Operating Systems CSE 411 CPU Management Dec. 4 2006 - Lecture Instructor: Bhuvan Urgaonkar.

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Presentation transcript:

Operating Systems CSE 411 CPU Management Dec Lecture Instructor: Bhuvan Urgaonkar

CPU Management Last class –Synchronization problem in concurrent programs Today –Continue from where we left –And meet some philosophers, may be!

Synchronization Problem (Shared Memory based IPC) Race condition –A situation where two or more processes are reading or writing some shared data and the final result depends on the exact ordering of events There has to be at least one writer (Why?) Critical Section – The part of the program where the shared memory is accessed How do we avoid race condition? –Mutual exclusion: If one process is using a shared variable or file, the other processes will be excluded from using the same resources i.e. no two processes are ever in their critical section at the same time

Mutual Exclusion do { entry section CRITICAL SECTION exit section REMAINDER SECTION } while (TRUE);

Requirements for Solving the the Critical Section Problem Mutual Exclusion: One process at a time gets in critical section  E.g., Only one car may pass at a 4-way stop sign at a given time Progress: If P i wants to get in critical section and no process is in critical section, then P i should be able to progress  E.g., If a car comes to a stop and there is no other car, then it should be able to proceed Bounded wait: P i should be able to get critical section with some upper waiting time  E.g., A car should get to proceed after some bounded amount of waiting at the stop sign What is the bound for this 4-way stop sign example?

Mutual Exclusion: Attempt 1 Can we get lock before get into critical section and after leave critical section, we release lock? Is there any trouble if we use a flag to lock? while (lock == 1) ; // Do nothing, just wait //position 1 lock = 1; // Critical Section..... lock = 0; If it is interrupted at position 1, the same trouble occurs The process was waiting until lock == 0. Before it announces its turn (i.e. set lock = 1;), the other process interrupts it and sees that lock is 0 It runs into critical section. When the CPU time comes back to the previous process, the latter process is still in critical section and unable to lock it since the previous process was on the stage of changing lock and has not done yet. The previous one will go ahead to get into critical section Now, two processes are in the critical section at the same time period! Doesn’t work!

Mutual Exclusion: Approach 1 Disable interrupts –Have each process disable all interrupts just after entering its critical region and re-enable them just before leaving it. –Disadvantage: Can lock out system: If there is only one process working and some problems occur then the system will stop and no other processes can interrupt in to use system.

Mutual Exclusion: Approach 2 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] initialized to [false, false] 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 P i is ready

Algorithm for Process P i while (true) { flag[i] = TRUE; turn = j; while ( flag[j] && turn == j); CRITICAL SECTION flag[i] = FALSE; REMAINDER SECTION }

Comments on Peterson’s Approach This is a purely software approach, and it does not require any support from hardware except atomic loads and stores However, it may cause "busy waiting” –Wastes CPU cycles

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

TestAndndSet Instruction Definition: boolean TestAndSet (boolean *target) { boolean rv = *target; *target = TRUE; return rv: }

Solution using TestAndSet Shared boolean variable lock, initialized to false Solution: while (true) { while ( TestAndSet (&lock )) ; /* do nothing // critical section lock = FALSE; // remainder section }

Swap Instruction Definition: void Swap (boolean *a, boolean *b) { boolean temp = *a; *a = *b; *b = temp: }

Solution using Swap Shared Boolean variable lock initialized to FALSE; Each process has a local Boolean variable key. Solution: while (true) { key = TRUE; while ( key == TRUE) Swap (&lock, &key ); // critical section lock = FALSE; // remainder section }

Semaphore Synchronization tool that does not require 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++; }

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 S; // initialized to 1 –wait (S); Critical Section signal (S);

Semaphore Implementation Must guarantee that no two processes can execute wait () and signal () on the same semaphore at the same time Thus, implementation becomes the critical section problem where the wait and signal code are placed in the crtical 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.

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.

Semaphore Implementation with no Busy waiting (Cont.) Implementation of wait: wait (S){ value--; if (value < 0) { add this process to waiting queue block(); } } Implementation of signal: Signal (S){ value++; if (value <= 0) { remove a process P from the waiting queue wakeup(P); } }

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 P 0 P 1 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.

Classical Problems of Synchronization Bounded-Buffer Problem Readers and Writers Problem Dining-Philosophers Problem

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

Bounded Buffer Problem (Cont.) The structure of the producer process while (true) { // produce an item wait (empty); wait (mutex); // add the item to the buffer signal (mutex); signal (full); }

Bounded Buffer Problem (Cont.) The structure of the consumer process while (true) { wait (full); wait (mutex); // remove an item from buffer signal (mutex); signal (empty); // consume the removed item }

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 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.

Readers-Writers Problem (Cont.) The structure of a writer process while (true) { wait (wrt) ; // writing is performed signal (wrt) ; }

Readers-Writers Problem (Cont.) The structure of a reader process while (true) { wait (mutex) ; readcount ++ ; if (readcount == 1) wait (wrt) ; signal (mutex) // reading is performed wait (mutex) ; readcount - - ; if (readcount == 0) signal (wrt) ; signal (mutex) ; }

Dining-Philosophers Problem Shared data –Bowl of rice (data set) –Semaphore chopstick [5] initialized to 1

Dining-Philosophers Problem (Cont.) The structure of Philosopher i: while (true) { wait ( chopstick[i] ); wait ( chopStick[ (i + 1) % 5] ); // eat signal ( chopstick[i] ); signal (chopstick[ (i + 1) % 5] ); // think }

Problems with Semaphores Correct use of semaphore operations: – signal (mutex) …. wait (mutex) – wait (mutex) … wait (mutex) – Omitting of wait (mutex) or signal (mutex) (or both)

Synchronization on Multi-processors

Monitors

System Bootstrap

Multi-processor Scheduling

Some History

UNIX

The POSIX Standard

Course Outline Resource Management (and some services an OS provides to programmers) CPU management  Memory management –I/O management (emphasis: Disk) Cross-cutting design considerations and techniques –Quality-of-service/fairness, monitoring, accounting, caching, software design methodology, security and isolation Advanced topics –Distributed systems –Data centers, multi-media systems, real-time systems, virtual machines