchronization (101010). The last 2 bits are 11 and alert the receiver that the next field is the destination address.

  After receiving 11 from the SFD field, the receiver or intermediate node collects the necessary information from the frame (DA, SA, Type, Data, CRC).

• T4-Q08

  Tutorial 4, Q8

  recallconceptual

  What are the advantages of dividing an Ethernet LAN with a bridge?

  • bridge
  • collision-domain
  • ethernet

  Show walkthrough

  Bridges have two main effects on an Ethernet LAN:

  1. They raise the bandwidth.
  2. They separate collision domains.

  Raising the bandwidth. With no bridge, a LAN of $n$ devices with a 10 Mbps data rate shares a total bandwidth of 10 Mbps among all stations.

  A bridge divides the network into two or more networks. Bandwidth-wise, each network is independent. Example: a bridge dividing a network into 2 segments, each with 3 stations, gives 10 Mbps per segment — 20 Mbps total, which is 2 times more than an unbridged network.

  Separating collision domains. With bridging, the collision domain becomes much smaller, and the probability of collision is reduced tremendously.

  Without bridging, all 12 stations (for example) contend for access to the medium. With bridging, only 3 stations plus 1 bridge contend for access to the medium in each segment.

• T4-Q09

  Tutorial 4, Q9

  recallconceptual

  What is the relationship between a switch and a bridge?

  • bridge
  • switch
  • switched-ethernet

  Show walkthrough

  • A two-layer switch is a bridge with many ports and a design that allows better (faster) performance.
  • The main reasons for using a bridge and a switch (Layer 2) are to increase bandwidth and reduce the collision domain.
  • Both operate at the physical and data link layers.
  • Both make filtering decisions based on the physical address of the frame they receive.
  • Both a bridge and a switch have a forwarding table of MAC addresses and port numbers.
  • A two-layer switch uses store-and-forward, cut-through, or fragment-free methods to forward a frame, while a bridge uses only the store-and-forward method.

• T4-Q10

  Tutorial 4, Q10

  analyseconceptual

  Why is there no need for CSMA/CD on a full-duplex LAN?

  • full-duplex
  • csma-cd
  • switched-ethernet

  Show walkthrough

  Background. One of the limitations of 10Base5 and 10Base2 is that communication is half-duplex. There is a need for the CSMA/CD method to avoid collision.

  Full-duplex LAN with a switch. In a full-duplex switched Ethernet, each station is connected to the switch via two separate links (Tx and Rx). Each link is a point-to-point dedicated path between the station and the switch.

  Why CSMA/CD is not needed. Each station or switch can send and receive independently without worrying about collision. There is no longer a need for carrier sensing, because there is no longer a need for collision detection. Since each link is dedicated and point-to-point, only two devices can communicate over each link at any time — one sending, one receiving — making collisions impossible.

• T4-Q11

  Tutorial 4, Q11

  applynumeric

  An Ethernet MAC layer receives 42 bytes of data from the upper layer. How many bytes of padding must be added to the data?

  • ethernet
  • frame-format
  • padding

  Show walkthrough

  Ethernet frame structure:

  • Minimum frame length in Ethernet is 64 bytes
  • Maximum frame length is 1518 bytes
  • 6 bytes for the destination address
  • 6 bytes for the source address
  • 2 bytes for length/Ethernet type
  • 46–1500 bytes for data
  • 4 bytes of CRC

  Worked example

  Padding required

  1. The data received is 42 bytes.

  2. The minimum data field size is 46 bytes.

  3. Append (46 - 42) = 4 bytes of padding data (zeros) to the data.

  Answer4 bytes

  Purpose. Padding ensures that the entire Ethernet frame is a detectable and processable length for the network hardware.

• T4-Q12

  Tutorial 4, Q12

  applynumeric

  An Ethernet MAC layer receives 1510 bytes of data from the upper layer. Can the data be encapsulated in one frame? If not, how many frames need to be sent? What is the size of the data in each frame?

  • ethernet
  • frame-format
  • frame-splitting

  Show walkthrough

  Maximum data size. The maximum data size in Standard Ethernet is 1500 bytes. The data of 1510 bytes, therefore, must be split between two frames.

  Worked example

  Frame distribution

  1. Data size for the first frame: 1500 bytes.

  2. Data size for the second frame: 10 bytes.

  Answer1500 bytes in the first frame, 10 bytes in the second

  Note on padding. The standard by default dictates that the first frame carries the maximum possible number of bytes (1500); the second frame then needs to carry only 10 bytes of data. The second frame may require padding (46 - 10 = 36 bytes of padding).

