Operating Systems BCS303

HKBK College of Engineering

Department of Artificial Intelligence & Machine Learning
SEMESTER – 3

Operating Systems – BCS303

Module III
Process Synchronization:
❖ Synchronization:
❖ The critical section problem;
❖ Peterson’s solution;
❖ Synchronization hardware;
❖ Semaphores;
❖ Classical problems of synchronization;
❖ Monitors.

Deadlocks:
❖ Deadlocks; System model;
❖ Deadlock characterization;
❖ Methods for handling deadlocks;
❖ Deadlock prevention;
❖ Deadlock avoidance; Deadlock detection and recovery from deadlock.

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MODULE 3

PROCESS SYNCHRONIZATION
A cooperating process is one that can affect or be affected by other processes executing
in the system. Cooperating processes can either directly share a logical address space
(that is, both code and data) or be allowed to share data only through files or messages.
Concurrent access to shared data may result in data inconsistency. To maintain data
consistency, various mechanisms is required to ensure the orderly execution of
cooperating processes that share a logical address space.

Producer- Consumer Problem

A Producer process produces information that is consumed by consumer process.
To allow producer and consumer process to run concurrently, A Bounded Buffer can
be used where the items are filled in a buffer by the producer and emptied by the
consumer.
The original solution allowed at most BUFFER_SIZE - 1 item in the buffer at the same
time. To overcome this deficiency, an integer variable counter, initialized to 0 isadded.
counter is incremented every time when a new item is added to the buffer and is
decremented every time when one item removed from thebuffer.

The code for the producer process can be modified as follows:

while (true) {

/* produce an item and put in nextProduced*/ while

(counter == BUFFER_SIZE)
; // do nothing
buffer [in] = nextProduced;
in = (in + 1) % BUFFER_SIZE;
counter++;
}

The code for the consumer process can be modified as follows:

while (true){
while (counter ==0)
; // donothing
nextConsumed =buffer[out];
out = (out + 1) % BUFFER_SIZE;
counter--;
/* consume the item in nextConsumed */
}

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Race Condition
When the producer and consumer routines shown above are correct separately, they
may not function correctly when executed concurrently.
Illustration:
Suppose that the value of the variable counter is currently 5 and that the producer and
consumer processes execute the statements "counter++" and "counter--" concurrently.
The value of the variable counter may be 4, 5, or 6 but the only correct result is counter
== 5, which is generated correctly if the producer and consumer execute separately.

The value of counter may be incorrect as shown below:

The statement counter++ could be implemented as
register1= counter
register1 = register1 + 1
counter =register1

The statement counter-- could be implemented as

register2 =counter
register2 = register2 – 1
count = register2

The concurrent execution of "counter++" and "counter--" is equivalent to a sequential

execution in which the lower-level statements presented previously are interleaved in some
arbitrary order. One such interleaving is

Consider this execution interleaving with “count = 5” initially:

S0: producer execute register1=counter {register1 = 5}
S1: producer execute register1 = register1+1 {register1 = 6}
S2: consumer execute register2=counter {register2 = 5}
S3: consumer execute register2 = register2-1 {register2 = 4}
S4: producer execute counter=register1 {count =6}
S5: consumer execute counter=register2 {count =4}

Note: It is arrived at the incorrect state "counter == 4", indicating that four buffers are
full, when, in fact, five buffers are full. If we reversed the order of the statements at T4
and T5, we would arrive at the incorrect state "counter==6".
Definition Race Condition: A situation where several processes access and
manipulate the same data concurrently and the outcome of the execution depends on
the particular order in which the access takes place, is called a RaceCondition.
To guard against the race condition, ensure that only one process at a time can be
manipulating the variable counter. To make such a guarantee, the processes are
synchronized in some way.

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The Critical Section Problems

• Consider a system consisting of n processes {Po, P1 , ... ,Pn-1}.

• Each process has a segment of code, called a critical section in which the process
may be changing common variables, updating a table, writing a file, and soon
• The important feature of the system is that, when one process is executing in its
critical section, no other process is to be allowed to execute in its critical section.
That is, no two processes are executing in their critical sections at the sametime.
• The critical-section problem is to design a protocol that the processes can use to
cooperate.

The general structure of a typical process Pi is shown in below figure.

Each process must request permission to enter its critical section. The section of code
implementing this request is the entry section.
The critical section may be followed by an exit section. The remaining code is the
reminder section.

Figure: General structure of a typical process Pi

A solution to the critical-section problem must satisfy the following three requirements:

1. Mutual exclusion:If process Pi is executing in its critical section, then no other

processes can be executing in their criticalsections.

2. Progress:If no process is executing in its critical section and some processes wish to
enter their critical sections, then only those processes that are not executing in their
remainder sections can participate in deciding which will enter its critical section next,
and this selection cannot be postponedindefinitely.

3. Bounded waiting:There exists a bound, or limit, on the number of times that other
processes are allowed to enter their critical sections after a process has made a request
to enter its critical section and before that request isgranted.

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PETERSON'S SOLUTION

• This is a classic software-based solution to the critical-section problem. There are

no guarantees that Peterson's solution will work correctly on modern computer
architectures
• Peterson's solution provides a good algorithmic description of solving the critical-
section problem and illustrates some of the complexities involved in designing
software that addresses the requirements of mutual exclusion, progress, and
bounded waiting.

Peterson's solution is restricted to two processes that alternate execution between theircritical
sections and remainder sections. The processes are numbered Po and P1 or Pi and Pj where j =
1-i
Peterson's solution requires the two processes to share two data items:
int turn;
boolean flag[2];

• turn: The variable turn indicates whose turn it is to enter its critical section. Ex:
if turn == i, then process Pi is allowed to execute in its criticalsection
• flag: The flag array is used to indicate if a process is ready to enter its critical
section. Ex: if flag [i] is true, this value indicates that Pi is ready to enter its critical
section.

do {
flag[i] = TRUE;
turn = j;
while (flag[j] && turn == j)
; // do nothing
critical section
flag[i] = FALSE;

remainder section

} while (TRUE);

Figure: The structure of process Pi in Peterson's solution

• To enter the critical section, process Pi first sets flag [i] to be true and then sets
turn to the value j, thereby asserting that if the other process wishes to enter the
critical section, it can doso.
• If both processes try to enter at the same time, turn will be set to both i and j at
roughly the same time. Only one of these assignments will last, the other will
occur but will be over written immediately.

