GATE | GATE-IT-2004 | Question 88

• Difficulty Level : Expert
• Last Updated : 15 Feb, 2018

Suppose that the maximum transmit window size for a TCP connection is 12000 bytes. Each packet consists of 2000 bytes. At some point of time, the connection is in slow-start phase with a current transmit window of 4000 bytes. Subsequently, the transmitter receives two acknowledgements. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?
(A) 4000 bytes
(B) 8000 bytes
(C) 10000 bytes
(D) 12000 bytes

Explanation:

In the figure shown above MTU (Maximum Transmission Unit) contains IP header, TCP header and the payload or the TCP MSS (maximum segment size)

In slow start phase transmission of the packets in TCP for every ACK( acknowledgement packet) received, the sender of the data increases the current transmitted window size by the MSS( Maximum Segment Size).

According to the question given every packet consists of 2000 bytes = maximum segment size.

So, after 2 ACKs received the current window size will now be increased by

4000+ 2000+2000

= 8000 bytes

This solution is contributed by Namita Singh.

My Personal Notes arrow_drop_up
Recommended Articles
Page :