Outline Please turn in Homework #2 Announcements Basic Synchronization Principles – continued Semaphores Bounded-buffer problem Read-writer’s problem Please turn in Homework #2

Announcements The second quiz will be at the end of class on this Thursday – October 9, 2003 It will be closed-book and closed-note It will cover the materials until today with emphasis on scheduling It requires to compute average wait time and turnaround time (or something like that) and you need a calculator 5/20/2019 COP4610

Announcements – cont. Demonstration of Lab 1 You can sign up to demonstrate your program tomorrow during the recitation session During the demo, Mr. Chen will ask you questions besides demonstration 5/20/2019 COP4610

Comments on Scheduling Algorithms Process Arrival Time Burst Time P1 0.0 7 P2 2.0 4 P3 4.0 1 P4 5.0 4 Wait time and turnaround time Round robin Where should we place the newly arrived ones? 5/20/2019 COP4610

Interacting Processes - review Independent process Cannot affect or be affected by the other processes in the system Does not share any data with other processes Interacting process Can affect or be affected by the other processes Shares data with other processes We focus on interacting processes through physically or logically shared memory 5/20/2019 COP4610

Bounded-Buffer – review 5/20/2019 COP4610

Bounded-Buffer - cont. Suppose we have one producer and one consumer, the variable counter is 5 Producer: counter = counter +1 P1: load counter, r1 P2: add r1, #1, r2 P3: store r2, counter Consumer: counter = counter - 1 C1: load counter, r1 C2: add r1, #-1, r2 C3: store r2, counter 5/20/2019 COP4610

Bounded-Buffer - cont. A particular execution sequence P1: load counter, r1 P2: add r1, #1, r2 --- Context switch ---- C1: load counter, r1 C2: add r1, #-1, r2 C3: store r2, counter P3: store r2, counter What is the value of counter? 5/20/2019 COP4610

Bounded-Buffer - cont. A particular execution sequence C1: load counter, r1 C2: add r1, #-1, r2 --- Context switch ---- P1: load counter, r1 P2: add r1, #1, r2 P3: store r2, counter C3: store r2, counter What is the value of counter this time? 5/20/2019 COP4610

Bounded-Buffer - cont. A particular execution sequence C1: load counter, r1 C2: add r1, #-1, r2 C3: store r2, counter --- Context switch ---- P1: load counter, r1 P2: add r1, #1, r2 P3: store r2, counter What is the value of counter this time? 5/20/2019 COP4610

Race Condition Race condition When several processes access and manipulate the same data concurrently, there is a race among the processes The outcome of the execution depends on the particular order in which the access takes place This is called a race condition 5/20/2019 COP4610

Traffic Intersections 5/20/2019 COP4610

A Semaphore 5/20/2019 COP4610

The Critical-Section Problem n processes all competing to use some shared data Each process has code segments, called critical section, in which the shared data is accessed. Problem – ensure that when one process is executing in its critical section, no other process is allowed to execute in its critical section, called mutual exclusion 5/20/2019 COP4610

The Critical-Section Problem – cont. Structure of process Pi repeat entry section critical section exit section reminder section until false; 5/20/2019 COP4610

Requirements for Critical-Section Solutions Mutual Exclusion. If process Pi is executing in its critical section, then no other processes can be executing in their critical sections. Progress If no process is executing in its critical section and there exist some processes that wish to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely. Bounded Waiting A bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted. 5/20/2019 COP4610

Solution through Disabling Interrupts In a uni-processor system, an interrupt causes the race condition The scheduler can be called in the interrupt handler, resulting in context switch and data inconsistency 5/20/2019 COP4610

Solution through Disabling Interrupts – cont. Process Pi repeat disableInterrupts(); //entry section critical section enableInterrupts(); //exit section reminder section until false; 5/20/2019 COP4610