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• The eventual value of turn determines which of the two processes is allowed to
enter its critical sectionfirst

To prove that solution is correct, then we need to show that

1. Mutual exclusion ispreserved
2. Progress requirement is satisfied
3. Bounded-waiting requirement is met

1. To prove Mutual exclusion

• Each pi enters its critical section only if either flag [j] == false or turn ==i.
• If both processes can be executing in their critical sections at the same time, then
flag [0] == flag [1]==true.
• These two observations imply that Pi and Pj could not have successfully executed their
while statements at about the same time, since the value of turn can be either 0 or 1 but
cannot be both. Hence, one of the processes (Pj) must have successfully executed the
while statement, whereas Pi had to execute at least one additional statement
("turn==j").
• However, at that time, flag [j] == true and turn == j, and this condition will persist as
long as Pi is in its critical section, as a result, mutual exclusion ispreserved.

2. To prove Progress and Bounded-waiting

• A process Pi can be prevented from entering the critical section only if it is stuck in
the while loop with the condition flag [j] ==true and turn=== j; this loop is the only
onepossible.
• If Pj is not ready to enter the critical section, then flag [j] ==false, and Pi can enter its
criticalsection.
• If Pj has set flag [j] = true and is also executing in its while statement, then either
turn === i or turn ===j.
▪ If turn == i, then Pi will enter the criticalsection.
▪ If turn== j, then Pj will enter the criticalsection.
• However, once Pj exits its critical section, it will reset flag [j] = false, allowing Pi to
enter its criticalsection.
• If Pj resets flag [j] to true, it must also set turn to i.
• Thus, since Pi does not change the value of the variable turn while executing the
while statement, Pi will enter the critical section (progress) after at most one entry by
Pj (boundedwaiting).

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SYNCHRONIZATIONHARDWARE

• The solution to the critical-section problem requires a simple tool-alock.

• Race conditions are prevented by requiring that critical regions be protected by
locks. That is, a process must acquire a lock before entering a critical section and it
releases the lock when it exits the critical section

do {
acquire lock
critical section
release lock
remainder section
} while (TRUE);

Figure: Solution to the critical-section problem using locks.

• The critical-section problem could be solved simply in a uniprocessor environment

if interrupts are prevented from occurring while a shared variable was being
modified. In this manner, the current sequence of instructions would be allowed to
execute in order without preemption. No other instructions would be run, so no
unexpected modifications could be made to the sharedvariable.
• But this solution is not as feasible in a multiprocessor environment. Disabling
interrupts on a multiprocessor can be time consuming, as the message is passed to
all the processors. This message passing delays entry into each critical section,
and system efficiency decreases.
TestAndSet ( ) and Swap( ) instructions
• Many modern computer systems provide special hardware instructions that allowto
test and modify the content of a word or to swap the contents of two words
atomically, that is, as one uninterruptibleunit.
• Special instructions such as TestAndSet () and Swap() instructions are used to solve
the critical-sectionproblem
• The TestAndSet () instruction can be defined as shown in Figure. The important
characteristic of this instruction is that it is executedatomically.

Definition:
booleanTestAndSet (boolean *target)
{
booleanrv = *target;
*target = TRUE;
return rv:
}
Figure: The definition of the TestAndSet () instruction.

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• Thus, if two TestAndSet () instructions are executed simultaneously, they will be

executed sequentially in some arbitrary order. If the machine supports the TestAndSet
() instruction, then implementation of mutual exclusion can be done by declaring a
Boolean variable lock, initialized to false.

do {
while ( TestAndSet (&lock ))
; // do nothing
// critical section
lock =FALSE;
// remaindersection
} while (TRUE);
Figure: Mutual-exclusion implementation with TestAndSet ()

• The Swap() instruction, operates on the contents of two words, it is defined as shown
below

Definition:
void Swap (boolean *a, boolean *b)
{
boolean temp = *a;
*a = *b;
*b = temp:
}
Figure: The definition of the Swap ( ) instruction
• Swap() it is executed atomically. If the machine supports the Swap() instruction, then
mutual exclusion can be provided as follows.
• A global Boolean variable lock is declared and is initialized to false. In addition, each
process has a local Boolean variable key. The structure of process Pi is shown in below

do {
key = TRUE;
while ( key == TRUE) Swap
(&lock, &key );

// critical section
lock =FALSE;
// remaindersection
} while (TRUE);

Figure: Mutual-exclusion implementation with the Swap() instruction

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• These algorithms satisfy the mutual-exclusion requirement, they do not

satisfy thebounded- waiting requirement.
• Below algorithm using the TestAndSet () instruction that satisfies all
the critical-section requirements. The common data structures are

boolean
waiting[
n];
boolean
lock;

These data structures are initialized to false.

do {
waiting[i] = TRUE;
key = TRUE;
while (waiting[i] && key)
key = TestAndSet(&lock);
waiting[i] = FALSE;

// critical section j

= (i + 1) % n;
while ((j != i) && !waiting[j])
j = (j + 1) % n;
if (j == i)
lock = FALSE;
else
waiting[j] = FALSE;
// remainder section
} while (TRUE);

Figure:Bounded-waiting mutual exclusion with TestAndSet ()

1. To prove the mutual exclusionrequirement

Note that process Pi can enter its critical section only if either waiting [i] == false or
key ==false.
The value of key can become false only if the TestAndSet( ) isexecuted.
The first process to execute the TestAndSet( ) will find key== false; all others must
wait.
The variable waiting[i] can become false only if another process leaves its critical
section; only one waiting[i] is set to false, maintaining the mutual-exclusion
requirement.

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2. To prove the progressrequirement

Note that, the arguments presented for mutual exclusion also apply here, since a process
exiting the critical section either sets lock to false or sets waiting[j] to false. Both allow a
process that is waiting to enter its critical section to proceed.

