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UNIT-1

PART-A

1) What is virtual machine?

The application program view everything under them in the hierarchy as though thesystem programs were part of the machine itself. This layered approach is taken as

logical conclusion in the concept of a virtual machine.

2) What is an operating system?

An operating system is a program that manages the computer hardware. It act as an

intermediate between a users of a computer and the computer hardware. It controls

and coordinates the use of the hardware among the various application programs for

the various users.

3) i!!erentiate Tightly couple" systems an" loosely couple" systems?

Loosely coupled systems Tightly coupled systems Each processor has its own local

memory ommon memory is shared by many processors Each processor cancommunicate with other all through communication lines !o need of any special

communication lines

PART-#

1) istinguish $et%een the client-server an" peer-to-peer mo"els o! "istri$ute"

systems.

AN&W'R(

The client-server mo"elfirmly distinguishes the roles of the client and server.

"nder this model# the client re$uests services that are provided by the server.

%erver systems can be broadly categori&ed as compute servers and fileservers'

The compute-server systemprovides an interface to which a client can send a

re$uest to perform an action (for eample# read data)* in response# the server

eecutes the action and sends back results to the client. A server running a

database that responds to client re$uests for data is an eample of such a

system.

The !ile-server system provides a file+system interface where clients can

create# update# read# and delete files. An eample of such a system is a web

server that delivers files to clients running web browsers.

The peer-to-peer mo"eldoesn,t have such strict roles. In fact# all nodes in the

system are considered peers and thus may act as either clients or servers + or both. A

node may re$uest a service from another peer# or the node may in fact provide such a

service to other peers in the system.

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-etermining what services are available is accomplished in one of two general

ways'

hen a node /oins a network# it registers its service with a centrali&ed lookup

service on the network. Any node desiring a specific service first contacts this

centrali&ed lookup service to determine which node provides the service. The

remainder of the communication takes place between the client and the service

provider.

A peer acting as a client must first discover what node provides a desired

service by broadcasting a re$uest for the service to all other nodes in the network. The

node (or nodes) providing that service responds to the peer making the re$uest. To

support this approach# a discovery protocol must be provided that allows peers to

discover services provided by other peers in the network.

0or eample# let,s consider a system of nodes that share cooking recipes.

"nder the client+server model# all recipes are stored with the server. If a client wishes

to access a recipe# it must re$uest the recipe from the specified server. "sing the peer+to+peer model# a peer node could ask other peer nodes for the specified recipe. The

node (or perhaps nodes) with the re$uested recipe could provide it to the re$uesting

node. !otice how each peer may act as both a client (i.e. it may re$uest recipes) and

as a server (it may provide recipes.)

) '*plain $rie!ly a$out the services an" !unctions provi"e" $y the operating

system+AN&W'R(

An operating system provides an environment for the eecution of programs. It provides

certain servicesto programs and to the users of those programs.

User inter!ace+ Almost all operating systems have a user inter!ace ("I). This

interface can take several forms. 1ne is a comman"-line inter!ace ,.I)/ which uses

tet commands and a method for entering them (say# a program to allow entering and

editing of commands). Another is a $atch inter!ace/ in which commands and

directives to control those commands are entered into files# and those files are

eecuted. ost commonly a graphical user inter!ace ,0UI) is used.

Program e*ecution+ The system must be able to load a program into memory and to

run that program. The program must be able to end its eecution# either normally orabnormally (indicating error).

I2 operations+ A running program may re$uire I31# which may involve a file or an

I31 device. 0or specific devices# special functions may be desired.

ile-system manipulation+ The file system is of particular interest. 1bviously#

programs need to read and write files and directories. They also need to create and

delete them by name# search for a given file# and list file information.

ommunications+ There are many circumstances in which one process needs to

echange information with another process. ommunications may be implemented

via shared memory or through message passing# in which packets of information are

moved between processes by the operating system.

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'rror "etection+ The operating system needs to be constantly aware of possible

errors. Errors may occur in the 4" and memory hardware (such as a memory error

or a power failure)# in I31 devices (such as a parity error on tape# a connection failure

on a network# or lack of paper in the printer)# and in the user program.

