1
GATE CSE 2024 Set 2
Numerical
+2
-0.66

Consider a TCP connection operating at a point of time with the congestion window of size 12 MSS (Maximum Segment Size), when a timeout occurs due to packet loss. Assuming that all the segments transmitted in the next two RTTs (Round Trip Time) are acknowledged correctly, the congestion window size (in MSS) during the third RTT will be _________

Your input ____
2
GATE CSE 2023
MCQ (More than One Correct Answer)
+2
-0.67

Suppose you are asked to design a new reliable byte-stream transport protocol like TCP. This protocol, named myTCP, runs over a 100 Mbps network with Round Trip Time of 150 milliseconds and the maximum segment lifetime of 2 minutes.

Which of the following is/are valid lengths of the Sequence Number field in the myTCP header?

A
30 bits
B
32 bits
C
34 bits
D
36 bits
3
GATE CSE 2021 Set 1
MCQ (More than One Correct Answer)
+2
-0.67

Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes, and between R and Q is 820 bytes.

A TCP segment of size 1400 bytes was transferred from P to Q through R, with IP identification value as 0 × 1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented i. e, Don't Fragment (DF) flag in the IP header is not set by P.

Which of the following statements is/are correct?

A
TCP destination port can be determined by analysing only the second fragement.
B
If the second fragment is lost, P is required to resend the whole TCP segment.
C
Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.
D
If the second fragment is lost, R will resend the fragment with the IP identification value 0 × 1234.
4
GATE CSE 2021 Set 1
Numerical
+2
-0.67

Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following:

1. The time taken for processing the data frame by the receiver is negligible.

2. The time taken for processing the acknowledgement frame by the sender is negligible.

3. The sender has infinite number of frames available for transmission.

4. The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits.

5. The link data rate in each direction is 1 Mbps (= 106 bits per second).

6. One way propagation delay of the link is 100 milliseconds.

The minimum value of the sender's window size in terms of the number of frames, (rounded to the nearest integer) needed to achieve a link utilization of 50% is ______

Your input ____
GATE CSE Subjects
Software Engineering
Web Technologies
EXAM MAP
Medical
NEET
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
CBSE
Class 12