mem alloc and process scheduling grogramming

8/19/2019 mem alloc and process scheduling grogramming

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Dynamic Storage-Allocation Problem

• First-ft: Allocate the frst hole that is bigenough

• Best-ft: Allocate the smallest hole that isbig enough; must search entire list, unlessordered by size – Produces the smallest leftover hole

• Worst-ft: Allocate the largest hole; mustalso search entire list –

Produces the largest leftover hole

How to satisfy a request of sizen from a list offree holes?

First-fit and best-fit better than worst-fit in terms

of speed and storage utilization


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Data Structures:

Free list & allocation list


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First Fit :

i/:! re"uired memory## $%

• First-ft: Allocate the frst hole that isbig enough


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After rst t


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'est Fit :

i/:! re"uired memory## (%

• Best-ft: Allocate the smallest holethat is big enough; must search entirelist, unless ordered by size


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After 'est Fit


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)orst Fit :

i/:! re"uired memory## (%

• Worst-ft: Allocate the largest hole;must also search entire list


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After )orst Fit


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De allocation

• De allocates the re"uested rocessfrom allocation list to free list

• i/:! rocess no

• *g: +


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After de allocation


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Process

• a rogram in eecution, -hich formsthe basis of all comutation

• rocess eecution must rogress inse"uential fashion

• Program is assive entity, rocess is

active – Program becomes rocess -hen

eecutable le loaded into memory


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Process Scheduling

• .aimize P0 use, "uic1ly s-itch rocesses

onto P0 for time sharing• Process scheduler selects among available

rocesses for net eecution on P0

• .aintains scheduling queues of rocesses – Job queue 2 set of all rocesses in the system

– Ready queue 2 set of all rocesses residing inmain memory, ready and -aiting to eecute

– Device queues 2 set of rocesses -aiting for an3/4 device

– Processes migrate among the various "ueues


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P0 Scheduler

• Selects from among the rocesses in ready "ueue,

and allocates the P0 to one of them – 5ueue may be ordered in various -ays

• P0 scheduling decisions may ta1e lace -hen arocess:

+6S-itches from running to -aiting state

76S-itches from running to ready state

86S-itches from -aiting to ready

96erminates

• Scheduling under + and 9 is nonreemtive

• All other scheduling is reemtive –.onsider access to shared data

–.onsider reemtion -hile in 1ernel mode

–.onsider interruts occurring during crucial 4S activities


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onreemtive scheduling

• 0nder nonreemtive scheduling,once the P0 has been allocated to arocess, – the rocess 1ees the P0 until it

releases the P0 either by terminatingor by s-itching to the -aiting state6


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Scheduling riteria• !P" utili#ation 2 1ee the P0 as busy as ossible

• $hroughut 2 < of rocesses that comlete theireecution er time unit

• $urnaround time 2 amount of time to eecute aarticular rocess

• Waiting time 2 amount of time a rocess has been-aiting in the ready "ueue

• Resonse time 2 amount of time it ta1es from-hen a re"uest -as submitted until the rstresonse is roduced, not outut =for time!sharing

environment>


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Scheduling Algorithm 4timizationriteria

• .a P0 utilization

• .a throughut

• .in turnaround time

• .in -aiting time

• .in resonse time


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First-!ome% First-Served &F!FS'Scheduling

Process 'urst ime

P1 79 P2 8

P3 8 • Suose that the rocesses arrive in the order:

P1 , P2 , P3 he ?antt hart for the schedule is:

• )aiting time for P1 @ %; P2 @ 79; P3 @ 7

• Average -aiting time: =% B 79 B 7>/8 @ +

P1 P2 P3

24 27 300


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FFS Scheduling =ont6>

Suose that the rocesses arrive in the order:

P2 , P3 , P1

• he ?antt chart for the schedule is:

• )aiting time for P1= C;

