The document discusses mechanisms for ensuring orderly execution of concurrent processes that share a logical address space. It describes concepts like concurrency, atomic operations, race conditions, and the critical section problem. It presents solutions to the critical section problem like mutual exclusion, progress, and bounded waiting. Algorithms like Peterson's and Dekker's are discussed for solving mutual exclusion between two processes. The use of semaphores as a synchronization tool that avoids busy waiting is also covered.

1. The OS as a gatekeeper

2. Objective
 To discuss various mechanisms to ensure
the orderly execution of cooperating
processes that share a logical address space
so that data is consistently maintained.

3. Concept of concurrency
 Existence of several simultaneous processes.
May be observed in situations like:
 Multi-user OS where different users execute
programs concurrently
 Single-user OS where input/output devices go
simultaneously with program execution
 Some processors whose activities are concurrent,
example, fetch next instruction is done
simultaneously with the execution of an instruction.

4. bounded buffer
PRODUCER CONSUMER
while (true) { while (true) {
/* produce an item and put in while (count == 0)
nextProduced */ do nothing
while (count == BUFFER_SIZE) nextConsumed = buffer [out];
do nothing;
out = (out + 1) % BUFFER_SIZE;
buffer [in] = nextProduced;
count--;
in = (in + 1) % BUFFER_SIZE;
/* consume the item in
count++; nextConsumed */
} }

5. Atomic Operation
 The statements count ++ and count -- must
be performed atomically.
 Atomic operation means an operation that
completes in its entirety without interruption.
 May not function correctly when executed
concurrently.

6. Atomic Operation
 If both the producer and consumer attempt to
update the buffer concurrently, the assembly
language may get interleaved
 Interleaving depends upon how the producer
and consumer processes are scheduled.

7. Atomic Operation
 count++ could be implemented as
MOV AX, COUNT register1 = count
INC AX register1 = register1 + 1
MOV COUNT, AX count = register1
 count-- could be implemented as
MOV BX, COUNT register2 = count
DEC BX register2 = register2 - 1
MOV COUNT, BX count = register2

8. Now consider this interleaving…
 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}

9. Race Condition
 Situation where several processes access and
manipulate shared data concurrently. The final
value of the shared data depends upon which
process finishes last.
 To prevent race conditions, concurrent
processes must be synchronized.

10. The Critical Section (CS) Problem
 Consider n processes all competing to use
some shared data.
 Each process has a code segment, called
critical section in which shared data is
accessed.
 Ensure that when one process is executing in
its critical section, no other process is allowed
to execute in its critical condition.

11. The Critical Section (CS) Problem
 Execution of critical sections by the processes
is mutually exclusive in time.
 Design protocol:
 Each process must request permission to enter its
CS (entry section)
 The CS is followed by an exit section
 The remaining code is in the remainder section.

12. Solution to Critical-Section
Problem
1.Mutual Exclusion
2.Progress
3.Bounded Waiting

13. Solution 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
CS
P1 P2 P
n

14. Solution to Critical-Section
Problem
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.
CS P
P 2
1
P

15. Solution to Critical-Section
Problem
1. Bounded Waiting
- No process should wait arbitrarily long to enter
its CS.
CS P
P 2
1
P
n

16. Note:
 No assumptions are made about relative
process speeds or number of CPU’s
 A process in one CS should not block others
in entering a different CS.

17. Sample Scenario – Milk Issue
Time Person A Person B
3:00 Look in fridge, Out of milk -------------
3:05 Leave for store -------------
3:10 Arrive at store Look in fridge, Out of Milk
3:15 Buy Milk Leave for store
3:20 Leave store Arrive at store
3:25 Arrive home, put milk away Buy milk
3:30 ------------- Leave Store
3:35 ------------- Arrive home, oops

18. Observation
 Someone gets milk, but not EVERYONE (too
much milk!)
 This shows that when cooperating processes are
not synchronized, they may face unexpected
“timing errors”
 Mutual Exclusion - ensures that only one
process (or person) is doing a thing at one time,
thus avoiding data inconsistency. All others
should be prevented from modifying shared data
until the current process finishes.
E.g. only one person buys milk at a time.

