1

SCS1501 - OPERATING
SYSTEM
UNIT – 3
SYNCHRONIZATION
AND DEADLOCK
Mrs. C.Kavitha
ASSISTANT PROFESSOR
DEPARTMENT OF CSE
SCHOOL OF COMPUTING
 UNIT 3
Syllabus
 Process Synchronization
 Background
 The Critical-Section Problem
 Peterson’s Solution
 Synchronization Hardware
 Mutex Locks
 Semaphores
 Classic Problems of Synchronization
 Monitors
 Synchronization Examples
 Alternative Approaches
 Objectives
 To present the concept of process synchronization.
 To introduce the critical-section problem, whose solutions
can be used to ensure the consistency of shared data
 To present both software and hardware solutions of the
critical-section problem
 To examine several classical process-synchronization
problems
 To explore several tools that are used to solve process
synchronization problems
 Background
 Processes can execute concurrently
 May be interrupted at any time, partially completing
execution
 Concurrent access to shared data may result in data
inconsistency
 Maintaining data consistency requires mechanisms to ensure
the orderly execution of cooperating processes
 Illustration of the problem:
Suppose that we wanted to provide a solution to the
consumer-producer problem that fills all the buffers. We can
do so by having an integer counter that keeps track of the
number of full buffers. Initially, counter is set to 0. It is
incremented by the producer after it produces a new buffer and
is decremented by the consumer after it consumes a buffer.
 Producer

while (true) {
/* produce an item in next produced */

while (counter == BUFFER_SIZE) ;

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

while (true) {
while (counter == 0)
; /* do nothing */
next_consumed = buffer[out];
out = (out + 1) % BUFFER_SIZE;
counter--;
/* consume the item in next consumed */
}
 Race Condition
 counter++ could be implemented as
register1 = counter
register1 = register1 + 1
counter = register1
 counter-- could be implemented as
register2 = counter
register2 = register2 - 1
counter = register2

 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 {counter = 6 }
S5: consumer execute counter = register2 {counter = 4}
 Critical Section Problem
 Consider system of n processes {p0, p1, … pn-1}
 Each process has critical section segment of code
 Process may be changing common variables, updating
table, writing file, etc
 When one process in critical section, no other may be in its
critical section
 Critical section problem is to design protocol to solve this
 Each process must ask permission to enter critical section in
entry section, may follow critical section with exit section,
then remainder section
 Critical Section

 General structure of process Pi

 Algorithm for Process Pi

do {

while (turn == j);

critical section
turn = j;

remainder section
} while (true);
 Solution to Critical-Section Problem
1. Mutual Exclusion - If process Pi is executing in its critical
section, then no other processes can be executing in their
critical sections
2. Progress - If no process is executing in its critical section and
there exist some processes that wish to enter their critical
section, then the selection of the processes that will enter the
critical section next cannot be postponed indefinitely
3. Bounded Waiting - A bound must exist 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 is granted
 Assume that each process executes at a nonzero speed
 No assumption concerning relative speed of the n
processes
 Critical-Section Handling in OS
Two approaches depending on if kernel is preemptive or non-
preemptive
 Preemptive – allows preemption of process when running
in kernel mode
 Non-preemptive – runs until exits kernel mode, blocks, or
voluntarily yields CPU
Essentially free of race conditions in kernel mode
 Peterson’s Solution
 Good algorithmic description of solving the problem
 Two process solution
 Assume that the load and store machine-language
instructions are atomic; that is, cannot be interrupted
 The two processes share two variables:
 int turn;
 Boolean flag[2]

