A storage system consists of a single disk drive The average

A storage system consists of a single disk drive. The average time to service an I/O request is 80 milliseconds. The I/O requests arrive to the disk system at a rate of 10 requests per second. Using an M/M/1 model for this system, determine the following: Average disk drive utilization. Probability of the system being idle. Probability of queueing Average number of jobs in the system. average number of jobs waiting in the queue. Mean response time Average waiting time.

Solution

For the given data:
Let arrival rate be L and service rate be U
here L=10 and U=1/0.08=12.5 sec
a)The average disk drive utilization=Arrival rate * Time(in sec)
(let p) =10 * 0.08(in sec)
               =0.8 sec
b)Prob of system being idle = 1-P
= 1-0.8
=0.2
c)Prob of queueing = 1-P^2
= 1-0.8^2
=0.36
d)Average no of jobs = P/1-P
in the system = 0.8 / 1-0.8
=0.8/0.2
= 4
e) AVerage no of jobs waiting
in the queue       = P^2 / 1-P
= 0.8^2 / 0.2
= 3.2
f)Mean Response Time = 1/U / 1-P
=12.5/0.2
= 62.5
g)Average waiting Time = P* (1/U) / (1-P)
=0.8*(62.5)
=50