Download Monitors

CENG334
Introduction to Operating Systems
Peterson’s Algorithm
Monitors, Condition variabless
Topics:
•Monitors
•Condition Variables
Erol Sahin
Dept of Computer Eng.
Middle East Technical University
Ankara, TURKEY
URL: http://kovan.ceng.metu.edu.tr/ceng334
1
Peterson’s Algorithm
int flag[2] = {0, 0};
int turn;
P0:
do{
flag[0] = 1;
turn = 1;
while (flag[1] == 1 && turn == 1)
{
// busy wait
}
// critical section
flag[0] = 0;
//remainder section
}while(1);
P1:
do{
flag[1] = 1;
turn = 0;
while (flag[0] == 1 && turn == 0)
{
// busy wait
}
// critical section
flag[1] = 0;
// remainder section
} while(1);
turn : indicates whose turn is it to enter critical section. If turn==i
process Pi is allowed to get in.
flag[2]: indicates if process Pi is ready to enter critical section. If
flag[i]is set, then Pi is ready to enter critical section.
2
Peterson’s Algorithm
int flag[2] = {0, 0};
int turn;
P0:
do{
flag[0] = 1;
turn = 1;
while (flag[1] == 1 && turn == 1)
{
// busy wait
}
// critical section
flag[0] = 0;
//remainder section
}while(1);
P1:
do{
flag[1] = 1;
turn = 0;
while (flag[0] == 1 && turn == 0)
{
// busy wait
}
// critical section
flag[1] = 0;
// remainder section
} while(1);
Mutual Exclusion: Only one process Pi (the one which set turn=i last) enters the critical
section.
3
Peterson’s Algorithm
int flag[2] = {0, 0};
int turn;
P0:
do{
flag[0] = 1;
turn = 1;
while (flag[1] == 1 && turn == 1)
{
// busy wait
}
// critical section
flag[0] = 0;
//remainder section
}while(1);
P1:
do{
flag[1] = 1;
turn = 0;
while (flag[0] == 1 && turn == 0)
{
// busy wait
}
// critical section
flag[1] = 0;
// remainder section
} while(1);
Progress: If process P1 is not in critical section then flag[1] = 0. Therefore while loop of P0
quits immediately and P0 can get into its critical section. And vice versa..
Bounded waiting: Process Pi keeps waiting in spinlocking only while the other process is in
its critical section.
4
Peterson’s Algorithm
int flag[2] = {0, 0};
int turn;
P0:
do{
flag[0] = 1;
turn = 1;
while (flag[1] == 1 && turn == 1)
{
// busy wait
}
// critical section
flag[0] = 0;
//remainder section
}while(1);
P1:
do{
flag[1] = 1;
turn = 0;
while (flag[0] == 1 && turn == 0)
{
// busy wait
}
// critical section
flag[1] = 0;
// remainder section
} while(1);
Uses spinlocking for waiting.
No strict alternation is required between processes. That is, P0,P0,P0,P1,P1 is doable.
Requires that processes alternate between critical and remainder sections.
Can be extended to n processes, only if n is known apriori (in advance). HOW?
5
Peterson’s Algorithm
int flag[2] = {0, 0};
int turn;
P0:
do{
flag[0] = 1;
turn = 1;
while (flag[1] == 1 && turn == 1)
{
// busy wait
}
// critical section
flag[0] = 0;
//remainder section
}while(1);
P1:
do{
flag[1] = 1;
turn = 0;
while (flag[0] == 1 && turn == 0)
{
// busy wait
}
// critical section
flag[1] = 0;
// remainder section
} while(1);
Prone to priority inversion:
Assume that P0 has a higher priority than P1.
When P1 is in its critical section, P0 may get scheduled to do spinlocking.
P1 never gets scheduled to finish its critical section and both processes end up waiting.
6
Issues with Semaphores
Much of the power of semaphores derives from calls to
down() and up() that are unmatched

See previous example!
Unlike locks, acquire() and release() are not always paired.
This means it is a lot easier to get into trouble with semaphores.

“More rope”
Would be nice if we had some clean, well-defined language
support for synchronization...

