What is the ideal TCP window size of a sender to fully utilize a link of…
2025
What is the ideal TCP window size of a sender to fully utilize a link of bandwidth 8192000bps and latency of 32ms?
- A.
65536 bytes
- B.
8192 bytes
- C.
16384 bytes
- D.
32768 bytes
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Correct answer: A
To fully utilize the link, the TCP window size must equal the Bandwidth-Delay Product (BDP). First, calculate the Round-Trip Time (RTT), which is twice the one-way latency: RTT = 2 × 32 ms = 64 ms (0.064 seconds). Next, calculate the BDP in bits: 8,192,000 bps × 0.064 s = 524,288 bits. Finally, convert to bytes: 524,288 / 8 = 65,536 bytes.