 The variable turn indicates whose turn it is to enter the critical

section
 The flag array is used to indicate if a process is ready to enter
the critical section. flag[i] = true implies that process Pi is
ready!
 Algorithm 1
do {
while (turn != i);
critical section
turn = j;
remainder section
}while (1);
 Algorithm 2
do
{
flag[i] =true;
while (flag[j]);
critical section
flag[i]= false;
remainder section
}while(1);
 Algorithm3

do {
flag[i] = true;
turn = j;
while (flag[j] && turn = = j);
critical section
flag[i] = false;
remainder section
} while (true);
 Peterson’s Solution (Cont.)
 Provable that the three CS requirement are met:
1. Mutual exclusion is preserved
Pi enters CS only if:
either flag[j] = false or turn = i
2. Progress requirement is satisfied
3. Bounded-waiting requirement is met
 Synchronization Hardware
 Many systems provide hardware support for implementing the
critical section code.
 All solutions below based on idea of locking
 Protecting critical regions via locks
 Uniprocessors – could disable interrupts
 Currently running code would execute without preemption
 Generally too inefficient on multiprocessor systems
 Operating systems using this not broadly scalable
 Modern machines provide special atomic hardware instructions
 Atomic = non-interruptible
 Either test memory word and set value
 Or swap contents of two memory words
 Solution to Critical-section Problem Using Locks

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

Definition:
boolean test_and_set (boolean *target)
{
boolean rv = *target;
*target = TRUE;
return rv:
}
1. Executed atomically
2. Returns the original value of passed parameter
3. Set the new value of passed parameter to “TRUE”.
 Solution using test_and_set()
 Shared Boolean variable lock, initialized to FALSE
 Solution:
do {
while (test_and_set(&lock))
; /* do nothing */
/* critical section */
lock = false;
/* remainder section */
} while (true);
 compare_and_swap Instruction
Definition:
int compare _and_swap(int *value, int expected, int new_value) {
int temp = *value;

if (*value == expected)
*value = new_value;
return temp;
}
1. Executed atomically
2. Returns the original value of passed parameter “value”
3. Set the variable “value” the value of the passed parameter “new_value”
but only if “value” ==“expected”. That is, the swap takes place only under
this condition.
 Solution using compare_and_swap
 Shared integer “lock” initialized to 0;
 Solution:
do {
while (compare_and_swap(&lock, 0, 1) != 0)
; /* do nothing */
/* critical section */
lock = 0;
/* remainder section */
} while (true);
 Bounded-waiting Mutual Exclusion with test_and_set

do {
waiting[i] = true;
key = true;
while (waiting[i] && key)
key = test_and_set(&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);
 Mutex Locks
 Previous solutions are complicated and generally inaccessible
to application programmers
 OS designers build software tools to solve critical section
problem
 Simplest is mutex lock
 Protect a critical section by first acquire() a lock then
release() the lock
 Boolean variable indicating if lock is available or not
 Calls to acquire() and release() must be atomic
 Usually implemented via hardware atomic instructions
 But this solution requires busy waiting
 This lock therefore called a spinlock
 acquire() and release()
 acquire() {
while (!available)
; /* busy wait */
available = false;
}
 release() {
available = true;
}
 do {
acquire lock
critical section
release lock
remainder section
} while (true);
 Semaphore
 Synchronization tool that provides more sophisticated ways (than Mutex locks) for
process to synchronize their activities.
 Semaphore S – integer variable
 Can only be accessed via two indivisible (atomic) operations
 wait() and signal()
 Originally called P() and V()
 Definition of the wait() operation
wait(S) {
while (S <= 0)
; // busy wait
S--;
}
 Definition of the signal() operation
signal(S) {
S++;
}
 Semaphore Usage
 Counting semaphore – integer value can range over an unrestricted
domain
 Binary semaphore – integer value can range only between 0 and 1
 Same as a mutex lock
 Can solve various synchronization problems
 Consider P1 and P2 that require S1 to happen before S2
Create a semaphore “synch” initialized to 0
P1:
S1;
signal(synch);
P2:
wait(synch);
S2;
 Can implement a counting semaphore S as a binary semaphore
 Semaphore Implementation
 Must guarantee that no two processes can execute the wait()
and signal() on the same semaphore at the same time
 Thus, the implementation becomes the critical section problem
where the wait and signal code are placed in the critical
section
 Could now have busy waiting in critical section
implementation
 But implementation code is short
 Little busy waiting if critical section rarely occupied
 Note that applications may spend lots of time in critical sections
and therefore this is not a good solution
 Semaphore Implementation with no Busy waiting

