GATE CS 2026 Set 2 — Question 65
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Computer Networks → Data Link Layer → Efficiency Numericals
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Question
It is necessary to design a link-layer protocol between two hosts that are directly connected over a lossless link of length 3000 kilometers. Assume that the link bandwidth is bits per second and that the propagation delay in the link is 5 nanoseconds per meter. Every transmitted data byte is assigned a unique sequence number.Let be the minimum number of bits needed for the sequence number field in the protocol header such thati. the sequence numbers do not wrap around before 60 seconds, and
ii. the maximum utilization of the link is achieved.The value of is ______. (answer in integer)
ii. the maximum utilization of the link is achieved.The value of is ______. (answer in integer)
Correct answer
30 to 30
Solution
We need to determine the minimum satisfying two conditions.Condition i: No wrap around before 60 seconds
Bandwidth bits/sec.
Since sequence numbers are assigned per byte, the byte rate is:Total bytes transmitted in 60 seconds:The sequence number space must be at least this size:Calculating powers of 2:
(too small)
(sufficient)
So, .Condition ii: Maximum utilization is achieved
For maximum utilization, the window size must cover the Bandwidth-Delay Product (BDP).
Propagation delay .
Round Trip Time .
BDP in bytes:The sequence number space must be larger than the window size (typically or ).
. This requires bits.Conclusion
The constraint from condition (i) () dominates the constraint from condition (ii) ().
Thus, the minimum value is .
Bandwidth bits/sec.
Since sequence numbers are assigned per byte, the byte rate is:Total bytes transmitted in 60 seconds:The sequence number space must be at least this size:Calculating powers of 2:
(too small)
(sufficient)
So, .Condition ii: Maximum utilization is achieved
For maximum utilization, the window size must cover the Bandwidth-Delay Product (BDP).
Propagation delay .
Round Trip Time .
BDP in bytes:The sequence number space must be larger than the window size (typically or ).
. This requires bits.Conclusion
The constraint from condition (i) () dominates the constraint from condition (ii) ().
Thus, the minimum value is .
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