Network Architecture

Question 1

What is a switched network? What is a cloud?

Answer 1

A switched network contains switches that store and forward data.
A cloud is a collection of switches.
Each cloud represents a network.

Question 2

What are the 2 types of switched networks?

Answer 2

Circuit-switched network (The stream of bits passing into a circuit emerges at the other end with the same pattern, delayed by a constant interval).
And
Packet switched network (breaks the stream of data into packets that are handled individually by the network).

Question 3

What are the requirements of a computer network?

Answer 3

• Connectivity: Every node in a network should be able to communicate with every other node.
• Scalability: After adding a new node, how are all the other nodes able to find it? (addressing).
• Performance: Latency (how long it takes to travel from one node to the other) and Bandwidth (the rate at which data can be transferred along a link).
• Reliability: Data might be corrupted by interference (error detecting)
• Cost-effective resource sharing: It’s not practical to have a point-to-point connection between every pair of nodes in a large network. How do we share a link between multiple flows? (Multiplexing).

Question 4

What problems does layering help solve?

Answer 4

With so many users, hardware, and software etc… how do hosts on a network communicate? Also within a host, how does everything interact

Question 5

What does layering do?

Answer 5

Divides different types of tasks into layers.
Abstraction: Hides the details of a particular layer’s implementation from the layers above and below.
Protocol: A particular implementation within a layer. It provides a communication service that higher level objects use to exchange messages.

Question 6

The OSI architecture

Answer 6

Application
Presentation (data formatting)
Session (associates multiple data streams)
Transport (delivery of messages)
Network (delivery of packets)
Data Link (delivery of frames)
Physical (delivery of bits)

Question 7

Encapsulation

Answer 7

The process of adding addresses and other information,

The RRP attaches a header (only a few bytes long) to each message. The rest of the message is called the body or payload

Question 8

Advantages of layering

Answer 8

The problem of building a network is broken down into a set of smaller and simpler problems.
Allows a modular design of a network. The addition of a new service may only require the modification of the functionality at one level.

Question 9

Addressing, routing and casting

Answer 9

Every node in a network must have an address. Then switches and routers are used to forward the message to a destination (unicast, multicast, broadcast).

Question 10

What problems does Multiplexing overcome? And how?

Answer 10

Multiplexing solves a lot of problems amongst
how all the hosts on a network communicate with each other.
And
How several hosts share the same link in the network at the same time?

By using multiplexing, a system resource is shared among multiple users.

Question 11

Methods of Multiplexing (STDM and FDM)

Answer 11

Synchronous time–division multiplexing (STDM): Divide time into quanta of equal size. Data is sent over the link in a round–robin manner.
Frequency division multiplexing (FDM): Transmit each flow of data over the physical link at a different frequency.

Question 12

What are the limitations of STDM and FDM in Multiplexing?

And what is the solution?

Answer 12

Inefficiency - If data is not being transmitted between a pair of hosts, then its share of the physical link is idle.
For both STDM and FDM, the number of data flows is fixed.

Solution:
Statistical multiplexing: Like STDM, statistical multiplexing operates in the time domain but unlike STDM, data is transmitted from each flow on demand rather than during a predetermined time slot.
It defines an upper bound on the size of the packet (block of data) that each flow is permitted to transmit at a given time.

Question 13

When does congestion occur?

Answer 13

When the switch receives packets at a higher rate than can be accommodated by the shared link.

This means some packets are stored in memory, and when the available memory runs out, packets must be deleted

Question 14

What is Bandwidth?

Answer 14

The number of bits that can be transmitted over the network in a given period of time (measured in bps).

Question 15

What is Latency?

Answer 15

length of time it takes to send a message from one end of the system to the other end.

Latency = Propagation + Transmit + Queue
Transmit = Size/Bandwidth (length of the message)
Propagation = Distance /Speed of light (time taken for the first bit of message to arrive at the destination).

Question 16

What is RTT (Round Trip Time)?

Answer 16

Double Propagation time.

Question 17

How many bits can be sent before the first one arrives at the receiver?

Answer 17

Suppose a link has a bandwidth of F Mbps and a latency (transmit plus propagation) of L seconds. Then the bandwidth is equal to F × 106 bits per second, and thus the number of bits transmitted in L seconds is F×L×106 bits or F×L×106 ×1/8 bytes of data can be stored.

Question 18

What is Throughput?

Answer 18

A measure of the performance of a channel that is achieved in practice.

Question 19

ASCII- scheme for encoding text characters as bytes

Answer 19

An upper-case letter to an ASCII code, find its position in the alphabet (A = 1, B = 2, etc.) and add 64.
A lower-case letter to an ASCII code, find its position in the alphabet (a = 1, b = 2, etc.) and add 96.
A digit to an ASCII code, add 48 to its value.
Space is 32.

Question 20

How does bit encoding work in network adapters?

Answer 20

Network adapters also contain a signaling component- encodes the bits into signals at the sending node, and decodes signals into bits at the receiving node.

Question 21

Non-return to zero encoding

Answer 21

Maps data value 0 to a low signal and 1 to a high signal.

Baud rate - the rate at which an encoder can make transitions.

Question 22

What are the problems with NRZ (non-return to zero encoding)?

Answer 22

Baseline wander - The receiver keeps an average of the signal that it has seen, and uses this value to discriminate between low and high signals. It is encoded as 0 if the signal is less than this average, and it is encoded as 1 if the signal is greater than this average. Too many consecutive 1s or 0s cause this average to change, making it difficult to detect a significant change in the signal.
Clock drift - Each network adaptor has a clock, and a signal is transmitted upon each clock tick. If the clocks of the sender and receiver are not synchronized, then the receiver may not be able to correctly recover the signals from the sender.

Question 23

What is Manchester Encoding?

Answer 23

Manchester encoding helps solve the problems with NRZ.
Transmits the exclusive–OR of the NRZ–encoded data and the clock.
The exclusive–OR of two logical variables p and q is
• True if p and q have different logical values
• False if p and q have the same logical values

Question 24

What are the disadvantages of Manchester Encoding?

Answer 24

The rate at which the signal changes (baud rate) is twice the rate of the transmitted bits:
• The receiver has only half the time to detect each pulse of the signal
• The bit rate is half the baud rate, and so Manchester coding is only 50% efficient

Question 25

A method for solving the framing problem: Byte-oriented protocols.

Answer 25

SYN, SYN, SOH, Header, STX, body, ETX, CRC
each are all 8 bits apart from the CRC which is 16

SYN are for synchronisation
SOH represents the start of the header
STX represents the start of the data
ETX represents the end of the data
CRC is for error detection

Question 26

What happens if the ETX or DLE (Data Link Escape) characters appear in the data portion of the frame? These are control or non-printing characters that do not represent written symbols.

Answer 26

Use character stuffing. Precede each character with a DLE character

Question 27

HDLC (High–level Data Link Control) Protocol

Answer 27

The beginning and end of a frame are defined by the sequence 01111110. Transmitted when the link is idle so that the sender and receiver clocks remain synchronised. Use bit stuffing if the string 01111110 appears in the data.

Bit–stuffing in HDLC Protocol
Sender inserts a 0 whenever five consecutive 1s have been transmitted If five consecutive 1s are sent, then the receiver acts as follows:
• If the next bit is 0, then it must have been stuffed, and it is therefore removed
• If the next bits are 10, then the end of frame marker has been received
• If the next bits are 11, then an error has been introduced and the whole frame is
deleted