Network Layer MCQ Quiz in मल्याळम - Objective Question with Answer for Network Layer - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

In class B, 16 bits are reserved for Network id and 16 for host id.

Explanation

6 bit is added due to subnetting

Number of subnest = 26 − 2= 62

Total Network Id = 16 + 6 = 22

Network Id + Host Id = 32 bits

Number of bit in host id = h = 32 - 22 = 10 bits

The number of available hosts = 2h − 2,

= 210 - 2 = 1024 - 2 = 1022

Therefore option 1 is the correct answer

2 is subtracted because subnet values consisting of all zeros and all ones (broadcast), reducing the number of available subnets by two in classic subnetting. In modern networks. Those two are counted leading to 64 subnets.

Answer: Option 2

Concept: 

In circuit-switched networks, the resources needed along a path (buffers, link transmission rate) to provide for communication between the end systems are reserved for the duration of the communication session between the end systems.

In packet-switched networks, these resources are not reserved, rather Simply in each packet mark the source address and destination and throw them in the network. These packets might not follow the same path.

ARPANET: 

The Advanced Research Projects Agency Network was established by the Advanced Research Projects Agency (ARPA) of the United States Department of Defense.

It was the first wide-area packet-switching network with distributed control.
It is one of the first networks to implement the TCP/IP protocol suite.

File size (data) = 5000 byte

Header size = 100 byte

Bandwidth (BW) = 2 × 10⁶ byte/second

First case,

\(Transmission\;time\left( {{T_t}} \right) = \frac{{data\; + \;header}}{{BW}}\;\)

\({T_t}\; = \;\frac{{5000 + 100}}{{2\; \times \;{{10}^6}}}\; = \;2.55\;ms\)

A → R1 → R2 → R3 → B

4 hops are needed

T1 = 4 × 2.55 ms

∴ T1 = 10.2 ms

Second case,

Data per packet = 5000 ÷ 20 = 250 B

Header size = 100 B

\(Transmission\;time\left( {{T_t}} \right) = \frac{{data\; + \;header}}{{BW}}\;\)

\({T_t}\; = \;\frac{{250 + 100}}{{2\; \times \;{{10}^6}}} = 0.175\;ms\)

X → R1 → R2 → R3 → Y

4 hops are needed

∴ time needed for 1^st packet = 4 × 0.175 ms = 0.7 ms

And time needed for remaining 19 packets = 19 × 0.175 = 3.325 ms

∴ T2 = 0.7 + 3.325 = 4.025 ms

Third case,

Data per packet = 5000 ÷ 40 = 125 B

Header size = 100 B

\(Transmission\;time\left( {{T_t}} \right)\; = \;\frac{{data\; + \;header}}{{BW}}\;\)

\({T_t} = \frac{{125 + 100}}{{2\; \times \;{{10}^6}}} = 0.1125\;ms\)

X → R1 → R2 → R3F → Y

4 hops are needed

∴ time needed for 1^st packet = 4 × 0.1125 ms = 0.45 ms

And time needed for remaining 19 packets = 39 × 0.1125 = 4.3875 ms

The correct answer is Ethernet networks

Key Points

• Ethernet networks:
  • In Ethernet networks, the MAC (Medium Access Control) protocol used is known as CSMA/CD (Carrier Sense Multiple Access with Collision Detection). This protocol employs a contention-based approach to access the shared communication medium.
  • Devices on the network listen for the presence of a carrier signal (carrier sense) before attempting to transmit data. If the medium is busy, they wait for a random amount of time before retrying.
  • In the case of a collision (when two devices transmit simultaneously and their signals overlap), CSMA/CD facilitates collision detection. Colliding devices stop transmitting, initiate a backoff period, and then retry.
  • So, contention-based MAC protocols, such as CSMA/CD, find common usage in Ethernet networks where multiple devices contend for access to the shared communication medium.

Additional Information

• Token Ring networks:
  • Token Ring networks, on the other hand, use a different approach. They utilize a token-passing protocol where a special token circulates around the network. Only the device possessing the token is allowed to transmit.
  • This is in contrast to contention-based protocols, as devices don't have to contend for access. The token is passed sequentially, ensuring orderly and fair access to the network.
• ATM networks:
  • ATM (Asynchronous Transfer Mode) networks typically use a combination of contention and reservation protocols. While some aspects may involve contention-based access, others may use reservation-based mechanisms to allocate bandwidth for specific connections.
• Point-to-Point networks:
  • In Point-to-Point networks, the need for contention is minimal as there are only two endpoints involved. The communication link is generally dedicated, and contention-based protocols like CSMA/CD are not commonly used in this context. Point-to-Point links often use protocols optimized for point-to-point communication without the need for contention resolution.