3. To prove the bounded-waitingrequirement

Note that, when a process leaves its critical section, it scans the array waiting in the
cyclic ordering (i + 1, i + 2, ... , n 1, 0, ... , i 1).
It designates the first process in this ordering that is in the entry section (waiting[j]
==true) as the next one to enter the critical section. Any process waiting to enter its
critical section will thus do so within n - 1 turns.

SEMAPHORE

A semaphore is a synchronization tool is used solve various synchronization problem

and can be implementedefficiently.
Semaphore do not require busywaiting.
A semaphore S is an integer variable that, is accessed only through two standard
atomic operations: wait () and signal (). The wait () operation was originally
termed P and signal() was calledV.
Definition of wait ():

wait (S) {
while S <= 0
; // no-op
S--;

Definition of signal ():

signal (S) {
S++;}

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All modifications to the integer value of the semaphore in the wait () and signal()
operations must be executed indivisibly. That is, when one process modifies the
semaphore value, no other process can simultaneously modify that same semaphore
value.

Binary semaphore
The value of a binary semaphore can range only between 0 and1.
Binary semaphores are known as mutex locks, as they are locks that provide mutual
exclusion. Binary semaphores to deal with the critical-section problem for multiple
processes. Then processes share a semaphore, mutex, initialized to1

Each process Pi is organized as shown in below figure

do {
wait (mutex);
// Critical Section
signal (mutex);
// remainder section
} while (TRUE);

Figure: Mutual-exclusion implementation with semaphores

Counting semaphore
The value of a counting semaphore can range over an unrestricteddomain.
Counting semaphores can be used to control access to a given resource consisting
of a finite number ofinstances.
The semaphore is initialized to the number of resources available. Each process
that wishes to use a resource performs a wait() operation on the semaphore. When
a process releases a resource, it performs a signal()operation.
When the count for the semaphore goes to 0, all resources are being used. After
that, processes that wish to use a resource will block until the count becomes
greater than 0.

Implementation
The main disadvantage of the semaphore definition requires busywaiting.
While a process is in its critical section, any other process that tries to enter its
critical section must loop continuously in the entry code.
This continual looping is clearly a problem in a real multiprogramming system,
where a single CPU is shared among many processes.

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Busy waiting wastes CPU cycles that some other process might be able to use
productively. This type of semaphore is also called a spinlock because the process
"spins" while waiting for thelock.

Semaphore implementation with no busy waiting

The definition of the wait() and signal() semaphore operations ismodified.
When a process executes the wait () operation and finds that the semaphore value
is not positive, it mustwait.
However, rather than engaging in busy waiting, the process can block itself. The
block operation places a process into a waiting queue associated with the
semaphore, and the state of the process is switched to the waiting state. Then control
is transferred to the CPU scheduler, which selects another process toexecute.
A process that is blocked, waiting on a semaphore S, should be restarted when some
other process executes a signal() operation. The process is restarted by a wakeup( )
operation, which changes the process from the waiting state to the ready state. The
process is then placed in the readyqueue.

To implement semaphores under this definition, we define a semaphore as a "C'

struct:

typedefstruct {
int value;
struct process *list;
} semaphore;
Each semaphore has an integer value and a list of processes list. When a process must
wait on a semaphore, it is added to the list of processes. A signal() operation removes
one process from the list of waiting processes and awakens that process.

The wait() semaphore operation can now be defined as:

wait(semaphore *S) {
S->value--;
if (S->value < 0) {
add this process to S-
>list; block();
}}

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The signal () semaphore operation can now be defined as

signal(semaphore *S) {
S->value++;
if (S->value <= 0) {
remove a process P
from S->list;
wakeup(P);
}
}

The block() operation suspends the process that invokes it. The wakeup(P)
operation resumes the execution of a blocked process P. These two operations are
provided by the operating system as basic systemcalls.
In this implementation semaphore values may be negative. If a semaphore value
is negative, its magnitude is the number of processes waiting on thatsemaphore.

Deadlocks and Starvation

The implementation of a semaphore with a waiting queue may result in a situation

where two or more processes are waiting indefinitely for an event that can be caused
only by one of the waiting processes. The event in question is the execution of a signal(
) operation. When such a state is reached, these processes are said to be deadlocked.
To illustrate this, consider a system consisting of two processes, Po and P1, each
accessing two semaphores, S and Q, set to the value 1

P0 P1
wait(S); wait(Q);
wait(Q); wait(S);
. .
. .
signal(S); signal(Q);
signal(Q); signal(S);

Suppose that Po executes wait (S) and then P1 executes wait (Q). When Poexecutes
wait (Q), it must wait until P1 executes signal (Q). Similarly, when P1 executes wait
(S), it must wait until Po executes signal(S). Since these signal() operations cam1ot
be executed, Po and P1 are deadlocked.

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Another problem related to deadlocks is indefinite blocking or starvation: A

situation in which processes wait indefinitely within the semaphore.
Indefinite blocking may occur if we remove processes from the list associated with
a semaphore in LIFO (last-in, first-out) order.

CLASSICAL PROBLEMS OF SYNCHRONIZATION

Bounded-BufferProblem
Readers and WritersProblem
Dining-PhilosophersProblem

Bounded-Buffer Problem
N buffers, each can hold one item
Semaphore mutexinitialized to the value 1
Semaphore full initialized to the value0
Semaphore empty initialized to the value N.

The structure of the producer process:

The structure of the consumerprocess:

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Readers-Writers Problem
A data set is shared among a number of concurrentprocesses
Readers – only read the data set; they do not perform anyupdates
Writers – can both read andwrite.
Problem – allow multiple readers to read at the same time. Only one single writer
can access the shared data at the sametime.
SharedData
Dataset
Semaphore mutexinitialized to 1.
Semaphore wrtinitialized to1.
Integer readcountinitialized to 0.

The structure of a writerprocess

The structure of a readerprocess

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Dining-Philosophers Problem

Consider five philosophers who spend their lives thinking and eating. The philosophers share
a circular table surrounded by five chairs, each belonging to one philosopher. In the center of
the table is a bowl of rice, and the table is laid with five singlechopsticks.