 With each semaphore there is an associated waiting queue

 Each entry in a waiting queue has two data items:
 value (of type integer)
 pointer to next record in the list
 Two operations:
 block – place the process invoking the operation on the appropriate
waiting queue
 wakeup – remove one of processes in the waiting queue and place it in
the ready queue
 typedef struct{
int value;
struct process *list;
} semaphore;
 Implementation with no Busy waiting (Cont.)

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

signal(semaphore *S) {
S->value++;
if (S->value <= 0) {
remove a process P from S->list;
wakeup(P);
}
}
 Deadlock and Starvation
 Deadlock – two or more processes are waiting indefinitely for an event
that can be caused by only one of the waiting processes
 Let S and Q be two semaphores initialized to 1
P0 P1
wait(S); wait(Q);
wait(Q); wait(S);
... ...
signal(S); signal(Q);
signal(Q); signal(S);

 Starvation – indefinite blocking

 A process may never be removed from the semaphore queue in which it is
suspended
 Priority Inversion – Scheduling problem when lower-priority process
holds a lock needed by higher-priority process
 Solved via priority-inheritance protocol
 Classical Problems of Synchronization
 Classical problems used to test newly-proposed synchronization
schemes
 Bounded-Buffer Problem
 Readers and Writers Problem
 Dining-Philosophers Problem
 Bounded-Buffer Problem

 n buffers, each can hold one item

 Semaphore mutex initialized to the value 1
 Semaphore full initialized to the value 0
 Semaphore empty initialized to the value n
 Bounded Buffer Problem (Cont.)

 The structure of the producer process

do {
...
/* produce an item in next_produced */
...
wait(empty);
wait(mutex);
...
/* add next produced to the buffer */
...
signal(mutex);
signal(full);
} while (true);
 Bounded Buffer Problem (Cont.)
 The structure of the consumer process

Do {
wait(full);
wait(mutex);
...
/* remove an item from buffer to next_consumed */
...
signal(mutex);
signal(empty);
...
/* consume the item in next consumed */
...
} while (true);
 Readers-Writers Problem
 A data set is shared among a number of concurrent processes
 Readers – only read the data set; they do not perform any updates
 Writers – can both read and write
 Problem – allow multiple readers to read at the same time
 Only one single writer can access the shared data at the same time
 Several variations of how readers and writers are considered – all
involve some form of priorities
 Shared Data
 Data set
 Semaphore rw_mutex initialized to 1
 Semaphore mutex initialized to 1
 Integer read_count initialized to 0
 Readers-Writers Problem (Cont.)

 The structure of a writer process

do {
wait(rw_mutex);
...
/* writing is performed */
...
signal(rw_mutex);
} while (true);
 Readers-Writers Problem (Cont.)
 The structure of a reader process
do {
wait(mutex);
read_count++;
if (read_count == 1)
wait(rw_mutex);
signal(mutex);
...
/* reading is performed */
...
wait(mutex);
read count--;
if (read_count == 0)
signal(rw_mutex);
signal(mutex);
} while (true);
 Readers-Writers Problem Variations
 First variation – no reader kept waiting unless writer has
permission to use shared object
 Second variation – once writer is ready, it performs the
write ASAP
 Both may have starvation leading to even more variations
 Problem is solved on some systems by kernel providing
reader-writer locks
 Dining-Philosophers Problem

 Philosophers spend their lives alternating thinking and eating

 Don’t interact with their neighbors, occasionally try to pick up 2
chopsticks (one at a time) to eat from bowl
 Need both to eat, then release both when done
 In the case of 5 philosophers
 Shared data
 Bowl of rice (data set)
 Semaphore chopstick [5] initialized to 1
 Dining-Philosophers Problem Algorithm
 The structure of Philosopher i:
do {
wait (chopstick[i] );
wait (chopStick[ (i + 1) % 5] );

// eat

signal (chopstick[i] );
signal (chopstick[ (i + 1) % 5] );

// think

} while (TRUE);
 What is the problem with this algorithm?
 Dining-Philosophers Problem Algorithm (Cont.)

