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Station A uses 32 byte packets to 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
asked in Computer Networks by Loyal (3.2k points) 4 39 | 26 views

3 Answers

+1 vote
Best answer

B = 64 Kbps

RTT = 40 ms 

No of bits which can be transmitted in 1 sec are

1 sec ---> 64 * 10³ bits 

No of bits which can be transmitted in RTT

40 * 10^(-3) ---> 64 * 40 bits 

Size of packet = 32 B

Size of window = (64*40)/(32*8) = 10 

Hence option 2) is correct

 

answered by Loyal (4.9k points) 2 34 80
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Option 2
answered by Loyal (2.5k points) 2 6 13
0 votes

2) 10

OWS=2a

a=Tp/Tt

2a=40/4

2a=10

answered ago by Junior (697 points) 1 5


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