A philosopher gets hungry and tries to pick up the two chopsticks that are closest to her
(the chopsticks that are between her and her left and right neighbors). A philosopher may
pick up only one chopstick at a time. When a hungry philosopher has both her chopsticks
at the same time, she eats without releasing the chopsticks. When she is finished eating,
she puts down both chopsticks and starts thinkingagain.
It is a simple representation of the need to allocate several resources among several
processes in a deadlock-free and starvation-freemanner.

Solution:One simple solution is to represent each chopstick with a semaphore. A

philosopher tries to grab a chopstick by executing a wait() operation on thatsemaphore.
She releases her chopsticks by executing the signal() operation on the appropriate
semaphores. Thus, the shared data are
semaphore chopstick[5];

where all the elements of chopstick are initialized to 1. The structure of

philosopher iis shown

Several possible remedies to the deadlock problem are replaced by:

• Allow at most four philosophers to be sitting simultaneously at thetable.

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• Allowaphilosophertopickupherchopsticksonlyifbothchopsticksareavailable.
• Use an asymmetric solution—that is, an odd-numbered philosopher picks up
first her left chopstick and then her right chopstick, whereas an even numbered
philosopher picks up her right chopstick and then her leftchopstick.

Problems with Semaphores

Correct use of semaphore operations:
signal (mutex) …. wait (mutex) : Replace signal with wait andvice-versa
wait (mutex) … wait(mutex)
Omitting of wait (mutex) or signal (mutex) (orboth)

Monitor
An abstract data type—or ADT—encapsulates data with a set of functions to operate
on that data that are independent of any specific implementation of the ADT.
A monitor typeis an ADT that includes a set of programmer defined operations that
are provided with mutual exclusion within the monitor. The monitor type alsodeclares
the variables whose values define the state of an instance of that type, alongwith the
bodies of functions that operate on those variables.
The monitor construct ensures that only one process at a time is active within the
monitor.

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To have a powerful Synchronization schemes a condition construct is added to the

Monitor. So synchronization scheme can be defined with one or more variables of type
condition Two operations on a conditionvariable:
Condition x, y
The only operations that can be invoked on a condition variable are wait() and signal().
The operation
x.wait () – a process that invokes the operation is suspended.
x.signal () – resumes one of processes (if any) that invoked x.wait ()

Fig: Monitor with Condition Variables

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Solution to Dining Philosophers

Each philosopher I invokes the operations pickup() and putdown() in the following

For each monitor, a semaphore mutex (initialized to 1) is provided. A process must

execute wait(mutex) before entering the monitor and must execute signal(mutex)
after leaving the monitor.
Since a signaling process must wait until the resumed process either leaves or
waits, an additional semaphore, next, is introduced, initialized to 0. The signaling
processes can use next to suspend themselves. An integer variable next_count is
also provided to count the number of processes suspended on next. Thus ,each
external function F is replaced by

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For each condition x, we introduce a semaphore x sem and an integer variable x

count, both initialized to 0. The operation x.wait() can now be implemented as

The operation x.signal() can be implementedas

Resuming Processes within a Monitor

If several processes are suspended on condition x, and an x.signal() operation is executed
by some process, then to determine which of the suspended processes should be resumed
next, one simple solution is to use a first-come, first-served (FCFS)ordering, so that the
process that has been waiting the longest is resumed first. For this purpose, the
conditional-wait construct can be used. This construct has theform
x.wait(c);
where c is an integer expression that is evaluated when the wait() operation is executed.
The value of c, which is called a priority number, is then stored with the name of the
process that is suspended. When x.signal() is executed, the process with the smallest
priority number is resumednext.

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The Resource Allocator monitor shown in the above Figure, which controls the
allocation of a single resource among competingprocesses.
A process that needs to access the resource in question must observe the following
sequence:
R.acquire(t);
...
access the resource;
...
R.release();
where R is an instance of type ResourceAllocator.
The monitor concept cannot guarantee that the preceding access sequence will be
observed. In particular, the following problems can occur:
A process might access a resource without first gaining access permission to the
resource.
A process might never release a resource once it has been granted access to the
resource.
A process might attempt to release a resource that it neverrequested.
A process might request the same resource twice (without first releasing the
resource).

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DEADLOCKS

A process requests resources, if the resources are not available at that time, the process enters a
waiting state. Sometimes, a waiting process is never again able to change state, because the
resources it has requested are held by other waiting processes. This situation is called aDeadlock.

SYSTEM MODEL

• A system consists of a finite number of resources to be distributed among a number of

competing processes. The resources are partitioned into several types, each consisting of
some number of identical instances. Memory space, CPU cycles, files, and I/0 devices are
examples of resource types.
• A process must request a resource before using it and must release the resource after using
it. A process may request as many resources as it requires carrying out its designated task.
The number of resources requested may not exceed the total number of resources available
in the system.

Under the normal mode of operation, a process may utilize a resource in only the following
sequence:
1. Request: The process requests the resource. If the request cannot be granted
immediately, then the requesting process must wait until it can acquire the resource.
2. Use: The process can operate on the resource.
3. Release: The process releases the resource.

A set of processes is in a deadlocked state when every process in the set is waiting for an event
that can be caused only by another process in the set. The events with which we are mainly
concerned here are resource acquisition and release. The resources may be either physical
resources or logical resources

To illustrate a deadlocked state, consider a system with three CD RW drives.

Suppose each of three processes holds one of these CD RW drives. If each process now
requests another drive, the three processes will be in a deadlocked state.
Each is waiting for the event "CD RW is released," which can be caused only by one of the
other waiting processes. This example illustrates a deadlock involving the same resource type.

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Deadlocks may also involve different resource types. For example, consider a system with one
printer and one DVD drive. Suppose that process Pi is holding the DVD and process Pj is holding
the printer. If Pi requests the printer and Pj requests the DVD drive, a deadlock occurs.

DEADLOCK CHARACTERIZATION

Necessary Conditions

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

1. Mutual exclusion: 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 requests that resource,
the requesting process must be delayed until the resource has been released.

2. Hold and wait: A process must be holding at least one resource and waiting to acquire
additional resources that are currently being held by other processes.

