Computer network congestion window


#1

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 timeout
occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.
(A) 8 MSS
(B) 14 MSS
© 7 MSS
(D) 12 MSS
gate2012 computer-netw


#2

[quote=“Gate_Aspirant1, post:1, topic:1036”]
here MSS size is 2KB i.e. every time 1MSS sent means 2 KB data is sent so either you write in terms of MSS by adding 1 or in terms of data adding 2KB

the question in the link has started with window size (which is 2MSS). The AIMD algorithm is applied when the segment being sent is lost or data size reaches threshold. We generally start from slow start phase i.e.

1st transmission 2MSS

2nd 4MSS

3rd 8MSS

now since it reached threshold

4th 9MSS (as per AIMD increasing by 1)

5th 10MSS (segment lost)

new threshold 10/2 i.e. 5MSS and transmission starts with 2MSS

6th 2MSS

7th 4MSS

8th 5MSS

9th 6MSS

10th 7MSS


#3

At

t=1,⇒2MSS
t=2,⇒4MSS
t=3,⇒8MSS
t=4,⇒9MSS (after threshold additive increase)

t=5,⇒10MSS (fails)

Threshold will be reduced by n/2i.e. 10/2=5
t=6,⇒1MSS

t=7⇒2MSS
t=8,⇒4MSS
t=9,⇒5MSS
t=10,⇒6MSS

So at the end of 10th10th sucessful transmission ,

the the congestion window size will be (6+1)=7MSS.


#4

Answer ©
Window size for 1st transmission,2nd transmission,3rd transmissions are 2MSS,4MSS,8MSS
threshold reached, increase linearly (according to AIMD) 4th transmission,5th transmissions are 9MSS,10MSS
time out occurs, resend 5th with window size starts with as slow start.6th transmission,7th transmission are 2MSS,4MSS
threshold reached, now increase linearly (according to AIMD)
Additive Increase: 5 MSS (since 8 MSS isn’t permissible anymore)
Window size for 8th transmission,9th transmission,10th transmissions are 5MSS,6MSS,7MSS.