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61 votes
61 votes

Suppose that the maximum transmit window size for a TCP connection is $12000$ $\text{bytes}$. Each packet consists of $2000$ $\text{bytes}$. At some point in time, the connection is in slow-start phase with a current transmit window of $4000$ $\text{bytes}$. Subsequently, the transmitter receives two acknowledgments. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?

  1. $4000$ $\text{bytes}$
  2. $8000$ $\text{bytes}$
  3. $10000$ $\text{bytes}$
  4. $12000$ $\text{bytes}$
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10 Answers

Best answer
75 votes
75 votes

In slow-start phase, for each ACK, the sender increases the current transmit window by Maximum Segment Size (MSS). In the question it is given a packet consists of $2000$ bytes and that can be taken as MSS. So, after two ACKs, current transmit window
$= 4000 + 2000 + 2000$
$= 8000$

http://www.ece.virginia.edu/~mv/edu/ee136/Lectures/congestion-control/tcp-congestion-control.pdf or archive

Correct Answer: $B$

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22 votes
22 votes

Since recvWind = 12000B and Packet Size = 2000B (or MSS )

 receiverWindow = 6 MSS and currentSenderWindow = 2MSS

Slow Start Threshold = receiverWindow/2 = 3 MSS (2 MSS < 3 MSS which implies transmission is in Slow Start Phase)

After receiving first ACK: currentSenderWindow should increase exponentially to 4 MSS but since threshold = 3 MSS, currentSenderWindow Size goes to threshold which is 3 MSS then 
After receiving second ACK: Since now it is in Congestion avoidance phase sender window size increases linearly which makes currentSenderWindow = 4 MSS = 4 * 1 MSS = 4 * 2000 B  = 8000 B

Thanks to @Nilesh Agrawal ji.

12 votes
12 votes
Packet size (MSS Maximum segment size)= 2000 B
Advertized window size (RWND)= 12000 B
Sender's Current window size (CWND)= 4000 B
CWND = 2 * MSS --- > This means that the sender can send upto 2 packets at this point and it did.
That is why it now recieves 2 Acks of those two packets sent
So it can now increase the CWND by 1 MSS for each packet acknowledged.
So Current Window Size = 4000 + 2* MSS = 8000 THis is the answer.
 
Now it can send upto 4 packets and if it receives 4 Acks for those packets sent then it will increment its window to 12000
 
This link will be of help !!!
7 votes
7 votes

In slow start phase, window size increases exponentially upon receiving 1 ack till threshold and after that collision avoidance phase start where window size is increased 1 mss after every acknowledgement till max window is reached.

Slow start : 1,2 4 ,8... threshold

Collision avoidance: threshold+1, threshold +2.....Max window size

Threshold = max transit window size/2

In given question Max window size is 12000B and therefore

Threshold = 6000

So the sequence of transit will be

1st 2000

2nd 4000 1 ack received

3rd 6000 1ack received

4th 8000

Hence, Answer is 8000 B

Answer:

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