GATE IT 2005

Question

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...

Answer 1

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.