GATE CS 2016 Set 2 Q55 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To achieve $100\%$ utilization, the sender must be able to transmit continuously during the round-trip time. This requires a minimum sender window size ($W_s$).

1. Calculate Transmission Delay ($T_t$):
Using $1 \text{ kilobyte} = 1024 \text{ bytes}$:
$$T_t = \frac{\text{Frame Size}}{\text{Bandwidth}} = \frac{1024 \times 8 \text{ bits}}{128 \times 10^3 \text{ bps}} = 64 \text{ ms}$$

2. Calculate Minimum Sender Window Size ($W_s$):
With propagation delay $T_p = 150 \text{ ms}$:
$$W_s \ge \frac{T_t + 2T_p}{T_t} = 1 + \frac{300}{64} = 5.6875$$
Since the window size must be an integer, the minimum $W_s = 6$.

3. Calculate Sequence Number Bits ($n$):
In the Selective Repeat protocol, the sequence number space size ($N$) must satisfy:
$$N \ge W_s + W_r$$
Since $W_s = W_r$ in Selective Repeat:
$$N \ge 2 W_s = 12$$
The minimum number of bits $n$ required is:
$$n = \lceil \log_2(12) \rceil = 4$$