Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

GATE | GATE-CS-2016 (Set 2) | Question 65

  • Last Updated : 05 Dec, 2018

Consider a 128×103 bits/second satellite communication link with one-way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgment. The minimum number of bits required for the sequence number field to achieve 100% utilization is ___________
(A) 2
(B) 4
(C) 6
(D) 8

Answer: (B)

Explanation: To achieve 100% efficiency, the number of frames that we should send = 1 + 2 * Tp / Tt
where, Tp is propagation delay, and Tt is transmission delay.
So, Number of frames sent = 1 + 4.687 = 5.687 (approx 6)
As the protocol used is Selective repeat, Receiver window size should be equal to Sender window size.
Then, Number of distinct sequence numbers required = 6 + 6 = 12
Number of bits required to represent 12 distinct numbers = 4
So, Answer is (B)

Quiz of this Question

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!