Java does!
Adapted from Matt Welsh’s (Harvard University) slides.
7
Monitors
A monitor is an object intended to be used safely
by more than one thread.
•
The defining characteristic of a monitor is that its
methods are executed with mutual exclusion.
•
•
also provide Condition Variables (CVs) for
threads to temporarily give up exclusive access,
in order to wait for some condition to be met,
•
•
That is, at each point in time, at most one thread may be
executing any of its methods.
before regaining exclusive access and resuming their task.
Use CVs for signaling other threads that such
conditions have been met.
8
Condition Variables
Conceptually a condition variable (CV) is a queue of threads,
associated with a monitor, upon which a thread may wait for
some assertion to become true.
Threads can use CV’s
•
to temporarily give up exclusive access, in order to wait for
some condition to be met,
•
•
before regaining exclusive access and resuming their task.
for signaling other threads that such conditions have been met.
9
Monitors
This style of using locks and CV's to protect access to a shared
object is often called a monitor

Think of a monitor as a lock protecting an object, plus a queue of waiting threads.
Shared data
Waiting threads
At most one thread
in the monitor at a
time
Methods accessing
shared data
How is this different than a lock???
Adapted from Matt Welsh’s (Harvard University) slides.
10
Monitors
unlocked
Shared data
Methods accessing
shared data
Adapted from Matt Welsh’s (Harvard University) slides.
11
Monitors
locked
Shared data
zzzz...
zzzz...
Methods accessing
shared data
Sleeping thread no longer “in” the monitor.
(But not on the waiting queue either! Why?)
Adapted from Matt Welsh’s (Harvard University) slides.
12
Monitors
locked
Monitor stays locked!
(Lock now owned by
different thread...)
Shared data
notify()
zzzz...
Adapted from Matt Welsh’s (Harvard University) slides.
Methods accessing
shared data
13
Monitors
locked
Shared data
notify()
Methods accessing
shared data
Adapted from Matt Welsh’s (Harvard University) slides.
14
Monitors
locked
Shared data
Methods accessing
shared data
No guarantee which order threads get into the monitor.
(Not necessarily FIFO!)
Adapted from Matt Welsh’s (Harvard University) slides.
15
Bank Example
monitor Bank{
int TL = 1000;
condition haveTL;
void withdraw(int amount) {
if (amount > TL)
wait(haveTL);
TL -= amount;
}
void deposit(int amount) {
TL += amount;
notify(haveTL);
}
}
16
Bank Example
monitor Bank{
int TL = 1000;
condition haveTL;
void withdraw(int amount) {
while (amount > TL)
wait(haveTL);
TL -= amount;
}
void deposit(int amount) {
TL += amount;
notifyAll(haveTL);
}
}
17
Hoare vs. Mesa Monitor Semantics
The monitor notify() operation can have two different meanings:
Hoare monitors (1974)


notify(CV) means to run the waiting thread immediately
Causes notifying thread to block
Mesa monitors (Xerox PARC, 1980)


notify(CV) puts waiting thread back onto the “ready queue” for the monitor
But, notifying thread keeps running
Adapted from Matt Welsh’s (Harvard University) slides.
18
Hoare vs. Mesa Monitor Semantics
The monitor notify() operation can have two different meanings:
Hoare monitors (1974)


notify(CV) means to run the waiting thread immediately
Causes notifying thread to block
Mesa monitors (Xerox PARC, 1980)


notify(CV) puts waiting thread back onto the “ready queue” for the monitor
But, notifying thread keeps running
What's the practical difference?


In Hoare-style semantics, the “condition” that triggered the notify()
will always be true when the awoken thread runs
 For example, that the buffer is now no longer empty
In Mesa-style semantics, awoken thread has to recheck the condition
 Since another thread might have beaten it to the punch
Adapted from Matt Welsh’s (Harvard University) slides.
19
Hoare Monitor Semantics
Hoare monitors (1974)


notify(CV) means to run the
waiting thread immediately
Causes notifying thread to block
The signaling thread must
wait outside the monitor (at
least) until the signaled
thread relinquishes
occupancy of the monitor
by either returning or by
again waiting on a
condition.
20
Mesa Monitor Semantics
Mesa monitors (Xerox PARC,
1980)


notify(CV) puts waiting thread back
onto the “ready queue” for the
monitor
But, notifying thread keeps running
Signaling does not cause the
signaling thread to lose
occupancy of the monitor.
Instead the signaled threads
are moved to the e queue.
21
Hoare vs. Mesa monitors
Need to be careful about precise definition of signal and wait.