3. No preemption: Resources cannot be preempted; that is, a resource can be released only
voluntarily by the process holding it, after that process has completed its task.

4. Circular wait: A set {P0, Pl, ... , Pn} of waiting processes must exist such that Po is waiting
for a resource held by P1, P1 is waiting for a resource held by P2, ... , Pn-1 is waiting for a
resource held by Pn and Pn is waiting for a resource held by Po.

Resource-Allocation Graph

Deadlocks can be described in terms of a directed graph called System Resource-Allocation

Graph

The graph consists of a set of vertices V and a set of edges E. The set of vertices V is
partitioned into two different types of nodes:
• P = {P1, P2, ...,Pn}, the set consisting of all the active processes in the system.
• R = {R1, R2, ..., Rm} the set consisting of all resource types in the system.

A directed edge from process Pi to resource type Rj is denoted by Pi → Rj it signifies that process
Pi has requested an instance of resource type Rj and is currently waiting for that resource.
A directed edge from resource type Rj to process Pi is denoted by Rj → Pi it signifies that an
instance of resource type Rj has been allocated to process Pi.
• A directed edge Pi → Rj is called a Request Edge.
• A directed edge Rj → Pi is called an Assignment Edge.

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Pictorially each process Pi as a circle and each resource type Rj as a rectangle. Since resource
type Rj may have more than one instance, each instance is represented as a dot within the
rectangle.
A request edge points to only the rectangle Rj, whereas an assignment edge must also designate
one of the dots in the rectangle.

When process Pi requests an instance of resource type Rj, a request edge is inserted in the
resource-allocation graph. When this request can be fulfilled, the request edge is instantaneously
transformed to an assignment edge. When the process no longer needs access to the resource, it
releases the resource; as a result, the assignment edge is deleted.

The resource-allocation graph shown in Figure depicts the following situation.

The sets P, K and E:

• P = {P1, P2, P3}
• R= {R1, R2, R3, R4}
• E = {Pl →Rl, P2 → R3, Rl → P2, R2 → P2, R2 → P1, R3 → P3 }

Resource instances:
• One instance of resource type R1
• Two instances of resource type R2
• One instance of resource type R3
• Three instances of resource type R4

Process states:
• Process P1 is holding an instance of resource type R2 and is waiting for an instance of
resource type R1.
• Process P2 is holding an instance of R1 and an instance of R2 and is waiting for an
instance of R3.
• Process P3 is holding an instance of R3.

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If the graph does contain a cycle, then a deadlock may exist.

• If each resource type has exactly one instance, then a cycle implies that a deadlock has
occurred. If the cycle involves only a set of resource types, each of which has only a
single instance, then a deadlock has occurred. Each process involved in the cycle is
deadlocked.
• If each resource type has several instances, then a cycle does not necessarily imply that
a deadlock has occurred. In this case, a cycle in the graph is a necessary but not a
sufficient condition for the existence of deadlock.

To illustrate this concept, the resource-allocation graph depicted in below figure:

Suppose that process P3 requests an instance of resource type R2. Since no resource instance is
currently available, a request edge P3 → R2 is added to the graph. At this point, two minimal
cycles exist in the system:
1. P1 →R1 → P2 → R3 → P3 → R2→P1
2. P2 →R3 → P3 → R2 → P2

Figure: Resource-allocation graph with a deadlock.

Processes P1, P2, and P3 are deadlocked. Process P2 is waiting for the resource R3, which is held
by process P3. Process P3 is waiting for either process P1 or process P2 to release resource R2.
In addition, process P1 is waiting for process P2 to release resource R1.

Consider the resource-allocation graph in below Figure. In this example also have a cycle:
P1→R1→P3→R2→P1

Figure: Resource-allocation graph with a cycle but no deadlock

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However, there is no deadlock. Observe that process P4 may release its instance of resource
type R2. That resource can then be allocated to P3, breaking the cycle.

METHODS FOR HANDLING DEADLOCKS

The deadlock problem can be handled in one of three ways:

1. Use a protocol to prevent or avoid deadlocks, ensuring that the system will never enter a
deadlocked state.
2. Allow the system to enter a deadlocked state, detect it, and recover.
3. Ignore the problem altogether and pretend that deadlocks never occur in the system.

To ensure that deadlocks never occur, the system can use either deadlock prevention or a
deadlock-avoidance scheme.

Deadlock prevention provides a set of methods for ensuring that at least one of the necessary
conditions cannot hold. These methods prevent deadlocks by constraining how requests for
resources can be made.

Deadlock-avoidance requires that the operating system be given in advance additional

information concerning which resources a process will request and use during its lifetime. With
this additional knowledge, it can decide for each request whether or not the process should
wait. To decide whether the current request can be satisfied or must be delayed, the system must
consider the resources currently available, the resources currently allocated to each process, and
the future requests and releases of each process

If a system does not employ either a deadlock-prevention or a deadlock avoidance algorithm,

then a deadlock situation may arise. In this environment, the system can provide an algorithm
that examines the state of the system to determine whether a deadlock has occurred and an
algorithm to recover from the deadlock.

In the absence of algorithms to detect and recover from deadlocks, then the system is in a
deadlock state yet has no way of recognizing what has happened. In this case, the undetected
deadlock will result in deterioration of the system's performance, because resources are being
held by processes that cannot run and because more and more processes, as they make requests
for resources, will enter a deadlocked state. Eventually, the system will stop functioning and will
need to be restarted manually.

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DEADLOACK PREVENTION

Deadlock can be prevented by ensuring that at least one of the four necessary conditions cannot
hold.

Mutual Exclusion
• The mutual-exclusion condition must hold for non-sharable resources. Sharable resources,
do not require mutually exclusive access and thus cannot be involved in a deadlock.
• Ex: Read-only files are example of a sharable resource. If several processes attempt to
open a read-only file at the same time, they can be granted simultaneous access to the file.
A process never needs to wait for a sharable resource.
• Deadlocks cannot prevent by denying the mutual-exclusion condition, because some
resources are intrinsically non-sharable.

