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Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the startof 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.
  • a)
    8 MSS
  • b)
    14 MSS
  • c)
    7 MSS
  • d)
    12 MSS
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider an instance of TCP’s Additive Increase Multiplicative D...
The Answer is correct , but method of solving is wrong .
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/2 i.e. 10/2 = 5.
t=6, =>2mss
t=7 =>4mss
t=8, =>5mss
t=9, =>6mss
t=10, =>7mss.
So at the end of 10th transmission congestion window size will be 8 mss.
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Most Upvoted Answer
Consider an instance of TCP’s Additive Increase Multiplicative D...
Solution:

Given data:

- Window size at the start of the slow start phase = 2 MSS
- Threshold at the start of the first transmission = 8 MSS
- A timeout occurs during the fifth transmission
- We need to find the congestion window size at the end of the tenth transmission

AIMD algorithm is used to control the congestion window size. It follows the following steps:

1. Slow start phase: In this phase, the congestion window size is increased exponentially until it reaches a threshold value.
2. Congestion avoidance phase: In this phase, the congestion window size is increased linearly until congestion is detected.

Let's calculate the congestion window size at the end of each transmission:

- At the start of transmission 1, the congestion window size is 2 MSS.
- At the start of transmission 2, the congestion window size is 4 MSS (due to slow start phase).
- At the start of transmission 3, the congestion window size is 8 MSS (due to slow start phase).
- At the start of transmission 4, the congestion window size is 8 MSS (due to slow start phase).
- At the start of transmission 5, the congestion window size is 8 MSS (due to slow start phase).
- During transmission 5, a timeout occurs. This indicates congestion, and the algorithm enters the congestion avoidance phase.
- After the timeout, the congestion window size is set to the threshold value (8 MSS).
- At the start of transmission 6, the congestion window size is 8 MSS (due to congestion avoidance phase).
- At the start of transmission 7, the congestion window size is 9 MSS (due to congestion avoidance phase).
- At the start of transmission 8, the congestion window size is 10 MSS (due to congestion avoidance phase).
- At the start of transmission 9, the congestion window size is 11 MSS (due to congestion avoidance phase).
- At the start of transmission 10, the congestion window size is 12 MSS (due to congestion avoidance phase).

Therefore, the congestion window size at the end of the tenth transmission is 8 MSS (option A).
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Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the startof 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.a)8 MSSb)14 MSSc)7 MSSd)12 MSSCorrect answer is option 'A'. Can you explain this answer?
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