TCP Sequence Number Wrap-Around Time

Given:

• Data rate = 2 Mbytes/sec
• Initial sequence number = 7000
• TCP sequence numbers are 32 bits, so the sequence number space is $2^{32} = 4{,}294{,}967{,}296$ bytes.

Step 1 — Bytes that must be sent before the number returns to 0

The sequence number wraps to 0 only after the entire 32-bit space is exhausted. Starting at 7000, the bytes still to be sent are:

$$N = 2^{32} - 7000 = 4{,}294{,}967{,}296 - 7000 = \mathbf{4{,}294{,}960{,}296 \text{ bytes}}$$

Step 2 — Convert the data rate to bytes/sec

Taking 1 Mbyte = $2^{20}$ bytes = 1,048,576 bytes:

$$R = 2 \times 1{,}048{,}576 = 2{,}097{,}152 \text{ bytes/sec}$$

Step 3 — Time = Data / Rate

$$T = \frac{4{,}294{,}960{,}296}{2{,}097{,}152} \approx 2047.997 \text{ s}$$

$$\boxed{T \approx 2048 \text{ s} \approx 34 \text{ min } 8 \text{ s}}$$

(Sanity check: $2^{32} / 2^{21} = 2^{11} = 2048$ s exactly; the initial 7000 bytes shave off only ~0.003 s, so the answer is essentially 2048 s.)

Alternative convention: if "Mbyte" is read as $10^6$ bytes, then $R = 2{,}000{,}000$ B/s and

$$T = \frac{4{,}294{,}960{,}296}{2{,}000{,}000} \approx 2147.5 \text{ s} \approx 35.8 \text{ min}$$

Final answer: ≈ 2048 seconds (about 34 minutes) for 1 Mbyte = $2^{20}$ bytes (≈ 2147.5 s if 1 Mbyte = $10^6$ bytes).

Note: this long wrap time is exactly why TCP may safely use a reduced window (e.g., window scaling / TCP timestamps) to disambiguate old from new segments — with wrap taking ~34 minutes, no segment can survive that long.