Solution through Disabling Interrupts – cont. 5/20/2019 COP4610

Solution through Disabling Interrupts – cont. This solution may affect the behavior of the I/O system Interrupts can be disabled for an arbitrarily long time The interrupts can be disabled permanently if the program contains an infinite loop in its critical section User programs are not allowed to enable and disable interrupts directly However, if the operating system can provide system calls, a user can use the system calls for synchronization Called semaphores and related system calls 5/20/2019 COP4610

Software Attempt to Solve the Problem Only 2 processes, P1 and P2 General structure of process Pi (other process Pj) repeat entry section critical section exit section reminder section until false; Processes may share some common variables to synchronize their actions. 5/20/2019 COP4610

Algorithm 1 Shared variables: Process Pi repeat var turn: (0..1); initially turn = 0 turn - i  Pi can enter its critical section Process Pi repeat while turn  i do no-op; critical section turn := j; reminder section until false; Satisfies mutual exclusion, but not progress 5/20/2019 COP4610

Algorithm 2 Shared variables repeat Process Pi var flag: array [0..1] of boolean; initially flag [0] = flag [1] = false. flag [i] = true  Pi ready to enter its critical section Process Pi repeat flag[i] := true; while flag[j] do no-op; critical section flag [i] := false; remainder section until false; Satisfies mutual exclusion, but not progress requirement. 5/20/2019 COP4610

Algorithm 2 – version 1 Shared variables repeat Process Pi var flag: array [0..1] of boolean; initially flag [0] = flag [1] = false. flag [i] = true  Pi ready to enter its critical section Process Pi repeat while flag[j] do no-op; flag[i] := true; critical section flag [i] := false; remainder section until false; Does not satisfy mutual exclusion 5/20/2019 COP4610

Algorithm 3 Combined shared variables of algorithms 1 and 2. Process Pi repeat flag [i] := true; turn := j; while (flag [j] and turn = j) do no-op; critical section flag [i] := false; remainder section until false; Meets all three requirements; solves the critical-section problem for two processes. 5/20/2019 COP4610

Bakery Algorithm Critical section for n processes Before entering its critical section, process receives a number. Holder of the smallest number enters the critical section. If processes Pi and Pj receive the same number, if i < j, then Pi is served first; else Pj is served first. The numbering scheme always generates numbers in increasing order of enumeration; i.e., 1,2,3,3,3,3,4,5... 5/20/2019 COP4610

Bakery Algorithm – cont. Notation < lexicographical order (ticket #, process id #) (a,b) < c,d) if a < c or if a = c and b < d max (a0,…, an-1) is a number, k, such that k  ai for i - 0, …, n – 1 Shared data var choosing: array [0..n – 1] of boolean; number: array [0..n – 1] of integer, Data structures are initialized to false and 0 respectively 5/20/2019 COP4610

Bakery Algorithm (Cont.) repeat choosing[i] := true; number[i] := max(number[0], number[1], …, number [n – 1])+1; choosing[i] := false; for j := 0 to n – 1 do begin while choosing[j] do no-op; while number[j]  0 and (number[j],j) < (number[i], i) do no-op; end; critical section number[i] := 0; remainder section until false; 5/20/2019 COP4610

Algorithm 4 Shared variables: Process P1 Process P2 repeat boolean lock; initially lock = FALSE lock is FALSE  Pi can enter its critical section Process P1 Process P2 repeat while (lock) do no-op; lock = TRUE; critical section lock = FALSE reminder section until false; repeat while (lock) do no-op; lock = TRUE; critical section lock = FALSE; reminder section until false; 5/20/2019 COP4610

Busy Wait Condition Code for p1 Code for p2 p2 p1 shared boolean lock = FALSE; shared double balance; Code for p1 Code for p2 /* Acquire the lock */ /* Acquire the lock */ while(lock) ; while(lock) ; lock = TRUE; lock = TRUE; /* Execute critical sect */ /* Execute critical sect */ balance = balance + amount; balance = balance - amount; /* Release lock */ /* Release lock */ lock = FALSE; lock = FALSE; p1 p2 at while Blocked lock = TRUE lock = FALSE Interrupt 5/20/2019 COP4610

