Suppose you are asked to design a new reliable byte-stream transport protocol…

Concept

In a byte-stream protocol, a k-bit sequence-number field labels bytes modulo 2k. To prevent a delayed duplicate from being accepted after the numbers wrap, the sequence-number space must cover more bytes than the sender can transmit during the maximum segment lifetime.

Thus the governing condition is 2k > R × M, where R is the byte rate and M is the maximum segment lifetime. RTT affects the window needed to keep the path busy, but it does not change this wraparound-within-MSL condition.

Application

Apply the sequence-space condition to myTCP.

1. Convert the link rate to bytes per second: R = 100×106 ÷ 8 = 12.5×106 bytes/s.

2. Convert the maximum segment lifetime: M = 2 minutes = 120 seconds.

3. Compute the bytes sent during one MSL: R × M = 12.5×106 × 120 = 1.5×109 bytes.

4. Compare powers of two: 230 = 1,073,741,824 < 1.5×109, while 231 = 2,147,483,648 > 1.5×109. Therefore the minimum safe integer length is 31 bits.

5. Among the offered values, 32 bits, 34 bits, and 36 bits are at least 31 bits; 30 bits is below the minimum.

Cross-check

• Using wrap time directly, 230 ÷ (12.5×106) ≈ 85.9 seconds, which is shorter than 120 seconds.

• For 32 bits, 232 ÷ (12.5×106) ≈ 343.6 seconds, which exceeds 120 seconds; 34-bit and 36-bit spaces take still longer to wrap.

Result

The valid offered sequence-number lengths are 32 bits, 34 bits, and 36 bits.