while (n==0) {
wait(not_empty); // If nothing, sleep
}
item = getItemFromArray();
// Get next item
Why didn’t we do this?
if (n==0) {
wait(not_empty); // If nothing, sleep
}
removeItemFromArray(val); // Get next item
Answer: depends on the type of scheduling


Hoare-style (most textbooks):
 Signaler gives lock, CPU to waiter; waiter runs immediately
 Waiter gives up lock, processor back to signaler when it exits critical section or if it
waits again
Mesa-style (Java, most real operating systems):
 Signaler keeps lock and processor
 Waiter placed on ready queue with no special priority
 Practically, need to check condition again after wait
22
Revisit: Readers/Writers Problem
Correctness Constraints:

Readers can access database when no writers

Writers can access database when no readers or writers

Only one thread manipulates state variables at a time
State variables (Protected by a lock called “lock”):

int NReaders: Number of active readers; initially = 0

int WaitingReaders: Number of waiting readers; initially = 0

int NWriters: Number of active writers; initially = 0

int WaitingWriters: Number of waiting writers; initially = 0

Condition canRead = NIL

Conditioin canWrite = NIL
23
Readers and Writers
Monitor ReadersNWriters {
int WaitingWriters, WaitingReaders,NReaders, NWriters;
Condition CanRead, CanWrite;
Void BeginWrite()
{
if(NWriters == 1 || NReaders > 0)
{
++WaitingWriters;
wait(CanWrite);
--WaitingWriters;
}
NWriters = 1;
}
Void EndWrite()
{
NWriters = 0;
if(WaitingReaders)
Signal(CanRead);
else
Signal(CanWrite);
}
Void BeginRead()
{
if(NWriters == 1 || WaitingWriters > 0)
{
++WaitingReaders;
Wait(CanRead);
--WaitingReaders;
}
++NReaders;
Signal(CanRead);
}
Void EndRead()
{
if(--NReaders == 0)
Signal(CanWrite);
}
24
Readers and Writers
Monitor ReadersNWriters {
int WaitingWriters, WaitingReaders,NReaders, NWriters;
Condition CanRead, CanWrite;
Void BeginWrite()
{
if(NWriters == 1 || NReaders > 0)
{
++WaitingWriters;
wait(CanWrite);
--WaitingWriters;
}
NWriters = 1;
}
Void EndWrite()
{
NWriters = 0;
if(WaitingReaders)
Signal(CanRead);
else
Signal(CanWrite);
}
Void BeginRead()
{
if(NWriters == 1 || WaitingWriters > 0)
{
++WaitingReaders;
Wait(CanRead);
--WaitingReaders;
}
++NReaders;
Signal(CanRead);
}
Void EndRead()
{
if(--NReaders == 0)
Signal(CanWrite);
}
25
Readers and Writers
Monitor ReadersNWriters {
int WaitingWriters, WaitingReaders,NReaders, NWriters;
Condition CanRead, CanWrite;
Void BeginWrite()
{
if(NWriters == 1 || NReaders > 0)
{
++WaitingWriters;
wait(CanWrite);
--WaitingWriters;
}
NWriters = 1;
}
Void EndWrite()
{
NWriters = 0;
if(WaitingReaders)
notify(CanRead);
else
notify(CanWrite);
}
Void BeginRead()
{
if(NWriters == 1 || WaitingWriters > 0)
{
++WaitingReaders;
Wait(CanRead);
--WaitingReaders;
}
++NReaders;
Signal(CanRead);
}
Void EndRead()
{
if(--NReaders == 0)
notify(CanWrite);
}
26
Readers and Writers
Monitor ReadersNWriters {
int WaitingWriters, WaitingReaders,NReaders, NWriters;
Condition CanRead, CanWrite;
Void BeginWrite()
{
if(NWriters == 1 || NReaders > 0)
{
++WaitingWriters;
wait(CanWrite);
--WaitingWriters;
}
NWriters = 1;
}
Void EndWrite()
{
NWriters = 0;
if(WaitingReaders)
notify(CanRead);
else
notify(CanWrite);
}
Void BeginRead()
{
if(NWriters == 1 || WaitingWriters > 0)
{
++WaitingReaders;
Wait(CanRead);
--WaitingReaders;
}
++NReaders;
notify(CanRead);
}
Void EndRead()
{
if(--NReaders == 0)
notify(CanWrite);
}
27
Understanding the Solution
A writer can enter if there are no other active writers and no readers
are waiting
28
Readers and Writers
Monitor ReadersNWriters {
int WaitingWriters, WaitingReaders,NReaders, NWriters;
Condition CanRead, CanWrite;
Void BeginWrite()
{
if(NWriters == 1 || NReaders > 0)
{
++WaitingWriters;
wait(CanWrite);
--WaitingWriters;
}
NWriters = 1;
}
Void EndWrite()
{
NWriters = 0;
if(WaitingReaders)
notify(CanRead);
else
notify(CanWrite);
}
Void BeginRead()
{
if(NWriters == 1 || WaitingWriters > 0)
{
++WaitingReaders;
Wait(CanRead);
--WaitingReaders;
}
++NReaders;
notify(CanRead);
}
Void EndRead()
{
if(--NReaders == 0)
notify(CanWrite);
}
29
Understanding the Solution
A reader can enter if

