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The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:
Correct answer is '12'. Can you explain this answer?
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The initial congestion window size over a TCP is 1. If slow strat algo...
To find the number of round-trip times (RTTs) it takes for the congestion window size to reach a certain value, we can use the formula for slow start:
Congestion Window Size = Initial Congestion Window Size + (Number of RTTs * MSS)
In this case, the initial congestion window size is 1 and it increments by 1 every RTT.
Let's say we want to find the number of RTTs it takes for the congestion window size to reach a value of N segments. We can set up the equation:
N = 1 + (Number of RTTs * 1)
Solving for the number of RTTs:
Number of RTTs = (N - 1) / 1
So, if the congestion window size reaches 2 segments after the first RTT, the number of RTTs is:
Number of RTTs = (2 - 1) / 1 = 1
Therefore, it takes 1 RTT for the congestion window size to reach 2 segments.
Congestion Window Size = Initial Congestion Window Size + (Number of RTTs * MSS)
In this case, the initial congestion window size is 1 and it increments by 1 every RTT.
Let's say we want to find the number of RTTs it takes for the congestion window size to reach a value of N segments. We can set up the equation:
N = 1 + (Number of RTTs * 1)
Solving for the number of RTTs:
Number of RTTs = (N - 1) / 1
So, if the congestion window size reaches 2 segments after the first RTT, the number of RTTs is:
Number of RTTs = (2 - 1) / 1 = 1
Therefore, it takes 1 RTT for the congestion window size to reach 2 segments.
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Community Answer
The initial congestion window size over a TCP is 1. If slow strat algo...
For every packet sent the window size increases by 2. So, when
Window size [WS=1] initially
⇒ After 1 RTT, window size = 2 and 1 segment is sent
⇒ After 2 RTT window size = 4 and 3 segment sent in total
⇒ After 3 RTT, window size = 8 and 7 segment sent in total
⇒ After ‘x’ RTTS window size 2x and 2x−1 segment are sent
Now, 2x−1 = 3999
⇒ 2x = 4000
⇒ x = log2(4000)
⇒ x = 12 RTT's
Alternately we can say that,
1st RTT- 1 segment sent.
2nd RTT- 2 new segments sent.
3rd RTT- 4 new segment sent.
4th RTT- 8 new segments sent.
5th RTT- 16 new segments sent.
6th RTT- 32 new segments sent.
7th RTT- 64 new segments sent.
8th RTT- 128 new segments sent.
9th RTT- 256 new segments sent.
10th RTT- 512 new segments sent.
11th RTT- 1024 new segments sent.
12th RTT- 2048 new segments sent.
Total segments sent = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048
=4095 > 3999
Hence, the correct answer is 12.
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The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:Correct answer is '12'. Can you explain this answer?
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The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:Correct answer is '12'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:Correct answer is '12'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:Correct answer is '12'. Can you explain this answer?.
The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:Correct answer is '12'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:Correct answer is '12'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The initial congestion window size over a TCP is 1. If slow strat algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first round trip time congestion window size is 2 segments. Assume connection never leaves slow start. Find the number of RTT’s to send 3999 segments:Correct answer is '12'. Can you explain this answer?.
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