GATE CS 2015 Set 3 Q28 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To utilize the network to its full capacity, the sender must transmit continuously without waiting for acknowledgments. This requires a sender window size ($W_s$) that can cover the entire round-trip time.

Step 1: Calculate Propagation Delay ($T_p$)
$$\text{Distance } (d) = 8000 \text{ Km} = 8 \times 10^6 \text{ m}$$
$$\text{Propagation Speed } (v) = 4 \times 10^6 \text{ m/s}$$
$$T_p = \frac{d}{v} = \frac{8 \times 10^6}{4 \times 10^6} = 2 \text{ seconds}$$

Step 2: Calculate Transmission Delay ($T_t$)
$$\text{Packet Size } (L) = 10^7 \text{ bits}$$
$$\text{Bandwidth } (B) = 500 \times 10^6 \text{ bps}$$
$$T_t = \frac{L}{B} = \frac{10^7}{500 \times 10^6} = 0.02 \text{ seconds}$$

Step 3: Calculate Sender Window Size ($W_s$)
For $100\%$ efficiency in Go-Back-$N$ (GBN):
$$W_s \ge 1 + 2a = 1 + \frac{2 T_p}{T_t}$$
$$W_s \ge 1 + \frac{2 \times 2}{0.02} = 1 + 200 = 201$$

Step 4: Calculate Minimum Sequence Numbers and Bits
In Go-Back-$N$, the total number of sequence numbers required is:
$$N_{\text{seq}} \ge W_s + 1 = 201 + 1 = 202$$
Let $k$ be the number of bits in the sequence number field:
$$2^k \ge 202 \implies k = \lceil \log_2(202) \rceil = 8 \text{ bits}$$