• T4-Q13

  Tutorial 4, Q13

  recallconceptual

  What is the difference between a BSS and an ESS?

  • wireless
  • bss
  • ess
  • access-point

  Show walkthrough

  Basic Service Set (BSS). The basic service set (BSS) is the building block of a wireless LAN.

  • A BSS without an AP is called an ad hoc architecture.
  • A BSS with an AP is sometimes referred to as an infrastructure network.

  Extended Service Set (ESS). An extended service set (ESS) is made up of two or more BSSs with APs. In this case, the BSSs are connected through a distributed system, which is usually a wired LAN.

• T4-Q14

  Tutorial 4, Q14

  recallconceptual

  Discuss the three types of mobility in a WLAN.

  • wireless
  • mobility
  • bss
  • ess

  Show walkthrough

  There are three types of stations based on their mobility in a wireless LAN:

  1. No-transition mobility
  2. BSS-transition mobility
  3. ESS-transition mobility

  No-transition mobility. A station with no-transition mobility is either stationary (not moving) or moving only inside a BSS.

  BSS-transition mobility. A station with BSS-transition mobility can move from one BSS to another, but the movement is confined inside one ESS.

  ESS-transition mobility. A station with ESS-transition mobility can move from one ESS to another.

• T4-Q15

  Tutorial 4, Q15

  recallconceptual

  What is the access method used by WLANs?

  • wireless
  • csma-ca
  • dcf
  • pcf
  • multiple-access

  Show walkthrough

  IEEE MAC sublayer protocols. Two protocols are defined by IEEE at the MAC sublayer:

  1. Distributed Coordination Function (DCF)
  2. Point Coordination Function (PCF)

  DCF — Distributed Coordination Function.

  • DCF uses CSMA/CA as the access method.
  • It is contention-based.

  Why wireless LANs cannot implement CSMA/CD:

  • In a wireless network, collisions do not add enough energy for stations to effectively detect them.
  • High costs and increased bandwidth requirements.
  • Collision may not be detected because of the hidden station problem.
  • Collision may be falsely detected because of the exposed node problem.
  • Signal fading could prevent a station at one end from hearing a collision at the other end.

  CSMA/CA operation in DCF:

  1. After finding an idle channel, the source waits for the Distributed Inter-Frame Space (DIFS).
  2. The source then sends a Request to Send (RTS) control frame to the destination.
  3. The destination, after receiving RTS, waits for the Short Inter-Frame Space (SIFS) and sends a Clear to Send (CTS) control frame to the source.
  4. The source waits for SIFS before sending the data.
  5. The destination waits another SIFS before sending an Acknowledgement (ACK).
  6. Collision Avoidance (CA) uses p-persistent methods; p randomly selects a Contention Window (CW), and decrements it.
  7. CSMA/CA waits for ACK before sending additional frames.

  Timings:

  • DIFS = 50 μs
  • SIFS = 20 μs
  • Time slot = 10 μs

  PCF — Point Coordination Function.

  • PCF is an optional access method that can be implemented in an infrastructure network (not in an ad hoc network).
  • PCF has a centralized, contention-free polling access method.
  • The AP performs polling for stations that are capable of being polled.
  • The stations are polled one after another, sending any data they have to the AP.

  Priority structure. To give priority to PCF over DCF, another set of inter-frame spaces has been defined: PIFS (Point Interframe Space), where SIFS < PIFS < DIFS.

  Managing PCF priority. Due to the priority of PCF over DCF, stations that only use DCF may not gain access to the medium. To prevent this, a beacon frame that is periodically transmitted is used to cover both contention-free (PCF) and contention-based (DCF) traffic. When the stations hear the beacon, they start their Network Allocation Vector (NAV) for the duration of the contention-free period.

• T4-Q16

  Tutorial 4, Q16

  recallconceptual

  What is the purpose of the NAV?

  • wireless
  • nav
  • csma-ca

  Show walkthrough

  Network Allocation Vector (NAV).

  How NAV works. The stations listening on the wireless medium read the Duration field and set their NAV, which is an indicator for a station of how long it must defer from accessing the medium — like a timer that counts down to zero.

  Functions of NAV:

  1. To prevent stations from accessing the medium and causing contention.
  2. To provide a kind of contention-free access in the contention-based access period (DCF).

