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GATE CS 2018 Q25 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Consider a long-lived TCP\text{TCP} session with an end-to-end bandwidth of 1  Gbps  (=1091\; \text{Gbps}\; (=10^9 bits-per-second).). The session starts with a sequence number of 12341234. The minimum time (in seconds, rounded to the closet integer) before this sequence number can be used again is _________.
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Answer: 34 : 35

To find the minimum time before a TCP sequence number can be reused (wrap-around time), we need to calculate how long it takes to consume the entire TCP sequence number space.

1. TCP Sequence Number Space

TCP sequence numbers are represented using a 32-bit field. Since TCP assigns a sequence number to every byte of data transmitted, the total sequence number space is:

Total Sequence Space=232 bytes=4,294,967,296 bytes\text{Total Sequence Space} = 2^{32} \text{ bytes} = 4,294,967,296 \text{ bytes}

2. Transmission Rate

The given bandwidth is 1 Gbps=109 bits per second1\text{ Gbps} = 10^9\text{ bits per second}.

Converting this rate into bytes per second:

Bandwidth in bytes/sec=1098 bytes/sec=1.25×108 bytes/sec\text{Bandwidth in bytes/sec} = \frac{10^9}{8} \text{ bytes/sec} = 1.25 \times 10^8 \text{ bytes/sec}

3. Wrap-around Time Calculation

The time TT required to consume the entire 2322^{32} sequence space is:

T=Total Sequence SpaceBandwidth in bytes/sec=2321.25×108T = \frac{\text{Total Sequence Space}}{\text{Bandwidth in bytes/sec}} = \frac{2^{32}}{1.25 \times 10^8} T=4,294,967,296125,000,000≈34.36 secondsT = \frac{4,294,967,296}{125,000,000} \approx 34.36 \text{ seconds}

4. Rounding

  • Rounding 34.3634.36 to the closest integer gives 3434.
  • If we approximate 232≈4.3×1092^{32} \approx 4.3 \times 10^9 or use other minor approximations, the result can be slightly higher, which is why the range 34 to 35 is accepted.

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