>For Q8, I still don't understand the equation 
>\alpha=pdf(t)/(1-CDF(t)) and don't know why the (b) is true. 
>I couldn't find this equation on the course slides, could you provide 
>some explanation of it?

See slide #25.


>For Q16, why we don't simply use D=1/\mu to obtain the response 
>time (where D is service time, and \mu is service rate)? 
>Since response time = wait time + service time and the wait time = 0 
>as \mu > \lambda, I think response time = 0 + D = 1/\mu. I know this result is 
>different than the answer, but I couldn't figure out what's wrong with my steps. 
>Could you help me and point out the issue of my steps?

Wait time is not zero even when \mu > \lambda because customers will wait in the queue 
while the server is busy servicing a customer. Wait time is zero only if there is 
always a free server available when a customer arrives such as an IS service center 
with an infinite number of servers.

Q16 is for M/M/1, so there is only a single server in the system and the wait 
time is certainly not zero.
 


>For Q17, why the (a) is not true, and (c) is true? 
>Since we allowing the repair, I think we cannot calculate the reliability.


"(a) we cannot calculate the reliability of the system since it has repair capabilities" is 
not true. An example is in your HW#2, Problem#2 for which you can use a Markov model 
for reliability calculation.
"(c) we can calculate the availability of the system using a fault tree model" is true 
because failure/repair of each component and therefore the availability of each component 
is independent of other components, so the TMR system availability is simply a 
function of individual component availabilities and can be simply solved using a 
non-state-based model such as a fault tree model. 
Of course, it can be solved by using a Markov model but the solution time complexity is higher.
 

>For Q20, why the (c) is not true? Since 

Q20 is about a M/M/3 system with three servers each with a service rate of \mu. 
Since the M/M/3 system does not have a queue size limitation, the throughput 
is equal to the arrival rate \lambda. so "(c) the throughput is (1-P0)*\lambda" 
is not true. You can also get 
throughput X = P1 * \mu + P2 * 2\mu + summation{i=3 to infinity} (Pi*3\mu) which is 
equal to \lambda after applying "local balance equations" (see slide #73 
about local balance equations).

Note that for a queuing system with queue size limitation (i.e., the 4th 
parameter is a number), the throughput is (1-rejection probability)* arrival rate. 
For Q20, the rejection probability is zero because there is no queue size limitation, 
so the throughput is equal to the arrival rate.