UNTI2N&(

Resource allocation+ hen there are multiple users or multiple /obs running at the

same time# resources must be allocated to each of them. any different types of

resources are managed by the operating system. %ome (such as 4" cycles# main

memory# and file storage) may have special allocation code# whereas others (such as

I31 devices) may have much more general re$uest and release code.

Accounting+ e want to keep track of which users use how much and what kinds of

computer resources. This record keeping may be used for accounting (so that users

can be billed) or simply for accumulating usage statistics.

Protection+ 4rotection involves ensuring that all access to system resources iscontrolled. %ecurity of the system from outsiders is also important. %uch security

starts with re$uiring each user to authenticate himself or herself to the system# usually

by means of a password# to gain access to system resources.

3) '*plain $rie!ly system calls an" their types+

AN&W'R(

&ystem callsprovide an interface to the services made available by an operating system.

Application developers often do not have direct access to the system calls# but can access

them through an application programming interface (A4I). The functions that are included in

the A4I invoke the actual system calls. 5y using the A4I# certain benefits can be gained'

4ortability' as long a system supports an A4I# any program using that A4I can

compile and run.

Ease of "se' using the A4I can be significantly easier then using the actual system

call.

&ystem all Parameters(

Three general methods eist for passing parameters to the 1%'

6. 4arameters can be passed in registers.

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2. hen there are more parameters than registers# parameters can be stored in a block

and the block address can be passed as a parameter to a register.

7. 4arameters can also be pushed on or popped off the stack by the operating system.

Types o! &ystem calls(

There are 8 different categories of system calls' process control# file manipulation# device

manipulation# information maintenance and communication.

Process ontrol( A running program needs to be able to stop eecution either

normally or abnormally. hen eecution is stopped abnormally# often a dump of

memory is taken and can be eamined with a debugger.

ile 4anagement( %ome common system calls are create# delete# read# write# repositi

on# or close. Also# there is a need to determine the file attributes 9 get and set file

attribute. any times the 1% provides an A4I to make these system calls.

evice 4anagement( 4rocess usually re$uire several resources to eecute# if these

resources are available# they will be granted and control returned to the user process.

These resources are also thought of as devices. %ome are physical# such as a video

card# and others are abstract# such as a file.

"ser programs re$uest the device# and when finished they release the device. %imilar

to files# we can read# write# and reposition the device.

In!ormation 4anagement( %ome system calls eist purely for transferring

information between the user program and the operating system. An eample of this

is time# or date. The 1% also keeps information about all its processes and provides

system calls to report this information. ommunication( There are two models of inter process communication# the

message+passing model and the shared memory model.

essage+passing uses a common mailbo to pass messages between processes.

%hared memory use certain system calls to create and gain access to create and

gain access to regions of memory owned by other processes. The two processes

echange information by reading and writing in the shared data.

5) Defne the essential properties o the ollowing types o operatingsystem:

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(i) Batch(ii) Real Time(iii) Time sharing(i) !l"stere#

#AT6(

:obs with similar needs are batched together and run through the computer as a group

by an operator or automatic /ob se$uencer. 4erformance is increased by attempting to keep

4" and I31 devices busy at all times through buffering# off+line operation# spooling# and

multi+programming. 5atch is good for eecuting large /obs that need little interaction* it can

be submitted and picked up later.

Real Time(

1ften used in a dedicated application# this system reads information from sensors and must

respond within a fied amount of time to ensure correct performance .

Time sharing( This systems uses 4" scheduling and multiprogramming to provide

economical interactive use of a system. The 4" switches rapidly from one user to another.

Instead of having a /ob defined by spooled card images# each program reads its net controlcard from the terminal# and output is normally printed immediately to the screen.

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lustere"(A clustered system combines multiple computers into a single system to perform

computational task distributed across the cluster.

i. ;andheld' A small computer system that performs simple tasks such as cal

endars# email# and web browsing. ;andheld systems differ from traditional

desktop systems with smaller memory and display screens and slower processors.

UNIT-5

PART-A1) What is the "i!!erence $et%een logical an" physical a""resses?

An address generated by the 4" is commonly referred to as a logical address# whereas an

address seen by the memory unit is commonly referred to as physical address.

2) What is e*ternal !ragmentation?

