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Bab 6: Sinkronisasi Sumber Utama: Silberschatz ed.8

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Page 1: 09 sinkronisasi proses

Bab 6: Sinkronisasi

Sumber Utama: Silberschatz ed.8

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Materi Bab 6: Sinkronisasi Proses

Background The Critical-Section Problem Peterson’s Solution Synchronization Hardware Semaphores Classic Problems of Synchronization Monitors Synchronization Examples Atomic Transactions

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Objectives

Setelah memelajari materi ini, mahasiswa mampu:

Memahami masalah ‘critical-section ‘ yang memilikiberbagai solusi yang dapat digunakan untuk menjaminkonsistensi ‘shared data’

Menyajikan berbagai solusi software dan hardware pada masalah ‘critical-section ‘

Memahami konsep dari suatu transaksi atomik danmenggambarkan mekanisme untuk menjaminatomisitas.

Memahami masalah-masalah klasik dari sinkronisasi

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Overview (1)

4

Proteksi OS: Independent process tidak terpengaruh atau dapat

mempengaruhi eksekusi/data proses lain.

“Concurrent Process” OS: mampu membuat banyak proses pada satu saat Proses-proses bekerja-sama: sharing data, pembagian task,

passing informasi dll Proses => mempengaruhi proses lain dalam menggunakan

data/informasi yang sengaja di-”share”

Cooperating process – sekumpulan proses yang dirancang untuk saling bekerja-sama untukmengerjakan task tertentu.

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Overview (2)

5

Keuntungan kerja-sama antar proses Information sharing: file, DB => digunakan bersama

Computation speed-up: parallel proses

Modularity: aplikasi besar => dipartisi dalam banyak proses.

Convenience: kumpulan proses => tipikal lingkungan kerja.

“Cooperating Process” Bagaimana koordinasi antar proses? Akses/Update data

Tujuan program/task: integritas, konsistensi data dapatdijamin

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Latar Belakang

6

Menjamin konsistensi data: Program/task-task dapat menghasilkan operasi yang

benar setiap waktu Deterministik: untuk input yang sama hasil harus

sama (sesuai dengan logika/algoritma program).

Contoh: Producer – Consumer Dua proses: producer => menghasilkan informasi;

consumer => menggunakan informasi Sharing informasi: buffer => tempat penyimpanan

data unbounded-buffer, penempatan tidak pada limit

praktis dari ukuran buffer bounded-buffer diasumsikan terdapat ukuran buffer

yang tetap

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Latar Belakang

Akses konkuren untuk ‘shared data’ bisa menghasilkan data yang inkonsisten (data inconsistency ).

Pengelolaan konsistensi data memerlukan mekanisme yang menjamin eksekusi proses-proses yang koorperasi (salingbekerja sama) secara terurut.

Andaikan bahwa kita hendak memberi sebuah solusi kepadamasalah consumer-produser yang mengisikan semua buffer. Kita dapat melakukan demikian dengan memiliki suatu bilanganhitungan integer(integer count) yang mencatat jumlah buffer yang penuh.

Awalnya, hitungan di-set ke 0. Bilangan dinaikkan oleh producer setelah ia menghasilkan sebuah buffer baru dan diturunkanoleh consumer setelah ia menkonsumsi sebuah buffer.

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Bounded Buffer (1)

8

Implementasi buffer: IPC: komunikasi antar proses melalui messages

membaca/menulis buffer

Shared memory: programmer secara eksplisit melakukan“deklarasi” data yang dapat diakses secara bersama.

Buffer dengan ukuran n => mampu menampung n data

Producer mengisi data buffer => increment “counter” (jumlah data)

Consumer mengambil data buffer => decrement “counter”

Buffer, “counter” => shared data (update oleh 2 proses)

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Bounded Buffer (2)

9

Shared data type item = … ;

var buffer array

in, out: 0..n-1;

counter: 0..n;

in, out, counter := 0;

Producer process

repeat…produce an item in nextp…while counter = n do no-op;buffer [in] := nextp;in := in + 1 mod n;counter := counter +1;until false;

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Bounded Buffer (3)

10

Consumer processrepeat

while counter = 0 do no-op;

nextc := buffer [out];

out := out + 1 mod n;

counter := counter – 1;

consume the item in nextc

until false;

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Bounded Buffer (4)

11

Apakah terdapat jaminan operasi akan benar jikaberjalan concurrent?

Misalkan: counter = 5 Producer: counter = counter + 1;

Consumer: counter = counter - 1;

Nilai akhir dari counter?

