Consider the data transfer using TCP over a 1 Gbps link. Assuming that the maximum segment lifetime (MSL) is set to 60 seconds, the minimum number of bits required for the sequence number field of the TCP header, to prevent the sequence number space from wrapping around during the MSL is _________.

To prevent sequence number wrap-around during the Maximum Segment Lifetime (MSL), the total number of unique byte sequence numbers available must be at least the maximum number of bytes that can be transmitted in time equal to MSL.

1. Link Bandwidth ($R$):

   $$R = 1\text{ Gbps} = 10^9\text{ bps} = \frac{10^9}{8}\text{ Bytes/s} = 1.25 \times 10^8\text{ Bytes/s}$$

2. Maximum bytes transmitted in MSL ($T = 60\text{ s}$):

   $$\text{Total Bytes} = R \times \text{MSL} = \left(\frac{10^9}{8}\right) \times 60 = 7.5 \times 10^9\text{ Bytes}$$

3. Number of bits ($n$) required:

   $$2^n \ge 7.5 \times 10^9$$

   Since $2^{32} \approx 4.295 \times 10^9 < 7.5 \times 10^9$ and $2^{33} \approx 8.590 \times 10^9 \ge 7.5 \times 10^9$, we need:

   $$n = 33\text{ bits}$$