GATE CSE
First time here? Checkout the FAQ!
x
+2 votes
33 views
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 44 | 33 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 Boss (5.2k points) 2 34 84
selected by
0 votes
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


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Top Users Oct 2017
  1. Arjun

    23488 Points

  2. Bikram

    17108 Points

  3. Habibkhan

    8682 Points

  4. srestha

    6314 Points

  5. Debashish Deka

    5458 Points

  6. jothee

    5068 Points

  7. Sachin Mittal 1

    4882 Points

  8. joshi_nitish

    4444 Points

  9. sushmita

    3996 Points

  10. Rishi yadav

    3868 Points


Recent Badges

Photogenic MiNiPanda
Reader Ali Jazib Mahmood
Good Question Ishrat Jahan
Nice Comment Arjun
Notable Question Kaifi7
Popular Question LORD ofKINGS
Great Question Rucha Shelke
Popular Question rahul sharma 5
Popular Question jothee
Popular Question Pankaj Joshi
27,351 questions
35,209 answers
84,265 comments
33,328 users