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Why can't I get answer by using formulae

Window size = 1+2a.....

Station A uses 32 byte packets to transmit messa

ges 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 Answer

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Bottleneck  bandwidth = 64 kbps=64x10bps

Round trip delay = 40 ms =40x10-3 sec

Total data 40x64x103 x10-3 =  bits  = 40x64 /8  bytes =320 bytes.

1 packet size = 32 byte

No. of packets 320/32  = 10

Hence (2 ) is the ans

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