1 Lecture : Concurrency: Mutual Exclusion and Synchronization Operating System Spring 2008.

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Lecture : Concurrency: Mutual Exclusion and Synchronization

Operating SystemSpring 2008

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Concurrency

An OS has many concurrent processes that run in parallel but share common access

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.

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Example for Race condition

Suppose a customer wants to book a seat on UAL 56. Ticket agent will check the #-of-seats. If it is greater than 0, he will grab a seat and decrement #-of-seats by 1.

UAL 56: #-of-seats=12Main memory

Terminal Terminal Terminal…Ticket Agent 1Ticket Agent 2 Ticket Agent n

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Example for Race condition(cont.)

Ticket Agent 1

P1: LOAD #-of-seatsP2: DEC 1P3: STORE #-of-seats

Ticket Agent 2

Q1: LOAD #-of-seatsQ2: DEC 1Q3: STORE #-of-seats

Ticket Agent 3

R1: LOAD #-of-seatsR2: DEC 1R3: STORE #-of-seats

Suppose, initially, #-of-seats=12Suppose instructions are interleaved as P1,Q1,R1,P2,Q2,R2,P3,Q3,R3The result would be #-of-seats=11, instead of 9

To solve the above problem, we must make sure that:P1,P2,P3 must be completely executed before we execute Q1 or R1, orQ1,Q2,Q3 must be completely executed before we execute P1 or R1, orR1,R2,R3 must be completely executed before we execute P1 or Q1.

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Critical Section Problem

Goal: To program the processes so that, at any moment of time, at most one of the processes is in its critical section.

Prefix0

CS0

Suffix0

P0

Prefix1

CS1

Suffix1

P1

Prefixn-1

CSn-1

Suffixn-1

Pn-1

…

Critical section: a segment of code in which the process may be changing common variables, updating a table, writing a file, and so on.

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Solution to Critical-Section Problem

Any facility to provide support for mutual exclusion should meet the following requirements:

1. Mutual exclusion must be enforced: Only one process at a time is allowed into its critical section

2. A process that halts in its noncritical section must do so without interfering with other processes.

3. A process waiting to enter its critical section cannot be delayed infinitely

4. When no process is in a critical section, any process that requests entry to its critical section must be permitted to enter without delay.

5. No assumption are made about the relative process speeds or the number of processors.

6. A process remains inside its critical section for a finite time only.

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Three Environments

1. There is no central program to coordinate the processes. The processes communicate with each other through global variable.

2. Special hardware instructions3. There is a central program to coordinate the

processes.

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Three Environments

1. There is no central program to coordinate the processes. The processes communicate with each other through global variable.

2. Special hardware instructions3. There is a central program to coordinate the

processes.

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1st AttemptStart with just 2 processes, P0 and p1

Global variable turn, initially turn=0

Prefix0

While (turn0) do {} CS0

turn=1 suffix0

Prefix1

While (turn1) do {} CS1

turn=0 suffix1

The processes take turn to enter its critical sectionIf turn=0, P0 entersIf turn=1, P1 enters

This solution guarantees mutual exclusion.

But the drawback is that the method is not fair, because P0 is priviledged.Worse yet, until P0 executed its CS, P1 is blocked.

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2st AttemptGlobal variable flag[0] and flag[1], initially flag[0] and flag[1] are both false

Prefix0

While (flag[1]) do {}flag[0]=true CS0

flag[0]=false suffix0

Prefix1

While (flag[0]) do {}flag[1]=true CS1

flag[1]= false suffix1

If P0 is in critical section, flag[0] is true; If P1 is in critical section, flag[1] is true

If one process leaves the system, it will not block the other process.

However, mutual exclusion is not guaranteed.P0 executes the while statement and finds that flag[1] is false;P1 executes the while statement and finds that flag[0] is false.P0 sets flag[0] to true and enters its critical section;P1 sets flag[1] to true and enters its critical section.

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3st AttemptGlobal variable flag[0] and flag[1], initially flag[0] and flag[1] are both false

Prefix0

flag[0]=trueWhile (flag[1]) do {} CS0

flag[0]=false suffix0

Prefix1

flag[1]=trueWhile (flag[0]) do {} CS1

flag[1]= false suffix1

If P0 is in critical section, flag[0] is true; If P1 is in critical section, flag[1] is true

Guarantees mutual exclusion.

But mutual blocking can occur.P0 sets flag[0] to be true;P1 sets flag[1] to be true;Both will be hung in the while loop.

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4st AttemptGlobal variable flag[0] and flag[1], initially flag[0] and flag[1] are both false

Prefix0

L0: flag[0]=trueIf (flag[1]) then {

flag[0]=false;goto L0}

CS0

flag[0]=false suffix0

Prefix1

L1: flag[1]=trueIf (flag[0]) then {

flag[1]=false;goto L1}}

CS1

flag[1]= false suffix1

Guarantees mutual exclusion.

mutual blocking can occur if they execute at the same speed.

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Correct Solution (Dekker’s Alg) – The first correct mutual exclusion alg (early 1960’s)

Initially, flag[0]=flag[1]=false; turn=0

Prefix0

flag[0]=truewhile (flag[1]) do { if (turn=1){

flag[0]=false;while(turn=1) do{}flag[0]=true;}

}CS0

turn=1flag[0]=falsesuffix0

Prefix1

flag[1]=truewhile (flag[0]) do { if (turn=0){

flag[1]=false;while(turn=0) do{}flag[1]=true;}

}CS1

turn=0flag[1]=falsesuffix1

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Peterson’s Algorithm for 2 processes – The simplest and most compact mutual exclusion alg.

