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. 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)
Attempted by 7 students.
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Correct answer: 30
Step-by-Step Solution
To find the minimum number of bits (N) for the sequence number field, we must satisfy two conditions: preventing wrap-around within 60 seconds and achieving maximum link utilization.
1. Calculate Link Parameters
First, determine the propagation delay and transmission time for a single bit or byte.
Link Length (L) = 3000 km = 3,000,000 meters
Propagation Speed (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
Since every byte (8 bits) gets a unique sequence number, the transmission time for one byte (T_trans) is:
T_trans = 8 bits / 10⁸ bps = 8 × 10⁻⁸ seconds
2. Condition for Maximum Utilization
To achieve maximum utilization (100%), the sender must be able to keep the pipe full. This requires the window size (W) to be at least the bandwidth-delay product in terms of bytes.
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
Thus, the window size W must be at least 375,000 bytes.
3. Condition for No Wrap-Around
The sequence numbers must not wrap around before 60 seconds. This means the total number of unique sequence numbers available (2^N) must be greater than the total number of bytes transmitted in 60 seconds.
Total bytes transmitted in 60 seconds = (Bandwidth × Time) / 8
Total bytes = (10⁸ bits/sec × 60 sec) / 8 bits/byte = 6,000,000,000 / 8 = 750,000,000 bytes
Therefore, 2^N > 750,000,000
4. Determine Minimum N
We need to find the smallest integer N such that 2^N > 750,000,000.
Let's check powers of 2:
2^29 = 536,870,912 (Too small)
2^30 = 1,073,741,824 (Greater than 750,000,000)
The condition 2^N > 750,000,000 is satisfied when N = 30.
Note: The window size requirement (375,000) is much smaller than the total sequence space required (750,000,000), so the wrap-around constraint is the dominant factor.
Final Answer
The minimum number of bits N is 30.