Eternal fragmentation eists when enough total memory space eists to satisfy a re$uest# butit is not contiguous* storage is fragmented into a large number of small holes.

7)What is the use o! vali"-invali" $its in paging?

hen the bit is set to valid# this value indicates that the associated page is in the process,s

logical address space# and is thus a legal page. If the bit is said to invalid# this value indicates

that the page is not in the process,s logical address space. "sing the valid+invalid bit traps

illegal addresses.

PART-#

1) iscuss in "etail the paging concept?

Ans%er(

Logical address space of a process can be non contiguous* process is

allocated physical memory whenever the latter is available.

-ivide physical memory into fied+si&ed blocks called !rames (si&e is

power of 2# between 862 bytes and eep

track of all free frames.

To run a program of si&e npages# need to find n free frames and load

program

%et up a page table to translate logical to physical addresses

A""ress Translation &cheme

Address generated by 4" is divided into

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Page num$er ,p)9 used as an inde into apage table which contains base

address of each page in physical memory

Page o!!set ,")9 combined with base address to define the physical memory

address that is sent to the memory unit0or given logical address space 2mand page si&e 2n

Paging 6ar"%are

Paging Model of Logical and Physical Memory

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Eample of the page table'

onsider the memory using a page si&e of ? bytes and a physical memory of 72

bytes.

3-$yte memory an" 5-$yte pages

Free Frames:

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hen a process arrives in the system to be eecuted# its si&e is epressed

in pages.

Each page of the process needs one frame. Thus if the process re$uires n

pages at least n frames must be available.

The first page of the is loaded into any one of the frame and frame

number is put in the table for this process.

The net page is allocated into another frame and its frame number is put

into the page table# and so on.

Implementation of page table'

4age table is kept in main memory

Page-ta$le $ase register ,PT#R)points to the page table

Page-ta$le length register ,PR.R) indicates si&e of the page table

In this scheme every data3instruction access re$uires two memoryaccesses. 1ne for the page table and one for the data3instruction.

The two memory access problem can be solved by the use of a special

fast+lookup hardware cache called associative memory or translation

loo7-asi"e $u!!ers ,T.#s)

%ome TL5s store a""ress-space i"enti!iers ,A&Is) in each TL5 entry

9 uni$uely identifies each process to provide address+space protection for

that processAssociative 4emory

Associative memory 9 parallel search

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Address translation (p# d)

If p is in associative register# get frame @ out

1therwise get frame @ from page table in memory

Paging 6ar"%are %ith T.#

'!!ective Access Time

Associative Lookup e time unit

Assume memory cycle time is 6 microsecond;it ratio 9 percentage of times that a page number is found in the associative

registers* ratio related to number of associative registers

;it ratio an '!!ective Access Time (EAT)

EAT (6 B e) a B (2 B e)(6 9 a)

2 B e 9 a.

,) What is eman" Paging iscuss?

A demand paging is similar to a paging system with swapping. hen we

want to eecute a process# we swap it into memory. Cather than swapping the

entire process into memory.

hen a process is to be swapped in# the pager guesses which pages will

be used before the process is swapped out again Instead of swapping in a whole

process# the pager brings only those necessary pages into memory. Thus# it

avoids reading into memory pages that will not be used in anyway# decreasing

the swap time and the amount of physical memory needed.

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;ardware support is re$uired to distinguish between those pages that are

in memory and those pages that are on the disk using the valid+invalid bit

scheme. here valid and invalid pages can be checked checking the bit and

marking a page will have no effect if the process never attempts to access the

pages. hile the process eecutes and accesses pages that are memory

resident# eecution proceeds normally.

0ig. Transfer of a paged memory to continuous disk space

Access to a page marked invalid causes a page+fault trap. This trap is the

result of the operating systemDs failure to bring the desired page into memory.

5ut page fault can be handled as following

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%teps in handling a page fault

6. e check an internal table for this process to determine whether the referencewas a valid or invalid memory access.

2. If the reference was invalid# we terminate the process. If .it was valid# but we

have not yet brought in that page# we now page in the latter.

7. e find a free frame.

?. e schedule a disk operation to read the desired page into the newly

allocated frame.

