Question

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 \(\underline{\hspace{1cm}}\).

Hint:

The sequence number space should be large enough to accommodate all the data transferred during the MSL, ensuring that the sequence numbers do not wrap around.

Answer 1

Correct Answer: 33

We are given the following data: - Link speed = 1 Gbps = \( 10^9 \, \text{bits/sec} \), - MSL = 60 seconds. The total number of bits that can be transferred in 60 seconds is: \[ \text{Total bits} = \text{Link speed} \times \text{MSL} = 10^9 \times 60 = 6 \times 10^{10} \, \text{bits}. \] To prevent the sequence number space from wrapping around during this time, the number of unique sequence numbers must be at least equal to the total number of bits transferred: \[ \text{Number of sequence numbers} \geq 6 \times 10^{10}. \] The sequence number field in the TCP header is 32 bits. The maximum number of sequence numbers that can be represented by 32 bits is: \[ 2^{32} = 4.294 \times 10^9. \] Now, the minimum number of bits required for the sequence number field can be calculated: \[ \text{Required bits} = \lceil \log_2(6 \times 10^{10}) \rceil. \] \[ \log_2(6 \times 10^{10}) = \log_2(6) + \log_2(10^{10}) = 2.585 + 33.219 = 35.804. \] Thus, the minimum number of bits required is: \[ \boxed{33 \, \text{bits}}. \]