TCP congestion control

Question

Ans given is 13.

Comments

1,2,4,8,12(new thresld),13 at 6th burst transmission and 14 after 6th burst

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MSS=1KB.APPLY AIMD

1,2,4,8,12(reaches threshhold) apply congestion avoidance 13,14,15,16......24.

at 6th transmission window size=13KB.

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But who asks for 6th transmission??

It should be considered for 7th

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all 6 transmission are succesful at that time what is window size,he is asking.?

it is mentioned that next 6 transmissions, starting from first.

Answer 1

When Time out occurs the $SSthresh=12KB$

Before 1 stsuccessfull transmission cwnd = 1KB  after cwnd = 2KB

Before 2 stsuccessfull transmission cwnd = 2KB  after cwnd = 4KB

Before 3 stsuccessfull transmission cwnd = 4KB  after cwnd = 8KB

Before 4 stsuccessfull transmission cwnd = 8KB  after cwnd = 12KB since  $16KB \gt 12KB$

Before 5 stsuccessfull transmission cwnd = 12KB  after cwnd = 13KB

Before 6 stsuccessfull transmission $cwnd\; = \;$13KB  after $cwnd\; = 14KB   $      

Thus answer is 14KB

Before 7 stsuccessfull transmission cwnd = 14KB  after cwnd = 15KB  and so on.

Comments

Actually, AIMD means Congestion Avoidance and Fast retransmit Phase (Knuth+Rosen Page 277)

It is given slow start AIMD .Thus we are using real congestion control algorithm having all 3 phases

Answer 2

Congetion Window Size = 24KB
Timeout Occurs . Now The New Threshold will be 50% of Congetion Window = 12KB
According to AIMD Upto 12KB Growth Will be Exponential . Then Linear

After Timeout 6 Transmissions are suucessfull and congetion size after 6th transmission

1. 1KB
2. 2KB
3. 4KB
4. 8KB
5. 12KB  (threshold)
6. 13KB
7. 14KB

After 6th  trasmission ,  Window Size  Ready to transmitted will be 14KB

Comments

@pC

For that Gate 2014 Q.

1 MSS = 1 RTT (Size at the end of RTT = 2 MSS)

2 MSS = 2 RTT (Size at the end of RTT = 4 MSS)

4 MSS = 3 RTT (Size at the end of RTT = 8 MSS)

8 MSS = 4 RTT (Size at the end of RTT = 9 MSS)

9 MSS = 5 RTT (Size at the end of RTT = 10 MSS)

10 MSS = 6 RTT (Size at the end of RTT = 11 MSS)

11 MSS = 7 RTT (Size at the end of RTT = 12 MSS)

12 MSS = 8 RTT (Size at the end of RTT = 13 MSS)

13 MSS = 9 RTT (Size at the end of RTT = 14 MSS)

14 MSS = 10 RTT (Size at the end of RTT = 15 MSS)

15 MSS = 11 RTT (Size at the end of RTT = 16 MSS) ( Hence, at the end of 15th RTT, it reaches 32 KB )

16 MSS = 12 RTT (Size at the end of RTT = 17 MSS) ( Here, at the start of 16th RTT, WS = 32 KB )

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PS: Like this same approach, I would tick 8 MSS in that GATE 2012 Q.

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Yeah it should start with 1 MSS..  I missed that

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And that's why I said, that answer for this one is 14 KB. Because every MSS is involved with RTT (It includes ACK.) , So, at the end means that completion of RTT :)