GATE CSE 2018 | Question: 14

Question

Consider the following statements regarding the slow start phase of the TCP congestion control algorithm. Note that cwnd stands for the TCP congestion window and MSS window denotes the Maximum Segments Size:

1. The cwnd increases by $2$ MSS on every successful acknowledgment
2. The cwnd approximately doubles on every successful acknowledgment
3. The cwnd increases by $1$ MSS every round trip time
4. The cwnd approximately doubles every round trip time

Which one of the following is correct?

• A. Only $\text{(ii)}$ and $\text{(iii)}$ are true
• B. Only $\text{(i)}$ and $\text{(iii)}$ are true
• C. Only $\text{(iv)}$ is true
• D. Only $\text{(i)}$ and $\text{(iv)}$ are true

Comments

Option 2 is false as -

The cwnd size increase by 1 on every ACK receives(i.e cwnd+=1)

Option 3 is false as-

cwnd increased by 2 MSS on every RTT

Option 1 is false as-

It is increased by 1 on every successful ACK

AND FOR EACH RTT: double cwnd(congestion window)

Correct answer is "C"

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@Ray Tomlinson option (ii) and option (iv) are not same ...

cwnd size increases by 1 MSS for each received acknowledgement and at the end of RTT, cwnd size increases exponentially. 

ex: 

RTT	Segments Sent	ACKs Received	cwnd (after RTT)	GrowthType
1	1	1	2 MSS	—
2	2	2	4 MSS	Exponential
3	4	4	8 MSS	Exponential
4	8	8	16 MSS	Exponential

 

option (ii) wrong - The cwnd approximately doubles on every successful acknowledgment

crct one - The cwnd approximately doubles on every RTT or The cwnd increases by  1 MSS on every successful acknowledgment

 

 

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This statement is also correct "The cwnd increases by 1 MSS on every successful acknowledgment"

Answer 1

Each time an $ACK$ is received by the sender, the congestion window is increased by $1$ segment: $CWND = CWND + 1$.

$CWND$ increases exponentially on every $RTT$.

Hence, correct answer is C.

https://www.utdallas.edu/~venky/acn/CongestionControl.pdf

Comments

1. The cwnd approximately doubles on every successful acknowledgment
2. why this option is incorrect

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In Slow Start Phase :

1. For each RTT, double the $cwnd$  $i.e, cwnd=2*cwnd$
2. For each ACK, increase the cwnd by 1.     $i.e,cwnd=cwnd+1$

In Congestion Avoidance Phase :

1. For each RTT, $cwnd$ increases by $1$
2. For each ACK, $cwnd$ increases by $1/cwnd$

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Refer this if anyone is confused: https://gateoverflow.in/333176/gate-cse-2020-question-55

Answer 2

Round trip time(RTT) is the length of time it takes for a signal to be sent plus the length of time it takes for an acknowledgement of that signal to be received.

Whenever a successful acknowledgement is received cwnd increases by 1. Now coming to RTT we are supposed to receive all acknowledgements within RTT. let number of segments sent is n, then in turn we will receive n ack. so now new cwnd size will be equal to last_cwnd_size(n) + number_of_ack_received(n) i.e. cwnd=n+n. which can also be said as cwnd doubles on every RTT.

now we can say that only (iv) option is the correct one

Comments

Very well explained!

Answer 3

In slow start phase, reception of each ACK adds 1 MSS to cwnd.

 

Generally, if you send, say 4 packets (cwnd = 4) then in the time approximately equal to an RTT you'll get 4 ACKs in return. These 4 ACKs would add a total of 4 MSS in the cwnd.

 

Hence approximately we double our cwnd every RTT in Slow Start Phase. Statement iv is True.

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Statement i is wrong as cwnd increases by 1 MSS on every ACK received. Hence, Statement ii is also wrong.

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Statement iii could be true with slight modifications.

If every RTT successfully delivers you an ACK, then in the time somewhat equal to RTT, cwnd will get doubled. "Somewhat equal" because we'll have to wait for ACKs to get delivered back-to-back.

Answer 4

 

• INITIALLY CWND = 1 
• CWND DOUBLES ON EVERY RTT
• CWND INCREMENTS BY ONE ON EVERY ACKNOLDEGEMENT

(i) False, because the cnwd increases by 1MSS(cnwd=cnwd+1) on every successful acknowledgement (when sender receive an acknowledgement).

(ii) False, because same reason as above.

(iii) False, because the cnwd increases exponentially on every RTT.

(iv) True, because the cnwd increases exponentially on every RTT so we can say that it approximately double the cnwd.

i hope the above picture clears everything.
 

Comments

This answer is really nice. Specially the diagram is very clearly depicting how 

• CWND DOUBLES ON EVERY RTT
• CWND INCREMENTS BY ONE ON EVERY ACKNOWLEDGEMENT

Answer 5

TCP uses sliding window protocol and that too sr protocol as it can receive out of window packets and with each ack received it it shift the window. In case of slow start with each ack congestion window is increased by 1 and effectively after each rtt congestion window has increased by 2 times. If an ack acknowledges 2 segment than also cwnd is still increased by 1 segment, enven if ack acknowlsedges a segment that is less than mss still cwnd is increased by 1.

In congestion avoidance phase increase cwnd as cwnd=cwnd+1/(floor(cwnd)). So cwnd segment is increased  if all the cwnd segment has been acknowledged.

Answer 6

in slow start algorithm sender first send one packet if it receives ack with in timeout the it send two packets in 2nd rtt tf it receives ack within timeout then it increase cwnd as exponentially.if it will not receive with in timeout the sender will decrease the size of cwnd to half.so in slow start 

The cwnd approximately doubles every round trip time .

so option c correct