 Deadlock handling
 Allow at most 4 philosophers to be sitting
simultaneously at the table.
 Allow a philosopher to pick up the forks only if both
are available (picking must be done in a critical
section.
 Use an asymmetric solution -- an odd-numbered
philosopher picks up first the left chopstick and then
the right chopstick. Even-numbered philosopher picks
up first the right chopstick and then the left chopstick.
 Problems with Semaphores

 Incorrect use of semaphore operations:

 signal (mutex) …. wait (mutex)

 wait (mutex) … wait (mutex)

 Omitting of wait (mutex) or signal (mutex) (or both)

 Deadlock and starvation are possible.

 Monitors
 A high-level abstraction that provides a convenient and effective
mechanism for process synchronization
 Abstract data type, internal variables only accessible by code within the
procedure
 Only one process may be active within the monitor at a time
 But not powerful enough to model some synchronization schemes

monitor monitor-name
{
// shared variable declarations
procedure P1 (…) { …. }

procedure Pn (…) {……}

Initialization code (…) { … }

}
}
Schematic view of a Monitor
 Condition Variables
 condition x, y;
 Two operations are allowed on a condition variable:
 x.wait() – a process that invokes the operation is
suspended until x.signal()
 x.signal() – resumes one of processes (if any) that
invoked x.wait()
 If no x.wait() on the variable, then it has no effect on
the variable
Monitor with Condition Variables
 Condition Variables Choices
 If process P invokes x.signal(), and process Q is suspended in
x.wait(), what should happen next?
 Both Q and P cannot execute in paralel. If Q is resumed, then P must
wait
 Options include
 Signal and wait – P waits until Q either leaves the monitor or it waits
for another condition
 Signal and continue – Q waits until P either leaves the monitor or it
waits for another condition
 Both have pros and cons – language implementer can decide
 Monitors implemented in Concurrent Pascal compromise
 P executing signal immediately leaves the monitor, Q is resumed
 Implemented in other languages including Mesa, C#, Java
 Monitor Solution to Dining Philosophers
monitor DiningPhilosophers
{
enum { THINKING; HUNGRY, EATING) state [5] ;
condition self [5];

void pickup (int i) {

state[i] = HUNGRY;
test(i);
if (state[i] != EATING) self[i].wait;
}

void putdown (int i) {

state[i] = THINKING;
// test left and right neighbors
test((i + 4) % 5);
test((i + 1) % 5);
}
 Solution to Dining Philosophers (Cont.)

void test (int i) {

if ((state[(i + 4) % 5] != EATING) &&
(state[i] == HUNGRY) &&
(state[(i + 1) % 5] != EATING) ) {
state[i] = EATING ;
self[i].signal () ;
}
}

initialization_code() {
for (int i = 0; i < 5; i++)
state[i] = THINKING;
}
}
 Solution to Dining Philosophers (Cont.)

 Each philosopher i invokes the operations pickup() and

putdown() in the following sequence:

DiningPhilosophers.pickup(i);

EAT

DiningPhilosophers.putdown(i);

 No deadlock, but starvation is possible

 Monitor Implementation Using Semaphores
 Variables

semaphore mutex; // (initially = 1)

semaphore next; // (initially = 0)
int next_count = 0;

 Each procedure F will be replaced by

wait(mutex);
…
body of F;
…
if (next_count > 0)
signal(next)
else
signal(mutex);

 Mutual exclusion within a monitor is ensured

 Monitor Implementation – Condition Variables

 For each condition variable x, we have:

semaphore x_sem; // (initially = 0)

int x_count = 0;

 The operation x.wait can be implemented as:

x_count++;
if (next_count > 0)
signal(next);
else
signal(mutex);
wait(x_sem);
x_count--;
 Monitor Implementation (Cont.)