19. Solution 1 – Attempt to computerize Milk
Buying
PROCESS A AND B
This solution works for some
If (NOmilk) people because the first
{ three lines are performed
if (NOnote) ATOMICALLY but does
{ not work otherwise.
Leave note;
Buy milk; What happens when the 2
Remove note; processes execute
} concurrently?
}

20. Solution 2 – Attempt to use two notes
PROCESS A PROCESS B
Leave (NOTEa); Leave (NOTEb);
If (NO_NOTEb) If (NO_NOTEa)
{ {
if (NOmilk) Buy Milk; if (NOmilk) Buy Milk;
} }
Remove (NOTEa); Remove (NOTEb);
WHAT CAN YOU SAY ABOUT THIS SOLUTION?

21. Solution 3:
PROCESS A PROCESS B
Leave (NOTEa); Leave (NOTEb)
If (NOnoteB) While (NOTEa) do
{ nothing;
if (NOmilk) Buy Milk;
} if (NOmilk) Buy Milk;
Remove (NOTEa); Remove (NOTEb);
In case of tie, who will buy first?

22. Critique for Solution 3
 Process B will always be the first to buy the
milk in case of a tie.
 Process B consumes CPU cycles while
waiting.
 More complicated if extended to many
processes.

23. Algorithm Template
do { ENTRY SECTION – Each
process must request
ENTRY_SECTION
permission to enter its CS
CRITICAL_SECTION
EXIT_SECTION
EXIT SECTION – Process
leaves the CS thereby
informing other processes
Remainder_Section that the CS is free for some
other process to enter
} while (TRUE)

24. Two Process Solution
Algo 1: Strict Alteration Shared var Turn <either a or b>
PROCESS A PROCESS B
repeat repeat
While (turn==B) do While (turn==A) do
nothing; nothing;
<critical section A>
<critical section B>
Turn = B;
Turn = A;
until false
until false
Which of the following requirement is violated?
Mutual Exclusion Progress Bounded Waiting

25. Algorithm 2
Shared Var pAinside, pBinside; FALSE
PROCESS A PROCESS B
While (TRUE) { While (TRUE) {
While (pBinside) do While (pAinside) do
nothing; nothing;
pAinside = True; pBinside = True;
<critical section A> <critical section B>
pAinside = False; pBinside = False;
} }
Which of the following requirement is violated?
Mutual Exclusion Progress Bounded Waiting

26. Algorithm 3
Shared Var pAtrying, pBtrying : FALSE
PROCESS A PROCESS B
While (TRUE) { While (TRUE) {
pAtrying = True; pBtrying = True;
While (pBtrying) do While (pAtrying) do
nothing; nothing;
<critical section A> <critical section B>
pAtrying = False; pBtrying = False;
} }
Which of the following requirement is violated?
Mutual Exclusion Progress Bounded Waiting

27. Dekker’s Algorithm - Elegant solution to
mutual exclusion problem
Shared var pAtrying pBtrying = False
PROCESS A PROCESS B
While (TRUE) { While (TRUE) {
pAtrying = True; pBtrying = True;
While (pBtrying) While (pAtrying)
if (turn==B) { if (turn==A) {
pAtrying = False; pBtrying = False;
while (turn==B) do nothing; while (turn==A) do nothing;
pAtrying = True pBtrying = True
} }
<critical section> <critical section>
turn==B; turn==A;
pAtrying = False; pBtrying = False;
} }

28. Peterson’s Algorithm: Much simpler than
Dekker’s Algorithm
Shared var pAtrying pBtrying = False
PROCESS A PROCESS B
While (TRUE) While (TRUE)
{ {
pAtrying = True; pBtrying = True;
turn= B; turn= A;
While (pBtrying && turn ==B) While (pAtrying && turn ==A)
do nothing; do nothing;
<critical section A> <critical section A>
pAtrying = False; pBtrying = False;
} }

29. Note
 Both Dekker’s and Peterson’s algorithms
are correct but they only work for 2 processes,
similar to the last solution of the “too-much-
milk” problem.