• T4-Q17

  Tutorial 4, Q17

  analyseconceptual

  Compare and contrast CSMA/CD with CSMA/CA.

  • csma-cd
  • csma-ca
  • wireless
  • ethernet

  Show walkthrough

  Aspect	CSMA/CD	CSMA/CA
  Used in	Wired networks (e.g. Ethernet)	Wireless networks (e.g. Wi-Fi)
  Function	Detects collisions after they occur	Avoids collisions before they occur
  Collision handling	Stops transmission and retries after a random backoff time	Uses backoff timers and waits for a clear channel before transmitting
  Efficiency	Good in low traffic, but performance degrades with high traffic	Essential for wireless communication, but overhead can reduce throughput

  Mechanism difference. In CSMA/CD, the protocol allows collisions to happen. If there is a collision, it is detected, transmission is terminated and the frame is resent. CSMA/CA uses an RTS/CTS mechanism to prevent collision.

  Why CSMA/CD cannot be used in wireless. CSMA/CD cannot be used in a wireless medium because:

  1. Wireless collisions do not add enough energy for effective detection.
  2. Collision may not be detected because of the hidden node problem.
  3. The distance between stations may be great, and signal fading could prevent a station at one end from hearing a collision at the other end.

  Conclusion. Collisions on wireless networks must be avoided because they cannot be detected. This is why CSMA/CA is used instead.

• T5-Q01

  Tutorial 5, Q1

  recallconceptual

  What are the differences between classful and classless addressing in IPv4?

  • classful
  • classless
  • cidr
  • ipv4
  • addressing

  Show walkthrough

  Internet Protocol version 4 (IPv4). IPv4 is the delivery mechanism used by the TCP/IP protocols in the Network Layer. An IP address (also known as a logical address) is a unique addressing scheme used to identify every host in the network (the internet). Each node using the TCP/IP protocol has a unique and universal 32-bit logical IP address — a total of $2^{32} \approx 4.3$ billion IP addresses in IPv4. Each IP address has two parts: the network number (netID) and host number (hostID).

  IP addresses are distributed based on two mechanisms:

  Classful addressing.

  • The address space is divided into five classes: A, B, C, D, E.
  • Assigns an organization a Class A, B, or C block of addresses.

  Classless addressing.

  • Now more commonly used to deal with address shortage.
  • No class definition.
  • The address contains a prefix length n, where n = 0, ..., 32, to classify which block an IP address belongs to.
  • Based on any IP address that belongs to a specific block, we can calculate the number of available addresses in that block, as well as the first and last IP addresses that belong to that block.
  • Example: based on 128.11.3.16/28, there is a total of $2^{(32-28)} = 16$ addresses in that block. The first address is 128.11.3.16, and the last address is 128.11.3.31.

• T5-Q02

  Tutorial 5, Q2

  analyseconceptual

  Explain why most of the addresses in class A are wasted. Explain why a medium or large size organization does not want a class of class C.

  • classful
  • class-a
  • class-c
  • addressing

  Show walkthrough

  Class A address waste. A block in class A is too large for almost any organisation. This means most of the addresses in class A are wasted and not used.

  Class C limitations. A block in class C is too small for many medium/large size organisations.

• T5-Q03

  Tutorial 5, Q3

  recallconceptual

  What is a mask in IPv4 addressing?

  • mask
  • ipv4
  • classful
  • addressing

  Show walkthrough

  Definition of mask. A mask is a 32-bit number in which the n leftmost bits are 1s and the (32 − n) rightmost bits are 0s.

  Classful addressing.

  • n = 8, 16, 24
  • Used to find the netid and hostid.
  • The concept of mask does not apply for class D and E.

• T5-Q04

  Tutorial 5, Q4

  applynumeric

  What is the network address in a block of addresses? How can we find the network address if one of the addresses in the block is known?

  • subnetting
  • mask
  • cidr
  • ipv4
  • addressing

  Show walkthrough

  Definition. The network address in a block of addresses is the first address.

  Finding the network address. The mask can be ANDed with any address in the block to find the network address.

  Example. Address is 205.16.37.34/28.

  Mask:       11111111 11111111 11111111 11110000
  Address:    11001101 00010000 00100101 00100010
  ANDed:      11001101 00010000 00100101 00100000 => 205.16.37.32 (Network address)

• T5-Q05

  Tutorial 5, Q5

  recallconceptual

  Briefly define subnetting. How does the subnet mask differ from a default mask in classful addressing?

