Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is _________.

Given:

• Frame size $L = 1\text{ KB} = 1024 \times 8\text{ bits} = 8192\text{ bits}$ (or $1000 \times 8 = 8000\text{ bits}$)

• Bandwidth $B = 1.5\text{ Mbps} = 1.5 \times 10^6\text{ bps}$

• Propagation delay $T_p = 50\text{ ms} = 50 \times 10^{-3}\text{ s}$

• Utilization $\eta = 60\% = 0.6$

1. Transmission time ($T_t$):

   $$T_t = \frac{L}{B} = \frac{8192}{1.5 \times 10^6} \approx 5.461\text{ ms}$$

   (Using $1\text{ KB} = 8000\text{ bits}$, $T_t = \frac{8000}{1.5 \times 10^6} \approx 5.333\text{ ms}$)

2. Parameter $a$:

   $$a = \frac{T_p}{T_t} = \frac{50}{5.461} \approx 9.155$$

3. Link utilization in sliding window:

   $$\eta = \frac{W_S}{1 + 2a} \ge 0.60$$

   $$W_S \ge 0.60 \times (1 + 2 \times 9.155) = 0.60 \times 19.31 = 11.586 \implies W_S = 12$$

4. Sequence numbers for Selective Repeat: For Selective Repeat, the sender window size is $W_S = W_R = N$. The minimum number of sequence numbers required is:

   $$\text{Total sequence numbers } M \ge W_S + W_R = 2N = 2 \times 12 = 24$$

5. Number of bits ($k$):

   $$2^k \ge 24 \implies k = \lceil \log_2(24) \rceil = 5$$

Thus, the minimum number of bits required is 5.