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Suppose host A is sending a large file to host B over a TCP connection.
The two end hosts are 10msec apart (20msec RTT) connected by a 1Gbps link.
Assume that they are using a packet size of 1000 bytes to transmit the file.
Also assume for simplicity that ACK packets are extremely small and can be ignored.

At least how big would the window size (in packets) have to be for the
channel utilization to be greater than 80%.

I am getting answer as 2001.

$\eta \gt0.8$

my final calculation gave me

$W_s \gt 2000.8$(Window Size)

And hence my answer came 2001. Is it correct?

0
i don't know about the correctness, but i'm also getting the same.
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by ur aproach as u r getting Ws>2000.8 , then suppose in question instead of efficiency greater then 80% they would have asked equal to 80%, then what would be ur window size ?

+1 vote
Bandwidth-delay product of A to B = 10^9*20*10^-3= 2*10^7

for 80% utilization , it will be 2*10^7*0.8 = 1.6*10^7

so , Number of packets = (1.6*10^7)/(8*1000) = 2000 packets + 1{1 more becoz here asked efficieny greater then 80%} = 2001 packets
edited
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Efficiency has to be to strictly greater than $80percent$. Inequality has to be maintained. Hence answer must be 2001
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It will be 2001. (Asking for Least windows size, So we have go for Ceil here)