It is necessary to design a link-layer protocol between two hosts that are…
2026
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 108 bits per second and that the propagation delay in the link is 5 nanoseconds per meter (that is, a signal takes 5 ns to travel each meter of the link). Every transmitted data byte is assigned a unique sequence number.
Let 𝑁 be the minimum number of bits needed for the sequence number field in the protocol header such that
i. the sequence numbers do not wrap around before 60 seconds, and
ii. the maximum utilization of the link is achieved.
The value of 𝑁 is ______. (answer in integer)
Answer: 30 — ConceptFor a sliding-window / Go-Back-N link protocol where every byte gets its own sequence number, the minimum sequence-number width N must satisfy TWO…
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Correct answer: 30
Concept
For a sliding-window / Go-Back-N link protocol where every byte gets its own sequence number, the minimum sequence-number width N must satisfy TWO independent constraints at once:
No-wrap-around: the total distinct sequence numbers, 2N, must exceed the total bytes transmitted during the required no-wrap interval — Bandwidth × Time / 8.
Maximum utilization: the sequence-number space must be large enough to number every byte in flight during one round-trip time, i.e. 2N must cover the bandwidth-delay product — Bandwidth × RTT / 8, where RTT = 2 × propagation delay.
Since both conditions must hold simultaneously, N is the LARGER of the two individually-required bit counts.
Application
1. Link parameters
Link Length (L) = 3000 km = 3,000,000 meters
Propagation delay per metre (v) = 5 ns/m = 5 × 10-9 s/m
Propagation Delay (T_prop) = L × v = 3,000,000 × 5 × 10-9 = 0.015 seconds
Bandwidth (B) = 108 bits per second
2. Maximum-utilization condition
Round Trip Time (RTT) = 2 × T_prop = 2 × 0.015 = 0.03 seconds
Bandwidth-Delay Product (in bits) = B × RTT = 108 × 0.03 = 3,000,000 bits
Bandwidth-Delay Product (in bytes) = 3,000,000 / 8 = 375,000 bytes
So the window size W must be at least 375,000 bytes, meaning the sequence-number space (2N − 1) must be at least 375,000.
219 = 524,288, which already exceeds 375,000 — so this condition alone needs only N = 19.
3. No-wrap-around condition
Total bytes transmitted in 60 seconds = (Bandwidth × Time) / 8 = (108 × 60) / 8 = 6,000,000,000 / 8 = 750,000,000 bytes
So we need 2N > 750,000,000.
229 = 536,870,912 (too small)
230 = 1,073,741,824 (greater than 750,000,000)
So this condition needs N = 30.
4. Combine the two conditions
N must satisfy BOTH conditions, so N = max(19, 30) = 30. The wrap-around requirement is the binding (dominant) constraint.
Cross-check
Verify N = 30 independently satisfies both requirements: 230 ≈ 1.074 × 109 bytes of sequence space, which is (a) greater than the 750,000,000 bytes sent in 60 seconds — no wrap — and (b) far greater than the 375,000-byte window needed for full utilization. A smaller value, N = 29, gives only 229 = 536,870,912 sequence numbers, which is less than 750,000,000 and would wrap around before 60 seconds — confirming 30 is the minimum.
Final Answer
The minimum number of bits N is 30.