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Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.
Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer?
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Let the size of congestion window of a TCP connection be 32 KB when a ...
TCP Congestion Window Size Calculation
Given:
- Congestion window size at timeout = 32 KB
- Round trip time (RTT) of connection = 100 ms
- Maximum segment size (MSS) used = 2 KB
To calculate the time taken by the TCP connection to get back to a congestion window size of 32 KB, we need to follow these steps:
Step 1: Calculate the threshold value
- At the time of timeout, the congestion window size is 32 KB, so the threshold value for the next round of Slow Start is half of this, which is 16 KB.
Step 2: Perform Slow Start
- Slow Start starts with the MSS size, which is 2 KB in this case.
- For each round trip time, the congestion window size is doubled.
- The congestion window size should not exceed the threshold value.
Slow Start sequence:
2KB|4KB|8KB|16KB
Step 3: Perform Additive Increase
- Once the threshold value is reached, Additive Increase starts.
- In Additive Increase, the congestion window size increases by one MSS for each round trip time.
Additive Increase sequence:
18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KB
Step 4: Calculate the total time taken
- Each round trip time is 100 ms, so the total number of round trip times required to reach 32 KB congestion window size is 8.
- Including the initial Slow Start phase, the total number of round trip times is 12.
- Therefore, the total time taken is 12 * 100 ms = 1200 ms.
Final Answer:
The time taken by the TCP connection to get back to a congestion window size of 32 KB is 1200 ms.
Given:
- Congestion window size at timeout = 32 KB
- Round trip time (RTT) of connection = 100 ms
- Maximum segment size (MSS) used = 2 KB
To calculate the time taken by the TCP connection to get back to a congestion window size of 32 KB, we need to follow these steps:
Step 1: Calculate the threshold value
- At the time of timeout, the congestion window size is 32 KB, so the threshold value for the next round of Slow Start is half of this, which is 16 KB.
Step 2: Perform Slow Start
- Slow Start starts with the MSS size, which is 2 KB in this case.
- For each round trip time, the congestion window size is doubled.
- The congestion window size should not exceed the threshold value.
Slow Start sequence:
2KB|4KB|8KB|16KB
Step 3: Perform Additive Increase
- Once the threshold value is reached, Additive Increase starts.
- In Additive Increase, the congestion window size increases by one MSS for each round trip time.
Additive Increase sequence:
18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KB
Step 4: Calculate the total time taken
- Each round trip time is 100 ms, so the total number of round trip times required to reach 32 KB congestion window size is 8.
- Including the initial Slow Start phase, the total number of round trip times is 12.
- Therefore, the total time taken is 12 * 100 ms = 1200 ms.
Final Answer:
The time taken by the TCP connection to get back to a congestion window size of 32 KB is 1200 ms.
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Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer?
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Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. 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 Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. 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 Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer?.
Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. 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 Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. 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 Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer?.
Solutions for Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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Here you can find the meaning of Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer?, a detailed solution for Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer? has been provided alongside types of Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of theconnection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.Correct answer is 'Ans: Given that at the time of Time Out, Congestion Window Size is 32 KB and RTT = 100 ms , When Time Out occurs, for the next round of Slow Start,Threshold = ,Threshold = 16KBSuppose we have a slow start ==>> 2KB|4KB|8KB|16KB (As the threshold is reached, Additive increase starts) 18KB|20KB|22KB|24KB|26KB|28KB|30KB|32KBHere | (vertical line) is representing RTT so the total number of vertical lines is 100*10ms ==>> and 1100m. sec so this is the answer...'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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