$13\mathrm{.}6$ Consider compressed video transmission in an ATM network. Suppose standard ATM cells must be transmitted through $5$ switches. The data rate is $43$ Mbps$.$

a$.$ What is the transmission time for one cell through one switch?

b$.$ Each switch may be transmitting a cell from other traffic all of which we assume to have lower $($non$-$preemptive for the cell$)$ priority. If the switch is busy transmitting a cell, our cell has to wait until the other cell completes transmission. If the switch is free, our cell is transmitted immediately. What is the maximum time from when a typical video cell arrives at the first switch $($and possibly waits$)$ until it is finished being transmitted by the fifth and last one? Assume that you can ignore propagation time, switching time, and everything else but the transmission time and the time spent waiting for another cell to clear a switch.

c$.$ Now suppose we know that each switch is utilized $60\%$ of the time with the other low$-$priority traffic. By this we mean that with probability $0\mathrm{.}6$ when we look at a switch it is busy. Suppose that if there is a cell being transmitted by a switch, the average delay spent waiting for a cell to finish transmission is one$-$half a cell transmission time. What is the average time from the input of the first switch to clearing the fifth?

d$.$ The measure of most interest is not delay, but jitter, which is the variability in the delay. Use parts $($b$)$ and $($c$)$ to calculate the maximum and average variability, respectively, in the delay.

In all cases assume that the various random events are independent of one another; for example, we ignore the burstiness typical of such traffic.