Question:medium

A new reliable byte-stream protocol (myTCP) runs over a 100 Mbps network with RTT = 150 ms and maximum segment lifetime (MSL) = 2 minutes. Which of the following are valid sequence number field lengths?

Show Hint

To design a sequence number field, compute the bandwidth–delay product over the MSL, then double it for wraparound safety. The field size must be the smallest power of 2 that exceeds this value.
Updated On: Feb 3, 2026
  • 30 bits
  • 32 bits
  • 34 bits
  • 36 bits
Show Solution

The Correct Option is B

Solution and Explanation

To determine the valid sequence number field lengths for the given network conditions in the question, we need to adhere to the concept of the sequence number space required by the TCP-like protocols. The sequence numbers must be large enough to ensure that data segments can be uniquely identified within the Maximum Segment Lifetime (MSL).

Key Formulas and Concepts:

  • The sequence number space should cover the amount of data sent over the network during the segment lifetime.
  • The calculation needs to fulfill the condition: 2^{\text{Sequence Number Field Length}} \geq \text{Bandwidth} \times \text{MSL}

Calculations:

  1. First, determine the amount of data that can be sent over the network during the MSL.
    The given network bandwidth = 100 \text{ Mbps}, and MSL = 2 minutes = 120 \text{ seconds}.
    Data sent during MSL = 100 \text{ Mbps} \times 120 \text{ s}
  2. Convert Mbps to bits, 100 \times 10^6 \text{ bits/second} \times 120 \text{ s} = 12 \times 10^9 \text{ bits}
  3. Sequence number field length must allow 12 billion unique positions.
    Therefore, we need to solve 2^n \geq 12 \times 10^9
  4. Calculate n:
    n \approx \log_2(12 \times 10^9).
    This approximates to n \approx 34 (since 2^{34} = 17.18 \times 10^9)

Evaluating the Options:

  • 30 bits: Fails since 2^{30} = 1.07 \times 10^9, which is less than 12 billion.
  • 32 bits: Insufficient since 2^{32} = 4.29 \times 10^9, less than required.
  • 34 bits: Sufficient as 2^{34} = 17.18 \times 10^9, which is more than needed.
  • 36 bits: While valid, it's larger than the minimal necessary size.

Conclusion:

According to the practical minimum for ensuring reliability in this MSL context, the minimum valid sequence number field length is 34 bits not 32 bits as provided in the correct answer prompt. This ensures the space is enough for a 100 Mbps network with 2-minute MSL. Therefore, careful reevaluation may suggest the correct solution should involve correcting the problem prompt to align with computed logic.

Was this answer helpful?
0


Questions Asked in GATE CS exam