Congestion Window 1.2

Question

Consider the effect of using slow start on a link with 5 millisec round trip time and no congestion. The receiver window is 36 KB and the maximum segment size = 2 KB. The time (in milliseconds) before the first full window can be sent is ________.

Answer 1

The question says - " Consider the effect of using slow start on a link with 5 millisec round trip time and no congestion " 

here No congestion is mentioned so no need of Threshold value..

With slow start, the first RTT sends out 1 segment (or 2KB),

the 2nd RTT sends out 2 segments (or 4KB),

the 3rd 4 segments (or 8KB),

the 4th 8 segments (or 16KB).

The 5th RTT would have sent out 16 segments (or 32KB),

6th sent out 32 segments ( 64 KB) 

however, it'll exceed the receiver's window. Therefore, the amont of time it takes BEFORE the 6th RTT (or full window, that is, 36 KB) is 5 * 5 = 25 msec.

2|4|8|16|32|64

total time needed = 5 * 5 msec = 25 msec  

Reference:

http://web.eecs.utk.edu/~qi/teaching/ece453f06/hw/hw7_sol.htm 

Problem 4 (10)

when Time-Out is mentioned see what to do , in that case only we need Threshold value : https://gateoverflow.in/1794/gate2014-1_27

Comments

@Bikram Sir, why don't we calculate RTT for last packet?

please explain sir.

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@ Shubhanshu

it'll exceed the receiver's window. See my answer again and see the question ..

Therefore, the amount of time it takes is up to that 6th RTT  . So we don't need to consider RTT of last packet.

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if slow start is not mentioned and their is no congestion.Then I think we can send a full window in one go.