30. Multiple Process Synchronization
Solution
 Hardware Support – special instructions
Many CPU’s today provide hardware
instructions to read, modify and store a word
atomically. Most common instructions with
this capability are:
 TAS – (Test and Set ) – Motorolla 68k
 CAS – (Compare and Swap) – IBM 370 and
Motorola 68k
 XCHG – (exchange) –x86

31. Mutual Exclusion with Test-&-Set
(TAS)
Shared data:
boolean TestAndSet boolean lock = false;
(boolean &target)
Process Pi
{ do {
boolean rv = target; while (TestAndSet(lock)) do
target = true; nothing( );
return rv;
} <critical section>
lock = false;
remainder section
}

32. Mutual Exclusion with SWAP and
XCHG
Atomically swap two Shared data (initialized to false):
variables.
boolean lock;
void Swap(boolean &a, boolean waiting[n];
boolean &b) Process Pi
do {
{ key = true;
boolean temp = a; while (key == true)
a = b; Swap(lock,key);
b = temp; critical section
} lock = false;
remainder section
}

33. Multiple Process Synchronization
Solution
 The basic idea is to be able to read the
contents of a variable (memory location) and
set it to something else all in one execution
cycle hence not interruptible.
 The use of these special instructions makes
the programming task easier and improves
efficiency.

34. Semaphores
 Synchronization tool that does not require
busy waiting
 Semaphore S – integer variable
 Two standard operations modify S: wait()
and signal()

35. A Semaphore…
 Essentially an integer S (initially S>0) and a
queue of processes, initially empty
 Synchronization variable (guard) that takes on
non-negative integer values with only two
atomic operations

36. Two Atomic Operations
Wait (S) PROBEREN:
 While s ≤ 0 do no_op; Probe/Test
 S--
Signal (S) VERHOGEN:
 S++ release
“make higher”

37. Solution to too much milk using
semaphore
Process A & B
Semaphore OkToBuyMilk = 1;
Wait (OkToBuyMilk);
If (NoMilk)
{
Buy Milk;
}
Signal (OkToBuyMilk)

38. Critical Section of n Processes
 Shared data: semaphore mutex (initially 1)
do{
wait (mutex);
< critical section >
signal (mutex);
remainder section
}while (true)

39. Critique: CS of n Processes
 The above algorithm requires busy-wait.
 To overcome busy waiting, the process will
block itself (placed in the waiting - queue
associated with the semaphore)
Define a semaphore as a record
typedef struct {
int value;
struct process *L;
} semaphore;

40. Example #1:
 There are 6 Synchronize the execution of the
processes waiting two CPUs:
in the ready queue CPU1: CPU2:
(P1,P2…P6) P1; P4;
 Let us assume that wait (s1);
 P5 can only be
P2; P5;
executed if P1 and
P2 have already signal (s1);
completed their
wait (s2);
tasks
 P3 will only be P3; P6;
executed if P6 is signal (s2);
finished

41. Exercise #2:
 Consider the following code which is being
executed by 3 CPU’s. This program computes
the value of X using the Quadratic formula.
 Synchronize the execution of the CPU’s by
using semaphore.

42. X = -B± √(B2 – 4 A C)
Exercise #2 Codes: 2A
CPU1: Readln (C); CPU2:
Readln (A); FourAC = 4*A*C;
TwiceA = A + A; If (FourAC > BSquare); Imaginary
Readln (B); SquareRoot = sqrt(abs(Bsquare – FourAC)
Bsquare = B * B; Root1 = (MinusB + SquareRoot)/TwiceA;
MinusB = -1 * B; Root2 = (MinusB - SquareRoot)/TwiceA;
CPU3:
If (Imaginary) writeln (“No real roots”)
Else writeln (“the roots are”, root1, root2”)

43. Solution to Exercise 2 X = -B± √(B2 – 4 A C)
2A
CPU1: Readln (C); CPU2:
Readln (A); Wait(S1);
Signal(S1); FourAC = 4*A*C;
TwiceA = A + A; Wait(S2);
Readln (B); If (FourAC > BSquare) then Imaginary =TRUE;
Bsquare = B * B; SquareRoot = sqrt(abs(Bsquare – FourAC)
Signal (S2); Wait(S3);
MinusB = -1 * B; Root1 = (MinusB + SquareRoot)/TwiceA;
Signal (S3); Root2 = (MinusB - SquareRoot)/TwiceA;
Signal (S4);
Wait (S4); CPU3
If (Imaginary) writeln (“No real roots”)
Else writeln (“the roots are”, root1, root2”)

