LN x 625 X 107 60 X J07 J J X Jo4 seconds

After the data frame has reached its intended destination, it must circulate through another (on average) N/2 hosts to return to the originating host, which then removes the frame from the network. When this occurs, the originating station then generates a new token onto the network. The time it takes for the frame to circulate through half of the network is then represented as [(A/2) (6.25 x 10-7 seconds)]. This time is then added to the time determined from the previous step, deriving the following:

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