Lecture16-17.ppt

 A cooperating process is one that can affect or be
affected by other processes executing in the
system.
 Concurrent access to shared data may result in
data inconsistency.

 Previously we studied producer-consumer
problem to understand cooperating processes.
 Our solution allowed at most BUFFER.SIZE - 1
items in the buffer at the same time (refer to old
slides).
 To remedy this deficiency we can add an integer
variable counter, initialized to 0.
 Counter is incremented every time we add a new
item to the buffer and is decremented every time
we remove one item from the buffer.

 producer-consumer problem
Shared Data:
 #define BUFFER_SIZE 10
 int buffer [BUFFER_SIZE] ;
 int in = 0 ,
 int out = 0 ;
 int count=0;

while (true)
/* produce an item and put in nextProduced
while (count == BUFFER_SIZE)
; // do nothing
buffer [in] = nextProduced;
in = (in + 1) % BUFFER_SIZE;
count++;
}

while (true)
{
while (count == 0)
; // do nothing
nextConsumed = buffer[out];
out = (out + 1) % BUFFER_SIZE;
count--;
/*consume the item in nextConsumed
}

 Although both producer and consumer routines are
correct separately, they may not function correctly
when executed concurrently.
 For example the variable counter is currently 5 and
that the producer and consumer processes execute
the statements "counter++" and "counter—"
concurrently.
 Following the execution of these two statements, the
value of the variable counter may be 4, 5, or 6!
.(correct result, though, is counter == 5)

 In machine language count++ could be implemented as
MOV R1, COUNTER (R1=5)
 INC R1 (R1=6)
 count-- could be implemented as
MOV R2, COUNTER (R2=5)
 DEC R2 (R2=4)
 Consider this execution interleaving with “count = 5” initially:
producer execute MOV COUNTER, R1{count = 6 }
consumer execute MOV COUNTER, R2{count = 4}
What will be the value of Count=?

 Notice that we have incorrect state "count == 4",
indicating that four buffers are full, when, in fact, five
buffers are full.
 This happened because we allowed both processes to
manipulate the variable count concurrently.
 When several processes access and manipulate the
same data concurrently and the outcome of the
execution depends on the particular order in which
the access takes place, is called a race condition.
 To guard against the race condition, we need to
ensure that only one process at a time can be
manipulating the variable counter, hence the need for
processes synchronization.

 In concurrent programming, a critical section is a
piece of code that accesses a shared resource (data
structure or device) that must not be concurrently
accessed by more than one thread of execution.
 The critical-section problem is to design a protocol
that the processes can use to cooperate.
 Each process must request permission to enter its
critical section. The section of code implementing this
request is the entry section. The critical section may
be followed by an exit section. The remaining code is
the remainder section.

A solution to the critical-section problem must satisfy the
following three requirements:
Mutual Exclusion - If process Pi is executing in its critical
section, then no other processes can be executing in their
critical sections.
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.

Bounded Waiting - There exists a bound, or limit, 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.

 A software-based solution to the critical-section problem
is known as Peterson's solution.
 It is restricted to two processes that alternate
execution between their CS and remainder sections.
 Peterson's solution may not work correctly on modern
architectures because of machine-language instructions
(like example of variable count in producer-consumer in
previous slides).
 We present the solution because it provides a good
algorithmic description of solving the critical-section
problem and addresses the requirements of mutual
exclusion, progress, and bounded waiting requirements.

 Processes are numbered P0 and P1.
 Peterson's solution requires two data items to be
shared between the two processes:
◦ 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!

 To enter the critical section, process Pi first sets
flag[i] to be true and then sets turn to the value j,
thereby asserting that if the other process wishes
to enter the critical section, it can do so.
 If both processes try to enter at the same time,
turn will be set to both i and j at roughly the same
time.
 Only one of these assignments will last; the other
will occur but will be overwritten immediately.

do {
flag[i] = TRUE;
turn = j;
while ( flag[j] && turn == j);
CRITICAL SECTION
flag[i] = FALSE;
REMAINDER SECTION
} while (TRUE);

do {
flag[0] = TRUE;
turn = 1;
while ( flag[1] && turn == 1);
CRITICAL SECTION
flag[0] = FALSE;
REMAINDER SECTION
} while (TRUE);
do {
flag[1] = TRUE;
turn = 0;
while ( flag[0] && turn == 0);
CRITICAL SECTION
flag[1] = FALSE;
REMAINDER SECTION
} while (TRUE);

 We now prove that this solution is correct. We need
to show that:
 1. Mutual exclusion is preserved?
 2. The progress requirement is satisfied?
 3. The bounded-waiting requirement is met?

Lecture16-17.ppt

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