  • subnetting
  • mask
  • cidr
  • classless
  • addressing

  Show walkthrough

  Subnetting definition. In subnetting, a large address block could be divided into several contiguous groups, and each group can be assigned to a smaller network called a subnet. With subnetting, the IP address now consists of network, subnet, and host.

  Subnet mask characteristics. The number of hosts per subnet is $2^x - 2$ (minus 2 because the first address is used as the network address and the last address is used as the broadcast address).

  Difference from the default mask.

  • The subnet mask is defined with a value of /n.
  • The subnet mask has more consecutive 1s than the corresponding default mask.

• T5-Q06

  Tutorial 5, Q6

  recallconceptual

  What is NAT? How can NAT help in address depletion?

  • nat
  • private-addressing
  • rfc1918
  • translation-table

  Show walkthrough

  The problem. Home users and small businesses may create small networks with several hosts and need an IP address for each host. Given the shortage of available addresses, this is a serious problem.

  The solution — NAT. A quick solution to this problem is network address translation (NAT). NAT is a mechanism that enables users to have a large set of addresses internally and one, or a small set of, addresses externally to communicate with the other networks (the internet).

  Using port addresses. Port addresses are used to specify the destination inside the private network. This allows multiple internal devices to share a single external IP address.

• T5-Q07

  Tutorial 5, Q7

  analyseconceptual

  What is the difference between connection oriented and connection less services?

  • network-layer
  • datagram-network

  Show walkthrough

  Aspect	Connection-oriented service	Connectionless service
  Connection	A virtual connection is established between the sender and the receiver before transferring data.	Each packet is independent from every other packet and can take a different path.
  Path	All packets follow the same path.	Packets may take different paths.
  Phases	Communication has three phases: setup, data transfer, and teardown.	Communication has only one phase: data transfer.
  Setup/teardown	Requires a setup and teardown process.	No setup and teardown process.
  Example	Virtual-circuit approach of packet switching.	Datagram approach of packet switching.

• T5-Q08

  Tutorial 5, Q8

  recallconceptual

  Define fragmentation and why it is needed?

  • fragmentation
  • mtu
  • offset
  • ipv4

  Show walkthrough

  Definition. Each data link layer protocol has a limited size of packet it can carry. When a datagram is encapsulated in a frame, the total size of the datagram must be less than this limit — otherwise, the datagram must be fragmented.

  Fragmentation offset. The fragmentation offset shows the relative position of the fragments with respect to the whole datagram (in units of 8 bytes).

• T5-Q09

  Tutorial 5, Q9

  analyseconceptual

  List transition strategies to move from IPv4 to IPv6. Explain the differences between tunneling and dual stack strategies during the transition period.

  • ipv6
  • dual-stack
  • tunneling
  • transition

  Show walkthrough

  Reason for migration. The main reason for migration from IPv4 to IPv6 is the small size of the address space in IPv4.

  Three strategies. Three strategies have been devised by the IETF to help the transition:

  1. Header translation technique
  2. Tunneling technique
  3. Dual stack technique

  Header translation. Necessary when the majority of the internet has moved to IPv6 but some systems still use IPv4. The sender wants to use IPv6, but the receiver does not understand IPv6, so the header format must be totally changed through header translation — the header of the IPv6 packet is converted to an IPv4 header.

  Tunneling. Used when two computers using IPv6 want to communicate with each other, and the packet must pass through a region that uses IPv4. The IPv6 packet is encapsulated in an IPv4 packet when it enters the IPv4 region.

  Dual stack. A station must run IPv4 and IPv6 simultaneously until all of the internet uses IPv6. If the DNS returns an IPv4 or IPv6 address, the source host sends an IPv4 or IPv6 packet, respectively.

  The difference. Tunneling encapsulates an IPv6 packet inside an IPv4 packet so it can cross an IPv4-only region. Dual stack instead has the host run both protocols natively and choose which one to send based on the DNS response — no encapsulation is involved.

• T5-Q10

  Tutorial 5, Q10

  recallconceptual

  What is the purpose of RIP?

  • rip
  • distance-vector
  • hop-count
  • routing

  Show walkthrough

  Routing Information Protocol (RIP). RIP is one of the most widely used intradomain routing protocols, based on the distance-vector routing algorithm. RIP uses hop count to determine the shortest path.

