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The data link layer uses a fixed-size sliding window protocol, where the window size for the connection is equal to twice the bandwidth-delay product of the network path. Consider the following three scenarios, in each of which only the given parameter changes as specified (no other parameters change). For each scenario, explain whether the throughput (not utilization) of the connection increases, decreases, remains the same, or cannot be determined:

  1. the packet loss rate $L$ decreases to $L/3$;
  2. the minimum value of the round trip time $R$ increases to $1.8R$;
  3. the window size $W$ decreases to $W/3$
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@Akash Ghosh

Window size is $1 + Tp/Tt$

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They asked for throughput.So,utilization means efficiency , what I think
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I think the you have written the formula wrong . In efficiency for the sliding window protocol we follow the formula n = N/(1+2a) where a = Tp/ Tt. Where N = The sliding window size in the sending side.

b) Now as RTT increases propagation time also increases , so the a increases and efficiency(n) decreases. We know Throughput = Efiiciency * Bandwidth. So as Efficiency decreases, Throughput also decreases.

c) As the window size(N) decreases so generally efficiency also decreases. Hence Throughput decreases.

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1 Answer

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  1. If the packet loss decreases then automatically throughput increases
  2. If Round trip time increases then link distance increases or propagation speed slows hence packet loss probablity increases thus throughput decreases.
  3. Its mentioned that “sliding window size equal to twice the bandwidth-delay product of the network path” BDP is directly proportional to round trip time. Hence is SW decreases then RTT decreases so link distance must decrease or speed increases so throughput increases. 

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