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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. $40$
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option B is correct
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10 Answers

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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.
https://www.youtube.com/watch?time_continue=812&v=bKHRbqwkMkg&feature=emb_title

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

BDP (Bandwidth Delay Product) = Link Bandwidth * Link Delay

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)

 

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The optimal window size of A is 10.

OPTION -B

Answer:

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