There are no writers active or waiting
So we can have many readers active all at once
Otherwise, a reader waits (maybe many do)
30
Readers and Writers
Monitor ReadersNWriters {
int WaitingWriters, WaitingReaders,NReaders, NWriters;
Condition CanRead, CanWrite;
Void BeginWrite()
{
if(NWriters == 1 || NReaders > 0)
{
++WaitingWriters;
wait(CanWrite);
--WaitingWriters;
}
NWriters = 1;
}
Void EndWrite()
{
NWriters = 0;
if(WaitingReaders)
notify(CanRead);
else
notify(CanWrite);
}
Void BeginRead()
{
if(NWriters == 1 || WaitingWriters > 0)
{
++WaitingReaders;
Wait(CanRead);
--WaitingReaders;
}
++NReaders;
notify(CanRead);
}
Void EndRead()
{
if(--NReaders == 0)
notify(CanWrite);
}
31
Understanding the Solution
When a writer finishes, it checks to see if any readers are waiting


If so, it lets one of them enter
That one will let the next one enter, etc…
Similarly, when a reader finishes, if it was the last reader, it lets a
writer in (if any is there)
32
Readers and Writers
Monitor ReadersNWriters {
int WaitingWriters, WaitingReaders,NReaders, NWriters;
Condition CanRead, CanWrite;
Void BeginWrite()
{
if(NWriters == 1 || NReaders > 0)
{
++WaitingWriters;
wait(CanWrite);
--WaitingWriters;
}
NWriters = 1;
}
Void EndWrite()
{
NWriters = 0;
if(WaitingReaders)
notify(CanRead);
else
notify(CanWrite);
}
Void BeginRead()
{
if(NWriters == 1 || WaitingWriters > 0)
{
++WaitingReaders;
Wait(CanRead);
--WaitingReaders;
}
++NReaders;
notify(CanRead);
}
Void EndRead()
{
if(--NReaders == 0)
notify(CanWrite);
}
33
Understanding the Solution
It wants to be fair



If a writer is waiting, readers queue up
If a reader (or another writer) is active or waiting, writers queue up
… this is mostly fair, although once it lets a reader in, it lets ALL waiting readers in all
at once, even if some showed up “after” other waiting writers
34
The Big Picture
The point here is that getting synchronization right is hard
How to pick between locks, semaphores, condvars, monitors???
Locks are very simple for many cases.