Hold and Wait

To ensure that the hold-and-wait condition never occurs in the system, then guarantee that,
whenever a process requests a resource, it does not hold any other resources.
• One protocol that can be used requires each process to request and be allocated all its
resources before it begins execution.
• Another protocol allows a process to request resources only when it has none. A process
may request some resources and use them. Before it can request any additional resources,
it must release all the resources that it is currently allocated.

Ex:
• Consider a process that copies data from a DVD drive to a file on disk, sorts the file, and
then prints the results to a printer. If all resources must be requested at the beginning of
the process, then the process must initially request the DVD drive, disk file, and printer.
It will hold the printer for its entire execution, even though it needs the printer only at the
end.
• The second method allows the process to request initially only the DVD drive and disk
file. It copies from the DVD drive to the disk and then releases both the DVD drive and
the disk file. The process must then again request the disk file and the printer. After
copying the disk file to the printer, it releases these two resources and terminates.

The two main disadvantages of these protocols:

1. Resource utilization may be low, since resources may be allocated but unused for a long
period.
2. Starvation is possible.

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No Preemption
The third necessary condition for deadlocks is that there be no preemption of resources that have
already been allocated.

To ensure that this condition does not hold, the following protocols can be used:
• If a process is holding some resources and requests another resource that cannot be
immediately allocated to it, then all resources the process is currently holding are
preempted.
• The preempted resources are added to the list of resources for which the process is waiting.
The process will be restarted only when it can regain its old resources, as wellas the new
ones that it is requesting.

If a process requests some resources, first check whether they are available. If they are, allocate
them.
If they are not available, check whether they are allocated to some other process that is waiting
for additional resources. If so, preempt the desired resources from the waiting process and
allocate them to the requesting process.
If the resources are neither available nor held by a waiting process, the requesting process must
wait. While it is waiting, some of its resources may be preempted, but only if another process
requests them.
A process can be restarted only when it is allocated the new resources it is requesting and recovers
any resources that were preempted while it was waiting.

Circular Wait
One way to ensure that this condition never holds is to impose a total ordering of all resource
types and to require that each process requests resources in an increasing order of enumeration.

To illustrate, let R = {R1, R2, ... , Rm} be the set of resource types. Assign a unique integer
number to each resource type, which allows to compare two resources and to determinewhether
one precedes another in ordering. Formally, it defined as a one-to-one function
F: R ->N, where N is the set of natural numbers.

Example: if the set of resource types R includes tape drives, disk drives, and printers, then the
function F might be defined as follows:
F (tape drive) = 1
F (disk drive) = 5
F (printer) = 12

Now consider the following protocol to prevent deadlocks. Each process can request resources
only in an increasing order of enumeration. That is, a process can initially request any number of
instances of a resource type -Ri. After that, the process can request instances of resource type Rj
if and only if F(Rj) > F(Ri).

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DEADLOCK AVOIDANCE

• To avoid deadlocks an additional information is required about how resources are to be

requested. With the knowledge of the complete sequence of requests and releases for each
process, the system can decide for each request whether or not the process should wait in
order to avoid a possible future deadlock
• Each request requires that in making this decision the system consider the resources
currently available, the resources currently allocated to each process, and the future
requests and releases of each process.
• The various algorithms that use this approach differ in the amount and type of information
required. The simplest model requires that each process declare the maximum number of
resources of each type that it may need. Given this a priori information, it is possible to
construct an algorithm that ensures that the system will never enter a deadlocked state.
Such an algorithm defines the deadlock-avoidance approach.

Safe State

• Safe state: A state is safe if the system can allocate resources to each process (up to its
maximum) in some order and still avoid a deadlock. A system is in a safe state only if
there exists a safe sequence.

• Safe sequence: A sequence of processes <P1, P2, ... , Pn> is a safe sequence for the current
allocation state if, for each Pi, the resource requests that Pi can still make can be satisfied
by the currently available resources plus the resources held by all Pj, with j <i.

In this situation, if the resources that Pi needs are not immediately available, then Pi can wait
until all Pj have finished. When they have finished, Pi can obtain all of its needed resources,
complete its designated task, return its allocated resources, and terminate. When Pi terminates,
Pi+1 can obtain its needed resources, and so on. If no such sequence exists, then the system state
is said to be unsafe.

A safe state is not a deadlocked state. Conversely, a deadlocked state is an unsafe state. Not all
unsafe states are deadlocks as shown in figure. An unsafe state may lead to a deadlock. As long
as the state is safe, the operating system can avoid unsafe states

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Figure: Safe, unsafe, and deadlocked state spaces.

Resource-Allocation-Graph Algorithm

• If a resource-allocation system has only one instance of each resource type, then a
variant of the resource-allocation graph is used for deadlock avoidance.
• In addition to the request and assignment edges, a new type of edge is introduced, called
a claim edge.
• A claim edge Pi ->Rj indicates that process Pi may request resource Rj at some time in
the future. This edge resembles a request edge in direction but is represented in the
graph by a dashed line.
• When process Pi requests resource Rj, the claim edge Pi ->Rj is converted to a request
edge. When a resource Rj is released by Pi the assignment edge Rj->Pi is reconverted to
a claim edge Pi->Rj.

Figure: Resource-allocation graph for deadlock avoidance.

Note that the resources must be claimed a priori in the system. That is, before process Pi starts
executing, all its claim edges must already appear in the resource-allocation graph.
We can relax this condition by allowing a claim edge Pi ->Rj to be added to the graph only if
all the edges associated with process Pi are claim edges.

Now suppose that process Pi requests resource Rj. The request can be granted only if
converting the request edge Pi ->Rj to an assignment edge Rj->Pi does not result in the
formation of a cycle in the resource-allocation graph.

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There is need to check for safety by using a cycle-detection algorithm. An algorithm for detecting
a cycle in this graph requires an order of n2 operations, where n is the number of processes in the
system.
• If no cycle exists, then the allocation of the resource will leave the system in a safe state.
• If a cycle is found, then the allocation will put the system in an unsafe state. In that case,
process Pi will have to wait for its requests to be satisfied.