Initially, flag[0]=flag[1]=false

Prefix0

flag[0]=trueturn=1while (flag[1] and turn=1) do{}CS0

flag[0]=falsesuffix0

Prefix1

flag[1]=trueturn=0while (flag[0] and turn=0) do{} CS1

flag[1]=falsesuffix1

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Solution for n processes Global Variable

1. Flag[0..n-1] – array of size n.

2. Turn. Initially, Turn=some no. between 0 and n-1

Idle if Pi is outside CsiWant-in if Pi wants to be in CSiin-CS if Pi is in CSi

Flag[i]=

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Solutions for n processes

PiPrefixi

Repeat Flag[i]=want-in; j=Turn; while ji do {if Flag[j]idle then j=Turn else j=(j+1) mod n} Flag[i]=in-CS j=0 while (j<n) and (j=i or Flag[j]in-CS) do {j=j+1}Until (jn) and (Turn=i or Flag[Turn]=idle)Turn=i;

CSi

j=(Turn+1)mod nWhile (jTurn) and (Flag[j]=idle) do{j=(i+1) mod n}Turn=jFlag[i]=idle

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Three Environments

1. There is no central program to coordinate the processes. The processes communicate with each other through global variable.

2. Special hardware instructions3. There is a central program to coordinate the

processes.

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Hardware SupportDisable interrupt CSEnable interrupt

Won’t work if we have multiprocessors

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Special Machine Instructions Modern machines provide special atomic hardware instructions

Atomic = non-interruptable Either test memory word and set value Or swap contents of two memory words

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TS – Test and SetBoolean TS(i)= true if i=0; it will also set i to 1 false if i=1

Initially, lock=0 Pi

Prefixi

While(¬ TS(lock)) do {} CSi

Lock=0 suffixi

It is possible that a process may starve if 2 processes enter the critical section arbitrarily often.

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Three Environments

1. There is no central program to coordinate the processes. The processes communicate with each other through global variable.

2. Special hardware instructions3. There is a central program to coordinate the

processes.

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Semaphores A variable that has an integer value upon

which 3 operations are defined. Three operations:

1. A semaphore may be initialized to a nonnegative value

2. The wait operation decrements the semaphore value. If the value becomes negative, then the process executing the wait is blocked

3. The signal operation increments the semaphore value. If the value is not positive, then a process blocked by a wait operation is unblocked.

Other than these 3 operations, there is no way to inspect or manipulate semaphores.

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Wait(s) and Signal(s) Wait(s) – is also called P(s)

{s=s-1;if (s<0) {place this process in a waiting queue}

} Signal(s) – is also called V(s)

{s=s+1;if(s0) {remove a process from the waiting

queue}}

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Semaphore as General Synchronization Tool

Counting semaphore – integer value can range over an unrestricted domain

Binary semaphore – integer value can range only between 0 and 1; can be simpler to implement

Also known as mutex locks Wait B(s) s is a binary semaphore

{ if s=1 then s=0 else block this process}

Signal B(s){ if there is a blocked process then unblock a process else s=1}

Can implement a counting semaphore S as a binary semaphore

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Note

The wait and signal primitives are assumed to be atomic; they cannot be interrupted and each routine can be treated as an indivisible step.

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Two classical examples

Producer and Consumer Problem Readers/Writers Problem

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Two classical examples

Producer and Consumer Problem Readers/Writers Problem

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Producer and Consumer Problem Producer can only put something in when there is an

empty buffer Consumer can only take something out when there is a

full buffer Producer and consumer are concurrent processes

…

0

N-1

N buffersproducer

consumer

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Producer and Consumer Problem(cont.) Global Variable

1. B[0..N-1] – an array of size N (Buffer)2. P – a semaphore, initialized to N3. C – a semaphore, initialized to 0

Local Variable1. In – a ptr(integer) used by the producer, in=0 initially2. Out – a ptr(integer) used by the consumer, out=0 initially

Producer Process

producer: produce(w) wait(p) B[in]=w in=(in+1)mod N signal(c) goto producer

Consumer Process

consumer: wait(c) w=B[out] out=(out+1)mod N signal(p) consume(w) goto consumer

W is a local buffer used by the producer to produce

W is a local buffer used by the consumer to store the item to be consumed

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Two classical examples

Producer and Consumer Problem Readers/Writers Problem

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Readers/Writers Problem

Suppose a data object is to be shared among several concurrent processes. Some of these processes want only to read the data object, while others want to update (both read and write)

Readers – Processes that read only Writers – processes that read and write

If a reader process is using the data object, then other reader processes are allowed to use it at the same time.

If a writer process is using the data object, then no other process (reader or writer) is allowed to use it simultaneously.

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Solve Readers/Writers Problem using wait and signal primitives(cont.)

Global Variable: Wrt is a binary semaphore, initialized to 1; Wrt is used by both readers and writers

For Reader Processes: Mutex is a binary semaphore, initialized to 1;Readcount is an integer variable,

initialized to 0 Mutex and readcount used by readers onlyReader Processes

Wait(mutex)Readcount=readcount+1If readcount=1 then wait(wrt)Signal(mutex);…Reading is performed…Wait(mutex)Readcount=readcount-1If readcount=0 then signal(wrt)Signal(mutex)

Writer Processes

Wait(wrt)…Writing is performed…Signal(wrt)

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End of lecture 6

Thank you!