Operasi concurrent P & C => Operasi dari high level language => sekumpulan instruksi

mesin: “increment counter”Load Reg1, CounterAdd Reg1, 1Store Counter, Reg1

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Bounded Buffer (5)

12

“decrement counter”Load Reg2, CounterSubtract Reg2, 1Store Counter, Reg2

Eksekusi P & C tergantung scheduler (dapat gantian) T0: Producer : Load Reg1, Counter (Reg1 = 5)

T1: Producer : Add Reg1, 1 (Reg1 = 6)

T2: Consumer: Loag Reg2, Counter (Reg2 = 5)

T3: Consumer: Subtract Reg1, 1 (Reg2 = 4)

T4: Producer: Store Counter, Reg1 (Counter = 6)

T5: Consumer: Store Counter, Reg2 (Counter = 4)

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Producer

while (true) {

/* produce an item and put in nextProduced*/

while (count == BUFFER_SIZE)

; // do nothing

buffer [in] = nextProduced;

in = (in + 1) % BUFFER_SIZE;

count++;

}

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Consumer

while (true) {

while (count == 0)

; // do nothing

nextConsumed = buffer[out];

out = (out + 1) % BUFFER_SIZE;

count--;

/* consume the item in nextConsumed

}

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Race Condition

15

Concurrent C & P Shared data “counter” dapat berakhir dengan nilai: 4, atau 5,

atau 6

Hasilnya dapat salah dan tidak konsisten

Race Condition: Keadaan dimana lebih dari satu proses meng-update data

secara “concurrent” dan hasilnya sangat bergantung dariurutan proses mendapat jatah CPU (run)

Hasilnya tidak menentu dan tidak selalu benar

Mencegah race condition: sinkronisasi proses dalam meng-update shared data

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Race Condition

Pada program producer/consumer dapat kita lihat terdapat perintah count++ dancount- - yang dapat diimplementasikan dengan bahasa mesin sebagai berikut:

count++ could be implemented as

register1 = countregister1 = register1 + 1count = register1

count-- could be implemented as

register2 = countregister2 = register2 - 1count = register2

Dapat dilihat jika perintahdari count+ + dan count - - dieksekusi secara bersama, maka akan sulit untuk m engetahui nilai count sebenarnya , sehingga nilai daricount itu akan menjadi tidak konsisten.

Marilah kita lihat contoh berikut:

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Consider this execution interleaving with “count = 5” initially:

S0: producer execute register1 = count {register1 = 5}S1: producer execute register1 = register1 + 1 {register1 = 6} S2: consumer execute register2 = count {register2 = 5} S3: consumer execute register2 = register2 - 1 {register2 = 4} S4: producer execute count = register1 {count = 6 } S5: consumer execute count = register2 {count = 4}

Pada contoh di atas dapat dilihat bahwa count memilki nilai dua nilai yaitu bernilai 5 (pada saat count + + dieksekusi) dan bernilai 4 (pada saat count- - dieksekusi).

Hal ini menyebabkan nilai dari count tsb inkonsisten.

Perhatikan bahwa nilai dari count akan bergantung pada perintah terakhir yang dieksekusi.

Oleh karenanya, kita membutuhkan sinkronisasi yang merupakan upaya yang dilakukan agar proses-proses yang saling bekerja bersama-sama dieksekusi secaraberaturan (orderly) demi mencegah timbulnya keadaan yang disebut Race Condition.

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Sinkronisasi

18

Sinkronisasi: Koordinasi akses ke shared data, misalkan hanya satu proses

yang dapat menggunakah shared var.

Contoh operasi terhadap var. “counter” harus dijamin di-eksekusi dalam satu kesatuan (atomik) :

counter := counter + 1;

counter := counter - 1;

Sinkronisasi merupakan “issue” penting dalam rancangan/implementasi OS (shared resources, data, dan multitasking).

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Problem Critical Section Problem ini karena adanya suatu race conditon pada suatu proses yang

dilakukan secara konkuren yang mengakibatkan tidak sinkron.

Nilai akhir tegantung pada proses mana yang terakhir dieksekusi.

Bagaimana cara mengatasi race condition?

Kuncinya adalah menemukan jalan untuk mencegah lebih dari suatu prosesmelakukan proses tulis atau baca kepada data atau berkas pada saat yang bersamaan.

