Computer Network

Question 1

List the five components of a data communication system.

Answer 1

Message, information that can be communicated
Sender, the device that sends the message
Receiver, the device that receives the message
Transmission medium, the physical path of which the message will travel from sender to receiver
Protocol, the rules that govern data communication

Question 2

2. What are the three most important network criteria that must be met for an effective and efficient network?

Answer 2

Performance, performance is measured by transmit time and response time. The performance of a network depends on the number of users, the type of transmission medium, the capacities of the connected hardware and, the efficiency of the software.
Reliability, reliability is measured by the frequency of failures.
Security, protection of data from unauthorized access.

Question 3

3. What are two types of line configurations?

Answer 3

Point-to-point, it provides a dedicated link between 2 devices, the capacity of the link is reserved for transmission between these 2 devices.
Multipoint, more than 2 devices share a link, the capacity of the link is shared spatially or temporarily.

Question 4

3 characteristics of effectiveness

Answer 4

1) delivery - the system must deliver the data to the correct destination
2) Accuracy - the system must deliver the data correctly
3) Timeliness - the system must deliver the data in a timely manner

Question 5

Data flow types

Answer 5

4) Simplex – only one can transmit, and the other receives
5) Half-duplex – both can transmit and receive but not at the same time
6) Full-duplex – both can transmit and receive simultaneously

Question 6

State the Physical topologies

Answer 6

Mesh topology
Star topology
Bus topology
Ring topology

Question 7

Describe Mesh Topology

Answer 7

each device has a point-to-point link to every other device resulting in
n(n-1)/2 links with n devices and (n-1) I/O ports per device.

Question 8

Describe Star Topology

Answer 8

each device has a point-to-point link to the hub, the hub acts as an exchange between nodes.

Question 9

Describe Bus Topology

Answer 9

it is a multipoint connection, A long cable acts as the backbone to link all devices in one network.

Question 10

Describe Ring Topology

Answer 10

Each device has a point-to-point connection with 2 devices on each side of it. The signal is passed along the ring in one direction, from device to device until it reaches its destination.

Question 11

advantages of mesh topology

Answer 11

The use of dedicated links eliminates traffic problems.
Privacy and security.
Easy fault identification and fault isolation.
Robustness. If one link becomes unusable, it does not incapacitate the entire topology.

Question 12

disadvantages of mesh topology

Answer 12

Big amount of cabling makes installation and reconnection difficult.
Wiring can be greater than available space (walls, ceiling, floors).
Big number of I/O ports required. Hardware required will be very expensive.
Mesh is implemented in a limited fashion

Question 13

advantages of star topology

Answer 13

Less expensive than mesh topology. Each device needs only one link and one I/O.
Easy to install and reconfigure.
Far less cabling needs to be housed compared with mesh topology and additions, moves, and deletions involve only the connection between a node and the hub.
Robustness.
Easy fault identification and fault isolation.

Question 14

advantages of bus topology

Answer 14

Ease of installation
Less cabling than mesh or star topology

Question 15

disadvantages of bus topology

Answer 15

Difficult to add devices
Signal reflection in taps can cause degradation in quality
A fault or break in the bus cable stops all transmission

Question 16

advantages of ring topology

Answer 16

Easy to install and reconfigure
The only constraints are media and traffic considerations
Fault isolation is simplified

Question 17

disadvantages of ring topology

Answer 17

In a simple ring (unidirectional), a disabled station can disable an entire network. This can be solved by using a dual ring or a switch capable of closing off the break

Question 18

pros and cons of hybrid (star of busses)

Answer 18

Pros: Better domain separation management, one hop delay.
Cons: better robustness than the “star”, but still vulnerable to Hub failure!

Question 19

What are Categories of Networks

Answer 19

A category of a network is determined by its size, its ownership, the distance it covers, and its physical architecture

Question 20

State the Categories of Networks

Answer 20

Local Area Netwok (LAN)
Metropolitan-Area Network (MAN)
Wide Area Network (WAN)

Question 21

what is The Internet

Answer 21

Huge number of interconnected Networks

Question 22

state the key elements of a protocol

Answer 22

Syntax: structure or format of the data
Semantics: Meaning of each portion of bits
Timing: When and how fast data should be sent

Question 23

what are standards

Answer 23

Standards provide guidelines to manufacturers, vendors, government agencies, and other providers to ensure the kind of interconnectivity necessary in today’s marketplace and in international communication.

