Consider a network connecting two systems located 8000 kilometers apart. The bandwidth of the network is $500 \times 10^6$ bits per second. The propagation speed of the media is $4 \times 10^6$ meters per second. It is needed to design a Go-Back-$N$ sliding window protocol for this network. The average packet size is $10^7$ bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be ________.

To utilize the network to its full capacity, the sender's window size $N$ in a sliding window protocol must satisfy:

where $a = \frac{T_p}{T_t}$.

1. Propagation time ($T_p$):

   $$\text{Distance } d = 8000\text{ km} = 8 \times 10^6\text{ m}$$

   $$\text{Propagation speed } v = 4 \times 10^6\text{ m/s}$$

   $$T_p = \frac{d}{v} = \frac{8 \times 10^6}{4 \times 10^6} = 2\text{ seconds}$$

2. Transmission time ($T_t$):

   $$\text{Packet size } L = 10^7\text{ bits}$$

   $$\text{Bandwidth } B = 500 \times 10^6\text{ bps}$$

   $$T_t = \frac{L}{B} = \frac{10^7}{500 \times 10^6} = \frac{1}{50} = 0.02\text{ seconds}$$

3. Ratio $a$ and Window Size ($W_s$):

   $$a = \frac{T_p}{T_t} = \frac{2}{0.02} = 100$$

   $$W_s \ge 1 + 2a = 1 + 2(100) = 201$$

4. Sequence Number Field Size ($k$ bits): In a Go-Back-$N$ protocol, the maximum window size allowed with $k$ bits is $2^k - 1$.

   $$2^k - 1 \ge W_s = 201$$

   $$2^k \ge 202$$

   Since $2^7 = 128 < 202$ and $2^8 = 256 \ge 202$, the minimum number of bits required in the sequence number field is 8.