Perlu adanya Mutual Exclusion yaitu suatu cara yang menjamin jika ada suatuproses yang menggunakan variabel atau berkas yang sama (digunakan juga olehproses lain), maka proses lain akan dikeluarkan dari pekerjaan yang sama.

Karena beberapa proses memiliki suatu segmen kode dimana jika segmen itudieksekusi, maka proses-proses itu dapat saling mengubah variabel, mengupdatesuatu tabel, menulis ke suatu file dsb.

Segmen kode ini dinamakan critical section.

Hal demikian, dapat membawa ke dalam bahaya race condition.

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Masalah Critical Section

20

n proses mencoba menggunakan shared data bersamaan

Setiap proses mempunyai “code” yang mengakses/ manipulasi shared data tersebut => “critical section”

Problem: Menjamin jika ada satu proses yang sedang

“eksekusi” pada bagian “critical section” tidak ada proses lain yang diperbolehkan masuk ke “code” critical section dari proses tersebut.

Structure of process Pi

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Solution to Critical-Section Problem

Solusi untuk memecahkan critical section adalah denganmendesain sebuah protokol di mana proses-proses dapatmenggunakannya secara bersama-sama.

Setiap proses harus ‘meminta izin’ untuk memasuki critical section-nya.

Bagian dari kode yang mengimplementasikan izin ini disebutentry section.

Akhir dari critical section disebut exit section.

Bagian kode selanjutnya disebut remainder section.

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Solusi Masalah Critical Section

22

Ide : Mencakup pemakaian secara “exclusive” dari shared

variable tersebut Menjamin proses lain dapat menggunakan shared

variable tersebut

Solusi “critical section problem” harus memenuhi:1. Mutual Exclusion: Jika proses Pi sedang “eksekusi”

pada bagian “critical section” (dari proses Pi) maka tidakada proses proses lain dapat “eksekusi” pada bagiancritical section dari proses-proses tersebut.

2. Progress: Jika tidak ada proses sedang eksekusi padacritical section-nya dan jika terdapat lebih dari satuproses lain yang ingin masuk ke critical section, makapemilihan siapa yang berhak masuk ke critical section tidak dapat ditunda tanpa terbatas.

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Solusi (cont.)

23

3. Bounded Waiting: Terdapat batasan berapalama suatu proses harus menunggu giliranuntuk mengakses “critical section” – jikaseandainya proses lain yang diberikan hak akseske critical section.

Menjamin proses dapat mengakses ke “critical section” (tidak mengalami starvation: prosesse-olah berhenti menunggu request akses kecritical section diperbolehkan).

Tidak ada asumsi mengenai kecepataneksekusi proses proses n tersebut.

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Solution to Critical-Section Problem Solusi dari masalah Critical-Section Problem harus memenuhi tiga syarat berikut:

1. Mutual Exclusion - If process Pi is executing in its critical section, then no other processes can be executing in their critical sections

2. 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

3. 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

Assume that each process executes at a nonzero speed

No assumption concerning relative speed of the N processes

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Pemecahan Masalah Critical Section: Solusi untuk dua proses

Ada dua jenis solusi masalah critical section, yaitu:

Solusi perangkat lunak

Dengan menggunakan algoritma-algoritma yang nilai kebenarannya tidaktergantung pada asumsi-asumsi lain, selain bahwa setiap proses berjalanpada kecepatan yang bukan nol

Solusi perangkat keras

Tergantung pada beberapa instruksi mesin tertentu, misalnya dengan me-nonaktifkan interupsi atau dengan mengunci suatu variabel tertentu.

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Solusi Sederhana : Kasus 2 proses

26

Hanya 2 proses

Struktur umum dari program code Pi dan Pj:

Software solution: merancang algoritma program untuk solusi critical section Proses dapat mengunakan “common var.” untuk menyusun algoritma tsb.

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Algoritma 1

27

Shared variables:

int turn;initially turn = 0

turn - i Pi dapat masuk ke criticalsection

Process Pi

do {

while (turn != i) ;

critical section

turn = j;

reminder section

} while (1);

Mutual exclusion terpenuhi, tetapi menentang progress

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Algoritma 2

28

Shared variables boolean flag[2];

initially flag [0] = flag [1] = false.

flag [i] = true Pi siap dimasukkan ke dalam critical section

Process Pi

do {

flag[i] := true;while (flag[j]) ;

critical section

flag [i] = false;

remainder section

} while (1);

Mutual exclusion terpenuhi tetapi progress belum terpenuhi.