Question 24

state the 2 types of standards

Answer 24

1. De facto: Standards that have not been approved by an organized body but have been adopted as standards through widespread use.
2. De Jure: Standards that have been legislated by an officially recognized body.

Question 25

concept of protocol layering

Answer 25

• When communication is simple, we only need one simple protocol
• When the communication is complex, we need a protocol at each layer, or protocol layering

Question 26

principles of protocol layering

Answer 26

1. In bidirectional communication, each layer has to be able to perform 2 opposite tasks, one in each direction.
2. the two objects under each layer at both sites should be identical.

Question 27

describe TCP/IP PROTOCOL SUITE

Answer 27

• It is a protocol suite which is used in the Internet today.
• Hierarchical protocol made up of interactive modules, each of which provides a specific functionality.
• Each upper level protocol is supported by the services provided by one or more lower level protocols.
• Original TCP/IP protocol suite was defined as four software layers built upon the hardware
• Today, TCP/IP is thought of as a five-layer model.

Question 28

Encapsulation at source

Answer 28

o The message is passed to the transport layer.
o The transport layer takes the message as the payload. It adds the transport layer header to the payload. The header contains the identifiers of source and destination application programs that want to communicate and additional information that is needed for the end-to-end delivery of the message. The transport layer passes the packet to the network layer.
o Network layers take the packet as payload and add its own header. The header contains addresses of source and destination hosts and additional information needed for error checking of the header, fragmentation information, etc. The resulting network layer packet is called a datagram. The network layer then passes the packet to the data link layer.
o The data link layer takes the network layer packet as a payload and adds its own header. The header contains link-layer addresses of the host or next hop. Result is the link-layer packet is called as a frame. Frame is passed to the physical layer for transmission.

Question 29

Decapsulation and Encapsulation at the Router

Answer 29

o After the set of bits are delivered to the data-link layer, this layer decapsulates the datagram from the frame and passes it to the network layer.
o The network layer inspects the source and destination addresses in the datagram header and consults the forwarding table to find the next hop to which the datagram is to be delivered. The datagram is then passed to the data-link layer of the next link
o Data-link layer of the next link encapsulates the datagram in a frame and passes it to the physical layer for transmission

Question 30

Decapsulation at the Destination

Answer 30

o At the destination host, each layer decapsulates the packet received, removes the payload, and delivers the payload to the next-higher layer protocol until the message reaches the application layer
o Decapsulation in the host involves error checking.

Question 31

the 7 layers in OSI model

Answer 31

Layer 7: Application
Layer 6: Presentation
Layer 5: Session
Layer 4: Transport
Layer 3: Network
Layer 2: Data link
Layer 1: Physical

Question 32

layers in TCP/IP protocol suite

Answer 32

Layer 5: Application
Layer 4: Transport
Layer 3: Network
Layer 2: Data link
Layer 1: Physical

Question 33

Where is the Transmission media Located?

Answer 33

located below the physical layer and directly controlled by the physical layer.

Question 34

how does the transmission media transmit signals?

Answer 34

Signals are transmitted from one device to another through electromagnetic energy propagated through transmission media.

Question 35

transmission media can be divided into two broad categories, which are?

Answer 35

Guided media include twisted-pair cable, coaxial cable, and fiber-optic cable.
Unguided media is wireless, such as radio wave, microwave and infrared

Question 36

describe transmission impairment

Answer 36

signals travel through transmission media, which are not perfect. The imperfection causes signal impairment.
the signal at the beginning of the medium is not the same as the signal at the end of the medium.

Question 37

List the causes of impairment in transmission

Answer 37

Attenuation
Distortion
Noise

Question 38

describe Attenuation

Answer 38

Attenuation means a loss of energy.
When a signal travels through a
medium, it loses some of its energy in overcoming the resistance of the medium.

Question 39

describe distortion

Answer 39

Distortion means that the signal changes its form or shape.
Distortion can occur in a composite signal made of different frequencies.
Each signal component has its own propagation speed through a medium and, therefore, its own delay in arriving at the final destination.
Differences in delay may create a difference in phase if the delay is not exactly the same as the period duration.

