@charul 1st the sender sends 1 segment and get an ack. So now congestion window increases by 1.
Now there are 2 segments. The sender sends these 2 segments and get ack. for each of them. Each segment increases by 1 segment so 2+2 =4 is the current window size.
Next these 4 segments are sent and 1 ack is received for each segment. So 4+4 =8.
I think what they meant by this---
congestion window increases by 1 segment each time it receives an acknowledgment,
is that the congestion window increases by 1 segment for each segment present in the current window. This is what happens in slow start phase. Tell me if i am wrong.