Frames arrive randomly at a 100 Mbps channel for transmission. If the channel is busy when a frame arrives, it waits its turn in a queue. Frame length is exponentially distributed with a mean of 10,000 bits/frame. For each of the following frame arrival rates, give the delay experienced by the average frame, including both queuing time and transmission time.
a. 90 frames/sec
b. 900 frames/sec
c. 9000 frames/sec
Please explain all work.© BrainMass Inc. brainmass.com March 4, 2021, 6:05 pm ad1c9bdddf
The standard formula for Markov queuing is T = 1/(mu*C - lambda), where
1/mu = mean frame length
C = Channel capacity = 100*10^6 bits/second
lambda = arrival rate
Here 1/mu ...
For each of the frame arrival rates, this solution provides the delay experienced by the average frame, including both queuing time and transmission time.