Question 40

Describe noise

Answer 40

Noise is another cause of impairment.
Several types of noise, such as thermal noise, induced noise, crosstalk, and impulse noise, may corrupt the signal.

Question 41

List the classes of transmission media

Answer 41

guided
1. Twisted-pair
2. Coaxial cable
3. Fiber-optic cable

unguided
1. Radio wave
2. Microwave
3. Infared

Question 42

describe guided media

Answer 42

A signal traveling along any guided media is directed and constrained by the physical limits of the medium.

Question 43

describe twisted-pair cable

Answer 43

A twisted-pair consist of two conductors (normally copper), each with its own plastic insulation, twisted together.
One of the wires is used to carry signals to the receiver, and
the other is used only as a ground reference. The receiver uses the difference between the two levels.
Interference may affect both wires and create unwanted signals. The receiver at the end, however, operates only on the difference between these unwanted signals.
If the two wires are affected by interference equally, the receiver is immune

Question 44

Describe Coaxial cable

Answer 44

Coaxial cable carries signals of higher frequency ranges than twisted-pair cable.
Coaxial cable (or coax) has a central core conductor of solid wire enclosed in an insulating sheath, which is, in turn, encased in an outer conductor of metal foil, braid, or a combination of the two.
The outer metallic wrapping serves both as a shield against noise and as the second conductor, which completes the circuit. This outer conductor is also enclosed in an insulating sheath, and the whole cable is protected by a plastic cover.

Question 45

State The Application of Coaxial Cable

Answer 45

Coaxial cable was used in analog telephone networks where a single coaxial network could carry 10000 voice signals.
Later it was used in digital telephone networks where a single coaxial cable could carry digital data up to 600 Mbps.
However, coaxial cable in telephone networks has largely been replaced today with fiber-optic cable.
Cable TV networks also use coaxial cables.
Another common application of coaxial cable is in traditional Ethernet LANs.

Question 46

Describe fiber-optic cable

Answer 46

Optical fibers use reflection to guide light through a channel. A glass or plastic case is surrounded by a cladding of less dense glass or plastic.
The difference in density of the two materials must be such that a beam of light moving through the core is reflected off the cladding instead of being refracted into it.

Question 47

Describe coaxial cable

Answer 47

The coaxial cable carries signals of higher frequency ranges than twisted-pair cable.
Coaxial cable (or coax) has a central core conductor of solid wire enclosed in an insulating sheath, which is, encased in an outer conductor of metal foil, braid, or a combination of the two.
The outer metallic wrapping serves both as a shield against noise and as the second conductor, which completes the circuit. This outer conductor is also enclosed in an insulating sheath, and the whole cable is protected by a plastic cover.

Question 48

describe fiber-optic cable

Answer 48

fiber-optic cable consists of 2 modes: a multimode and a single mode.
multimode has 2 forms: step-index and graded-index.
In the step-index, the density of the core remains constant from the center to the edges.
A beam of light moves through this constant density in a straight line until it reaches the interface of the core and the cladding.
At the interface, there is an abrupt change to a lower density that alters the angle of the beam’s motion.
A graded-index fiber is one with varying densities. Density is highest at the center of the core and decreases gradually to its lowest at the edge.
single-mode fiber: Single-mode uses step-index fiber and a highly focused source of light that limits beams to a small range of angles. it is manufactured with a much smaller diameter than that of multimode fiber, and with substantially lower density (index of refraction).

Question 49

state the application of fiber-optic cable

Answer 49

Fiber-optic cable is often found in backbone networks because its wide bandwidth is cost-effective.
Some cable TV companies use a combination of optical fiber and coaxial cable, thus creating a hybrid network.
Local area networks such as 100Base-FX network (Fast Ethernet) and 1000Base-X also use fiber-optic cable.

Question 50

Describe ungiuded media

Answer 50

Unguided media transport electromagnetic waves without using a physical conductor. this type of communication is known as wireless communication.

Question 51

describe radio waves

Answer 51

Radio waves can travel long distances. This makes radio waves a good candidate for long-distance broadcasting such as AM radio.
Radio waves of low and medium frequencies can penetrate walls.

