Congestion avoidance in TCP

Question

During Congestion avoidance(Additive Increase) in tcp when 3 duplicate ACKs arrived then during that arrival phase by which algorithm did the  sender window adopt increment procedure?
a) Sender window uses additive increase
b)Sender window uses slow start
which option is correct?

Answer 1

From forouzan :: TCP/IP protocol suite:

If three duplicate ACKs are received, there is a weaker possibility of congestion; a segment may have been dropped but some segments after that have arrived safely since three duplicate ACKs are received. This is called fast transmission and fast recovery. In this case, TCP has a weaker reaction as shown below:
(A). It sets the value of the threshold to half of the current window size.
(B). It sets cwnd to the value of the threshold (some implementations add three segment sizes to the threshold).
(C). It starts the congestion avoidance phase.
 

Comments

Is the Value of Sender Window(SW) is correct during 3 duplicate Acks arrival?

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In AIMD phase, if there are 3 duplicate ACK , then there is a strong possibility of congestion (not weak possibility)

Moreover there should be a threshhold value. Sender Window value cannot go upto infinity

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3 duplicate acknowledgement means mild congestion.

Answer 2

Option a is correct .

Slow start mechanism occurs whenever time out occurs

Answer 3

3 duplicate ack means congestion is less so sender opts for additive increase rather than slow start which occurs when congestion is severe and time out happens.

Hence correct option is a.

Comments

U mean to say During the phase when 3 duplicate acks are coming sender opts for additive increase and After the 3 duplicate Acks arrived sender opts for slow start??

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No, i am saying whenever timeout occurs it means congestion is severe slow start mechanism happens.

 After the 3 duplicate Acks arrived sender opts for slow start(This is wrong)

Answer 4

b option is correct

Comments

3 duplicate packet means week posiblities of congestion and use congestion avoidance algorithm.

Option a is correct.