  Purpose of RIP.

  • To forward packets, one of the available routes is chosen by the network layer; routing protocols handle this function.
  • If more than one possible route exists to reach a subnet, RIP picks the best route based on hop count.
  • In case of a link failure, it can forward packets via a different route.
  • Advertises routing information about IP subnets to neighboring routers.

• T5-Q11

  Tutorial 5, Q11

  recallconceptual

  What are the functions of a RIP message?

  • rip
  • distance-vector
  • ripv1
  • ripv2
  • routing-table

  Show walkthrough

  RIP message types. RIP has two types of messages:

  Request messages. Sent by a router that has just come up, or by a router that has some time-out entries.

  Response messages. A response (or update) message can be either solicited or unsolicited.

  • Solicited: sent only in answer to a request message.
  • Unsolicited: sent periodically, every 30 seconds, or on a change to the forwarding table.

  States of a network using RIP.

  Initialization state. At the beginning, each router shares its routing table with the routers directly connected to it (its neighbor routers). Node A does not know about node E; node C does. Node D does not know about node C and B; node A does.

  Sharing and updating state. Node A updates its routing table after receiving the partial table from node C.

  Final state. A stable connection; each node knows how to reach any other node and the cost.

• T5-Q12

  Tutorial 5, Q12

  analyseconceptual

  Why do OSPF messages propagate faster than RIP messages?

  • ospf
  • rip
  • link-state
  • distance-vector

  Show walkthrough

  Open Shortest Path First (OSPF). OSPF uses Link State (LS) routing to find the shortest path. Whenever a link cost changes, the new link cost must be sent to all nodes (flooding). When the flooding of LSPs is complete, each router can create its own shortest-path tree and forwarding table; convergence is fairly quick.

  Routing Information Protocol (RIP). RIP uses Distance Vector (DV) routing to find the shortest path. When link costs change, the DV propagates the results to one node attached to that link at a time. The DV can converge slowly and can have routing loops while the algorithm is converging. DV also suffers from the count-to-infinity problem.

  The count-to-infinity problem in RIP.

  • At the beginning, both nodes A and B know how to reach node X.
  • The link between A and X fails and node A changes its table.
  • If A can send its table to B immediately, everything is fine.
  • But if B sends its routing table to A before receiving A’s routing table, A receives the update and assumes that B has found a way to reach X.
  • A also sends its new update to B. Now B thinks that something has changed around A and updates its routing table.
  • Infinite loop: node A thinks the route to X is via B; node B thinks the route to X is via A.

  Why OSPF propagates faster. OSPF propagates faster because link-state messages flood to all nodes at once, providing immediate global network knowledge. RIP relies on periodic distance-vector updates that propagate hop-by-hop, leading to slow convergence and potential routing loops.

• T5-Q13

  Tutorial 5, Q13

  recallconceptual

  What is the purpose BGP?

  • bgp
  • autonomous-system
  • as-path
  • policy-routing

  Show walkthrough

  Border Gateway Protocol (BGP). BGP is an interdomain routing and gateway protocol. It connects individual autonomous systems together using a path vector routing mechanism.

  Primary purpose. The primary purpose of a BGP system is to exchange routing information among the autonomous systems on the internet.

  BGP autonomous system classification. In BGP, ASs are divided into three categories:

  1. Stub AS: has only one connection to another AS. Data traffic can be either initiated or terminated in a stub AS; it cannot pass through it.
  2. Multihomed AS: can have more than one connection to other ASs, but does not allow data traffic to pass through it.
  3. Transient AS: connected to more than one other AS and also allows traffic to pass through.

  Path vector routing. Path vector routing proved useful for interdomain routing. The speaker node in an AS creates a routing table and advertises it to speaker nodes in neighboring ASs. A speaker node advertises the path, not the metric of the nodes, in its AS or other ASs.

  BGP operation states.

  Initialization. At the beginning, each speaker node knows only the reachability of nodes inside its own AS.

  Sharing. In path vector routing, a speaker in an AS shares its table with its immediate neighbors.

  Updating (speaker A1 of AS1). If router A1 receives a packet for node A3, it knows the path is inside AS1 (the packet is “at home”). If it receives a packet for D1, it knows the packet should go from AS1, to AS2, and then to AS3.

  Updating (speaker D1 of AS4). If node D1 in AS4 receives a packet for node A2, it knows it should go through AS4, AS3, and AS1.

