1 Lecture 8: Concurrency: Mutual Exclusion and Synchronization Advanced Operating System Fall 2012
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Lecture 8: Concurrency: Mutual Exclusion and Synchronization
Advanced Operating SystemFall 2012
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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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End of lecture 8
Thank you!
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