Question 52

Advantages and disadvantages of radio waves

Answer 52

advantage:
Radio waves, can travel long distances.
an AM radio can receive signals inside a building
The omnidirectional characteristics of radio waves make them useful for multicasting, in which there is one sender but many receivers.
disadvantage:
We cannot isolate communication just inside or outside a building.
The radio waves transmitted by one antenna are susceptible to interference by another antenna that sends signals using the same frequency or band.

Question 53

describe microwaves

Answer 53

Electromagnetic waves having frequencies between 1 and 300 GHz are called microwaves.
Microwaves are unidirectional. When an antenna transmits microwave waves, they can be narrowly focused.
This means that the sending and receiving antennas need to be aligned.
Microwave propagation is line-of-sight. Since the towers with the mounted antennas need to be in direct sight of each other, towers that are far apart need to be very tall.

Question 54

state the application of microwaves

Answer 54

Microwaves, due to their unidirectional properties, are very useful when unicast (one-to-one) communication is needed between the sender and the receiver. They are used in cellular phones , satellite networks, and wireless LANs

Question 55

State the application of infrared

Answer 55

the infrared band, almost 400 THz has an excellent potential for data transmission.
Such a wide bandwidth can be used to transmit digital data with a very high data rate. Some manufacturers provide a
special port called the Infrared Data Association (IrDA) port
that allows a wireless keyboard to communicate with a PC.

Question 56

describe infrared

Answer 56

Infrared signals, with frequencies from 300 GHz to 400 THz (wavelengths from 1 mm to 770 nm), can be used for short-range communication.
Infrared signals, having high frequencies, cannot penetrate walls.
This advantageous characteristic prevents interference between one system and another. However, this same characteristic makes infrared signals useless for long-range communication.
we cannot use infrared waves outside a building because the sun’s rays contain infrared waves that can interfere with communication.

Question 57

State the Services provided by Data Link Layer

Answer 57

Farming
- A packet at the data-link layer is called a frame
- The data-link layer at each node must encapsulate the datagram (packet received from the network layer) in a frame before sending it to the next node.
- The node must also decapsulate the datagram from the frame received on the logical channel.
Flow control
- Different data-link-layer protocols use different strategies for flow control:
- Drop the frames if its buffer is full
- Send feedback to the sending data-link layer to stop or slow down
Error control
- Error need to be detected and corrected at the receiver node or discarded and retransmitted by the sending node
Congestion control
- Link may be congested with frames, which may result in frame loss.

Question 58

Describe framing

Answer 58

A packet at the data-link layer is called a frame
The data-link layer at each node must encapsulate the datagram (packet received from the network layer) in a frame before sending it to the next node.
The node must also decapsulate the datagram from the frame received on the logical channel.

Question 59

state the 2 categories of links

Answer 59

Point-to-Point link
- Link is dedicated to two devices
- Data-link layer that uses the whole capacity of the medium
Broadcast link
- Link is shared between several pairs of devices
- Data-link layer that uses only part of the capacity of the link.

Question 60

state the 2 sublayers

Answer 60

Data link Control (DLC) sub layer
- Deal with all issues common to both point-to-point and broadcast links
Media access Control sublayer
- Deal with issues specific to broadcast links

Question 61

state the 3 types of addresses

Answer 61

Unicast Address (IPv4 and IPv6)
- One-to-one communication
- A frame with a unicast address destination is destined for only one entity in the link
- Each host or each interface of a router is assigned a unicast address
Multicast Address (IPv4 and IPv6)
- One-to-many communication
Broadcast Address (IPv4)
- One-to-all communication
- A frame with a destination broadcast address is sent to all entries in the link

Question 62

Define the type of the following destination addresses:
a. 4A:30:10:21:10:1A b. 47:20:1B:2E:08:EE
c. FF:FF:FF:FF:FF:FF

Answer 62

look at the second hexadecimal digit. If it is odd, the address is multicast. If all digits are F’s, the address is broadcast. Therefore, we have the following:
a. This is a unicast address because A in binary is 1010.
b. This is a multicast address because 7 in binary is 0111.
c. This is a broadcast address because all digits are F’s.

Question 63

Show how the address 47:20:1B:2E:08:EE is sent out on line.