 The operation x.signal can be implemented as:

if (x_count > 0) {
next_count++;
signal(x_sem);
wait(next);
next_count--;
}
 Resuming Processes within a Monitor

 If several processes queued on condition x, and x.signal()

executed, which should be resumed?
 FCFS frequently not adequate
 conditional-wait construct of the form x.wait(c)
 Where c is priority number
 Process with lowest number (highest priority) is
scheduled next
 Single Resource allocation
 Allocate a single resource among competing processes using
priority numbers that specify the maximum time a process
plans to use the resource

R.acquire(t);
...
access the resurce;
...

R.release;

 Where R is an instance of type ResourceAllocator

A Monitor to Allocate Single Resource
monitor ResourceAllocator
{
boolean busy;
condition x;
void acquire(int time) {
if (busy)
x.wait(time);
busy = TRUE;
}
void release() {
busy = FALSE;
x.signal();
}
initialization code() {
busy = FALSE;
}
}
 Deadlocks
 System Model
 Deadlock Characterization
 Methods for Handling Deadlocks
 Deadlock Prevention
 Deadlock Avoidance
 Deadlock Detection
 Recovery from Deadlock
 Objectives

 To develop a description of deadlocks, which prevent

sets of concurrent processes from completing their
tasks
 To present a number of different methods for
preventing or avoiding deadlocks in a computer
system
 System Model
 System consists of resources
 Resource types R1, R2, . . ., Rm
CPU cycles, memory space, I/O devices
 Each resource type Ri has Wi instances.
 Each process utilizes a resource as follows:
 request
 use
 release
 Deadlock Characterization
Deadlock can arise if four conditions hold simultaneously.

 Mutual exclusion: only one process at a time can use a

resource
 Hold and wait: a process holding at least one resource is
waiting to acquire additional resources held by other
processes
 No preemption: a resource can be released only voluntarily
by the process holding it, after that process has completed
its task
 Circular wait: there exists a set {P0, P1, …, Pn} of waiting
processes such that P0 is waiting for a resource that is held
by P1, P1 is waiting for a resource that is held by P2, …, Pn–1
is waiting for a resource that is held by Pn, and Pn is waiting
for a resource that is held by P0.
 Deadlock with Mutex Locks
 Deadlocks can occur via system calls, locking, etc.
 See example box in text page 318 for mutex deadlock
 Resource-Allocation Graph
A set of vertices V and a set of edges E.
 V is partitioned into two types:
 P = {P1, P2, …, Pn}, the set consisting of all the processes
in the system

 R = {R1, R2, …, Rm}, the set consisting of all resource

types in the system

 request edge – directed edge Pi  Rj

 assignment edge – directed edge Rj  Pi

 Resource-Allocation Graph (Cont.)
 Process

 Resource Type with 4 instances

 Pi requests instance of Rj

Pi
Rj
 Pi is holding an instance of Rj

Pi
Rj
Example of a Resource Allocation Graph
Resource Allocation Graph With A Deadlock
Graph With A Cycle But No Deadlock
 Basic Facts

 If graph contains no cycles  no deadlock

 If graph contains a cycle 
 if only one instance per resource type, then deadlock
 if several instances per resource type, possibility of
deadlock
 Methods for Handling Deadlocks
 Ensure that the system will never enter a deadlock
state:
 Deadlock prevention
 Deadlock avoidence
 Allow the system to enter a deadlock state and then
recover
 Ignore the problem and pretend that deadlocks never
occur in the system; used by most operating systems,
including UNIX
 Deadlock Prevention
Restrain the ways request can be made

