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2 votes
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Station $A$ uses $32-byte$ packets to transmit messages to Station $B$ using a sliding window protocol.
The round trip time delay between $A$ and $B$ is $40\,ms$ and the bottleneck bandwidth on the path
$A$ and $B$ is $64\, kbps.$What is the optimal window size that $A$ should use?

  1. $5$
  2. $10$
  3. $40$
  4. $80$
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2 Answers

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Transmission delay $T_t=\frac{32\times8}{64\times10^{3}}=$ 4 msec.

Assuming 100% bandwidth utilization, we get $1=\frac{N\times T_t}{RTT}$
$\rightarrow1=\frac{N\times 4}{40}\rightarrow N=10.$
Answer:

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