Answer 63

The address is sent left-to-right, byte by byte; for each byte, it is sent right-to-left, bit by bit, as shown below:

11100010 00000100 11011000 01110100 00010000 01110111

Question 64

A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the following formula. Each signal level is represented by 3 bits.

Answer 64

Number of bits per level = log2(8) = 3

Question 65

describe latency

Answer 65

The latency defines how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source.

Question 66

describe jitter

Answer 66

jitter is a problem if different packets of data encounter different delays and the application using the data at the receiver site is time-sensitive

Question 67

State the problem that can occur if different packets of data arrive at different times at the receiving end.

Answer 67

jitter

Question 68

state the solution to critical-section problem

Answer 68

Mutual Exclusion - If a process is executing in its critical section, then no other processes can be executing in their critical sections
Progress - If no process is executing in its critical section and there exist some processes that wish to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely
Bounded Waiting - A bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted

Question 69

what is the propose of semaphore implementation

Answer 69

Must guarantee that no two processes can execute wait () and signal () on the same semaphore at the same time

Question 70

state the type of semaphores

Answer 70

Counting semaphore – integer value can range over an unrestricted domain. it is used to control access to a resource that has multiple instances
Binary semaphore (Mutex lock) – integer value can range only between 0 and 1; can be simpler to implement

Question 71

state the types of errors

Answer 71

single-bit error: only 1 bit in the data unit is changed from 1 to 0 or from 0 to 1
burst error: 2 or more bits in the data unit have changed from 1 to 0 or from 0 to 1.

Question 72

Detection vs Correction

Answer 72

The correction of errors is more difficult than the detection.
In error detection, we only look to see if any error has occurred.
We are not interested in the number of corrupted bits. A single-bit error is the same for us as a burst error.
In error correction, we need to know the exact number of corrupted bits
and, more importantly, their location in the message.

Question 73

Process of error detection in block coding

Answer 73

the sender sends the data.
then, the generator generates a redundant bit
the receiver receives the data and gives the data to the checker
the checker decides whether to discard or extract the data

Question 74

describe cyclic codes

Answer 74

Cyclic codes are special linear block codes with one extra property.

Question 75

describe flow control

Answer 75

Flow control refers to a set of procedures used to restrict the amount of data that the sender can send
before waiting for an acknowledgment.

Question 76

describe error control

Answer 76

Error control is both error detection and error correction. It allows the receiver to inform the sender of any frames lost or damaged in transmission and coordinates the retransmission of those frames by the sender.

Question 77

how does Stop-and-Wait ARQ work

Answer 77

Error correction in Stop-and-Wait ARQ is done by keeping a copy of the sent frame and retransmitting the frame when the timer expires

Question 78

Taxonomy of protocols

Answer 78

for noiseless channel
- Simplest
- channel
for noisy channel
- Stop-and-Wait ARQ
- Go-Back-N ARQ
- Selective Repeat ARQ

Question 79

describe redundancy

Answer 79

The central concept in detecting or correcting errors is redundancy.
the sender adds redundant bits to our data and the receiver removes them.
Their presence allows the receiver to detect or correct corrupted bits.

Question 80

state the hamming distance of d(000, 011)

Answer 80

2 because (000 xor 011) = two 1s

Question 81

advantage of two-dimensional parity check

Answer 81

increases the likelihood of detecting burst errors.

Question 82

describe CRC encoder

Answer 82

In the encoder, the dataword has k bits (4 here); the codeword has n bits (7 here).
The size of the dataword is augmented by adding n − k (3 here) 0s to the right-hand side of the word.
The n-bit result is fed into the generator.
The generator uses a divisor of size n − k + 1 (4 here).
The generator divides the augmented dataword by the divisor.
The quotient of the division is discarded.
the remainder (r2r1r0) is appended to the dataword to create the codeword.

Question 83

describe CRC decoder

Answer 83

The decoder receives the codeword.
A copy of all n bits is fed to the checker, which is a replica of the generator.
The remainder produced by the checker is a syndrome of n − k (3 here) bits, which is
fed to the decision logic analyzer.
The analyzer has a simple function. If the syndrome bits are all 0s, the 4 left- most bits of the codeword are accepted as the dataword (interpreted as no error); otherwise, the 4 bits are discarded (error).

