+1 vote
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Q. Compute the approximate optimal window size  when packet size 53 byte and round  trip time (RTT) is 60 ms and bandwidth (BW)155 mbps (in byte) using sliding window protocol.
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+2

@joshi_nitish @just_bhavana

Check it once!

0
For optimal window size exhaust your bandwidth in given RTT so sender in never idle (Thats the basic idea).

so bandwidth = 155Mbps

and RTT = 60 ms

so in 1 sec we may send 155MB

in RTT it should be 155 MB * 60 ms = 9300 KB

packet size is 53 B

win size = 9300 KB / 53 B

=175.47169 K or 175471 (taking floor as what we have just calculated is the maximum limit so cant go further).
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@G.K.T

Bandwidth is given in BytesI I guess?
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Yeah thanks for pointing .
+2
Yes, it's correct @saxena0612
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correct.
0
why you are not using
w= 1 + 2a
0

For maximum eff = 100% = 1

1 = $\frac{n*TT}{TT + 2PT}$

where n = window size.

TT = 2.73 sec

PT = 30

$\frac{n*2.73}{2.73 + 60}$ = 1

n = 22.97

taking floor value it is 22.

If we take ceil value then it will be more than 100%.

Hope you will get this...:)

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