• T5-Q14

  Tutorial 5, Q14

  recallconceptual

  What is an autonomous system?

  • autonomous-system
  • igp
  • egp
  • routing

  Show walkthrough

  Autonomous system (AS). An autonomous system is a group of networks and routers controlled by a single administrator. A network can therefore be seen as a large collection of autonomous systems.

  Key characteristic. Routers within the same AS run the same routing algorithm.

  Router classification. Routers connecting different ASs are called gateway routers.

  Routing protocol levels.

  • A routing algorithm running within an autonomous system is called an intradomain (intra-autonomous system) routing protocol — examples: RIP and OSPF.
  • A routing algorithm used by gateway routers between autonomous systems is called an interdomain (inter-autonomous system) routing protocol — example: BGP.

• T5-Q15

  Tutorial 5, Q15

  analyseconceptual

  Compare and contrast distance vector routing method against link state routing method.

  • distance-vector
  • link-state
  • routing
  • routing-metrics

  Show walkthrough

  Aspect	Distance vector routing	Link state routing
  Information sharing	Each node shares its routing table with its immediate neighbors periodically and when there is a change.	Each node in the domain has the entire topology of the domain.
  Message format	The update messages have a very simple format and are sent only to neighbors.	The link-state messages in OSPF have a somewhat complex format.
  Traffic impact	They do not normally create heavy traffic.	For a large area, messages may create heavy traffic.
  Convergence	Slow convergence.	Convergence is fairly quick.
  Failure impact	Corruption or failure in one router affects other routers as seriously.	Corruption or failure in one router does not affect other routers.
  Stability	Distance vector routing is subject to instability if there are more than a few hops in the domain of operation.	Link state routing needs a huge number of resources to calculate routing tables.
  Example	RIP	OSPF

• T5-Q16

  Tutorial 5, Q16

  applynumeric

  Workout with your tutor the answers for the following slides 28 and 45 from the lectures.

  • network-layer

  Show walkthrough

  The source solution is incomplete for this question. What follows is only what the tutorial actually gives.

  The unit did not release a solution for this question — it is worked through with the tutor in the session itself. The entire extent of what survives in the source material is the line:

  SOLUTION NOT RELEASED by the unit.

  The question names two lecture slides and nothing else, so there is nothing here to work from, and no safe way to fill in an answer without guessing at what those slides contain.

  Check the tutorial recording, or ask in class, for the worked answers to slides 28 and 45.

• T5-Q17

  Tutorial 5, Q17

  applynumeric

  Workout with your tutor the answers for the Lecture 5, slides 30 and 31.

  • subnetting
  • cidr
  • mask
  • ipv4
  • addressing

  Show walkthrough

  The unit’s released solution for this worksheet question works through three subnetting calculations. That is everything the tutorial gives for it; nothing beyond these three is reconstructed here.

  1. A block of addresses is granted to a small organization. We know that one of the addresses is 205.16.37.36/29. What is the first, the last, and the total number of addresses of this block?

  29 bits for network address.

  IP Address:       11001101 00100000 00100101 00100100 (205.16.37.36)
  Mask:             11111111 11111111 11111111 11111000 (255.255.255.248)
  Subnet Address:   11001101 00100000 00100101 00100000 (205.16.37.32)
  
  1st Address:      11001101 00100000 00100101 00100001 (205.16.37.33)
  2nd Address:      11001101 00100000 00100101 00100010 (205.16.37.34)
  3rd Address:      11001101 00100000 00100101 00100011 (205.16.37.35)
  4th Address:      11001101 00100000 00100101 00100100 (205.16.37.36)
  5th Address:      11001101 00100000 00100101 00100101 (205.16.37.37)
  6th Address:      11001101 00100000 00100101 00100110 (205.16.37.38)
  Broadcast Address: 11001101 00100000 00100101 00100111 (205.16.37.39)

  2. A block of addresses is granted to a small organization. We know that one of the addresses is 205.16.37.36/30. What is the first, the last, and the total number of useable addresses of this block?

  Mask:             11111111 11111111 11111111 11111100 (255.255.255.252)
  Subnet Address:   11001101 00100000 00100101 00100100 (205.16.37.36)
  
  1st Address:      11001101 00100000 00100101 00100101 (205.16.37.37)
  2nd Address:      11001101 00100000 00100101 00100110 (205.16.37.38)
  Broadcast Address: 11001101 00100000 00100101 00100111 (205.16.37.39)

  In total, there are only two usable addresses, which is $2^2 - 2 = 2$.