File size (data) = 1500 byte

Header size = 100 byte

Bandwidth (BW) = 10⁷ byte/second

First case,

\(Transmission\;time\left( {{T_t}} \right) = \frac{{data\; + \;header}}{{BW}}\;\)

\({T_t} = \frac{{1500 + 100}}{{{{10}^7}}} = 160\;\mu s\)

X → R1 → R2 → R3 → R4 → Y

5 hops are needed

∴ total time needed = 5 × 160 μs = 800 μs

Second case,

Data per packet = 1500 ÷ 15 = 100 B

Header size = 100 B

\(Transmission\;time\left( {{T_t}} \right) = \frac{{data\; + \;header}}{{BW}}\;\)

\({T_t}\; = \;\frac{{100 + 100}}{{{{10}^7}}}\; = \;20\;\mu s\)

X → R1 → R2 → R3 → R4 → Y

5 hops are needed

∴ time needed for 1^st packet = 5 × 20 μs = 100 μs

And time needed for remaining 14^th packet = 14 × 20 μs = 280 μs

∴ total time = 100 + 280 = 380 μs

The correct answer is option 4.

Concept:

A single cable links all of the included nodes in a bus topology. The primary cable serves as the network's backbone. If the common cable breaks, the entire system will be brought to a halt.

Every device in a ring network has exactly two neighbors for communication purposes. It's called ring topology because it's shaped like a ring. Every computer in this topology is linked to another computer. A single computer failure can disrupt the entire network.

As a result, bus and ring topologies have difficulties, such as the possibility of the entire network being deactivated if one connection fails.

Hence the correct answer is Bus and Ring Topology.

Additional InformationStar Topology: 

All of the computers in a star topology communicate through a hub. This cable is known as a central node, and it is used to link all other nodes. The attached nodes are deactivated if the hub or concentrator fails.

Mesh Topology:

A mesh topology is a network configuration in which each computer and network device is linked to the others. Most communications can be dispersed using this topology even if one of the links fails. It is a wireless network topology that is extensively used.

Answer: Option 4

Concept: 

Classful Addressing : 

In class A: 

The First byte should be in the range of 0 -126 

The First byte only contains network part and (Second, Third, Fourth) byte will be the host part.

In class B:

The First byte should be in the range of 128 - 191.

Network part includes (First byte and Second byte) and host part includes (Third and Fourth byte).

In class C: 

The First byte should be in the range of 192 - 223.

Network part includes (First byte, Second byte, Third byte) and host part includes (Third and Fourth byte).

In class D: 

The First byte should be in the range of 192 - 223.

These are addresses are used for multicast networking.

In class E: 

These addresses are reserved for Future and experimental purposes.

The process of modifying IP address information in IP packet headers while in transit across a traffic routing device is called Network address translation (NAT).

CONCEPT:

Private Network uses the private IP address and public IP address is used to access the Internet.

Network Address Translation(NAT) is a process to translate one or more Private IP address into one or more public IP address so that localhost can access the Internet.

The idea of NAT is to allow multiple devices to access the Internet through a single public IP address.

Hence, the correct answer is "option 2".

Additional Information

Option 1:

Port Address Translation(PAT) translates the Private IP address into Public IP address using port numbers.

More than one device can use a single IP address to access the Internet using different port numbers.

Option 3:

Address mapping is used to map the IP address into different forms.

Option 4:

Port mapping is the process of redirecting the communication request from one IP address and port number combination to another IP address and port number combination.

Three types of NAT are:

1. Static NAT: In static NAT, a single private IP address is translated into a single public IP address.

2. Dynamic NAT: In dynamic NAT, multiple IP addresses are mapped to a pool of public IP addresses.

3. Port Address Translation(PAT): PAT is also called NAT overload.

ICMP error message should NOT be sent after receiving when:

• After another ICMP error message
• After an IGMP message of any kind.
• After any fragment other than the first fragment from a fragmented IP datagram

ICMP error message should be sent after receiving when:

• If DF( do not fragment) bit is set and a router needs to fragment the datagram to send it further then the ICMP error message must be sent to the sender.
• If TTL is exceeded then also it should report to the sender by sending the ICMP error message.

Options 1 and 3 are the correct answer.

The correct answer is (i) False, (ii) False.

Key Points

• HTTP stands for Hyper Text Transfer Protocol.
  • It is the foundation of the World Wide Web.
  • It is an application layer protocol designed to transfer information between networked devices.
  • It runs on top of other layers of the network protocol stack.
• HTTP is not a markup language. Hence, statement 1 is false.
• A valid IP address must be in the form of A.B.C.D, where A, B, C and D are numbers from 0-255.
  • A valid IP address must be in the form of xxx.xxx.xxx.xxx.
• Hence, statement 2 is false.

Additional Information

• An HTTP request is a way internet communications platforms such as web browsers ask for the information they need to load a website.
• Each HTTP request made across the Internet carries with it a series of encoded data that carries different types of information.
• A typical HTTP request contains:
  • HTTP version type
  • a URL
  • an HTTP method
  • HTTP request headers
  • Optional HTTP body