To illustrate this algorithm, consider the resource-allocation graph as shown above. Suppose that
P2 requests R2. Although R2 is currently free, we cannot allocate it to P2, since this action will
create a cycle in the graph.
A cycle, indicates that the system is in an unsafe state. If P1 requests R2, and P2 requests R1,
then a deadlock will occur.

Figure: An unsafe state in a resource-allocation graph

Banker's Algorithm

The Banker’s algorithm is applicable to a resource allocation system with multiple instances of
each resource type.
• When a new process enters the system, it must declare the maximum number of instances
of each resource type that it may need. This number may not exceed the total number of
resources in the system.
• When a user requests 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.

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To implement the banker's algorithm the following data structures are used.

Let n = number of processes, and m = number of resources types

Available: A vector of length m indicates the number of available resources of each type. If
available [j] = k, there are k instances of resource type Rj available.

Max: An n x m matrix defines the maximum demand of each process. If Max [i,j] = k,
thenprocess Pi may request at most k instances of resource type Rj

Allocation: An n x m matrix defines the number of resources of each type currently allocated
toeach process. If Allocation[i,j] = k then Pi is currently allocated k instances of Rj

Need: An n x m matrix indicates the remaining resource need of each process. If Need[i,j] = k,
then Pi may need k more instances of Rj to complete its task.

Need [i,j] = Max[i,j] – Allocation [i,j]

Safety Algorithm

The algorithm for finding out whether or not a system is in a safe state. This algorithm can be
described as follows:

1. Let Work and Finish be vectors of length m and n, respectively. Initialize:

Work = Available
Finish [i] = false for i = 0, 1,…,n- 1

2. Find an index i such that both:

(a) Finish[i] = false
(b) Needi Work
If no such i exists, go to step 4

3. Work = Work + Allocationi

Finish[i] = true
go to step 2

4. If Finish [i] == true for all i, then the system is in a safe state

This algorithm may require an order of m x n2 operations to determine whether a state is safe.

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Resource-Request Algorithm

The algorithm for determining whether requests can be safely granted.

Let Requesti be the request vector for process Pi. If Requesti [j] == k, then process Pi wants k
instances of resource type Rj. When a request for resources is made by process Pi, the following
actions are taken:

1. If RequestiNeedigo to step 2. Otherwise, raise error condition, since process has exceeded
its maximum claim

2. If RequestiAvailable, go to step 3. Otherwise Pi must wait, since resources are not available

3. Have the system pretend to allocate requested resources to Pi by modifying the state as
follows:
Available = Available – Request;
Allocationi= Allocationi +
Requesti;Needi=Needi – Requesti;

If safe  the resources are allocated to Pi

If unsafe  Pi must wait, and the old resource-allocation state is restored

Example

Consider a system with five processes Po through P4 and three resource types A, B, and C.
Resource type A has ten instances, resource type B has five instances, and resource type C
hasseven instances. Suppose that, at time T0the following snapshot of the system has been
taken:

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The content of the matrix Need is defined to be Max - Allocation

The system is currently in a safe state. Indeed, the sequence <P1, P3, P4, P2, P0> satisfies the
safety criteria.

Suppose now that process P1 requests one additional instance of resource type A and two
instances of resource type C, so Request1 = (1,0,2). Decide whether this request can be
immediately granted.

Check that Request  Available

(1,0,2)  (3,3,2)  true

Then pretend that this request has been fulfilled, and the following new state is arrived.

Executing safety algorithm shows that sequence <P1, P3, P4, P0, P2> satisfies safety
requirement.

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DEADLOCK DETECTION
If a system does not employ either a deadlock-prevention or a deadlock avoidance algorithm,
then a deadlock situation may occur. In this environment, the system may provide:
• An algorithm that examines the state of the system to determine whether a deadlock has
occurred
• An algorithm to recover from the deadlock

Single Instance of Each Resource Type

• If all resources have only a single instance, then define a deadlock detection algorithm
that uses a variant of the resource-allocation graph, called a wait-for graph.
• This graph is obtained from the resource-allocation graph by removing the resource nodes
and collapsing the appropriate edges.
• An edge from Pi to Pj in a wait-for graph implies that process Pi is waiting for process Pj
to release a resource that Pi needs. An edge Pi → Pj exists in a wait-for graph if and only if
the corresponding resource allocation graph contains two edges P i →Rq and Rq→Pi for some
resource Rq.

Example: In below Figure, a resource-allocation graph and the corresponding wait-for graph is
presented.

Figure: (a) Resource-allocation graph. (b) Corresponding wait-for graph.

• A deadlock exists in the system if and only if the wait-for graph contains a cycle. To
detect deadlocks, the system needs to maintain the wait-for graph and periodically
invoke an algorithm that searches for a cycle in the graph.
• An algorithm to detect a cycle in a graph requires an order of n2 operations, where n is
the number of vertices in the graph.

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Several Instances of a Resource Type

A deadlock detection algorithm that is applicable to several instances of a resource type. The
algorithm employs several time-varying data structures that are similar to those used in the
banker's algorithm.

• Available: A vector of length m indicates the number of available resources of

eachtype.
• Allocation: Ann x m matrix defines the number of resources of each type
currentlyallocated to each process.
• Request: An n x m matrix indicates the current request of each process. If Request[i][j]
equals k, then process P; is requesting k more instances of resource type Rj.

Algorithm:

1. Let Work and Finish be vectors of length m and n, respectively Initialize:

(a) Work = Available
(b) For i = 1,2, …, n, if Allocationi 0, then Finish[i] = false;
otherwise, Finish[i] = true

2. Find an index isuch that both:

(a) Finish[i] == false
(b)RequestiWork

If no such i exists, go to step 4

3. Work = Work + Allocationi

Finish[i] = true
go to step 2

4. If Finish[i] == false, for some i, 1 in, then the system is in deadlock state. Moreover, if
Finish[i] == false, then Pi is deadlocked

Algorithm requires an order of O(m x n2) operations to detect whether the system is in
deadlocked state

Example of Detection Algorithm

Consider a system with five processes Po through P4 and three resource types A, B, and C.
Resource type A has seven instances, resource type B has two instances, and resource type
Chas six instances. Suppose that, at time T0, the following resource-allocation state:

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After executing the algorithm, Sequence <P0, P2, P3, P1, P4> will result in Finish[i] = true for
all i
Suppose now that process P2 makes one additional request for an instance of type C. The
Request matrix is modified as follows:

The system is now deadlocked. Although we can reclaim the resources held by process Po, the
number of available resources is not sufficient to fulfill the requests of the other processes.
Thus, a deadlock exists, consisting of processes P1, P2, P3, and P4.