  3. A block of addresses is granted to a small organization. We know that one of the addresses is 192.168.10.64/28. What is the first, the last, and the number of useable addresses of this block?

  Mask:             11111111 11111111 11111111 11110000 (255.255.255.240)
  Subnet Address:   11000000 10101000 00001010 01000000 (192.168.10.64)
  
  1st Address:      11000000 10101000 00001010 01000001 (192.168.10.65)
  Last Address:     11000000 10101000 00001010 01001110 (192.168.10.78)
  Broadcast Address: 11000000 10101000 00001010 01001111 (192.168.10.79)

  In total, there are only 14 usable addresses, which is $2^4 - 2 = 14$.

• T5-Q18

  Tutorial 5, Q18

  analysediagram

  Can Router R1 in the figure below receive a packet with destination address 140.24.7.194. How is the packet routed to its final destination?

  • routing
  • routing-table
  • forwarding
  • network-layer

  Show walkthrough

  Analysis. Router R1 has four interfaces. Consider the possibility of a packet with destination 140.24.7.194 arriving from each of these interfaces, and how it would be routed.

  Result. The packet can arrive from one of the interfaces m0, m1, or m2, because one of the computers in organization 1, 2, or 3 could have sent this packet.

  Applying the /26 prefix to the address gives the network address 140.24.7.192/26. Since none of the network addresses/masks at R1 matches this result, the packet is sent to the default router R2.

  Additional note. The packet cannot arrive at router R1 from interface m3, because that would mean it had arrived from interface m0 of router R2 — which is impossible. If this packet arrives at router R2 (from any interface), applying the prefix length /26 to the destination address gives the network address/mask 140.24.7.192/26, and the packet is sent out from interface m1, directed to organization 4, and never reaches router R1.

• T5-Q19

  Tutorial 5, Q19

  applydiagram

  Assume Router R2 in the figure below receive a packet with destination address 140.24.7.42. How is the packet routed to its final destination?

  • routing
  • routing-table
  • forwarding
  • network-layer

  Show walkthrough

  Routing process. The packet is sent to router R1 and eventually to organization 1, as follows:

  Step 1. Router R2 applies the mask /26 to the address (extracting the leftmost 26 bits), giving the network address/mask 140.24.7.0/26, which does not match the first entry in the forwarding table.

  Step 2. Router R2 applies the mask /24 to the address (extracting the leftmost 24 bits), giving the network address/mask 140.24.7.0/24, which matches the second entry in the forwarding table. The packet is sent out from interface m0 to router R1.

  Step 3. Router R1 applies the mask /26 to the address (extracting the leftmost 26 bits), giving the network address/mask 140.24.7.0/26, which matches the first entry in the forwarding table. The packet is sent out from interface m0 to organization 1.

• T5-Q20

  Tutorial 5, Q20

  applynumeric

  In IPv4 datagram, the value of total-length field is (00A0)₁₆ and the value of the header length (HLEN) is (5)₁₆. How many bytes of payload are being carried by the datagram? What is this datagram’s efficiency (ratio of the payload length to the total length)?

  • ipv4
  • header
  • datagram

  Show walkthrough

  Given.

  • Total-length field = (00A0)₁₆
  • Header length (HLEN) = (5)₁₆

  Calculation. The total length of the datagram is (00A0)₁₆ = 0000 0000 1010 0000 = 160 bytes (128 + 32).

  The header length is 5 × 4 = 20 bytes. (The receiver needs to multiply the value of HLEN by a constant 4 to find the total header length.)

  The size of the payload is then 160 − 20 = 140 bytes.

  Efficiency. Efficiency = payload length / total length = 140 / 160 = 87.5%.

• T5-Q21

  Tutorial 5, Q21

  applynumeric

  A packet has arrived in which the offset value is 100, the value of HLEN is 5, and the total length value is 100. What are the numbers of the first and last bytes?

  • fragmentation
  • offset
  • ipv4

  Show walkthrough

  Given.

  • Offset value = 100
  • HLEN = 5
  • Total length value = 100

  Calculation. The first byte number is offset × 8 = 100 × 8 = 800.

  The total length is 100 bytes, and the header length is 5 × 4 = 20 bytes, which means there are 80 bytes of data in this datagram.

  If the first byte number is 800, the last byte number is 800 + 80 − 1 = 879.