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Algoritma 3

29

Kombinasi shared variables dari algoritma 1 and 2. Process Pi

do {

flag [i]:= true;turn = j;while (flag [j] and turn = j) ;

critical section

flag [i] = false;

remainder section

} while (1);

Ketiga kebutuhan terpenuhi, solusi masalah critical section pada dua proses

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Algoritma Bakery

30

Critical section untuk n proses

Sebelum proses akan masuk ke dalam “critical section”, maka proses harus mendapatkan “nomor” (tiket).

Proses dengan nomor terkecil berhak masuk ke critical section.

Jika proses Pi dan Pj menerima nomor yang sama, jika i < j, maka Pi dilayani pertama; jika tidak Pj dilayani pertama

Skema penomoran selalu dibuat secara berurutan, misalnya 1,2,3,3,3,3,4,5...

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Algoritma Bakery (2)

31

Notasi < urutan lexicographical (ticket #, process id #)

(a,b) < c,d) jika a < c atau jika a = c and b < d

max (a0,…, an-1) dimana a adalah nomor, k, seperti pada k ai untuk i - 0, …, n – 1

Shared data

var choosing: array [0..n – 1] of boolean

number: array [0..n – 1] of integer,

Initialized: choosing =: false ; number => 0

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Algoritma Bakery (3)

32

do {

choosing[i] = true;

number[i] = max(number[0], number[1], …, number [n –1])+1;

choosing[i] = false;

for (j = 0; j < n; j++) {

while (choosing[j]) ;

while ((number[j] != 0) && (number[j,j] < number[i,i])) ;

}critical section

number[i] = 0;

remainder section

} while (1);

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Sinkronisasi Hardware

33

Memerlukan dukungan hardware (prosesor)

Dalam bentuk “instruction set” khusus: test-and-set

Menjamin operasi atomik (satu kesatuan): test nilai dan ubah nilai tersebu

Test-and-Set dapat dianalogikan dengan kode:

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Test-and-Set (mutual exclusion)

34

Mutual exclusion dapat diterapkan: Gunakan shared data,

variabel: lock: boolean (initially false)

lock: menjaga critical section

Process Pi:do {

while (TestAndSet(lock)) ;

critical section

lock = false;

remainder section

}

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Semaphore

35

Perangkat sinkronisasi yang tidak membutuhkan busy waiting

Semaphore S – integer variable

Dapat dijamin akses ke var. S oleh dua operasi atomik: wait (S): while S ≤ 0 do no-op;

S := S – 1;

signal (S): S := S + 1;

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Contoh : n proses

36

Shared variables var mutex : semaphore

initially mutex = 1

Process Pido {

wait(mutex);critical section

signal(mutex);remainder section

} while (1);

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Implementasi Semaphore

37

Didefinisikan sebuah Semaphore dengan sebuah record

typedef struct {

int value;struct process *L;

} semaphore;

Diasumsikan terdapat 2 operasi sederhana : block menhambat proses yang akan masuk

wakeup(P) memulai eksekusi pada proses P yang di block

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Implementasi Semaphore (2)

38

Operasi Semaphore-nya menjadi :

wait(S):S.value--;

if (S.value < 0) {

add this process to S.L;block;

}

signal(S): S.value++;

if (S.value <= 0) {

remove a process P from S.L;wakeup(P);

}

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Masalah Klasik Sinkronisasi

39

Bounded-Buffer Problem

Readers and Writers Problem

Dining-Philosophers Problem

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Bounded-Buffer Problem

40

Shared data

semaphore full, empty, mutex;

Initially:

full = 0, empty = n, mutex = 1

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Bounded-Buffer Problem : Producer-Consumer

41

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Readers-Writers Problem

42

Shared data

semaphore mutex, wrt;

Initially

mutex = 1, wrt = 1, readcount = 0

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Readers-Writers Problem (2)

43

Writters Process

wait(wrt);

writing is performed

signal(wrt);

Readers Processwait(mutex);

readcount++;

if (readcount == 1)

wait(rt);

signal(mutex);

reading is performed

wait(mutex);

readcount--;

if (readcount == 0)

signal(wrt);

signal(mutex):

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Dining-Philosophers Problem

44

Shared data

semaphore chopstick[5];

Semua inisialisasi bernilai 1

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Dining-Philosophers Problem

45

Philosopher i:

do {

wait(chopstick[i])

wait(chopstick[(i+1) % 5])

eat

signal(chopstick[i]);

signal(chopstick[(i+1) % 5]);

think

} while (1);

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Solusi Tingkat Tinggi

46

Motif:

Operasi wait(S) dan signal(S) tersebar pada code program => manipulasi langsung struktur data semaphore

Bagaimana jika terdapat bantuan dari lingkungan HLL (programming) untuk sinkronisasi ?

