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TDC 311

TDC 311. Process Synchronization. Processes. Whenever processes share things, there is a chance for screwing up things For example Two processes sharing one printer Two processes updating one database record Two processes sharing a single counter. Producer / Consumer Problem.

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TDC 311

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  1. TDC 311 Process Synchronization

  2. Processes • Whenever processes share things, there is a chance for screwing up things • For example • Two processes sharing one printer • Two processes updating one database record • Two processes sharing a single counter

  3. Producer / Consumer Problem • Two processes – one a producer of something, and one a consumer of something • The producer produces something and places the something into a buffer • The consumer consumes the something, taking the something out of the buffer • Once again two processes are sharing something – the buffer

  4. repeat repeat : while counter = 0 do no-op; produce an item in nextp (buffer is empty) : nextc := buffer[out]; while counter = n do no-op; out := out + 1 mod n; (buffer is full) counter := counter – 1; buffer[in] := nextp; : in := in + 1 mod n; consume the item in nextc counter := counter + 1; : until false; until false; PRODUCER CONSUMER

  5. Producer and Consumer • Note that both the producer and the consumer reference variable counter • In the producer: counter := counter + 1 • which in machine language is: • register1 := counter; • register1 := register1 + 1; • counter := register1;

  6. Producer and Consumer • In the consumer: counter := counter - 1 • Which in machine language is: • register2 := counter • register2 := register2 - 1 • counter := register2 • counter could be 4, 5 or 6 depending upon where the code is interrupted • For example:

  7. Producer and Consumer • counter is at 5 and producer starts: • register1 := counter; register1 = 5 • register1 := register1 + 1; register1 = 6 • the producer is swapped out and the consumer starts anew • register2 := counter; register2 = 5 • register2 := register2 - 1; register2 = 4 • the consumer is swapped out and the producer returns • counter := register1; counter = 6 • the producer is done, and the consumer returns • counter := register2; counter = 4 • counter should be 5!

  8. Critical Sections • Thus, the accessing of counter is a critical section • What happens above is called a race condition • Must ensure than ONLY one process at a time references counter • We need process synchronization

  9. Critical Section • To solve critical section problem, three things are needed • Mutual exclusion – only one at a time • Progress – process moves through event • Bounded waiting – everyone has to get a turn

  10. Semaphores • A simple, clean way to support process synchronization • Supported by many OSs and even some processors • A semaphore is nothing more than an integer. • But, this integer can only be accessed (except for initialization) through two atomic operations:

  11. Wait and Signal Semaphores • wait(s): if s = 0 then go to sleep; s := s – 1; • signal(s): s := s + 1; wakeup sleeping process;

  12. Semaphores • To use semaphores, create a structure that looks like the following: repeat : wait(semaphore) : (critical section) : signal(semaphore) : until false;

  13. Database Example • Database update in which you want to lock out all but one concurrent user procedure UpdateDatabase; var s: semaphore (:=1); begin : (prompt user for data) : wait(s); Update(record); signal(s); : end;

  14. CD-ROM Reader Example • CD-ROM reader which allows 5 concurrent users procedure ReadCD; var t: semaphore (:=5); begin : (prompt user for data) : wait(t); ReadCD(buffer); signal(t); : end;

  15. Expanded Producer/Consumer Problem • Let’s re-visit the producer/consumer problem and solve the problem using semaphores (on the whiteboard)

  16. var s: semaphore (:=1); n: semaphore (:=0); e: semaphore (:=sizeofbuffer); procedure producer; begin repeat produce item for buffer; wait (e); wait(s); append to buffer; signal(s); signal(n); forever; end; procedure consumer; begin repeat wait(n); wait(s); take from buffer; signal(s); signal(e); consume item forever; end;

  17. Deadlock • When two or more processes have conflicting needs for resources, we have deadlock • Example: P1 holds the disk and wants the tape; P2 holds the tape and wants the disk • Example: P1 has 80K or memory and wants 60K more; P2 has 70K of memory and wants 80K more; there is only 150K total of memory

  18. Deadlock • Three conditions must first exist for deadlock to occur: • Mutual exclusion • Hold and wait • No preemption • Then the following must happen: • Circular wait • How do you avoid deadlock?

  19. Deadlock Prevention • Indirect methods – stop one of the conditions 1-3 from happening (not desirable) • Direct methods – stop occurrence of condition 4 from happening • For example: Order all resources from 0 to N. If you have resource m, you can only ask for a resource > m, where Disk=5; Tape=7; Printer=12

  20. Deadlock Avoidance • Really is prevention, but not as restrictive • Avoidance allows the 3 conditions but makes dynamic decisions whether a current resource allocation request will lead to deadlock, such as Bankers’ Algorithm

  21. Deadlock Detection • Use an algorithm that can detect cycles in directed graphs • Costly, as OS must maintain graphs and check for cycles

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