GATE CS 2022 Q50 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To prevent the TCP sequence number space from wrapping around during the Maximum Segment Lifetime ($\text{MSL}$), the total number of unique sequence numbers available must be at least equal to the maximum number of bytes that can be transmitted during the $\text{MSL}$.

Step 1: Calculate the data transmission rate in bytes per second
In computer networks, bandwidth is measured in powers of $10$.
$$\text{Bandwidth} = 1 \text{ Gbps} = 10^9 \text{ bits/second}$$ Since TCP sequence numbers are assigned per byte (octet): $$\text{Transmission Rate} = \frac{10^9 \text{ bits/second}}{8 \text{ bits/byte}} = 1.25 \times 10^8 \text{ bytes/second}$$

Step 2: Calculate the total bytes transmitted during the MSL
Given $\text{MSL} = 60 \text{ seconds}$: $$\text{Total Bytes} = 1.25 \times 10^8 \text{ bytes/second} \times 60 \text{ seconds} = 7.5 \times 10^9 \text{ bytes}$$

Step 3: Determine the minimum number of bits ($n$) for the sequence number
We need the sequence number space $2^n$ to be at least the total bytes transmitted: $$2^n \ge 7.5 \times 10^9$$ Let's evaluate powers of $2$:

• For $n = 32$: $2^{32} \approx 4.29 \times 10^9$ (Insufficient)
• For $n = 33$: $2^{33} \approx 8.59 \times 10^9$ (Sufficient)