Pemrograman tingkat tinggi disediakan sintaks-sintaks khusus untuk menjamin sinkronisasi antar proses, thread

Misalnya: Monitor & Condition

Conditional Critical Region

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Monitor

47

Monitor mensinkronisasi sejumlah proses: suatu saat hanya satu yang aktif dalam monitor dan yang lain

menunggu

Bagian dari bahasa program (mis. Java). Tugas compiler menjamin hal tersebut terjadi dengan

menerjemahkan ke “low level synchronization” (semphore, instruction set dll)

Cukup dengan statement (deklarasi) suatu section/fungsi adalah monitor => mengharuskan hanya ada satu proses yang berada dalam monitor (section) tsb

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Monitor (2)

48

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Monitor (3)

49

Proses-proses harus disinkronisasikan di dalam monitor:

Memenuhi solusi critical section.

Proses dapat menunggu di dalam monitor.

Mekanisme: terdapat variabel (condition) dimana proses dapat menguji/menunggu sebelum mengakses “critical section”

var x, y: condition

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Monitor (4)

50

Condition: memudahkan programmer untuk menulis code pada monitor.

Misalkan : var x: condition ;

Variabel condition hanya dapat dimanipulasi dengan operasi: wait() dan signal()

x.wait() jika dipanggil oleh suatu proses maka proses tsb. akan suspend - sampai ada proses lain yang memanggil: x. signal()

x.signal() hanya akan menjalankan (resume) 1 proses saja yang sedang menunggu (suspend) (tidak ada proses lain yang wait maka tidak berdampak apapun)

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Skema Monitor

51

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Pemecahan Masalah Critical Section: Peterson’s Solution

Two process solution

Assume that the LOAD and STORE instructions are atomic; that is, cannot be interrupted.

The two processes share two variables:

int turn;

Boolean flag[2]

The variable turn indicates whose turn it is to enter the critical section.

The flag array is used to indicate if a process is ready to enter the critical section. flag[i] = true implies that process Pi is ready!

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PROCESS SYNCHRONIZATION

Here’s an example of a simple piece of code containing the components required in a critical section.

do {while ( turn ^= i );/* critical section */turn = j;/* remainder section */

} while(TRUE);

Two ProcessesSoftware

Entry Section

Critical Section

Exit Section

Remainder Section

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Algorithm for Process Pi

do {

flag[i] = TRUE;

turn = j;

while (flag[j] && turn == j);

critical section

flag[i] = FALSE;

remainder section

} while (TRUE);

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Synchronization Hardware

Many systems provide hardware support for critical section code

Uniprocessors – could disable interrupts

Currently running code would execute without preemption

Generally too inefficient on multiprocessor systems Operating systems using this not broadly scalable

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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Solution to Critical-section Problem Using Locks

do {

acquire lock

critical section

release lock

remainder section

} while (TRUE);

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TestAndSet Instruction

Definition:

boolean TestAndSet (boolean *target)

{

boolean rv = *target;

*target = TRUE;

return rv:

}

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Solution using TestAndSet

Shared boolean variable lock., initialized to false.

Solution:

do {

while ( TestAndSet (&lock ))

; // do nothing

// critical section

lock = FALSE;

// remainder section

} while (TRUE);

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Swap Instruction

Definition:

void Swap (boolean *a, boolean *b)

{

boolean temp = *a;

*a = *b;

*b = temp:

}

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Solution using Swap Shared Boolean variable lock initialized to FALSE; Each process has a

local Boolean variable key

Solution:

do {

key = TRUE;

while ( key == TRUE)

Swap (&lock, &key );

// critical section

lock = FALSE;

// remainder section

} while (TRUE);

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Bounded-waiting Mutual Exclusion with TestandSet()

do {

waiting[i] = TRUE;

key = TRUE;

while (waiting[i] && key)

key = TestAndSet(&lock);

waiting[i] = FALSE;

// critical section

j = (i + 1) % n;

while ((j != i) && !waiting[j])

j = (j + 1) % n;

if (j == i)

lock = FALSE;

else

waiting[j] = FALSE;

// remainder section

} while (TRUE);

