1. Single server model
Queue
Server
I/O Controller
And
device
Arrivals

2. Timesys = Timeq + Times

3. Times = Average time to server
task
Timeq = Average time per task in
the queue
Timesys= Aver time /task
Arrival Rate = λ = Average number of
arriving tasks
Lengths = Average number of task in service
Lengthq = Average length of queue
Lengthsys= Lengths +Lengthq

4. Server Utilization
= Arrival Rate x Times

5. Example#1 Problem Statement
Suppose an I/O system with a single disk gets on average 100 I/O requests/second. Assume that average time for a disk to service an I/O request is 5ms. What is the utilization of the I/O system?

6. Solution Time for an I/O request = 5ms =0.005sec
Server utilization = 100 x 0.005
= 0.5

7. Poisson Distribution Probability (k)= (e-k x ak) /k!
a= Rate of events x Elapsed time

8. Variance
C2 =
(Arithmetic mean time)2

9. Average Residual Service Time
= ½ x weighted mean time x (1+C2)

10. All the tasks in the queue ahead of the new task must be completed.
Each task takes on average Times.
A task on server takes average residual service time to complete.
The chance that the server will be busy is called server utilization.

11. Timeq = Lengthq x Times
+ (Server utilization
x Average residual service time )
Suppose:
Server utilization = β
Average residual = λ
service time

12. Timeq = server utilization x times
+ (Arrival time x Time q) x Times
Timeq = β x Times
+ (Arrival time x Timeq) x Times
Timeq = (Times x β ) / (1 – β)

13. Example#2
For the system of previous example having server utilization of 0.5, what is the mean number of I/O requests in the queue?

14. Solution (Server utilization )2 Lengthq = (1- Server utilization )

15. Example#3
Suppose a processor sends 10 disk I/O per second, these requests are exponentially distributed, and the average service time of an older disk is 10ms. Answer the following questions:
What is the number of requests in the queue?
What is the average time a spent in the queue?
What is the average response time for a disk request?

16. Solution Average number of arriving tasks/second = 20
Average disk time = 10ms = 0.01sec
Sever utilization = 20 x 0.01=0.2
Timeq = 10ms x 0.2/(1-0.2) = 2.5ms
Average response time = =22.5ms

17. M/M/m Model of Queuing Theory
Arrival Rate x Time
Utilization =
Nserver

18. M/M/m Model of Queuing Theory
Lengthq = Arrival Rate x Timeq
(Time s x (Ptasks>= Ns))
Timeq =
Ns x (1- utilization)

19. M/M/m Model of Queuing Theory
Ns x utilization
Probtasks>= Ns = x Prob0task
Ns! x (1-utilization)

20. Example#4
Suppose instead of a new, faster disk, we add a second slow disk, and duplicate the data so that read can be serviced by either disk. Let’s assume that the requests are all reads. Recalculate the answers to the earlier questions, this time using an M/M/m queue.

21. Solution The average utilization of the two disks is given as;
Arrival rate x Times
Server utilization = Ns
= (40 x 0.02) / 2
= 0.4

22. Probability of no tasks in the queue
(2 x utilization)2
Prob0tasks =
2! x (1- utilization)
(2 x utilization)n n! -1

23. (2x 0.4) 2 Prob0tasks = 1 + ---------------- + (2 x 0.4) 2! X (1- 0.4)
= ( ) -1 = -1

24. (2 x utilization)2
Prob tasks >=Ns = x Prob0tasks
2! x (1- utilization)
(2x 0.4) 2
= x
2! x (1- 0.4) =

25. Time waiting in the queue:
Prob tasks >=Ns
Time q = Times x
Ns x (1- utilization)
= x 0.229/ ( 2 x 0.6)
= 3.8msec
Average response time = 20msec + 3.8msec
= 23.8msec

26. Bench Mark=C: Complex query OLTP
Data Size (GB) =
Performance metric = New order transaction per minute
Date of 1st result = 1992

27. TPC-C TPC-C is measured In transaction per min (tpm).
In price of system, including hardware, and software.

28. Synchronous I/O The operating system requests the data
Then switches to another process
When requested data arrives, then operating switches back to the requesting process.

29. Asynchronous I/O Allows the processor to make continuous requests.
Shares the same philosophy as caches in out-of- order CPUs.