Background

1
Background

• Concurrent access to shared data may result in
  data inconsistency
• Maintaining data consistency requires mechanisms
  to ensure the orderly execution of cooperating
  processes (i.e., processes share memory or use
  message passing to cooperate).
• Example
• Recall Producer Consumer problem in Chapter 4
  (could only use n-1 elements of the buffer).
• New solution adds a variable to track number of
  items in the buffer.

2
Race Conditions

• global shared int counter 0, BUFFER_SIZE 10
  
• Producer
• while (1)
• while (counter BUFFER_SIZE) // do
  nothing
• // produce an item and put in
  nextProduced
• bufferin nextProduced
• in (in 1) BUFFER_SIZE
• counter

3
Race Conditions

• Consumer
• while (1)
• while (counter 0) // do nothing
• nextConsumed bufferout
• out (out 1) BUFFER_SIZE
• counter--
• // consume the item

4

• Executed Separately work as expected.
• Execute concurrently, causes race condition
• Two or more processes (threads) access shared
  data concurrently and the outcome of the
  execution depends on the particular order in
  which the access takes place.

5

• Consider this execution interleaving of
• counter and counter--
• S0 producer execute register1 counter
• register1 5S1 producer execute register1
  register1 1
• register1 6 S2 consumer execute register2
  counter
• register2 5 S3 consumer execute register2
  register2 - 1
• register2 4 S4 producer execute counter
  register1
• counter 6 S5 consumer execute counter
  register2
• counter 4 //final value for this execution
  sequence

6
Critical Section Problem

• The part of a processes (threads) code that
  modifies shared data is termed its critical
  section.
• The critical section problem is to design a
  protocol that the processes can use to handle
  shared data without causing race conditions.
• Each process must request permission to enter
  into its critical section (entry section). The
  protocol may also require an exit section.
• The rest of the program code is termed the
  remainder section. We are not concerned with this
  code.

7
Program Model

• while(1)
• ltentry sectiongt
• ltcritical sectiongt
• ltexit sectiongt
• ltremaindergt

8
Consumer

• while (1)
• while (counter 0) // Entry Section
• nextConsumed bufferout
• out (out 1) BUFFER_SIZE
• counter-- //Critical section
• // consume the item (remainder section)

9
Producer

• while (1)
• while (counter BUFFER_SIZE) //
  entry section
• // produce an item and put in nextProduced
• bufferin nextProduced
• in (in 1) BUFFER_SIZE
• counter //Critical section

10
Correct Solutions to Critical-Section Problem

1. Mutual Exclusion - If process Pi is executing in
   its critical section, then no other processes can
   be executing in their critical sections.
2. No assumptions may be made about speed or number
   of CPUs.
3. No process executing in its remainder section can
   block another process from entering into its
   critical section.
4. No process should have to wait forever to enter
   its critical section.

11
Attempt 1

• shared int turn 0
• while (1) while(1)
• while (turn !0)
  while(turn !1) //entry section
• critical section critical
  section
• turn 1 turn
  0 //exit section
• ltremaindergt ltremaindergt

12

• shared int turn 0
• P0 P1
• while (1) while(1)
• while (turn !0) while(turn !1)
• critical_region() critical_region()
• turn 1 turn 0
• ltremaindergt ltremaindergt
• P0 accesses CR sets turn to 1.
• P1 accesses CR sets turn to 0. In remainder
  section reads binary equivalent of War and Peace.
• P0 enters CR sets turn to 1.
• P0 attempts to enter CR but cannot (until after
  book report due).

13
Does it meet requirements?

• Mutual Exclusion?
• Violates rule three.

14
Attempt 2 Using flag to indicate interest

• global shared int flag2 0,0
• while(1) while(1)
• flag0 1 flag1 1
• while(flag1 1) while(flag0
  1 ) //entry section
• ltcritical sectiongt ltcritical
  sectiongt
• flag0 0 flag1 0 //exit
  section
• ltremainder ltremaindergt

15
Attempt 2 Using flag to indicate interest

• global shared int flag2 0,0
• while(1) while(1)
• flag0 1 flag1 1
• while(flag1 1) while(flag0
  1 ) //busy wait
• ltcritical sectiongt ltcritical
  sectiongt
• flag0 0 flag1 0
• ltremainder ltremaindergt
• P0 sets flag0 1 gets preempted.
• P1 sets flag1 1 gets preempted.

