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Consider the efffect of using slow start on a line with 10msec RTT.The receiver window and size of congestion window are 38KB and 36KB respectively.Sender side threshold is set to 18KB.After 8 transmission timeout occurs,after timeout the time taken to send first full window of 18KB is ________________(in msec).Assume window size at the start of slow start phase is 2KB

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For congestion control policy , the condition which is needed is :

congestion window size  <  receiver window size

which holds here as maximum congestion window size < receiver window size..

Initially window size  =  2 KB ( 1 MSS )   i.e. at t = 1 , window size  =  2 KB

                                                                  at t = 2 , window size  = 4 KB

                                                                  at t = 3 , window size  = 8 KB

                                                                  at t = 4 , window size  = 16 KB

                                                                  at t = 5 , window size  = 18 KB [threshold reached , after this congestion avoidance]

                                                                  at t = 6 , window size  =  20 KB

                                                                  at t = 7 , window size  =  22 KB

                                                                  at t = 8 , window size  =  24 KB

                                                                  at t = 9 , window size  =  26 KB

During this transmission(means same as after 8 transmissions) , timeout occurs and hence transmission at t = 9 fails..

So applying multiplicative decrease (as timeout means strong possibility of congestion).

So now new value of  threshold   =  Current window size / 2      =    13 KB and slow start phase starts again

                                                        Now   at  t  =  10 , window size  =  2 KB

                                                                 at  t  =  11 , window size  =  4 KB

                                                                at   t  =  12 , window size  =  8 KB

                                                                at   t  =  13 , window size  = 13 KB (threshold value reached)

                                                                at   t  =   14 , window size  =  15 KB

                                                                at   t  =  15 , window size   = 17 KB

                                                                at   t   =  16 , window size   = 19 KB

Hence number of RTTs involved to send data of 18 KB   =  7 RTT (as till t = 15 , we can send only 17 KB but we need to send 18 KB so we need one more RTT and timeout occurs during t = 9 , so we need to see the time needed after timeout which is at t = 9 , so we need to consider time interval between t = 9   to   t = 16).

Hence time taken        =                           7 RTT      =    7 *  10

                                                                             =    70 ms

                  

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