GATE CS 2014 Set 1 Q28 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To find the minimum number of bits required for the sequence number field in Selective Repeat (SR), we first calculate the transmission time ($T_t$) and the round-trip time ($RTT$).

1. Calculate Transmission Time ($T_t$):
Frame size, $L = 1 \text{ KB} = 1024 \times 8 \text{ bits} = 8192 \text{ bits}$
Bandwidth, $B = 1.5 \text{ Mbps} = 1.5 \times 10^6 \text{ bps}$
$$T_t = \frac{L}{B} = \frac{8192}{1.5 \times 10^6} \approx 5.46 \text{ ms}$$

2. Calculate Round-Trip Time ($RTT$):
One-way latency ($T_p$) = $50 \text{ ms}$
$$RTT = 2 \times T_p = 100 \text{ ms}$$

3. Determine Sender Window Size ($W_s$):
For a link utilization of $\eta = 60\% = 0.6$:
$$\eta = \frac{W_s \times T_t}{T_t + 2T_p} \ge 0.6$$ $$W_s \ge 0.6 \times \left(1 + \frac{2T_p}{T_t}\right) = 0.6 \times \left(1 + \frac{100}{5.46}\right) \approx 0.6 \times 19.31 = 11.59$$ Since the window size must be an integer, the minimum sender window size is $W_s = 12$.

4. Calculate Sequence Number Bits ($n$):
In the Selective Repeat protocol, to avoid ambiguity, the sender window size $W_s$ and receiver window size $W_r$ must satisfy:
$$W_s + W_r \le 2^n$$ Since $W_s = W_r$ in Selective Repeat:
$$2 W_s \le 2^n \implies 2^{n-1} \ge W_s$$ $$2^{n-1} \ge 12 \implies n - 1 \ge 4 \implies n \ge 5$$

Thus, the minimum number of bits required to represent the sequence number field is 5.