44. Possible uses of semaphores
 Mutual Exclusion
– Initializes the semaphore to one (1)
 Example:
Wait (Mutex);
< Critical section; >
Signal (Mutex);

45. Possible uses of semaphores
 Synchronization of cooperating processes
(signalling)
– initializes the semaphore to zero.
 Example
Pi: Pj
… …
… …
A wait (flag)
Signal (flag) B

46. Possible uses of semaphore
 Managing multiple instances of a resource
- Initializes the semaphore to the number of
instances.

47. Types of semaphores
 Binary
– semaphore with an integer value of 0 or 1
 Counting
– semaphore with an integer value ranging
between 0 and an arbitrarily large number.
- Initial value represents the number of units of
the critical resources that are available.
- Also known as general semaphore.

48. Classical Problems of
Synchronization

49. Bounded-Buffer Problem:
 Two processes share a common, fixed size
buffer. One of them, the producer puts
information into the buffer, and the other, the
consumer, takes out information from the
buffer. When the producer wants to put a new
item in the buffer but it is already full then allow
the producer to sleep to be awakened when the
consumer has removed one or more items.
Similarly to the consumer, it goes to sleep if the
buffer is empty until the producer puts
something in the buffer.

50. shared data
semaphore full = 0, empty = n, mutex
= 1;
Producer: Consumer:
do { do {
… wait ( full )
produce an item in nextp wait ( mutex );
… …
remove item from buffer to
wait ( empty ); nextc
wait ( mutex ); …
… signal ( mutex );
add nextp to buffer signal ( empty );
… …
signal ( mutex ); consume item in nextc
signal ( full ); …
} while (TRUE); } while (TRUE);

51. Readers-Writers Problem
 This problem models access to a database with
many competing processes wishing to read and
write into it (imagine an airline reservation
system). It is possible to have many processes
reading the database at the same time (for
query purposes). But if one process is writing
(modifying), no other process shall access to
the database, not even readers. With this
condition in mind, how will you program the
reader and writers?

52. Illustration: Readers and Writers
Writers Area
wrt
Readers
Area Critical section
wrt

53. Readers-Writers: Solution
Types
 Algorithm 1:
Readers have higher priority than writers
 Algorithm 2:
Writers have higher priority than readers

54. Algorithm #1:
 Readers have higher priority than writers
If a writer appears while several readers are in
the database, the writer must wait. If new
readers keep appearing so that there is always
at least one reader in the database, the writer
must keep waiting until no more readers are
interested in the database.

55. Solution to Readers/Writers:
Algorithm #1
shared data semaphore mutex = 1, wrt = 1;
int readcount = 0; Readers:
Writers: do {
wait ( mutex );
do { readcount + +;
wait(wrt); if ( readcount == 1 ) wait ( wrt );
… signal ( mutex );
…
writing is performed reading is performed
… …
signal(wrt); wait ( mutex );
readcount - -;
} while (TRUE) if (readcount == 0 ) signal ( wrt );
signal ( mutex ):
} while (TRUE)

56. Algorithm 2:
 How about if we prioritize Writers over
readers? Reprogram the above codes.
 Seatwork….
Writers have higher priority than readers

57. Dining-Philosophers
Problem
 Five philosophers spend their lives thinking and
eating.
 Philosophers share a common circular table
surrounded by five chairs, each belonging to
one philosopher.
 At the center of the table, there is a bowl of rice,
and the table is laid with five single chopsticks.

58. Dining-Philosophers cont…
 When a philosopher thinks, he does not interact
with his colleagues.
 When a philosopher gets hungry, he tries to
pick up two chopsticks that are closest to him.
 Philosopher could pick up only one chopstick at
a time.
 Obviously, he cannot pick up a chopstick that is
already in the hand of his neighbor.
 When a hungry philosopher has his both
chopsticks, he eats without releasing them.
 When he is finished, he puts down the
chopsticks and starts thinking again.

