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 $10^8$ 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)

To determine the minimum number of bits $N$ for the sequence number field, we evaluate both constraints:

1. Condition (i): Sequence numbers do not wrap around before 60 seconds

• Bandwidth $B = 10^8\text{ bits/second} = \frac{10^8}{8} = 1.25 \times 10^7\text{ bytes/second}$.

• In $60\text{ seconds}$, the number of bytes transmitted is:

  $$\text{Total bytes} = 1.25 \times 10^7 \times 60 = 7.5 \times 10^8\text{ bytes}$$

• Since each byte is assigned a unique sequence number, the sequence number space must be at least $7.5 \times 10^8$ to prevent wrap-around within 60 seconds:

  $$2^{N_1} \ge 7.5 \times 10^8$$

  $$N_1 = \lceil \log_2(7.5 \times 10^8) \rceil = 30\text{ bits}$$

  *(Since $2^{29} = 536,870,912 < 7.5 \times 10^8$ and $2^{30} = 1,073,741,824 > 7.5 \times 10^8$)*

2. Condition (ii): Maximum link utilization is achieved

• Link length $d = 3000\text{ km} = 3 \times 10^6\text{ m}$.

• Propagation delay $T_p = 3 \times 10^6\text{ m} \times 5 \times 10^{-9}\text{ s/m} = 15 \times 10^{-3}\text{ s} = 15\text{ ms}$.

• Round Trip Time $\text{RTT} = 2 \times T_p = 30\text{ ms} = 0.03\text{ s}$.

• Bandwidth-Delay Product (BDP) in bytes:

  $$\text{BDP} = B \times \text{RTT} = 1.25 \times 10^7 \times 0.03 = 3.75 \times 10^5\text{ bytes}$$

• To achieve maximum utilization, the sequence number space must support at least the window size (even for Selective Repeat where sequence space is $2 \times \text{Window}$, $2 \times 3.75 \times 10^5 = 7.5 \times 10^5$, which requires $\lceil \log_2(7.5 \times 10^5) \rceil = 20\text{ bits}$).

Taking the minimum bits that satisfy both conditions: