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A TCP sender uses a window of 64 KB (65,536 bytes) and the round-trip time is 100 ms. Ignoring all other delays, the maximum throughput is…

1981Medium DSSSB TGT Computer Science 0 replies

A TCP sender uses a window of 64 KB (65,536 bytes) and the round-trip time is 100 ms. Ignoring all other delays, the maximum throughput is about

  • A52.4 Mbps
  • B0.52 Mbps
  • C5.24 Mbps
  • D8.19 Mbps
Correct answer: C

Explanation

At most one window can be sent per round trip, so throughput = 65,536 × 8 bits / 0.1 s ≈ 5.24 Mbps.
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