16
Problem?

• Mutual Exclusion?
• Progress?
• Bounded Wait?

17
Attempt 2.1 Using flag to indicate interest

• global shared int flag2 0,0
• while(1) while(1)
• while(flag1 1 ) while(flag0 1)
  
• flag0 1 flag1 1
• ltcritical sectiongt ltcritical sectiongt
• flag0 0 flag1 0
• ltremainder ltremaindergt

18
Attempt 2.1 Using flag to indicate interest

• global shared int flag2 0,0
• while(1) while(1)
• while(flag1 1 ) while(flag0 1)
  
• flag0 1 flag1 1
• ltcritical sectiongt ltcritical
  sectiongt
• flag0 0 flag1 0
• ltremainder ltremaindergt
• P0 reads flag1 0 Gets preempted.
• P1 reads flag0 0 Gets preempted.
• P0 sets flag0 1 and enters critical_region.
• P1 sets flag1 1 and enters critical_region.

19
Problem?

• Mutual Exclusion?
• Progress?
• Bounded Wait?

20
Correct for Two Processes

• while(1) while(1)
• flag0 1 flag1 1
• turn 1 turn 0
• while(flag1turn1) while(flag0turn0)
  
• ltcritical sectiongt ltcritical sectiongt
• flag0 0 flag1 0

21
Correct for Two Processes

• while(1) while(1)
• flag0 1 flag1 1
• turn 1 turn 0
• while(flag1turn1) //entry
  while(flag0turn0) //entry section
• ltcritical sectiongt ltcritical sectiongt
• flag0 0 // exit
  flag1 0 // exit section.
• do_boring_things() //Remainder do_interesting_th
  ings() // Remainder section
• P0 sets flag to 1 and turn to 1. Gets preempted.
• e.g., reg1 1 // ready to store value of 1 to
  turn.
• P1 sets flag to 1 and turn to 0. Gets preempted.
• e.g., reg1 0 //ready to store value of 0 to
  turn.
• What happens if P0 stores value of turn 1 and
  falls into while loop??
• What happens if P1 then stores value of turn 0 ?

22
Goal of Critical Sections (Regions)

• Mutual exclusion using critical regions

23
Hardware Solutions

• Many systems provide hardware support to critical
  section code.
• Simple Disable interrupts during critical
  section.
• Would this provide mutual exclusion??
• Problems

24
Hardware Solutions

• Many systems provide hardware support to critical
  section code.
• Simple Disable interrupts during critical
  section.
• Would this provide mutual exclusion??
• Problems
• Do not want to do this at user level.
• Does not scale to multiprocessor system.
• OK for OS to do so for short periods of time, not
  users.

25

• Most machines provide special atomic hardware
  instructions
• Atomic non-interruptable
• Either test memory word and set value
• Or swap contents of two memory words

26
Test and Set Instruction (Busy Waiting)

• This is a software description of the
  TestandSet hardware instruction.
• int lock 0 // global integer var initially set
  to 0.
• int TestAndSet (int lock)
• return_val lock
• lock 1
• return (return_val)

27
Usage

• shared int lock 0
• while(TestAndSet(lock)) //Entry code
• ltcritical sectiongt
• lock 0 //Required for correctness
• Do_something_else() //Remainder Section

28
Solutions that minimize busy waiting

• Producer-consumer problem (Revisited)

29
Deadlock?

30
Deadlock?

Producer If (count N) //assume condition
true. Preempted Consumer N-- If (count
N-1) wakeup(producer) Producer Sleep()
//taking the dirt nap
31
Semaphore

• Synchronization tool that does not require busy
  waiting (spin lock)
• Semaphore S integer variable
• Two standard operations modify S wait() and
  signal()
• Originally called P() and V(), called up and down
  in text.
• Can only be accessed via two indivisible (atomic)
  operations

32

• typedef struct
• int count // for mutual exclusion between 2
  processes initialize to 1.
• PCB Sem_Q
• Semaphore

33
Use of Semaphores

• struct Semaphore exclusion 1
• while (1)
• wait(exclusion) //Note Not a busy wait!
• ltcritical sectiongt
• signal(exclusion)
• ltremainder sectiongt