P2 @ %

;P3= 8

• Average -aiting time: =C B % B 8>/8 @ 8

• .uch better than revious case

• !onvoy e(ect ! short rocess behind long rocess – onsider one P0!bound and many 3/4!bound rocesses

P1P3P2

63 300


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Shortest!ob!First =SF> Scheduling

• Associate -ith each rocess the

length of its net P0 burst – 0se these lengths to schedule the

rocess -ith the shortest time

• SF is otimal 2 gives minimumaverage -aiting time for a given set

of rocesses – he diEculty is 1no-ing the length of the

net P0 re"uest

– ould as1 the user


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*amle of SF ProcessArr iva e'urst ime

P1 %6% C

P2 76%

P3 96%

P4 $6% 8• SF scheduling chart

• Average -aiting time @ =8 B +C B ( B %> / 9 @

P4 P3P1

3 160 9

P2

24


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Determining Gength of et P0 'urst

• an only estimate the length 2

should be similar to the reviousone

– hen ic1 rocess -ith shortestredicted net P0 burst

• Preemtive version called

shortest-remaining-time-frst


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*amle of Shortest-remaining-time-frst

• o- -e add the concets of varying arrival times andreemtion to the analysis

ProcessA arri Arrival ime 'urst ime

P1 %

P2 + 9

P3 7 ( P4 8 $

• Preemptive SF ?antt hart

• Average -aiting time @ C6$ msec

P1 P1P2

1 170 10

P3

265

P4

am e o or es rema n ng


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am e o or es -rema n ng-time-frst ProcessA arri Arrival ime 'urst ime

P1 %

P2 + 9

P3

7 (

P4 8 $

• Heeat the follo-ing until cl1 I sumburst

• For rocess + to n

•

hec1 PJiK6atimeIcl1 and JiK6rtimeL% – oy the rocess JiK into ready "ueue r"JMK

– Select min from ready "ueue6= for sMf min=burst> forriority min=ri>

–

3f reemting the rocess in eecution rint ganttchart

– Do rocess scheduling• Decrement r"JMK6rtime of rocessing rocess

• 3ncrement r"JILMK6 -time of other rocesses in r"

• 3ncrement cl1


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Priority Scheduling

• A riority number =integer> is associated -ith eachrocess

• he P0 is allocated to the rocess -ith the highestriority =smallest integer ≡ highest riority> –

Preemtive – onreemtive

• *"ual riority rocesses are scheduled in FFS order

• Problem ≡ Starvation 2 lo- riority rocesses may

never eecute

• Solution ≡ Aging 2 as time rogresses increase theriority of the rocess


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*amle of Priority Scheduling

ProcessA arri 'urst ime Priority

P1 +% 8 P2 + +

P3 7 9

P4 + $

P5$ 7• Priority scheduling ?antt hart

• Average -aiting time @ 67 msec

P2 P3P5

1 180 16

P4

196

P1

H d H bi =HH>


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Hound Hobin =HH>

• *ach rocess gets a small unit of P0 time =timequantum ">, usually +%!+%% milliseconds6 Afterthis time has elased, the rocess is reemtedand added to the end of the ready "ueue6

• 3f there are n rocesses in the ready "ueue andthe time "uantum is q, then each rocess gets

+/n of the P0 time in chun1s of at most q timeunits at once6 o rocess -aits more than =n!+>qtime units6

• imer interruts every "uantum to schedule net

rocess• Performance

– q large ⇒ F3F4

– q small ⇒ q must be large -ith resect to

contet s-itch, other-ise overhead is too high


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*amle of HH -ith ime 5uantum @ 9

Process 'urst ime

P179

P2 8

P3 8

• he ?antt chart is:

• yically, higher average turnaround than SF, but betterresponse

• " should be large comared to contet s-itch time

• " usually +%ms to +%%ms, contet s-itch I +% usec

P1 P2 P3 P1 P1 P1 P1 P1

0 4 7 10 14 18 22 26 30

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mem alloc and process scheduling grogramming