Issues: Maybe not the most efficient solution
For example, can't allow multiple readers but one writer inside a standard lock.
Condition variables allow threads to sleep while holding a lock

Just be sure you understand whether they use Mesa or Hoare semantics!
Semaphores provide pretty general functionality

But also make it really easy to botch things up.
Adapted from Matt Welsh’s (Harvard University) slides.
35
Barbershop problem
A barber shop consists of a waiting
room with N chairs, and the barber
room containing the barber chair.
If there are no customers to be
served, the barber goes to sleep.
If a customer enters the barber shop
and all chairs are occupied, then
the customer leaves the shop.
If the barber is busy, but chairs are
available, then the customer sits in
one of the free chairs.
If the barber is asleep, the customer
wakes up the barber.
36
Monitor template
monitor BarberShop{
condition waitingForCustomers,
waitingForBarbers;
int waiting = 0; // number of waiting customers in chairs
void barber(){
………………….
cutHair();
……………….
}
void customer(){
………………….
getHairCut(); // may not be executed if all chairs are full.
………………
}
}
void barberThread(){
while(1)
BarberShop.barber();
}
void customerThread(){
BarberShop.customer();
}
37
Semaphore template
semaphore customer = 0; // number of customers waiting for service
semaphore barber = 0; // number of barbers waiting for customers
semaphore mutex = 1; // for mutual exclusion
int waiting = 0; //customers who are sitting in chairs
void barberThread(){
while (1) {
……………….
cutHair();
…………….
}
}
void customerThread(){
………………..
getHairCut(); // may not be executed if all chairs are full.
………………
}
38
Checking your code
• B: Does the Barber sleep when there are no customers.
• BC: Does the first customer wake up the sleeping barber and
have his haircut.
• BCC: Does the second customer waits for the barber while he is
giving a haircut to the first customer.
• BCCCCCCC: Does the 7th customer (first customer having a
haircut, the next 5 customers waiting), exits without getting a
haircut?
• BCCCCC: Does the barber wake up a waiting customer after
finishing the haircut of the first customer?
• Finally, is the solution efficient? Sending more notify signals than
needed, or using more variables is not good practise.
39
CENG334
Introduction to Operating Systems
Deadlocks
Topics:
Deadlocks
•Dining philosopher problem
•
Erol Sahin
Dept of Computer Eng.
Middle East Technical University
Ankara, TURKEY
URL: http://kovan.ceng.metu.edu.tr/~erol/Courses/ceng334
40
What’s a deadlock?
41
Deadlock
A deadlock happens when


Two (or more) threads waiting for each other
None of the deadlocked threads ever make progress
Mutex
1
holds
Thread 1
waits for
waits for
Mutex
2
Adapted from Matt Welsh’s (Harvard University) slides.
holds
Thread 2
42
Deadlock Definition
Two kinds of resources:


Preemptible: Can take away from a thread
 e.g., the CPU
Non-preemptible: Can't take away from a thread
 e.g., mutex, lock, virtual memory region, etc.
Why isn't it safe to forcibly take a lock away from a thread?
Starvation

A thread never makes progress because other threads are using a resource it needs
Deadlock

A circular waiting for resources
 Thread A waits for Thread B
 Thread B waits for Thread A
Starvation ≠ Deadlock
Adapted from Matt Welsh’s (Harvard University) slides.
43
Dining Philosophers
Classic deadlock problem



Multiple philosophers trying to lunch
One chopstick to left and right of each philosopher
Each one needs two chopsticks to eat
Adapted from Matt Welsh’s (Harvard University) slides.
44
Dining Philosophers
What happens if everyone grabs the chopstick to their right?


Everyone gets one chopstick and waits forever for the one on the left
All of the philosophers starve!!!
Adapted from Matt Welsh’s (Harvard University) slides.
45
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, …, P0} 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
•
P0 is waiting for a resource that is held by P0.
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
46
Deadlock Prevention
Restrain the ways request can be made to ensure that at least one of
the four conditions DO NOT HOLD!
Mutual Exclusion
•
not required for sharable resources;
•
must hold for non-sharable resources,
•
such as a printer.
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.
•
low resource utilization;
•
starvation possible.
47
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.
Can be applied to resources whose state can be saved such as CPU, and memory.
Not applicable to resources such as printer and tape drives.
Circular Wait
•
•
impose a total ordering of all resource types, and
require that each process requests resources in an increasing order of
enumeration.
48
Circular Wait - 1
Each resource is given an ordering:

F(tape drive) = 1

F(disk drive) = 2

F(printer) = 3

F(mutex1) = 4

F(mutex2) = 5

…….
Each process can request resources only in increasing order of
enumeration.
A process which decides to request an instance of Rj should first
release all of its resources that are F(Ri) >= F(Rj).
49
Circular Wait - 2
For instance an application program may use ordering among all of its
synchronization primitives:




F(semaphore1) = 1
F(semaphore2) = 2
F(semaphore3) = 3
…….
After this, all requests to synchronization primitives should be made only
in the increasing order:


Correct use:
 down(semaphore1);
 down(semaphore2);
Incorrect use:
 down(semaphore3);
 down(semaphore2);
Keep in mind that it’s the application programmer’s responsibility to
obey this order.
50
Methods for Handling Deadlocks
How should we handle deadlocks



Ensure that the system will never enter a
deadlock state.
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.
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
51
Dining Philosophers
How do we solve this problem??

(Apart from letting them eat with forks.)
Adapted from Matt Welsh’s (Harvard University) slides.
52
How to solve this problem?
Solution 1: Don't wait for chopsticks




Grab the chopstick on your right
Try to grab chopstick on your left
If you can't grab it, put the other one back down
Breaks “no preemption” condition – no waiting!
Solution 2: Grab both chopsticks at once


Requires some kind of extra synchronization to make it atomic
Breaks “multiple independent requests” condition!
Solution 3: Grab chopsticks in a globally defined order



Number chopsticks 0, 1, 2, 3, 4
Grab lower-numbered chopstick first
 Means one person grabs left hand rather than right hand first!
Breaks “circular dependency” condition
Solution 4: Detect the deadlock condition and break out of it


Scan the waiting graph and look for cycles
Shoot one of the threads to break the cycle
Adapted from Matt Welsh’s (Harvard University) slides.
53
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.
•
•
Is this possible at all?
When should the algorithm be called?
Resource-allocation state is defined by the number of
available and allocated resources, and the maximum
demands of the processes.
54
System Model
Resource types R1, R2, . . ., Rm

CPU,

memory,

I/O devices

disk

network
Each resource type Ri has Wi instances.

For instance a quad-core processor has

4 CPUs
Each process utilizes a resource as follows:

request

use

release
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
55
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 P1  Rj
assignment edge – directed edge Rj  Pi
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
56
Resource Allocation Graph With A Deadlock
If there is a deadlock => there is a
cycle in the graph.
However the reverse is not true!
i.e. If there is a cycle in the graph
=/> there is a deadlock
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
57
Resource Allocation Graph With A Cycle But No Deadlock
However the existence of a
cycle in the graph does not
necessarily imply a
deadlock.
Overall message:
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.
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
58
Resource-Allocation Graph Algorithm
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.
When a resource is released by a process,
assignment edge reconverts to a claim edge.
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
59
Resource-Allocation Graph Algorithm
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.
When a resource is released by a process,
assignment edge reconverts to a claim edge.
Cycle => Unsafe
Resources must be claimed a priori in the system.
Note that the cycle detection algorithm does not work
with resources that have multiple instances.
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
60
Safe, unsafe and deadlock states
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.
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
61
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 safe sequence of all
processes.
Sequence <P1, P2, …, Pn> is safe if 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.
 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.
Adapted from Operating System Concepts (Silberschatz, Galvin, Gagne) slides.
62
Banker’s Algorithm
While giving credits, a banker should
ensure that it never allocates all of its
cash in such a way that none of its
creditors can finish their work and pay
back the loan.
63
Example
The system has three
processes and 12
tape drives.
t=t0
P0
P1
P2
Maximum Needs
10
4
9
Current Needs
5
2
2
The system at t0 is safe since the sequence <P1,P0,P2>
exists.
64
Example
The system has three
processes and 12 tape
drives.
t=t0
P0
P1
P2
Maximum Needs
10
4
9
Current Needs
5
2
2
P2 requests one more drive
t=t1
P0
P1
P2
Maximum Needs
10
4
9
Current Needs
5
2
3
The system at t1 is no longer safe since
• P1 requests 2 more tape drives, finishes and releases 4 drives.
• However 4 drives are not sufficient for P0 or P2 complete its operation and would
result in a deadlock.
65