Question

On a TCP connection, current congestion window size is Congestion Window = 4 KB. The window size advertised by the received is Advertise Window = 6 KB. The last byte sent by the sender is Last Byte Sent = 10240 and the last byte acknowledged by the receiver is Last Byte Acked = 8192. The current window size at the sender is:
1. 2048 bytes
2. 4096 bytes
3. 6144 bytes
4. 8192 bytes

Answer 1

Correct Answer - Option 2 : 4096 bytes

Unacknowledged Bytes = 10240 - 8192 = 2048 Bytes

Receiver Available Window Size = Receiver Advertised WindowSize - Unacknowledged Bytes = 6KB - 2KB = 4KB

Sender Current Window Size = min( 4KB, 4KB) = 4KB) i.e min(Congestion Window Size, Receiver Available WindowSize)