Question

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 acknowledgement. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?

• A. 10000 bytes
• B. 12000 bytes
• C. 8000 bytes
• D. 4000 bytes

Answer 1

The correct option is C 8000 bytes

In slow start phase, the sender increase the size of current transmission window by MSS for each
ACK.
Here Packet size = 2000 B,
Data to be send =4000 B
So, # packet 2 be sent$=\frac{4000B}{2000B}=2$

So, Maximum value of transmit window
$=4000\times (2\times 2000)$
$=4000+4000$
$=8000B$