59. Dining-Philosophers
diagram
0 4
1 3
2

60. Dining Philosophers: Initial Solution
Shared data semaphore chopstick[5]; {Initially all values are 1}
Philosopher i:
do {
wait ( chopstick [ i ] ) { his left chopstick}
wait ( chopstick [ ( i + 1) % 5 ] ) { his right chopstick}
…
eat
…
signal ( chopstick [ i ] ); 0 4
signal ( chopstick [ ( i + 1) % 5 ] ); 0
… 1 4
think
… 1 2 3 3
} while (TRUE);
2
Question: What is wrong with this solution?

61. Solution for Dining Philosophers…
Shared data:
state [ 5 ] of (thinking, hungry, eating);
// init to all thinking//
self [ 5 ]: semaphore; { init to all 0 }
mutex: semaphore; { init to 1 }
left = (i + 4) % 5; { left neighbor }
right = (i + 1) % 5; { right neighbor }

62. Solution for Dining Philosophers…
Philosopher ( i ) {
void test ( int i ) {
if ( (state [ left ] ! = eating) &&
state [ i ] = hungry; (state [ i ] == hungry) &&
wait ( mutex ); (state [ right ] != eating)) {
test [ i ]; state [ i ] = eating;
signal (mutex); signal (self [ i ] );
if (state [ i ] != eating ) }
wait ( self [ i ] ); }
void putdown ( int i ) {
… eating …
state [ i ] = thinking;
wait ( mutex ); wait ( self [ i ] );
putdown [ i ];
signal ( mutex ); // test left and right neighbors
} test ( left );
test ( ( right );
}

63. Sleeping – Barbers
Problem:
 A barber shop consists of a waiting room with
5 chairs and the barber room containing the
barber chair.
 If no customer to serve, the barber goes to
sleep to be awakened by the customer.
 If a customer enters the barber shop but all
chairs are occupied, the customer leaves the
shop; otherwise he sits in one of the free
chairs.
 Write the program to coordinate the actions of
the barber and the customer.

64. Sleeping – Barbers Solution
Shared data: customer = 0, barber = 0, mutex = 1;
int waiting = 0; CHAIR = 5;
Barber: Customer:
while (TRUE) { while ( TRUE ) {
wait ( customer ); wait ( mutex );
wait ( mutex ); if ( waiting < CHAIR) {
waiting - -; waiting + +;
signal ( barber ); signal ( customer );
signal ( mutex ); signal ( mutex );
.... wait ( barber );
cut hair(); …..
} get haircut();
} else signal ( mutex);
}

65. Exercise #1:
 Discuss the correctness of the Use_resource( )
function below for two process solution:
int free = TRUE;
use_resource ( )
{
while ( ! free ); /* wait */
free = TRUE;
/* use the resource – CRITICAL SECTION */
free = FALSE;
}

66. Exercise #2:
 Discuss the correctness of the Use_resource( )
function below for two process solution:
char buffer [ n ];
int full, empty; { initially full = empty = TRUE }
empty_buffer( ) fill_buffer( )
{ {
while ( ! full ) do nothing( ); while ( ! empty ) do nothing( );
full = TRUE; empty = TRUE;
/* critical section */ /* critical section */
empty ( buffer ); fill (buffer );
empty = FALSE; full = FALSE;
} }

67. Exercise #3: Crossing Baboons
Problem
 There are two groups of baboons, one from the
east and the other from the west.
 A one-way bridge is attached between the east
cave where the east baboons live and west
cave where the west baboons reside.
 Whoever group reaches the foot of the bridge
first, will cross the bridge first signaling the
other group not to cross.
 With this condition in mind, how will you
synchronize the action of the two groups of
baboons to cross the bridge?

68. Exercise #4:Cigarette-Smoker
Problem
 There are three smoker processes and one agent
process.
 Each smoker continuously rolls a cigarette and then
smokes it. But to roll and smoke a cigarette, the
smoker needs three ingredients: tobacco, paper, and
matches.
 One of the smoker processes has paper, another has
tobacco, and the third has matches.
 Agent has an infinite supply of all three materials. He
places two of the ingredients on the table.
 The smoker who has the remaining ingredient then
makes and smokes a cigarette, signaling the agent on
completion.
 The agent then puts out another two of the three
ingredients and the cycle repeats. Write a program to
synchronize the agent and the smokers.

69. End of Presentation