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Semaphore

Sinkronisasi adalah alat bantu yang tidak memerlukan busy waiting

Semaphore S – integer variable

Two standard operations modify S: wait() and signal()

Originally called P() and V()

Less complicated

Can only be accessed via two indivisible (atomic) operations

wait (S) {

while S <= 0

; // no-op

S--;

}

signal (S) {

S++;

}

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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

Can implement a counting semaphore S as a binary semaphore

Provides mutual exclusion

Semaphore mutex; // initialized to 1

do {

wait (mutex);

// Critical Section

signal (mutex);

// remainder section

} while (TRUE);

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Semaphore Implementation Harus menjamin bahwa tidak ada duap roses dapat mengeksekusi wait () and

signal () pada semaphore yang sama dan waktu yang sama

Thus, implementasi menjadi masalah critical section dimana kode wait dansignal ditempatkan pada critical section

Could now have busy waiting in critical section implementation

But implementation code is short

Little busy waiting if critical section rarely occupied

Note that applications may spend lots of time in critical sections and therefore this is not a good solution.

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Semaphore Implementation with no Busy waiting

With each semaphore there is an associated waiting queue. Each entry in a waiting queue has two data items:

value (of type integer)

pointer to next record in the list

Two operations:

block – place the process invoking the operation on the appropriate waiting queue.

wakeup – remove one of processes in the waiting queue and place it in the ready queue.

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Semaphore Implementation with no Busy waiting (Cont.)

Implementation of wait:

wait(semaphore *S) {

S->value--;

if (S->value < 0) {

add this process to S->list;

block();

}

}

Implementation of signal:

signal(semaphore *S) {

S->value++;

if (S->value <= 0) {

remove a process P from S->list;

wakeup(P);

}

}

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Deadlock and Starvation 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

wait (S); wait (Q);

wait (Q); wait (S);

. .

. .

. .

signal (S); signal (Q);

signal (Q); signal (S);

Starvation – indefinite blocking. A process may never be removed from the semaphore queue in which it is suspended

Priority Inversion - Scheduling problem when lower-priority process holds a lock needed by higher-priority process

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Classical Problems of Synchronization

Bounded-Buffer Problem

Readers and Writers Problem

Dining-Philosophers Problem

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Bounded-Buffer Problem

N buffers, each can hold one item

Semaphore mutex initialized to the value 1

Semaphore full initialized to the value 0

Semaphore empty initialized to the value N.

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Bounded Buffer Problem (Cont.)

The structure of the producer process

do {

// produce an item in nextp

wait (empty);

wait (mutex);

// add the item to the buffer

signal (mutex);

signal (full);

} while (TRUE);

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Bounded Buffer Problem (Cont.)

The structure of the consumer process

do {

wait (full);

wait (mutex);

// remove an item from buffer to nextc

signal (mutex);

signal (empty);

// consume the item in nextc

} while (TRUE);

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Readers-Writers Problem

A data set is shared among a number of concurrent processes

Readers – only read the data set; they do not perform any updates

Writers – can both read and write

Problem – allow multiple readers to read at the same time. Only one single writer can access the shared data at the same time

Shared Data

Data set

Semaphore mutex initialized to 1

Semaphore wrt initialized to 1

Integer readcount initialized to 0

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Readers-Writers Problem (Cont.)

The structure of a writer process

do {

wait (wrt) ;

// writing is performed

signal (wrt) ;

} while (TRUE);

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Readers-Writers Problem (Cont.) The structure of a reader process

do {

wait (mutex) ;

readcount ++ ;

if (readcount == 1)

wait (wrt) ;

signal (mutex)

// reading is performed

wait (mutex) ;

readcount - - ;

if (readcount == 0)

signal (wrt) ;

signal (mutex) ;

} while (TRUE);

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Dining-Philosophers Problem

Shared data

Bowl of rice (data set)

Semaphore chopstick [5] initialized to 1

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Dining-Philosophers Problem (Cont.)

