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Station $A$ uses $32$ byte packets t transmit messages to station $B$ using sliding window protocol. The round trip delay between $A$ and $B$ is $40$ milliseconds and the bottleneck bandwidth on the path between $A$ and $B$ is $64$ kbps. The optimal window size of $A$ is

1. $20$
2. $10$
3. $30$
4. $40$

1 comment

option B is correct

If anyone is following the NPTEL course(Computer Networks and Internet Protocol), the course discusses the concept of BDP(Bandwidth Delay product) that can also be used to calculate the window size.

Given, RTT = 40ms, therefore one side time = 40/2 = 20ms

Hence,

BDP = 64kbps * 20ms = 1280 bits

packet size = 32byte = 256 bits,

number_of_packets = 1280/256 = 5

WindowSize = 2*no_of_packets + 1

w = 2*5 + 1 = 11 (we add +1 as the ACK is sent only when the first first segment is received)

Hence, (b) 10 is the correct option (closest option)

The optimal window size of A is 10.

OPTION -B