TCP MCQ Quiz in தமிழ் - Objective Question with Answer for TCP - இலவச PDF ஐப் பதிவிறக்கவும்

RFC 1122

Application Layer of TCP/IP corresponds to the Application layer, Session layer, the Presentation layer of OSI.

Therefore, the TCP/IP model does not have Session and  Presentation layers but the Application layer includes the required functions of these layers.

Concept:

In a sliding window protocol,

window size of sender = W_s = 2⁶ bits = 64

window size of sender = window size of receiver

Calculation:

∴ Ws +Wr   =  n

Ws = Wr

∴ 2Ws    =  n 

∴ n = 2 × 64  =  128

1st frame → 0 (sequence number)

128th frame → 127

129th frame → 0

180th frame → 51

The correct answer is Fregmentation offset

Key Points

• Fragmentation Offset:
  • This is not a field in the TCP header; rather, it is a field in the IP header. The IP protocol allows large packets to be fragmented into smaller fragments for transmission over networks with smaller Maximum Transmission Unit (MTU) sizes.
  • The fragmentation offset field in the IP header indicates the position of a fragment in the original, unfragmented packet.

Additional Information

• Sequence Number:
  • This field in the TCP header is used to identify the order of the bytes sent in a TCP connection. It is crucial for ensuring that the data is received in the correct order and for handling retransmissions.
• Checksum:
  • The checksum field in the TCP header is used for error-checking purposes. It helps verify the integrity of the TCP header and data. The sender calculates the checksum based on the contents of the TCP header and data, and the receiver verifies it to detect any transmission errors.
• Window Size:
  • The window size field in the TCP header is used to manage flow control in a TCP connection. It indicates the amount of data, in bytes, that can be sent by the sender before an acknowledgment is required from the receiver. Adjusting the window size allows for efficient data transfer and congestion control.

Key Points

 Option 1: FTP is primarily designed for large file transfers with the ability to resume downloads after a download has been interrupted.

True, FTP is a TCP/IP protocol that transfers files between FTP servers and clients. FTP was designed specifically for transferring large files. All you need to do to start using the protocol is an FTP client. The good thing about FTP is that what it lacks in security it makes up for with its file management capabilities. File transfer protocol is a way to download, upload, and transfer files from one location to another on the internet and between computer systems.

Option 2: TCP establishes a connection at both ends before any data can flow.

True, TCP uses a three-way handshake to establish a reliable connection. The connection is full-duplex, and both sides synchronize (SYN) and acknowledge (ACK) each other. It establishes a connection at both ends before any data can flow. The exchange of these four flags is performed in three steps—SYN, SYN-ACK, and ACK.

Hence the correct answer is (i) - True, (ii) - True.

Answer: Option 4

Concept: 

• In TCP, for Connection Establishment 3-way handshake protocol is used.
• FIN bit : FIN bit is set while terminating the connection.
• Syn bit : Syn bit is used for initiating the request for connection establishment.
• Ack bit : Ack bit is used for indicating that the segment contains the acknowledgment.
• Sequence Number : In TCP , Sequence number used for counting each byte transferred in a particular connection and in each Segment Header, Sequence number field contains the Sequence number of first byte of data part of the Segment.
• Ack Number: In TCP , Ack number indicates the Sequence number of Byte expected next.

Explanation:

GIven: 

Next segment will be a piggybacked Acknowledgement segment So Ack bit = 1 and  Q will increment the Sequence number ( which it received from P) put this into Ack Number field. Syn = 1 as Q will also establish the connection from Q to P and Sequence number will be Y(As given in Question).

So , SYN bit = 1 SEQ Number = Y ACK bit = 1 ACK number = X+1 FIN bit = 0. Matches with Option 4.​

The correct answer is Network layer.

Important Points

Data:

TCP message contains 1024 bytes of data

TCP header = 20 bytes

Total TCP segment length = 1024 + 20 = 1044 Bytes

Payload length from 0 to 1043.

CASE1:

First network uses MTU of 512 bytes.

IP payload = 512 – 20 = 492 bytes

492 is not a multiple of 8. So, we will take 488.

So, first fragment will go from 0 to 487

Second from 488 to 975.

Third fragment from 976 to 1043.

Since MTU of other networks is less it is the bottleneck.

CASE 2:

MTU size = 256 bytes

Payload length = 256 – 20 = 236 bytes

We will consider 232, because 236 is not a multiple of 8.

1^st fragment from 0 to 231

2^nd fragment from 232 to 463

3^rd fragment from 464 to 695

4^th fragment from 696 to 927

5^th fragment from 928 to 1043

So, total 5 fragments are required.

4 Gbps → 1 second

\(1b \to \frac{1}{{4 \times {{10}^9}\;}}\;seconds\)

\(\frac{1}{8}\;B \to \frac{1}{{500 \times {{10}^6}\;}}\;seconds\)

\(1 \; B \rightarrow \frac{8}{{4 \times {{10}^9}\;}}\;seconds\)

Sequence number = 32 bits

\({2^{32}}\;B \rightarrow \frac{8 \times 2^{32}}{{4 \times {{10}^9}\;}}\;seconds\)

\({2^{32}}B \to 8.5899\;seconds\)

The minimum time before this sequence number can be used again is 8.5899  ≈ 9 seconds

In the OSI-reference model, the networking system is decided into 7 layers.

Each layer has its own data unit in which they communicate.

BIND:

bind () function call binds a unique local name to the socket with descriptor socket. After calling socket (), a descriptor does not have a name associated with it.

LISTEN:

It causes a bound TCP socket to enter the listening state. It waits for the connections to socket.

ACCEPT:

It accepts a connection on a socket.

RECV:

It is used to read incoming data on connection-oriented sockets or connectionless sockets.