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