A. Frank - P. Weisberg Operating Systems The Critical-Section Problem

A. Frank - P. Weisberg

Operating Systems

The Critical-Section Problem

2 A. Frank - P. Weisberg

Cooperating Processes

• Introduction to Cooperating Processes

• Producer/Consumer Problem

• The Critical-Section Problem

• Synchronization Hardware

• Semaphores


The Critical-Section Problem

• n processes competing to use some shared data.

• No assumptions may be made about speeds or the number of CPUs.

• Each process has a code segment, called Critical Section (CS), in which the shared data is accessed.

• Problem – ensure that when one process is executing in its CS, no other process is allowed to execute in its CS.


CS Problem Dynamics (1)

• When a process executes code that manipulates shared data (or resource), we say that the process is in it’s Critical Section (for that shared data).

• The execution of critical sections must be mutually exclusive: at any time, only one process is allowed to execute in its critical section (even with multiple processors).

• So each process must first request permission to enter its critical section.


CS Problem Dynamics (2)

• The section of code implementing this request is called the Entry Section (ES).

• The critical section (CS) might be followed by a Leave/Exit Section (LS).

• The remaining code is the Remainder Section (RS).

• The critical section problem is to design a protocol that the processes can use so that their action will not depend on the order in which their execution is interleaved (possibly on many processors).


General structure of process Pi (other is Pj)

do {entry section

critical sectionleave section

remainder section} while (TRUE);

• Processes may share some common variables to synchronize their actions.


Solution to Critical-Section Problem

• There are 3 requirements that must stand for a correct solution:

1. Mutual Exclusion

2. Progress

3. Bounded Waiting

• We can check on all three requirements in each proposed solution, even though the non-existence of each one of them is enough for an incorrect solution.


Solution to CS Problem – Mutual Exclusion

1. Mutual Exclusion – If process Pi is executing in its critical section, then no other processes can be executing in their critical sections.

• Implications: Critical sections better be focused and short. Better not get into an infinite loop in there. If a process somehow halts/waits in its critical

section, it must not interfere with other processes.


Solution to CS Problem – Progress

2. 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 process that will enter the critical section next cannot be postponed indefinitely:

• If only one process wants to enter, it should be able to.

• If two or more want to enter, one of them should succeed.


Solution to CS Problem – Bounded Waiting

3. Bounded Waiting – A bound must exist 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.

• Assume that each process executes at a nonzero speed.

• No assumption concerning relative speed of the n processes.


Types of solutions to CS problem

• Software solutions –– algorithms who’s correctness does not rely on any

other assumptions.

• Hardware solutions –– rely on some special machine instructions.

• Operating System solutions –– provide some functions and data structures to the

programmer through system/library calls.

• Programming Language solutions – – Linguistic constructs provided as part of a language.


Software Solutions

• We consider first the case of 2 processes:– Algorithm 1 and 2/3 are incorrect.– Algorithm 4 is correct (Peterson’s algorithm).

• Then we generalize to n processes:– The Bakery algorithm.

• Initial notation:– Only 2 processes, P0 and P1

– When usually just presenting process Pi (Larry, I, i), Pj (Jim, J, j) always denotes other process (i != j).


Initial Attempts to Solve Problem

• General structure of process Pi (other is Pj) – do {

entry sectioncritical section

leave sectionremainder section

} while (TRUE);• Processes may share some common variables

to synchronize their actions.


Algorithm 1 – Larry/Jim version

• Shared variables: – string turn; initially turn = “Larry” or “Jim” (no matter)– turn = “Larry” Larry can enter its critical section

• Process Larrydo {

while (turn != “Larry”);critical section

turn = “Jim”;remainder section

} while (TRUE);• Jim’s version is similar but “Larry”/“Jim” reversed.


