Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
633 views
in Computer by (95.4k points)
closed by
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 _________.

1 Answer

0 votes
by (95.2k points)
selected by
 
Best answer

Data:

Maximum Segment Size (MSS) = 2 KB

Round trip time (RTT) = 100 ms

Congestion window Size (CWS) = 32 KB

Formula:

\({\rm{Threshold}} = \frac{{{\rm{CWS}}}}{{{\rm{MSS}}}}\)

Calculation

\({\rm{Threshold}} = \frac{{{\rm{CWS}}}}{{{\rm{MSS}}}} = \frac{{32}}{2} = 16\;KB\)

 

 

Transmission

segment

size (KB)

 

 

Slow start phase

1

1

2

2

2

4

3

4

8

4

8

16 (Threshold)

 

 

congestion avoidance phase

5

9

18

6

10

20

7

11

22

8

12

24

9

13

26

10

14

28

11

15

30

12

16

32

 

Number of round trips = 12 – 1 = 11

Total time taken = 11 × 100 ms = 1100

The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is 1100 ms.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...