8. hen the disk read is complete# we modify the internal table kept with the

process and the page table to indicate that the page is now in memory.

. e restart the instruction that was interrupted by the illegal address trap. The

process can now access the page as though it had always been memory.

Therefore# the operating system reads the desired page into memory and restarts

the process as though the page had always been in memory.

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The page replacement is used to make the frame free if they are not in used. If

no frame is free then other process is called in.

A"vantages o! eman" Paging(

6. Large virtual memory.

2. ore efficient use of memory.

7. "nconstrained multiprogramming. There is no limit on degree of

multiprogramming.

isa"vantages o! eman" Paging(

?. !umber of tables and amount of processor over head for handling page

interrupts are greater than in the case of the simple paged managementtechni$ues.

8. -ue to the lack of an eplicit constraints on a /obs address space si&e.

,3) '*plain in "etail a$out segmentation %ith paging+

%egmentation is a memory+management scheme that supports this user

view of memory. A logical address space is a collection of segments.

Each segment has a name and a length.

The addresses specify both the segment name and the offset within the

segment.

The user therefore specifies each address by two $uantities' a segment

name and an offset.

0or simplicity of implementation# segments are numbered and are referred to by

a segment number# rather than by a segment name. Thus# a logical address

consists of a two tuple:

F segment+number# offset G.

!ormally# the user program is compiled# and the compiler automatically

constructs segments reflecting the input program.

A compiler might create separate segments for the following'

6. The code

2. Hlobal variables

7. The heap# from which memory is allocated

?. The stacks used# by each thread

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8. The standard library

The user can now refer to ob/ects in the program by a two+dimensional address#

the actual physical memory is still# of course# a one+dimensional se$uence of

bytes. Thus# we must define an implementation to map two dimensional user

defined addresses into one+dimensional physical addresses.

This mapping is effected by a segment table. Each entry in the segment

table has asegment base and asegment limit.

The segment base contains the starting physical address where the

segment resides in memory# whereas the segment limit specifies the

length of the segment.

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A logical address consists of two parts' a segment number#s, and an

offset into that segment# d.

The segment number is used as an inde to the segment table. The offset

d of the logical address must be between and the segment limit.

If it is not# we trap to the operating system (logical addressing attempt

beyond# end of segment). hen an offset is legal# it is added to the segment base to produce the

address in physical memory of the desired byte.

'*ample(

e have five segments numbered from through ?. The segments are

stored in physical memory as shown.

The segment table has a separate entry for each segment# giving the

beginning address of the segment in physical memory (or base) and the

length of that segment (or limit).

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,5) '*plain in "etail the structure o! the page ta$le

The most common techni$ues for structuring the page table.

;ierarchical 4aging

;ashed 4age Tables

Inverted 4age Tables

6ierarchical Paging

ost modern computer systems support a large logical address space (272

to 2?). In such an environment# the page table itself becomes ecessively large.

0or eample# consider a system with a 72+bit logical address space.

If the page si&e in such a system is ? >5 (262

)# then a page table mayconsist of up to 6 million entries (2723262).

1ne simple solution to this problem is to divide the page table into

smaller pieces. A simple techni$ue is a two+level page table.

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T%o-.evel Paging '*ample

A logical address (on 72+bit machine with 6> page si&e) is divided into'

a page number consisting of 22 bits

a page offset consisting of 6 bits

%ince the page table is paged# the page number is further divided into'

a 62+bit page number

a 6+bit page offset

Thus# a logical address is as follows' wherepi is an inde into the outer

page table# andp2 is the displacement within the page of the outer page

table

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Address-Translation Scheme

6ashe" Page Ta$les

ommon in address spaces G 72 bits

The virtual page number is hashed into a page table

This page table contains a chain of elements hashing to the same location

Jirtual page numbers are compared in this chain searching for a match

If a match is found# the corresponding physical frame is etracted

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Inverte" Page Ta$le

1ne entry for each real page of memory

Entry consists of the virtual address of the page stored in that real

memory location# with information about the process that owns that page

-ecreases memory needed to store each page table# but increases time

needed to search the table when a page reference occurs

"se hash table to limit the search to one K or at most a few K page+table

entries

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UNIT-8

PART-A

1) What is a Resource-Allocation 0raph?

