Lecture : Concurrency: Mutual Exclusion and Synchronization

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

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.

UAL 56: #-of-seats=12 Main memory … Terminal Terminal Terminal Ticket Agent 1 Ticket Agent n Ticket Agent 2 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.

Example for Race condition(cont.) Ticket Agent 1 P1: LOAD #-of-seats P2: DEC 1 P3: STORE #-of-seats Ticket Agent 2 Q1: LOAD #-of-seats Q2: DEC 1 Q3: STORE #-of-seats Ticket Agent 3 R1: LOAD #-of-seats R2: DEC 1 R3: STORE #-of-seats Suppose, initially, #-of-seats=12 Suppose instructions are interleaved as P1,Q1,R1,P2,Q2,R2,P3,Q3,R3 The 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, or Q1,Q2,Q3 must be completely executed before we execute P1 or R1, or R1,R2,R3 must be completely executed before we execute P1 or Q1.

P0 P1 Pn-1 Prefix0 Prefix1 Prefixn-1 … CS0 CS1 CSn-1 Suffix0 Suffix1 Suffixn-1 Critical Section Problem Critical section: a segment of code in which the process may be changing common variables, updating a table, writing a file, and so on. Goal: To program the processes so that, at any moment of time, at most one of the processes is in its critical section.

Solution to Critical-Section Problem • Any facility to provide support for mutual exclusion should meet the following requirements: • Mutual exclusion must be enforced: Only one process at a time is allowed into its critical section • A process that halts in its noncritical section must do so without interfering with other processes. • A process waiting to enter its critical section cannot be delayed infinitely • When no process is in a critical section, any process that requests entry to its critical section must be permitted to enter without delay. • No assumption are made about the relative process speeds or the number of processors. • A process remains inside its critical section for a finite time only.

Three Environments • There is no central program to coordinate the processes. The processes communicate with each other through global variable. • Special hardware instructions • There is a central program to coordinate the processes.

1st Attempt Start with just 2 processes, P0 and p1 Global variable turn, initially turn=0 Prefix1 While (turn1) do {} CS1 turn=0 suffix1 Prefix0 While (turn0) do {} CS0 turn=1 suffix0 The processes take turn to enter its critical section If turn=0, P0 enters If 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.

2st Attempt Global variable flag[0] and flag[1], initially flag[0] and flag[1] are both false Prefix1 While (flag[0]) do {} flag[1]=true CS1 flag[1]= false suffix1 Prefix0 While (flag[1]) do {} flag[0]=true CS0 flag[0]=false suffix0 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.

3st Attempt Global variable flag[0] and flag[1], initially flag[0] and flag[1] are both false Prefix1 flag[1]=true While (flag[0]) do {} CS1 flag[1]= false suffix1 Prefix0 flag[0]=true While (flag[1]) do {} CS0 flag[0]=false suffix0 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.

4st Attempt Global variable flag[0] and flag[1], initially flag[0] and flag[1] are both false Prefix1 L1: flag[1]=true If (flag[0]) then { flag[1]=false; goto L1}} CS1 flag[1]= false suffix1 Prefix0 L0: flag[0]=true If (flag[1]) then { flag[0]=false; goto L0} CS0 flag[0]=false suffix0 Guarantees mutual exclusion. mutual blocking can occur if they execute at the same speed.

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]=true while (flag[1]) do { if (turn=1){ flag[0]=false; while(turn=1) do{} flag[0]=true; } } CS0 turn=1 flag[0]=false suffix0 Prefix1 flag[1]=true while (flag[0]) do { if (turn=0){ flag[1]=false; while(turn=0) do{} flag[1]=true; } } CS1 turn=0 flag[1]=false suffix1

Peterson’s Algorithm for 2 processes – The simplest and most compact mutual exclusion alg. Initially, flag[0]=flag[1]=false Prefix0 flag[0]=true turn=1 while (flag[1] and turn=1) do{} CS0 flag[0]=false suffix0 Prefix1 flag[1]=true turn=0 while (flag[0] and turn=0) do{} CS1 flag[1]=false suffix1

Solution for n processes • Global Variable • Flag[0..n-1] – array of size n. • Turn. Initially, Turn=some no. between 0 and n-1 Idle if Pi is outside Csi Want-in if Pi wants to be in CSi Flag[i]= in-CS if Pi is in CSi

Solutions for n processes Pi Prefixi Repeat Flag[i]=want-in; j=Turn; while ji 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 (jn) and (Turn=i or Flag[Turn]=idle) Turn=i; CSi j=(Turn+1)mod n While (jTurn) and (Flag[j]=idle) do{j=(i+1) mod n} Turn=j Flag[i]=idle

Hardware Support Disable interrupt CS Enable interrupt Won’t work if we have multiprocessors

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

TS – Test and Set Boolean 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.

Semaphores • A variable that has an integer value upon which 3 operations are defined. • Three operations: • A semaphore may be initialized to a nonnegative value • The wait operation decrements the semaphore value. If the value becomes negative, then the process executing the wait is blocked • 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.

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(s0) {remove a process from the waiting queue} }

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

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.

Two classical examples • Producer and Consumer Problem • Readers/Writers Problem

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 consumer producer N buffers N-1

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 consumer to store the item to be consumed W is a local buffer used by the producer to produce Producer and Consumer Problem(cont.) • Global Variable • B[0..N-1] – an array of size N (Buffer) • P – a semaphore, initialized to N • C – a semaphore, initialized to 0 • Local Variable • In – a ptr(integer) used by the producer, in=0 initially • Out – a ptr(integer) used by the consumer, out=0 initially

Two classical examples • Producer and Consumer Problem • Readers/Writers Problem

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.

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

End of lecture 6 Thank you!

Lecture : Concurrency: Mutual Exclusion and Synchronization