4330/6310 first assignment spring 2015 jehan-françois pâris
DESCRIPTION
AN EXAMPLETRANSCRIPT
4330/6310 FIRST ASSIGNMENTSpring 2015
Jehan-François Pâ[email protected]
The modelWe haveOne quad-core CPUOne diskOne input deviceThree queues
CPU queue "ready queue"
Disk queueDevice queue
RQ
DQ
Disk
CCCC
IQ
Input
AN EXAMPLE
P0 begins at t = 0
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30
RQ
DQ
Disk
CCCC
IQ
Input
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
P0 gets core until t = 0 + 200 = 200
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30
RQ
DQ
Disk
CCCC
IQ
Input
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
What's next?
RQ
DQ
Disk
CCCC
IQ
Input
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
P1 arrives at t = 100NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P1 gets core until t = 100 + 30 = 130NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
What's next?NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P1 waits for input until t = 130 +800 = 930 NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
What's next? NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P0 waits for input device NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
What's next?NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P1 gets core until t = 930 + 40 = 970P0 waits for input until t = 930 + 900 = 1830
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
What's next?
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P1 terminates at t = 970
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
Your program will display (I)
Process 1 terminates at t = 970Core 0 is IDLECore 1 is IDLECore 2 is IDLECore 3 is IDLEDisk is IDLEAverage number of busy cores:
Ready queue contains: --Disk queue contains: --
Your program will display (II)
Process ID Start time CPU time Status0 0 200 WAITING1 100 70TERMINATED
How to compute CPU utilization
Keep track of total time for all CPU requests:200 + 30 + 40 = 270 ms
Divide by elapsed time:270/970 = 0.278 (rounded)
Since there are four cores, the maximum CPU utilization is 4.0
What's next?
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P0 gets core until t = 1830 +10 = 1840
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
What's next?
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P0 gets disk until t = 1840 + 10 = 1850
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
What's next?
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P0 gets CPU until t = 1850 + 30 =1880
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
What's next?
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
P0 terminates at t = 1880
NEW 0START 0CPU 200INPUT 900CPU 10I/O 10CPU 30NEW 1START 100CPU 30INPUT 800CPU 40
RQ
DQ
Disk
CCCC
IQ
Input
Your program will display (I)
Process 0 terminates at t = 1880Core 0 is IDLECore 1 is IDLECore 2 is IDLECore 3 is IDLEDisk is IDLEAverage number of busy cores: 0.174
Ready queue contains: --Disk queue contains: --
Your program will display (II)
Process ID Start time CPU time Status0 0 240TERMINATED
How to compute CPU utilization
Keep track of total time for all CPU requests:200 + 10 + 30 + 30 + 40 = 310 ms
Divide by elapsed time:310/1880 = 0.165 (rounded)
Since there are four cores, the maximum CPU utilization is 4.0
HANDLINGZERO-DELAYDISK ACCESSES
Zero-delay disk accesses
Represent disk requests that can be satisfied without actually accessing the disk:Data can be read from the I/O bufferData are written to the I/O buffer
A process performing a zero-delay I/O request will:Skip the diskStart without further delay its next processing
step
An example (I)…CPU 5I/O 0CPU 20…
RQ
DQ
Disk
CCCC
IQ
Input
Assume next stepfor red process isI/O 0
An example (II)…CPU 5I/O 0CPU 20…
RQ
DQ
Disk
CCCC
IQ
Input
Red process willimmediatelyrequest a core(and get one becauseone or more coreswere idle)
ENGINEERING THE SIMULATION
Simulating time
Absolutely nothing happens to our model between two successive "events"
Events areArrival of a new processStart of a computing stepCompletion of a computing step
We associate an event routine with each event
Arrival event routine
Process first request of process It will always be a CPU request
CPU request routine
current time is clockrequest time is crt
if a core is free :mark core busy until clock + crt
add crt to corebusytimeselse :
enter process in ready queue
CPU request completion routine
if ready queue is empty :mark core idle
else:pick first process P' in ready queuecrt' is request time for P'mark core busy until clock + crt'
add crt' to corebusytimes proceed with next request for completing
process
Disk request routine
current time is clockrequest time is drt
if drt == 0: proceed with next process request
if disk is free :mark disk busy until clock + drt
add drt to diskbusytimeselse :
enter process request in disk queue
Disk request completion routine
if disk queue is empty :mark disk idle
else :pick first process request P' in disk queuedrt' is request time for P'mark disk busy until clock + drt'
add drt' to diskbusytimes proceed with next request for completing
process
Input request routine
current time is clockrequest time is irt
if input device is free :mark input device busy until clock + irt
else :enter process request in device queue
Input request completion routine
if input queue is empty :mark input device idle
else :pick first process request P' in
input queueirt' is request time for P'mark disk busy until clock + irt'
proceed with next request for completing process
The simulation scheduler
1. Find next event by looking at: CPU request completion times Disk request completion time time Input request completion time Arrival time of next process
2. Set current time to event time3. Process event routine4. Repeat until all processes are done
Organizing our program (I)
Most steps of simulation involve scheduling future completion events
Associate with each completion event an event noticeTime of eventDevice Process sequence number
Organizing our program (II)
Do the same with process startsTime of eventProcess start Process sequence number
Organizing our program (III)
Process all event notices in chronological orderRelease
disk247
Releasecore250
Newprocess
245
Newprocess
270
Newprocess
310
First notice tobe processed
Organizing our program (IV)
Keep the event list sorted (priority queue)
Releasedisk247
Releasecore250
Newprocess
245
Newprocess
270
Newprocess
310
First notice to be processed is head of the list
Organizing our program (V)
Overall organization of main program
schedule first event # will be a process startwhile event list is not empty :
process next event in listprint simulation results
Organizing our event list
Priority queue Two kinds of entries
Computational steps completion times: Created and inserted "on the fly"
Process arrivals: Created during input phase Already sorted
An implementation
My main data structures would be:Data tableProcess tableDevice table
The data table
Stores the input data Line indices are used
in process table
Operation Parameter
NEW 5
CPU 10
INPUT 0
CPU 20
NEW 20
CPU 50
… …
The process table (I)
StartTime
First Line
Last Line
CurrentLine
5 0 3 varies
20 4 … …
… … … …
The process table (I)StartTime
First Line
Last Line
CurrentLine
5 0 3 varies
20 4 … …
… … … …
The process table (II)
One line per processLine index is process sequence number!
First column has start time of process First line, last line and current line respectively
identify first line, last line and current line of the process in the input table
Last column is for processes waiting for end of an INPUT step
The device table (I)DeviceDevice Process Completion time
Core 0Core 0 P0 15Core1Core1Core 2Core 2Core 3Core 3 - -DIskDIskInputInput - -
First column is not needed:use row index to specify device
The device table (II)
One line per device (core, disk and Input) Contains
A busy/free flagA process sequence number if device is busyA completion time
Zero (or a very large value) if device is free
Finding the next event (I)
If you use a priority list for your events, you should start by "seeding" your priority list with the arrival times of all processes
After that, you add successive events to the list each time you can allocate a resource to a process.
Your simulation will end once the list is empty (and you cannot add anything to it).
Finding the next event (II)
If you do not use a priority list for your events, you can find the next event to process by searching the lowest value in The process start times in the process tableCompletion times of INPUT stepsThe completion times in the device table
A full list implementation
INT…0
INT…5
RUN10
RUN20
… …
…
Very elegant but harder to debug
Reading your input
You must use I/O redirectionassign1 < input_file
AdvantagesVery flexibleProgrammer writes her code as if it was read
from standard input No need to mess with fopen(), argc and argcv