+1 vote
168 views
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.
| 168 views
+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).
0
@G.K.T

Bandwidth is given in BytesI I guess?
0
Yeah thanks for pointing .
+2
Yes, it's correct @saxena0612
0
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...:)

by Active (1.5k points)