# ISI2018-PCB-CS9

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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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1. loss rate decreases ie, efficiency  increases, throughput = efficiency * bandwidth,so throughput increases.

2.RTT  increases means Tp increases, so windowsize =1 + Tt/Tp decreases , as throughput=L/1+2a ,so throughput increases.

3.Windowsize decreases ,so throughput will increase.considering Throughput =Length of the packet/window size)

Correct me if I am wrong.
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There is an ambiguity in the Question. Throughput is also Called Bandwidth Utilization. They haven't mentioned which Utilization it is, Line Utilization or Bandwidth Utilization

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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
1

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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