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Help with Jacobson Algorithm _ TCP

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My answer is
2|4|8|16|  20  |21|22|23|24|25|26|27|28|29|30|31|32

 %%% At time 5, the threshold is reached. Slow-Start  ends and threshold/2 and cwnd= 1 MSS   %%%


#of RTT = 17
17*RTT= 17*25=  425 msec
File size/the receiver window size=300/32= 6.25*425= 2762.5 msec

Am I right?
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Receiver window=32kB/2kB=16

Thresh hold=20kB/2kB=10

No of MSS = 200/2=100

 

RTT

1

2

3

4

5

6

7

8

9

10

11

12

CW

1

2

4

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10

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12

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16

4

TH

10

10

10

10

10

10

10

10

10

10

10

10

 

No of RTT= 12

Total time=12 × 25=300 msec

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