Synchronization Hardware Test and modify the content of a word atomically (indivisibly) The procedure cannot be interrupted until it has completed the routine Implemented as a test-and-set instruction boolean Test-and-Set (boolean target) { boolean tmp tmp = target; target = true; return tmp; } 5/20/2019 COP4610

Test and Set Instruction – cont. TS(m): [Reg_i = memory[m]; memory[m] = TRUE;] FALSE m Primary Memory … R3 Data Register CC Before Executing TS TRUE =0 (b) After Executing TS 5/20/2019 COP4610

Mutual Exclusion with Test-and-Set Shared data: boolean lock ; Initially false Process Pi repeat while Test-and-Set (lock) do no-op; critical section lock = false; remainder section until false; 5/20/2019 COP4610

Semaphores Semaphore S – integer variable can only be accessed via two indivisible (atomic) operations As if they were a single machine instruction wait (S): while S 0 do no-op; S := S – 1; signal (S): S := S + 1; 5/20/2019 COP4610

Two Types of Semaphores 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. Can implement a counting semaphore S using binary semaphores. 5/20/2019 COP4610

Example: Critical Section of n Processes Shared variables var mutex : semaphore initially mutex = 1 Process Pi repeat wait(mutex); critical section signal(mutex); remainder section until false; 5/20/2019 COP4610

Shared Account Balance Problem Proc_0() { proc_1() { . . . . . . /* Enter the CS */ /* Enter the CS */ P(mutex); P(mutex); balance += amount; balance -= amount; V(mutex); V(mutex); } } semaphore mutex = 1; fork(proc_0, 0); fork(proc_1, 0); 5/20/2019 COP4610

Sharing Two Variables semaphore s1 = 0; semaphore s2 = 0; proc_A() { while(TRUE) { <compute section A1>; update(x); /* Signal proc_B */ V(s1); <compute section A2>; /* Wait for proc_B */ P(s2); retrieve(y); } semaphore s1 = 0; semaphore s2 = 0; fork(proc_A, 0); fork(proc_B, 0); proc_B() { while(TRUE) { /* Wait for proc_A */ P(s1); retrieve(x); <compute section B1>; update(y); /* Signal proc_A */ V(s2); <compute section B2>; } 5/20/2019 COP4610

Bounded Buffer Problem Empty Pool Producer Consumer Full Pool 5/20/2019 COP4610

Bounded Buffer Problem – cont. producer() { buf_type *next, *here; while(TRUE) { produce_item(next); /* Claim an empty */ P(empty); P(mutex); here = obtain(empty); V(mutex); copy_buffer(next, here); release(here, fullPool); /* Signal a full buffer */ V(full); } semaphore mutex = 1; semaphore full = 0; /* A general (counting) semaphore */ semaphore empty = N; /* A general (counting) semaphore */ buf_type buffer[N]; fork(producer, 0); fork(consumer, 0); consumer() { buf_type *next, *here; while(TRUE) { /* Claim full buffer */ P(full); P(mutex); here = obtain(full); V(mutex); copy_buffer(here, next); release(here, emptyPool); /* Signal an empty buffer */ V(empty); consume_item(next); } 5/20/2019 COP4610

Bounded Buffer Problem – cont. producer() { buf_type *next, *here; while(TRUE) { produce_item(next); /* Claim an empty */ P(empty); P(mutex); here = obtain(empty); V(mutex); copy_buffer(next, here); release(here, fullPool); /* Signal a full buffer */ V(full); } semaphore mutex = 1; semaphore full = 0; /* A general (counting) semaphore */ semaphore empty = N; /* A general (counting) semaphore */ buf_type buffer[N]; fork(producer, 0); fork(consumer, 0); consumer() { buf_type *next, *here; while(TRUE) { /* Claim full buffer */ P(full); P(mutex); here = obtain(full); V(mutex); copy_buffer(here, next); release(here, emptyPool); /* Signal an empty buffer */ V(empty); consume_item(next); } 5/20/2019 COP4610

