The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
0 votes

On a TCP connection, current congestion window size is Congestion Window = 4 KB. The window size advertised by the receiver is Advertise Window = 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is

(A) 2048 bytes
(B) 4096 bytes
(C) 6144 bytes
(D) 8192 bytes

I think it will be option A, I was doing from geeks from geeks website but the person who answered is not sure and he has calculated as B. I think first round congestion window is 4096 bytes after 10240 bytes send and receiver received only 8192,so it can consume only till 8192 till its window overflow and 10240-8192 = 2048 ,this will be advertised again back to sender so min of 4096 and 2048 will be 2048.Please explain a bit also thanks...

asked in Computer Networks by Active (3.7k points) | 90 views

1 Answer

0 votes
Actually current window size is minimum of congestion window size and receiver window size. so minimum here will be 4kB which in turn 4096 B.
answered by Active (3.1k points)

Related questions

0 votes
0 answers
asked Mar 13, 2017 in DS by Bhargav Zantye 1 (15 points) | 211 views
0 votes
0 answers
asked Sep 11, 2018 in Mathematical Logic by Shiv kumari (9 points) | 12 views
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
50,108 questions
53,221 answers
70,462 users