The structure of Philosopher i:

do {

wait ( chopstick[i] );

wait ( chopStick[ (i + 1) % 5] );

// eat

signal ( chopstick[i] );

signal (chopstick[ (i + 1) % 5] );

// think

} while (TRUE);

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Problems with Semaphores

Incorrect use of semaphore operations:

signal (mutex) …. wait (mutex)

wait (mutex) … wait (mutex)

Omitting of wait (mutex) or signal (mutex) (or both)

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Monitors A high-level abstraction that provides a convenient and effective mechanism

for process synchronization

Only one process may be active within the monitor at a time

monitor monitor-name

{

// shared variable declarations

procedure P1 (…) { …. }

procedure Pn (…) {……}

Initialization code ( ….) { … }

}

}

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Schematic view of a Monitor

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Condition Variables

condition x, y;

Two operations on a condition variable:

x.wait () – a process that invokes the operation is

suspended.

x.signal () – resumes one of processes (if any) that

invoked x.wait ()

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Monitor with Condition Variables

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Solution to Dining Philosophers

monitor DP

{

enum { THINKING; HUNGRY, EATING) state [5] ;

condition self [5];

void pickup (int i) {

state[i] = HUNGRY;

test(i);

if (state[i] != EATING) self [i].wait;

}

void putdown (int i) {

state[i] = THINKING;

// test left and right neighbors

test((i + 4) % 5);

test((i + 1) % 5);

}

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Solution to Dining Philosophers (cont)

void test (int i) {

if ( (state[(i + 4) % 5] != EATING) &&

(state[i] == HUNGRY) &&

(state[(i + 1) % 5] != EATING) ) {

state[i] = EATING ;

self[i].signal () ;

}

}

initialization_code() {

for (int i = 0; i < 5; i++)

state[i] = THINKING;

}

}

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Solution to Dining Philosophers (cont)

Each philosopher I invokes the operations pickup()

and putdown() in the following sequence:

DiningPhilosophters.pickup (i);

EAT

DiningPhilosophers.putdown (i);

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Monitor Implementation Using Semaphores

Variables semaphore mutex; // (initially = 1)semaphore next; // (initially = 0)int next-count = 0;

Each procedure F will be replaced by

wait(mutex);…

body of F;

…if (next_count > 0)

signal(next)else

signal(mutex);

Mutual exclusion within a monitor is ensured.

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Monitor Implementation For each condition variable x, we have:

semaphore x_sem; // (initially = 0)

int x-count = 0;

The operation x.wait can be implemented as:

x-count++;

if (next_count > 0)

signal(next);

else

signal(mutex);

wait(x_sem);

x-count--;

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Monitor Implementation The operation x.signal can be implemented as:

if (x-count > 0) {

next_count++;

signal(x_sem);

wait(next);

next_count--;

}

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A Monitor to Allocate Single Resource

monitor ResourceAllocator

{

boolean busy;

condition x;

void acquire(int time) {

if (busy)

x.wait(time);

busy = TRUE;

}

void release() {

busy = FALSE;

x.signal();

}

initialization code() {

busy = FALSE;

}

}

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Synchronization Examples

Solaris

Windows XP

Linux

Pthreads

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Solaris Synchronization

Implements a variety of locks to support multitasking, multithreading (including real-time threads), and multiprocessing

Uses adaptive mutexes for efficiency when protecting data from short code segments

Uses condition variables and readers-writers locks when longer sections of code need access to data

Uses turnstiles to order the list of threads waiting to acquire either an adaptive mutex or reader-writer lock

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Windows XP Synchronization

Uses interrupt masks to protect access to global resources on uniprocessor systems

Uses spinlocks on multiprocessor systems

Also provides dispatcher objects which may act as either mutexes and semaphores

Dispatcher objects may also provide events An event acts much like a condition variable

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Linux Synchronization

Linux:

Prior to kernel Version 2.6, disables interrupts to implement short critical sections

Version 2.6 and later, fully preemptive

Linux provides:

semaphores

spin locks

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Pthreads Synchronization

Pthreads API is OS-independent

It provides:

mutex locks

condition variables

Non-portable extensions include:

read-write locks

spin locks

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Atomic Transactions

System Model

Log-based Recovery

Checkpoints

Concurrent Atomic Transactions

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System Model

Assures that operations happen as a single logical unit of work, in its entirety, or not at all

Related to field of database systems

Challenge is assuring atomicity despite computer system failures

Transaction - collection of instructions or operations that performs single logical function

Here we are concerned with changes to stable storage – disk

Transaction is series of read and write operations

Terminated by commit (transaction successful) or abort (transaction failed) operation

Aborted transaction must be rolled back to undo any changes it performed

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Types of Storage Media

Volatile storage – information stored here does not survive system crashes

Example: main memory, cache

Nonvolatile storage – Information usually survives crashes

Example: disk and tape

Stable storage – Information never lost

Not actually possible, so approximated via replication or RAID to devices with independent failure modes

Goal is to assure transaction atomicity where failures cause loss of information on volatile storage