Bounded Buffer Problem – cont. consumer() { buf_type *next, *here; while(TRUE) { /* Claim full buffer */ P(mutex); P(full); here = obtain(full); V(mutex); copy_buffer(here, next); release(here, emptyPool); /* Signal an empty buffer */ V(empty); consume_item(next); } producer() { buf_type *next, *here; while(TRUE) { produce_item(next); /* Claim an empty */ P(empty); P(mutex); here = obtain(empty); V(mutex); copy_buffer(next, here); release(here, fullPool); /* Signal a full buffer */ V(full); } semaphore mutex = 1; semaphore full = 0; /* A general (counting) semaphore */ semaphore empty = N; /* A general (counting) semaphore */ buf_type buffer[N]; fork(producer, 0); fork(consumer, 0); 5/20/2019 COP4610

Readers-Writers Problem 5/20/2019 COP4610

Readers-Writers Problem (2) Shared Resource 5/20/2019 COP4610

Readers-Writers Problem (3) Shared Resource 5/20/2019 COP4610

Readers-Writers Problem (4) Shared Resource 5/20/2019 COP4610

Readers-writers with active readers 5/20/2019 COP4610

First and Second Policy 5/20/2019 COP4610

First Solution First reader competes with writers while(TRUE) { <other computing>; P(mutex); readCount++; if(readCount == 1) P(writeBlock); V(mutex); /* Critical section */ access(resource); readCount--; if(readCount == 0) V(writeBlock); } resourceType *resource; int readCount = 0; semaphore mutex = 1; semaphore writeBlock = 1; fork(reader, 0); fork(writer, 0); writer() { while(TRUE) { <other computing>; P(writeBlock); /* Critical section */ access(resource); V(writeBlock); } First reader competes with writers Last reader signals writers 5/20/2019 COP4610

First Solution (2) First reader competes with writers while(TRUE) { <other computing>; P(mutex); readCount++; if(readCount == 1) P(writeBlock); V(mutex); /* Critical section */ access(resource); readCount--; if(readCount == 0) V(writeBlock); } resourceType *resource; int readCount = 0; semaphore mutex = 1; semaphore writeBlock = 1; fork(reader, 0); fork(writer, 0); writer() { while(TRUE) { <other computing>; P(writeBlock); /* Critical section */ access(resource); V(writeBlock); } First reader competes with writers Last reader signals writers Any writer must wait for all readers Readers can starve writers “Updates” can be delayed forever May not be what we want 5/20/2019 COP4610

Writer Precedence 4 3 2 1 reader() { while(TRUE) { <other computing>; P(readBlock); P(mutex1); readCount++; if(readCount == 1) P(writeBlock); V(mutex1); V(readBlock); access(resource); readCount--; if(readCount == 0) V(writeBlock); } int readCount = 0, writeCount = 0; semaphore mutex = 1, mutex2 = 1; semaphore readBlock = 1, writeBlock = 1, writePending = 1; fork(reader, 0); fork(writer, 0); writer() { while(TRUE) { <other computing>; P(mutex2); writeCount++; if(writeCount == 1) P(readBlock); V(mutex2); P(writeBlock); access(resource); V(writeBlock); P(mutex2) writeCount--; if(writeCount == 0) V(readBlock); } 4 3 2 1 5/20/2019 COP4610

Writer Precedence (2) 4 3 2 1 reader() { while(TRUE) { <other computing>; P(writePending); P(readBlock); P(mutex1); readCount++; if(readCount == 1) P(writeBlock); V(mutex1); V(readBlock); V(writePending); access(resource); readCount--; if(readCount == 0) V(writeBlock); } int readCount = 0, writeCount = 0; semaphore mutex = 1, mutex2 = 1; semaphore readBlock = 1, writeBlock = 1, writePending = 1; fork(reader, 0); fork(writer, 0); writer() { while(TRUE) { <other computing>; P(mutex2); writeCount++; if(writeCount == 1) P(readBlock); V(mutex2); P(writeBlock); access(resource); V(writeBlock); P(mutex2) writeCount--; if(writeCount == 0) V(readBlock); } 4 3 2 1 5/20/2019 COP4610