-eadlocks can be described more precisely in terms of a directed graph calleda system resource allocation graph. This graph consists of a set of vertices J and a set

of edges E. The set of vertices J is partitioned into two different types of nodes* 4 the

set consisting of all active processes in the system and C the set consisting of all

resource types in the system.

) What are the metho"s !or 6an"ling ea"loc7s?

The deadlock problem can be dealt with in one of the three ways'

"se a protocol to prevent or avoid deadlocks# ensuring that the system will

never enter a deadlock state.

Allow the system to enter the deadlock state# detect it and then recover.

Ignore the problem all together# and pretend that deadlocks never occur in the

system.

3) What are a &a!e &tate an" an Unsa!e &tate?

A state is safe if the system can allocate resources to each process in some order and

still avoid a deadlock. A system is in safe state only if there eists a safe se$uence. A

se$uence of processes F46#42#.4nG is a safe se$uence for the current allocation state if# for

each 4i# the resource that 4i can still re$uest can be satisfied by the current available resource

plus the resource held by all the 4/# with /Fi. if no such se$uence eists# then the system state

is said to be unsafe.

PART-#

1) '*plain a$out "ea"loc7 prevention $rie!ly?

,or)

6o% can %e prevent the occurrence o! "ea"loc7s? iscuss in $rie!+

AN&W'R(

0or a deadlock occurrence# each of the four necessary conditions must hold. 5y

ensuring that at least one of these conditions cannot hold# we can prevent the occurrence of a

deadlock.

5y eamining each of the four necessary conditions'6) utual Eclusion

2) ;old and ait

7) !o 4reemption

?) ircular ait

4utual '*clusion(

The mutual+eclusion condition must hold for non+sharable resources. 0or eample# a

printer cannot be simultaneously shared by several processes. %harable resources# in contrast#

do not re$uire mutually eclusive access and thus cannot be involved in a deadlock. Cead+

only files are a good eample of a sharable resource. If several processes attempt to open aread+only file at the same time# they can be granted simultaneous access to the file. A process

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never needs to wait for a sharable resource. In general# however# we cannot prevent deadlocks

by denying the mutual+eclusion condition.

6ol" an" Wait(

To ensure that the hold+and+wait condition never occurs in the system# we must

guarantee that# whenever a process re$uests a resource# it does not hold any other resources.eans protocol allows a process to re$uest resources only when it has none.

Another protocol is that can be used re$uires each process to re$uest and be allocated

all its resources before it begins eecution.

The difference between these two protocols# we consider a process that copies data

from a -J- drive to a file on disk# sorts the file# and then prints the results to a printer. If all

resources must be re$uested at the beginning of the process# then the process must initially

re$uest the -J- drive#

disk file# and printer. It will hold the printer for its entire eecution# even though it needs the

printer only at the end.

The second method allows the process to re$uest initially only the -J- drive and

disk file. It copies from the -J- drive to the disk and then releases both the -J- drive andthe disk file. The process must then again re$uest the disk file and the printer. After copying

the disk file to the printer# it releases these two resources and terminates.

5oth protocols have some disadvantages'

6) Cesource utili&ation may be low# since resources may be allocated but unused for a long

period.

2) %tarvation' A process that needs several resources may have to wait indefinitely# because

at least one of the resources that it needs is always allocated to some other process.

No Preemption(

If a process is holding some resources and re$uests another resource that cannot be

immediately allocated to it (that is# the process must wait)# then all resources currently being

held are preempted. In other words# these resources are implicitly released. The preempted

resources are added to the list of resources for which the process is waiting.

