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

Given Parameters:

• Transmission Rate ($R$) = 1 Gbps = $10^9$ bits per second
• Maximum Segment Lifetime ($MSL$) = 60 seconds

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Step 1: Calculate the total data volume transmitted in one MSL

The sequence number space must be large enough so that no sequence number is reused within the time a packet might still be alive in the network (the MSL). First, we find the total number of bits sent:

$Total\ Bits = R \times MSL$ $Total\ Bits = 10^9 \text{ bits/sec} \times 60 \text{ sec} = 6 \times 10^{10} \text{ bits}$

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Step 2: Determine the required Sequence Number Space

TCP sequence numbers traditionally count bytes, not bits. To prevent wrap-around, we convert the total bits to bytes:

$Total\ Bytes = \frac{6 \times 10^{10} \text{ bits}}{8 \text{ bits/byte}} = 7.5 \times 10^9 \text{ bytes}$

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Step 3: Calculate the minimum bit-width ($n$)

The number of bits $n$ required to represent $7.5 \times 10^9$ unique values is found using the ceiling of the base-2 logarithm:

$2^n \geq 7,500,000,000$ $n \geq \log_2(7.5 \times 10^9)$ $n \geq \log_2(7.5) + \log_2(10^9) \approx 2.906 + 29.897 = 32.803$

Rounding up to the nearest whole number, we get $n = 33$.

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Final Answer:

The minimum number of bits required for the sequence number field to avoid wrap-around is: 33 bits