Question 84

Advantages of Cyclic Codes

Answer 84

Cyclic codes have a very good performance in detecting single-bit errors, double errors, an odd number of errors, and burst errors.

They can easily be implemented in hardware and software.

They are fast when implemented in hardware.

This makes cyclic codes a good candidate for many networks.

Question 85

describe checksum

Answer 85

Checksum is an error-detecting technique that can be applied to a message of any length.

Question 86

list the steps of checksum

Answer 86

At the source, the message is first divided into m-bit units.
The generator then creates an extra m-bit unit called the checksum, which is sent with the message.
At the destination, the checker creates a new checksum from the combination of the message and sent checksum.
If the new checksum is all 0s, the message is accepted; otherwise, the message is discarded

Question 87

Describe Forward Error Correction Process

Answer 87

The incoming signal is demodulated at the receiver to produce a bit string similar to the original codeword but may contain errors. This block is passed through an FEC decoder, with one of four possible outcomes:

No errors: If there are no bit errors, the input to the FEC decoder is identical to the original codeword, and the decoder produces the original data block as output.

Detectable, correctable errors: For certain error patterns, the decoder can detect and correct those errors. Thus, even though the incoming data block differs from the transmitted codeword, the FEC decoder can map this block into the original data block.

Detectable, not correctable errors: For certain error patterns, the decoder can detect but not correct the errors. In this case, the decoder simply reports an uncorrectable error.

Undetectable errors: typically rare error patterns, the decoder does not detect the error and maps the incoming n-bit data block into a k-bit block that differs from the original k-bit block.

Question 88

Describe piggybacking

Answer 88

piggybacking is a method to combine a data frame with an acknowledgment.

Piggybacking can save bandwidth because the overhead from a data frame and an ACK frame can be combined into just one frame.

Question 89

describe Bidirectional Transmission

Answer 89

we can have bidirectional transmission if the two parties have 2 separate channels for full-duplex transmission or share the same channel for half-duplex transmission

Question 90

HDLC defines three types of frames, describe them

Answer 90

I-frames are used to transport user data and control information relating to user data (piggybacking).
S-frames are used only to transport control information.
U-frames are reserved for system management.

Question 91

state the type of multiple-access protocols

Answer 91

Random-access protocols
Controlled-access protocols
Channelization protocols

Question 92

describe random-access

Answer 92

no station is superior to another station and
none is assigned control over another.
At each instance, a station with data to send uses a procedure defined by the protocol to decide whether to send.
This decision depends on the state of the medium (idle or busy).

Question 93

state the methods of random-access protocols

Answer 93

ALOHA
CSMA/CD
CSMA/CA

Question 94

describe the nonpersistent strategy in CSMA

Answer 94

a station senses the line. If the line is idle, the station sends the frame immediately.
If the line is not idle, the station waits for a random period of time and then senses the line again.

Question 95

describe the 1-persistent strategy in CSMA

Answer 95

In a persistent strategy, a station continuously senses the line. If the line is idle, the station sends a frame. This method has two variations: 1-persistent and p-persistent.
In the 1-persistent method, if the station finds the line idle, the station sends its frame immediately (with a probability of 1).

Question 96

describe the p-persistent strategy in CSMA

Answer 96

The p-persistent is used if the channel has time slots with a slot duration equal to or greater than the propagation time.
In this method, if the station finds the line idle, the station may or may not send it.
It sends with probability p and refrains from sending with probability (1-p).

Question 97

describe ALOHA

Answer 97

ALOHA transmits data whenever they have data to be sent.
No carrier sensing.
No collision detection
when the station sends data it waits for acknowledgment.
when data from the two stations collide and become garbled, retransmit the data after the collision

Question 98

State the Differences between ALOHA and CSMA/CD

Answer 98

In ALOHA, the station sends the frame without sensing it. In CSMA/CD, the station senses the channel before sending a frame.
In ALOHA, we first transmit the entire frame and then wait for an acknowledgment.
In CSMA/CD, transmission and collision detection are simultaneous processes.
In ALOHA, there is no collision detection. In CSMA/CD, the station sends a short jamming signal to make sure that all other stations become aware of the collision.