MadeEasy Subject Test: Computer Networks - Congestion Control

Question

Assume a scenario where the size of congestion window of a TCP connection be 40 KB when a timeout occurs. The maximum segment size (MSS) be 2 KB. Let the propagation delay be 200 msec. The time taken by the TCP connection to get back to 40 KB congestion window is _________ msec.

Comments

I am getting 5600 msec  but ans. given is 6000 msec

Answer 1

comment if i am wrong.

Comments

https://gateoverflow.in/1794/gate2014-1-27  you can refer here too.

---

yes its correct ....i took MSS as 1 ...and landed up with wrong answer...thanks a lot

---

why after 16 u are getting 18 it should be 20 ryt

---

no after thresold ....we have linear increment of MSS....as 2,4,8,16 are there after that 32 is there but its above the thresold ....hence we go with linear increment  of MSS which is 2...

---

I am saying like 2 | 4 | 8 | 16 | 20| 22 | 24 | 26 | 28 | 30 | 32 | 34 | 36 | 38 | 40

14 verical lines hence 5600

Threshold=20 kb after that linear increment would take place

---

after 16 next should be 32...but we cant allow as thresold is 20....hence after 16 we start MSS of 2kb...you can refer to above link...

---

No I didn't think it to be correct ...please verify

---

See they are talking about tcp connection congestion window size....

now whenever we talk about congestion window, it actually the amount of data sender can send. Now there are totally 14 round trip after timeout to reach current congestion window.

14*400=5600msec.  

some say about 15 round trip time i.e 15*400=6000msec. After 15 round trip time congestion window will become 42KB not 40kb.

---

In theory we learn that 1stly it increase exponentially bt after thresold value it increase linearly  then  in answer why we don't use this concepts???

---

I think it should be 5600.

It will increase exponentially till 16 then it will not go till 32 due to the constraint of 20KB being the congestion window size. So it will become 20 and after that linear phase start.

It has 14 lines so 14*2*200 = 5600.

---

after 16 .....20kb will not come as after 16kb linear phase will start..... so 1 by 1 segment it will increase....and i segment size is 2kb so

16KB 18KB 20KB....

15 RTT'S WILL COME hence answer should be 15*400=6000ms

---

Any references please........ would be a great help

---

@pranav ray

Please add reference regarding your approach.

---

https://www.utdallas.edu/~venky/acn/CongestionControl.pdf ......pg no 4 and 5 ...hope so it may help u...

---

I am also getting answer as 14RTT=14*400=5600

I think answer given is wrong and it should be 5600 only