Topic: Computer Networks
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
Answer: A. 65536 bytes
Explanation: To fully utilize a link, the TCP window size should be at least equal to the Bandwidth-Delay Product (BDP).
BDP = Bandwidth $\times$ Round Trip Time (RTT).
Given Bandwidth = $8,192,000$ bps.
Latency (one-way delay) = 32 ms. Therefore, RTT = $2 \times 32 = 64$ ms = $0.064$ seconds.
Window size in bits = $8,192,000 \times 0.064 = 524,288$ bits.
Window size in bytes = $524,288 / 8 = 65,536$ bytes.