Alternatively# if a process re$uests some resources# we first check whether they are

available. If they are# we allocate them. If they are not# we check whether they are allocated

to some other process that is waiting for additional resources. If so# we preempt the desired

resources from the waiting process and allocate them to the re$uesting process. If the

resources are neither available

nor held by a waiting process# the re$uesting process must wait. hile it is waiting# some of

its resources may be preempted# but only if another process re$uests them. A process can berestarted only when it is allocated the new resources it is re$uesting and recovers any

resources that were pre+empted while it was waiting.

ircular Wait(

Each process re$uests resources in an increasing order of enumeration. To illustrate#

we letR {R0, Ri, ...#Rm} be the set of resource types. e assign to each resource type a

uni$ue integer number# which# allows us to compare two resources and to determine whether

one precedes another in our ordering. 0ormally# we define a one+to+one function 0'R > N,

whereN is the set of natural numbers. 0or eample# if the set of resource typesR includes

tape drives# disk drives# and printers# then the function 0 might be defined as follows'0(tape drive) 6

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0(disk drive)8

0 (printer) 62

Each process can re$uest resources only in an increasing order of enumeration. That is# a

process can initially re$uest any number of instances of a resource typeKsay# C. After that#

the process can re$uest instances of resource type C* if and only if 0(C*) G 0(C#).

) '*plain in "etail the $an7er9s algorithm?

,or)

AN&W'R

0or avoiding deadlocks is to re$uire additional information about how resources are to

be re$uested. The various algorithms that use this approach differ in the amount and type of

information re$uired. The simplest and most useful model re$uires that each process declare

the maimum number of resources of each type that it may need. Hiven this a priori#

information# it is possible to construct an algorithm that ensures that the system will never

enter a deadlocked state. %uch an algorithm defines the deadlock+avoidance approach.hen# a new process enters the system# it must declare the maimum number of

instances of each resource type that it may need. This number may not eceed the total

number of resources in the system. hen a user re$uests a set of resources# the system must

determine whether the allocation of these resources will leave the system in a safe state. If it

will# the resources are allocated* otherwise# the process must wait until some other process

releases enough resources.

%everal data structures must be maintained to implement the bankerDs algorithm.

Availa$le+ A vector of length m indicates the number of available resources of each

type. If AvailableM/N e$uals k# there are k instances of resource type C/ available.

4a*+ An n m matri defines the maimum demand of each process. If aMiN

M/Ne$uals k# then process 4i may re$uest at most k instances of resource type C/.

Allocation. An n m matri defines the number of resources of each type currently

allocated to each process. If AllocationMiNM/N e$uals k# then process 4i is currently

allocated k instances of resource type C/.

Nee"+ An n m matri indicates the remaining resource need of each process. If

!eedMiNM/N e$uals k# then process 4i may need k more instances of resource type C/ to

complete its task. !ote that !eedMiNM/N e$uals aMiNM/N+ AllocationMiNM/N.

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&a!ety algorithm

The algorithm for finding out whether or not a system is in a safe state.

6) Let ork and 0inish be vectors of length in and n# respectively. Initiali&e

ork Available and 0inishMiN false for i#6#.. n+6.

2. 0ind an i such that both

a. 0inishMiN falseb. !eediF ork

If no such 3 eists# go to step ?.

7. ork ork B Allocation#

0inishMiN true

Ho to step 2.

?. If 0inishMiN true for all. I# then the system is in a safe state.

Resource-Re:uest Algorithm

The algorithm which determines if re$uests can be safely granted.

Let Ce$uestibe the re$uest vector for process 4i If Ce$uestiM /N k# then process 4iwants k instances of resource type C/ hen a re$uest for resources is made by process 4i the

following actions are taken'

6. If Ce$uestiF!eedi go to step 2. 1therwise# raise an error condition# since the process

has eceeded its maimum claim.

2. If Ce$uestiF Available# go to step 7. 1therwise# 4i must wait# since the resources are

not available.

7. ;ave the system pretend to have allocated the re$uested resources to process 4 iby

modifying the state as follows'

Available Available 9 Ce$uesti*

Allocationi AllocationiB Ce$uesti*

!eedi !ecdi9 Ce$uesti*

If the resulting resource+allocation state is safe# the transaction is completed# and process 4 iis

allocated its resources.

';A4P.'(

onsider a system with five processes 4through 4? and three resource types A# 5#

and . Cesource type A has 6 instances# resource type 5 has 8 instances# and resource type

has O instances.

proce

sses

ALL1A

TI1!