 Mutual Exclusion – not required for sharable resources

(e.g., read-only files); must hold for non-sharable resources
 Hold and Wait – must guarantee that whenever a process
requests a resource, it does not hold any other resources
 Require process to request and be allocated all its
resources before it begins execution, or allow process
to request resources only when the process has none
allocated to it.
 Low resource utilization; starvation possible
 Deadlock Prevention (Cont.)
 No Preemption –
 If a process that is holding some resources requests
another resource that cannot be immediately allocated to
it, then all resources currently being held are released
 Preempted resources are added to the list of resources
for which the process is waiting
 Process will be restarted only when it can regain its old
resources, as well as the new ones that it is requesting
 Circular Wait – impose a total ordering of all resource types,
and require that each process requests resources in an
increasing order of enumeration
 Deadlock Example
/* thread one runs in this function */
void *do_work_one(void *param)
{

pthread_mutex_lock(&first_mutex);
pthread_mutex_lock(&second_mutex);
/** * Do some work */
pthread_mutex_unlock(&second_mutex);
pthread_mutex_unlock(&first_mutex);
pthread_exit(0);
}
/* thread two runs in this function */
void *do_work_two(void *param)
{

pthread_mutex_lock(&second_mutex);
pthread_mutex_lock(&first_mutex);
/** * Do some work */
pthread_mutex_unlock(&first_mutex);
pthread_mutex_unlock(&second_mutex);
pthread_exit(0);
}
 Deadlock Example with Lock Ordering
void transaction(Account from, Account to, double amount)
{
mutex lock1, lock2;
lock1 = get_lock(from);
lock2 = get_lock(to);
acquire(lock1);
acquire(lock2);
withdraw(from, amount);
deposit(to, amount);
release(lock2);
release(lock1);
}

Transactions 1 and 2 execute concurrently. Transaction 1 transfers $25

from account A to account B, and Transaction 2 transfers $50 from account
B to account A
 Deadlock Avoidance
Requires that the system has some additional a priori information
available
 Simplest and most useful model requires that each process
declare the maximum number of resources of each type
that it may need
 The deadlock-avoidance algorithm dynamically examines
the resource-allocation state to ensure that there can never
be a circular-wait condition
 Resource-allocation state is defined by the number of
available and allocated resources, and the maximum
demands of the processes
 Safe State

 When a process requests an available resource, system must

decide if immediate allocation leaves the system in a safe state
 System is in safe state if there exists a sequence <P1, P2, …, Pn>
of ALL the processes in the systems such that for each Pi, the
resources that Pi can still request can be satisfied by currently
available resources + resources held by all the Pj, with j < I
 That is:
 If Pi resource needs are not immediately available, then Pi can
wait until all Pj have finished
 When Pj is finished, Pi can obtain needed resources, execute,
return allocated resources, and terminate
 When Pi terminates, Pi +1 can obtain its needed resources, and
so on
 Basic Facts

 If a system is in safe state  no deadlocks

 If a system is in unsafe state  possibility of deadlock

 Avoidance  ensure that a system will never enter an

unsafe state.
Safe, Unsafe, Deadlock State
 Avoidance Algorithms
 Single instance of a resource type
 Use a resource-allocation graph

 Multiple instances of a resource type

 Use the banker’s algorithm
 Resource-Allocation Graph Scheme
 Claim edge Pi  Rj indicated that process Pj may request
resource Rj; represented by a dashed line
 Claim edge converts to request edge when a process requests
a resource
 Request edge converted to an assignment edge when the
resource is allocated to the process
 When a resource is released by a process, assignment edge
reconverts to a claim edge
 Resources must be claimed a priori in the system
Resource-Allocation Graph
Unsafe State In Resource-Allocation Graph
 Resource-Allocation Graph Algorithm

 Suppose that process Pi requests a resource Rj

 The request can be granted only if converting the
request edge to an assignment edge does not result
in the formation of a cycle in the resource allocation
graph
 Banker’s Algorithm
 Multiple instances

