Question

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?

• A. 30 bits
• B. 32 bits
• C. 34 bits
• D. 36 bits

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.

Answer 1

The Correct Option is B

Step 1: Bandwidth–delay product.
Throughput = \(100 \,\text{Mbps} = 100 \times 10^6 \,\text{bits/sec} = 12.5 \times 10^6 \,\text{bytes/sec}\).
Over 2 minutes = \(120 \,\text{sec}\), maximum outstanding bytes that can still exist in the network: \[ 12.5 \times 10^6 \times 120 = 1.5 \times 10^9 \,\text{bytes}. \] Step 2: Sequence number space requirement.
The sequence number space must be at least twice this value to avoid ambiguity (wraparound shouldn’t reuse sequence numbers within MSL). \[ \text{Required space} \geq 2 \times 1.5 \times 10^9 = 3.0 \times 10^9. \] Step 3: Convert to bits.
We need \[ 2^k \geq 3 \times 10^9. \] - For \(k=30\): \(2^{30} = 1.07 \times 10^9<3 \times 10^9\) (Not sufficient).
- For \(k=32\): \(2^{32} = 4.29 \times 10^9>3 \times 10^9\) (Sufficient).
- For \(k=34\): \(2^{34} = 1.72 \times 10^{10}\) (More than enough).
- For \(k=36\): \(2^{36} = 6.87 \times 10^{10}\) (Also valid).
Step 4: Conclusion.
Valid sequence number lengths are \(\boxed{32, 34, 36 \,\text{bits}}\).