A

P

AJAILA

5LE

A 5 A 5

A 5

4 6 O 8

7

7 7 2

46 2 7 2

2

42 7 2 =

2

47 2 6 6 2 2

2

4? 2 ? 7

7

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!ow we have to calculate the need. !eedMiNM/N aMiNM/N+AllocationMiNM/N

4rocesse

s

!eed

A 5

4 O ? 746 6 2 2

42

47 6 6

4? ? 7 6

1ur work available# so work 7 7 2

!ow check for need weather !eedFwork then p not satisfied then go for p6# which satisfies

the need then p6 will starts its eecution. After p6 completes its eecution it releases its

allocated resources then availability is 8 7 2.

Then again check for need of resources of other processes then p 7 satisfies after that itreleases its resources then availability is O ? 7. Then net p#p2# p?.%o the safe se$uence is

Fp6# p7# p# p2# p?G.

3) 6o% to recover !rom "ea"loc7? '*plain

AN&W'R(

hen a detection algorithm determines that a deadlock eists# several alternatives are

available. 1ne possibility is to inform the operator that a deadlock has occurred and to let the

operator deal with the deadlock manually. Another possibility is to let the system recover

from the deadlock automatically. There are two options for breaking a deadlock. 1ne is

simply to abort one or more processes to break the circular wait. The other is to preempt

some resources from one or more of the deadlocked processes.

Process Termination

To eliminate deadlocks by aborting a process# we use one of two methods. In both

methods# the system reclaims all resources allocated to the terminated processes.

A$ort all "ea"loc7e" processes+ This method clearly will break the deadlock cycle#

but at great epense* the deadlocked processes may have computed for a long time#

and the results of these partial computations must be discarded and probably will have

to be recomputed later.

A$ort one process at a time until the "ea"loc7 cycle is eliminate"+ This method

incurs considerable overhead# since# after each process is aborted# a deadlock+

detection algorithm must be invoked to determine whether any processes are still

deadlocked.

Resource Preemption

To eliminate deadlocks using resource preemption# we successively pre+empt some resources

from processes and give these resources to other processes until the deadlock cycle is broken.

If preemption is re$uired to deal with deadlocks# then three issues need to be addressed'

1+ &electing a victim. hich resources and which processes are to be preemptedQ As in

process termination# we must determine the order of preemption to minimi&e cost.

ost factors may include such parameters as the number of resources a deadlocked

process is holding and the amount of time the process has thus far consumed duringits eecution.

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+ Roll$ac7. If we preempt a resource from a process# what should be done with that

processQ learly# it cannot continue with its normal eecution* it is missing some

needed resource. e must roll back the process to some safe state and restart it from

that state. %ince# in general# it is difficult to determine what a safe state is# the simplest

solution is a total rollback' Abort the process and then restart it.

3+ &tarvation+;ow do we ensure that starvation will not occurQ That is how can we

guarantee that resources will not always be preempted from the same processQ

5) e!ine "ea"loc7? What are the !our con"itions that hol" simultaneously in a

system !or the "ea"loc7 situation to arise?

AN&W'R(

A process re$uests resources* and if the resources are not available at that time# the

process enters a waiting state. %ometimes# a waiting process is never again able to change

state# because the resources it has re$uested are held by other waiting processes. Thissituation is called a "ea"loc7+

A process may utili&e a resource in only the following se$uence'

1+ Re:uest+ If the re$uest cannot be granted immediately (for eample# if the resource is

being used by another process)# then the re$uesting process must wait until it can

ac$uire the resource.

+ Use# The process can operate on the resource (for eample# if the resource is a printer#

the process can print on the printer)

3+ Release+The process releases the resource.

Necessary on"itions(

A deadlock situation can arise if the following four conditions hold simultaneously in

a system'

4utual e*clusion+ At least one resource must be held in a non+sharable mode* that is#

only one process at a time can use the resource. If another process re$uests that

resource# the re$uesting process must be delayed until the resource has been released.

6ol" an" %ait+ A process must be holding at least one resource and waiting to

ac$uire additional resources that are currently being held by other processes.

No preemption+ Cesources cannot be preempted* that is# a resource can be releasedonly voluntarily by the process holding it# after that process has completed its task.

ircular %ait. A set R4# 46# ...# 4nS of waiting processes must eist such that 4 is

waiting for a resource held by 46# 46is waiting for a resource held by 42# # 4n+6 is

waiting for a resource held by 4n# and 4n is waiting for a resource held by 4.