 Each process must a priori claim maximum use

 When a process requests a resource it may have to wait

 When a process gets all its resources it must return them in a

finite amount of time
 Data Structures for the Banker’s Algorithm

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

 Available: Vector of length m. If available [j] = k, there are k

instances of resource type Rj available

 Max: n x m matrix. If Max [i,j] = k, then process Pi may request at

most k instances of resource type Rj

 Allocation: n x m matrix. If Allocation[i,j] = k then Pi is currently

allocated k instances of Rj

 Need: n x m matrix. 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
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 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
 Resource-Request Algorithm for Process Pi

Requesti = request vector for process Pi. If Requesti [j] = k then

process Pi wants k instances of resource type Rj
1. If Requesti  Needi go 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. Pretend to allocate requested resources to Pi by modifying the
state as follows:
Available = Available – Requesti;
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 of Banker’s Algorithm

 5 processes P0 through P4;

3 resource types:
A (10 instances), B (5instances), and C (7 instances)
 Snapshot at time T0:
Allocation Max Available
ABC ABC ABC
P0 010 753 332
P1 200 322
P2 302 902
P3 211 222
P4 002 433
 Example (Cont.)
 The content of the matrix Need is defined to be Max – Allocation

Need
ABC
P0 743
P1 122
P2 600
P3 011
P4 431

 The system is in a safe state since the sequence < P1, P3, P4, P2, P0>
satisfies safety criteria
 Example: P1 Request (1,0,2)
 Check that Request  Available (that is, (1,0,2)  (3,3,2)  true
Allocation Need Available
ABC ABCABC
P0 010 743 230
P1 302 020
P2 302 600
P3 211 011
P4 002 431

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

 Can request for (3,3,0) by P4 be granted?

 Can request for (0,2,0) by P0 be granted?

 Deadlock Detection

 Allow system to enter deadlock state

 Detection algorithm

 Recovery scheme
 Single Instance of Each Resource Type

 Maintain wait-for graph

 Nodes are processes
 Pi  Pj if Pi is waiting for Pj

 Periodically invoke an algorithm that searches for a cycle in the

graph. If there is a cycle, there exists a deadlock

 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
Resource-Allocation Graph and Wait-for Graph

Resource-Allocation Graph Corresponding wait-for graph

 Several Instances of a Resource Type
 Available: A vector of length m indicates the number of
available resources of each type
 Allocation: An n x m matrix defines the number of resources
of each type currently allocated to each process
 Request: An n x m matrix indicates the current request of
each process. If Request [i][j] = k, then process Pi is
requesting k more instances of resource type Rj.
 Detection 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 i such that both:

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

If no such i exists, go to step 4

 Detection Algorithm (Cont.)
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
 Five processes P0 through P4; three resource types
A (7 instances), B (2 instances), and C (6 instances)

 Snapshot at time T0:

Allocation Request Available
ABC ABC ABC
P0 010 000 000
P1 200 202
P2 303 000
P3 211 100
P4 002 002

 Sequence <P0, P2, P3, P1, P4> will result in Finish[i] = true for all i
 Example (Cont.)

 P2 requests an additional instance of type C

Request
ABC
P0 000
P1 202
P2 001
P3 100
P4 002

 State of system?
 Can reclaim resources held by process P0, but insufficient
resources to fulfill other processes; requests
 Deadlock exists, consisting of processes P1, P2, P3, and P4
 Detection-Algorithm Usage
 When, and how often, to invoke depends on:
 How often a deadlock is likely to occur?
 How many processes will need to be rolled back?
 one for each disjoint cycle

 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.
 Recovery from Deadlock: Process Termination

 Abort all deadlocked processes

 Abort one process at a time until the deadlock cycle is eliminated

 In which order should we choose to abort?

1. Priority of the process
2. How long process has computed, and how much longer to
completion
3. Resources the process has used
4. Resources process needs to complete
5. How many processes will need to be terminated
6. Is process interactive or batch?
 Recovery from Deadlock: Resource Preemption

 Selecting a victim – minimize cost

 Rollback – return to some safe state, restart process for that

state

 Starvation – same process may always be picked as victim,

include number of rollback in cost factor
End