Algorithm 1 – Pi/Pj version

• Shared variables: – int turn; initially turn = 0– turn = i Pi can enter its critical section

• Process Pi

do {while (turn != i);

critical sectionturn = j;


• Satisfies mutual exclusion and bounded waiting, but not progress.



• Shared variables– boolean flag-larry, flag-jim;

initially flag-larry = flag-jim = FALSE– flag-larry= TRUE Larry ready to enter its critical section


while (flag-jim);flag-larry = TRUE;

critical sectionflag-larry = FALSE;




• Shared variables– boolean flag[2]; initially flag [0] = flag [1] = FALSE– flag [i] = TRUE Pi ready to enter its critical section

• Process Pi

do { while (flag[j]);

flag[i] = TRUE;critical section

flag [i] = FALSE;remainder section

} while (TRUE);• Satisfies progress, but not mutual exclusion and

bounded waiting requirements.



• Shared variables– boolean flag-larry, flag-jim;

initially flag-larry = flag-jim = FALSE– flag-larry= TRUE Larry ready to enter its critical section

• Process Larrydo {flag-larry = TRUE;

while (flag-jim); critical section

flag-larry = FALSE; remainder section

} while (TRUE);



• Shared variables– boolean flag[2]; initially flag [0] = flag [1] = FALSE– flag [i] = TRUE Pi wants to enter its critical section

• Process Pi

do {flag[i] = TRUE;while (flag[j]);

critical sectionflag [i] = FALSE;


• Satisfies mutual exclusion, but not progress and bounded waiting (?) requirements.



• Combined shared variables of algorithms 1 and 2/3.• Process Larry

do {flag-larry = TRUE;turn = “Jim”;while (flag-jim and turn == “Jim”);





• Combined shared variables of algorithms 1 and 2/3.• Process Pi

do {flag [i] = TRUE;turn = j;while (flag [j] and turn == j);

critical sectionflag [i] = FALSE;


• Meets all three requirements; solves the critical-section problem for two processes.



• Like Algorithm 4, but with the first 2 instructions of the entry section swapped – is it still a correct solution?


turn = “Jim”;flag-larry = TRUE;while (flag-jim and turn == “Jim”);




Bakery Algorithm (1)

• Critical Section for n processes:– Before entering its critical section, a process

receives a number (like in a bakery). Holder of the smallest number enters the critical section.

– The numbering scheme here always generates numbers in increasing order of enumeration; i.e., 1,2,3,3,3,3,4,5...

– If processes Pi and Pj receive the same number, if i < j, then Pi is served first; else Pj is served first (PID assumed unique).


Bakery Algorithm (2)

• Choosing a number:– max (a0,…, an-1) is a number k, such that k ai for

i = 0, …, n – 1

• Notation for lexicographical order (ticket #, PID #) – (a,b) < (c,d) if a < c or if a == c and b < d

• Shared data:

boolean choosing[n];

int number[n];

Data structures are initialized to FALSE and 0, respectively.


Bakery Algorithm for Pi

do { choosing[i] = TRUE;number[i] = max(number[0], …, number[n – 1]) +1;choosing[i] = FALSE;for (j = 0; j < n; j++) {

while (choosing[j]) ; while ((number[j] != 0) &&

((number[j],j) < (number[i],i))) ;}

critical sectionnumber[i] = 0;



What about process failures?

• If all 3 criteria (ME, progress, bounded waiting) are satisfied, then a valid solution will provide robustness against failure of a process in its remainder section (RS).– since failure in RS is just like having an infinitely

long RS.

• However, no valid solution can provide robustness against a process failing in its critical section (CS).– A process Pi that fails in its CS does not signal that

fact to other processes: for them Pi is still in its CS.


Drawbacks of software solutions

• Software solutions are very delicate .

• Processes that are requesting to enter their critical section are busy waiting (consuming processor time needlessly).– If critical sections are long, it would be more

efficient to block processes that are waiting.

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A. Frank - P. Weisberg Operating Systems The Critical-Section Problem