Detection-Algorithm Usage

The detection algorithm can be invoked on two factors:

1. How often is a deadlock likely to occur?
2. How many processes will be affected by deadlock when it happens?

If deadlocks occur frequently, then the detection algorithm should be invoked frequently.
Resources allocated to deadlocked processes will be idle until the deadlock can be broken.

If detection algorithm is invoked arbitrarily, there may be many cycles in the resource graph and
so we would not be able to tell which of the many deadlocked processes “caused” the deadlock.

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RECOVERY FROM DEADLOCK

The system recovers from the deadlock automatically. There are two options for breaking a
deadlock one 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, use one of two methods. In both methods, the
system reclaims all resources allocated to the terminated processes.

1. Abort all deadlocked processes: This method clearly will break the deadlock cycle, but
at great expense; 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.
2. Abort one process at a time until the deadlock cycle is eliminated: 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.

If the partial termination method is used, then we must determine which deadlocked process (or
processes) should be terminated. Many factors may affect which process is chosen, including:

1. What the priority of the process is

2. How long the process has computed and how much longer the process will compute
before completing its designated task
3. How many and what types of resources the process has used.
4. How many more resources the process needs in order to complete
5. How many processes will need to be terminated?
6. Whether the process is interactive or batch

Resource Preemption

To eliminate deadlocks using resource preemption, we successively preempt some resources

from processes and give these resources to other processes until the deadlock cycle is broken.
If preemption is required to deal with deadlocks, then three issues need to be addressed:

1. Selecting a victim. Which resources and which processes are to be preempted? As in

process termination, we must determine the order of preemption to minimize cost. Cost
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 during its execution.
2. Rollback. If we preempt a resource from a process, what should be done with that
process? Clearly, it cannot continue with its normal execution; it is missing some needed
resource. We must roll back the process to some safe state and restart it from that state.
Since it is difficult to determine what a safe state is, the simplest solution is a total rollback:
abort the process and then restart it.

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3. Starvation. How do we ensure that starvation will not occur? That is, how can we
guarantee that resources will not always be preempted from the same process?

QUESTION BANK
1. What are semaphores? Explain two primitive semaphore operations. What are its advantages?
2. Explain any one synchronization problem for testing newly proposed sync scheme
3. Explain three requirements that a solution to critical –section problem must satisfy.
4. State Dining Philosopher’s problem and give a solution using semaphores. Write structure of
philosopher.
5. What do you mean by binary semaphore and counting semaphore? With C struct, explain
implementation of wait() and signal. Semaphore as General Synchronization Tool.
6. Describe term monitor. Explain solution to dining philosophers.
7. What are semaphores? Explain solution to producer-consumer problem using semaphores
8. What is critical section? Explain the various methods to implement process synchronization.
9. Explain the various classical synchronization problems.
10. What are deadlocks? What are its characteristics? Explain the necessary conditions forits
occurrence.
11. Define Semaphores. Explain its usage and implementation.
12. Explain Reader-Writer problem with semaphore in detail.
13. What are monitors? Explain dining Philospher's solution using monitor.
14. Illustrate how Reader's-Writer's problem can be solved by using semaphores.
15. Explain the process of recovery from deadlock.
16. Describe RAG:
i) With deadlock
ii) With a cycle but no deadlock
17. What is Resource Allocation Graph (RAG)? Explain how RAG is very useful indescribing
deadly embrace (dead lock ) by considering your own example.
18. With the help of a system model, explain a deadlock and explain the necessaryconditions that
must hold simultaneously in a system for a deadlock to occur.
19. Explain how deadlock can be prevented by considering four necessary conditions cannothold.
20. Using Banker's algorithm determines whether the system is in a safe state.
21. How is a system recovered from deadlock? Explain the different methods used torecover from
deadlock.
22. Explain deadlock detection with algorithm and example

Additional Quesitions:
1. Define the terms: safe state and safe sequence. Give an algorithm to find whether or nota
system is in a safe state.
2. Find need matrix and calculate the safe sequence by using Banker's algorithm. Mention the
above system is safe or not safe.
Jobs Allocation Maximum Available
R1 R2 R3 R1 R2 R3 R1 R2 R3
J1
0 1 0 7 5 3
J2 2 0 0 3 2 2 3 3 2
J3 3 0 2 9 0 2

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J4 2 1 1 2 2 2
J5 0 0 2 4 3 3

3. Given the memory partitions of 100 K, 500 K, 200 K, 300 K and 600 K apply first fit, best fit
and worst fit algorithms to place 212K, 417K, 112K and 426K.
4. Explain the structure of page table.
5. Assume that there are 5 processes PO through P4 and 4 types of resources. At time T0 we
have the following state:
Process Allocation Maximum Available

A B C D A B C D A B C D
P0
0 0 1 2 0 0 1 2 1 5 2 0
P1 1 0 0 0 1 7 5 0
P2 1 3 5 4 2 3 5 6
P3 0 6 3 2 0 6 5 2
P4 0 0 1 4 0 6 5 6
P5 0 0 2 4 3 3
Apply Banker's algorithm to answer the following:
i) What is the content of need matrix?
ii) Is the system in a safe state?
iii) If a request from a process P1(0, 4, 2, 0) arrives, can it be granted?

6. Consider the following snapshot of a system

Process Allocation Maximum Available

A B C A B C A B C
P0
0 0 2 0 0 4
P1 1 0 0 2 0 1
P2 1 3 5 1 3 7 1 0 2
P3 6 3 2 8 4 2
P4 1 4 3 1 5 7
Find the need matrix and calculate safe sequence using Banker's algorithm. Mention the above
system is safe or not safe.

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