The Sleepy Barber Barber can cut one person’s hair at a time Other customers wait in a waiting room Entrance to Waiting Room (sliding door) Shop Exit Entrance to Barber’s Room (sliding door) Waiting Room 5/20/2019 COP4610

Sleepy Barber customer() { while(TRUE) { customer = nextCustomer(); if(emptyChairs == 0) continue; P(chair); P(mutex); emptyChairs--; takeChair(customer); V(mutex); V(waitingCustomer); } semaphore mutex = 1, chair = N, waitingCustomer = 0; int emptyChairs = N; fork(customer, 0); fork(barber, 0); barber() { while(TRUE) { P(waitingCustomer); P(mutex); emptyChairs++; takeCustomer(); V(mutex); V(chair); } 5/20/2019 COP4610

Implementing Semaphores Minimize effect on the I/O system Processes are only blocked on their own critical sections (not critical sections that they should not care about) If disabling interrupts, be sure to bound the time they are disabled 5/20/2019 COP4610

Implementing Semaphores Using Interrupts class semaphore { int value; public: semaphore(int v = 1) { value = v;}; P(){ disableInterrupts(); while(value == 0) { enableInterrupts(); } value--; }; V(){ value++; 5/20/2019 COP4610

Using the TS Instruction boolean s = FALSE; . . . while(TS(s)) ; <critical section> s = FALSE; semaphore s = 1; . . . P(s) ; <critical section> V(s); 5/20/2019 COP4610

Implementing the General Semaphore struct semaphore { int value = <initial value>; boolean mutex = FALSE; boolean hold = TRUE; }; shared struct semaphore s; P(struct semaphore s) { while(TS(s.mutex)) ; s.value--; if(s.value < 0) ( s.mutex = FALSE; while(TS(s.hold)) ; } else V(struct semaphore s) { while(TS(s.mutex)) ; s.value++; if(s.value <= 0) ( while(!s.hold) ; s.hold = FALSE; } s.mutex = FALSE; 5/20/2019 COP4610

Semaphore Implementation – cont. Semaphores implemented using interrupt disabling and test-and-set require busy waiting This type of semaphores is often called spinlock 5/20/2019 COP4610

Semaphore Implementation – cont. Define a semaphore as a structure typedef struct { int value; queue L; } semaphore; Assume two simple operations: block suspends the process that invokes it. wakeup(P) resumes the execution of a blocked process P. 5/20/2019 COP4610

Semaphore Implementation - cont. Semaphore operations now defined as P(S): S.value = S.value – 1; if S.value < 0 then begin add this process to S.L; block; end; 5/20/2019 COP4610

Semaphore Implementation - cont. V(S): S.value = S.value + 1; if S.value  0 then begin remove a process P from S.L; wakeup(P); end; 5/20/2019 COP4610

Semaphore Implementation - cont. Semaphores are resources in the above implementation 5/20/2019 COP4610

Issues Using Semaphores Deadlock two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes. Let S and Q be two semaphores initialized to 1 P0 P1 P(S); P(Q); P(Q); P(S);   V(S); V(Q); V(Q) V(S); Starvation – indefinite blocking A process may never be removed from the semaphore queue in which it is suspended. 5/20/2019 COP4610

Active vs. Passive Semaphores A process can dominate the semaphore Performs V operation, but continues to execute Performs another P operation before releasing the CPU Called a passive implementation of V Active implementation calls scheduler as part of the V operation. Changes semantics of semaphore! Cause people to rethink solutions 5/20/2019 COP4610

Summary Processes that share data need to be synchronized Otherwise, a race condition may exist Semaphores are the basic mechanism underlying synchronization and can be used to solve different synchronization problems Critical section problem Bounded-buffer problem Readers-writers problem 5/20/2019 COP4610