34

• Semaphore Sem 1 //Initialize counter
  variable
• wait(Sem)
• Sem.count--
• if (Sem.count lt 0)
• Place process on SemQ
• Block process

35

• Signal(Sem)
• Sem.count
• if (Sem.count lt 0 ) //someone is sleeping on
  S.Q
• Remove process from SemQ
• Place on ready queue

36
Example

• Semaphore Sem_Baby 1
• Process Operation Sem.count Result
• PO wait 0 Passes through
• P1 wait -1 Blocked
• P 0 signal 0 removes P1 from SemQ and it
  passes through the semaphore.

semq
37
Ensuring Mutual Exclusion of Sem Code.

• shared int lock 0
• Semaphore Sem 1 //Initialize count
• wait(Sem)
• while(TestAndSet(lock)) //busy loop. Why
  not a problem??
• Sem.count--
• if (Sem.count lt 0)
• Place process on SemQ
• Block process and set lock to 0
• else
• lock 0

38

• Signal(Sem)
• while (TestAndSet(lock)) //busy wait
• Sem.count
• if (Sem.count lt 0 )
• Remove process from SemQ
• Place on ready queue
• lock 0

39
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

40

• Semaphore Q, S 1,1
• P1 P2
• wait(S) wait(Q)
• wait(Q) wait(S)
• ltCSgt ltCSgt
• signal S signal(Q)
• signal Q signal(S)

41

• Semaphore Q, S 1,1
• P0 wait(S) // S.count 0
• P1 wait(Q) //Q.count 0
• P0 wait(Q) // Q.count -1 recall if
  Q.count lt 0 put on Q.semqueue
• P1 wait(S) // S.count -1 recall if
  S.count lt 0 put on S.semqueue

42
Semaphores for Synchronization

• Want Statement S1 in P0 to execute before
  statement S2 in P1.
• Semaphore Syncher 0 // What does this do?
• P1
• wait(Syncher) // Syncher.count -1
• S2

43
Semaphores for Synchronization

• Want Statement S1 in P0 to execute before
  statement S2 in P1.
• Semaphore Syncher 0 // What does this do?
• P0 P1
• S1 wait(Syncher)
• signal (Syncher) S2

44
Semaphores for Synchronization

• Case 1 P1 executes wait first
• P1
• wait(Syncher) //Syncher.count -1, (put on
  Syncher.queue)
• P0
• S1
• signal(Syncher) //
  Syncher.count 0 take P1 off
  //Syncher.queue and // place on run
  queue.
• // Synchronization accomplished!!

45
Semaphores for Synchronization

• Case 2 P0 executes wait first
• P0
• S1
• signal (Syncher) //Syncher.count 1, (put on
  Syncher.queue)
• P1
• wait (Syncher) //Syncher.count 0. P1
  does not have to wait and
• // executes statement S2. Synchronization
  // accomplished!!

46
Semaphores for Synchronization

• Want Statement S1 in P0 to execute before
  statement S2 in P1 before S3 in P2.
• Semaphore Syncher 0
• Semaphore Synch1 0
• P0 P1 P2
• S1 wait(Syncher) wait(Synch1)
• signal (Syncher) S2 S3
• signal(Synch1)

47
global shared int counter 0, BUFFER_SIZE 10
Producer while (1) while
(counter BUFFER_SIZE) // do nothing
// produce an item and put in
nextProduced bufferin nextProduced in
(in 1) BUFFER_SIZE counter
48
Consumer while (1) while
(counter 0) // do nothing nextConsumed
bufferout out (out 1)
BUFFER_SIZE counter-- // consume the item

49

• How can this busy waiting be (largely) avoided?

50

• How can this busy waiting be (largely) avoided?
• Use semaphores.
• How many semaphores?

51

• How can this busy waiting be (largely) avoided?
• Use semaphores.
• How many semaphores?
• 2 One for producer and one for consumer (call
  them full and empty)
• To what should they be initialized
• That is, how many times can a process fall
  through the semaphore before it becomes a
  problem?

52
Struct semaphore full ? Struct semaphore empty
?
53
struct semaphore full 10 struct semaphore
empty 0 Think about how you would use these
semaphores in the bounded buffer problem.
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