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Log-Based Recovery Record to stable storage information about all modifications by a transaction

Most common is write-ahead logging

Log on stable storage, each log record describes single transaction write operation, including

Transaction name

Data item name

Old value

New value

<Ti starts> written to log when transaction Ti starts

<Ti commits> written when Ti commits

Log entry must reach stable storage before operation on data occurs

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Log-Based Recovery Algorithm Using the log, system can handle any volatile memory errors

Undo(Ti) restores value of all data updated by Ti

Redo(Ti) sets values of all data in transaction Ti to new values

Undo(Ti) and redo(Ti) must be idempotent

Multiple executions must have the same result as one execution

If system fails, restore state of all updated data via log

If log contains <Ti starts> without <Ti commits>, undo(Ti)

If log contains <Ti starts> and <Ti commits>, redo(Ti)

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Checkpoints

Log could become long, and recovery could take long

Checkpoints shorten log and recovery time.

Checkpoint scheme:

1. Output all log records currently in volatile storage to stable storage

2. Output all modified data from volatile to stable storage

3. Output a log record <checkpoint> to the log on stable storage

Now recovery only includes Ti, such that Ti started executing before the most recent checkpoint, and all transactions after Ti All other transactions already on stable storage

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Concurrent Transactions

Must be equivalent to serial execution – serializability

Could perform all transactions in critical section

Inefficient, too restrictive

Concurrency-control algorithms provide serializability

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Serializability

Consider two data items A and B

Consider Transactions T0 and T1

Execute T0, T1 atomically

Execution sequence called schedule

Atomically executed transaction order called serial schedule

For N transactions, there are N! valid serial schedules

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Schedule 1: T0 then T1

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Nonserial Schedule

Nonserial schedule allows overlapped execute

Resulting execution not necessarily incorrect

Consider schedule S, operations Oi, Oj

Conflict if access same data item, with at least one write

If Oi, Oj consecutive and operations of different transactions & Oi and Oj

don’t conflict

Then S’ with swapped order Oj Oi equivalent to S

If S can become S’ via swapping nonconflicting operations

S is conflict serializable

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Schedule 2: Concurrent Serializable Schedule

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Locking Protocol

Ensure serializability by associating lock with each data item

Follow locking protocol for access control

Locks

Shared – Ti has shared-mode lock (S) on item Q, Ti can read Q but not write Q

Exclusive – Ti has exclusive-mode lock (X) on Q, Ti can read and write Q

Require every transaction on item Q acquire appropriate lock

If lock already held, new request may have to wait

Similar to readers-writers algorithm

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Two-phase Locking Protocol Generally ensures conflict serializability

Each transaction issues lock and unlock requests in two phases

Growing – obtaining locks

Shrinking – releasing locks

Does not prevent deadlock

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Timestamp-based Protocols Select order among transactions in advance – timestamp-ordering

Transaction Ti associated with timestamp TS(Ti) before Ti starts

TS(Ti) < TS(Tj) if Ti entered system before Tj

TS can be generated from system clock or as logical counter incremented at each entry of transaction

Timestamps determine serializability order

If TS(Ti) < TS(Tj), system must ensure produced schedule equivalent to serial schedule where Ti appears before Tj

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Timestamp-based Protocol Implementation

Data item Q gets two timestamps

W-timestamp(Q) – largest timestamp of any transaction that executed write(Q) successfully

R-timestamp(Q) – largest timestamp of successful read(Q)

Updated whenever read(Q) or write(Q) executed

Timestamp-ordering protocol assures any conflicting read and write executed in timestamp order

Suppose Ti executes read(Q)

If TS(Ti) < W-timestamp(Q), Ti needs to read value of Q that was already overwritten

read operation rejected and Ti rolled back

If TS(Ti) ≥ W-timestamp(Q)

read executed, R-timestamp(Q) set to max(R-timestamp(Q), TS(Ti))

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Timestamp-ordering Protocol Suppose Ti executes write(Q)

If TS(Ti) < R-timestamp(Q), value Q produced by Ti was needed previously and Ti

assumed it would never be produced

Write operation rejected, Ti rolled back

If TS(Ti) < W-tiimestamp(Q), Ti attempting to write obsolete value of Q

Write operation rejected and Ti rolled back

Otherwise, write executed

Any rolled back transaction Ti is assigned new timestamp and restarted

Algorithm ensures conflict serializability and freedom from deadlock

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Schedule Possible Under Timestamp Protocol

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