1
$\begingroup$

Consider an instance of TCP’s Additive Increase Multiplicative Decrease(AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a time out occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

I am bit confuse in this concept.
What I know :

Slow start and congestion avoidance are different.
Slow start is used in conjunction with congestion avoidance.

Window size for 1st transmission = 2 MSS
Window size for 2nd transmission = 4 MSS
Window size for 3rd transmission = 8 MSS threshold reached, increase linearly (according to AIMD)
Window size for 4th transmission = ?? MSS Window size for 5th transmission = ?? MSS
time out occurs, resend 5th with window size reduced to half (according to AIMD)
Window size for 5th transmission = ??? MSS

What will be the Window size for 4th transmission ? 9 MSS or 10 MSS
It means i want to ask you after threshold window size should increase linearly means it should increase by 1 MSS or 2 MSS ( 2 MSS is given in example ) ?

What will be window size for 5th transmission ? Will it be 1 MSS or 2 MSS ?

Please clear my concept. I am so confused.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

This is a home work problem. By the way the problem is ambiguous. For concept see 3.6 pg259 of kurose 6th edition.

As for the solution normally what AIMD do is after the time out the it set window size to 1 in the next slow start phase In sort the window increase in this manner 2,4,8,9,10,1,2,4,5,6 So after 10th transmission window size will be 6 But in question says the slow start phase start with 2 MSS So to consider that the window will Grow this way 2,4,8(threshold reached additive increase),9,10(time out& set threshold to 5),2,4,5(threshold reached additive increase),6,7

Improve this answer
$\endgroup$
0
$\begingroup$

Window size for 4th transmission is 9 MSS
Window size for 5th transmision is 10 MSS, here timeout happens
then we have to reduce the threshold to the half of the current window (i.e new threshold becomes 10/2 = 5 MSS) and window is re-set to 1 MSS and slow start phase is started again. so, Window size for 5th transmission is 1 MSS. In AIMD the window size is increased by 1 MSS only.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.