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GATE | Gate IT 2005 | Question 73
  • Difficulty Level : Medium
  • Last Updated : 28 Dec, 2015

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


Answer: (B)

Explanation: Congestion Window: It is the factor which tells how much maximum byte can be sent through network in TCP connection. It is dynamic in TCP.

Receiver Advertised Window: It is the factor which tells what is the maximum capacity of receiver to receive the data in bytes.

Total amount of maximum bytes that can be send by receiver at any time is called receiver window size.
At any time receiver can send minimum of Congestion Window and Receiver Advertised Window.
Thus current window size=min(congestion window ,receiver advertised window)
=min(4KB,6KB)
=4KB
=4*1024 bytes=4096bytes

But this is confusing question In my opinion 4KB will be for first time.
After that Receiver Window Size will be =  buffered data(unacknowledged) + min(congestion window ,receiver advertised window)
=
Unacknowledged data= (10240-8192)bytes=2048 bytes.
= (4096+2048) bytes=6144 bytes



 

This solution is contributed by Abhishek Kumar.

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