Chapter 3 The Critical Section Problem 1 Producer

Chapter 3 The Critical Section Problem 1

Producer - Consumer Problem Producer Process Consumer Process Produce Get from buffer Put in buffer Consume BUFFER n Buffer is shared (ie. , it is a shared variable) 2

Progress in time…. . Producer p 1 1 p 2 2 p 3 3 Buffer p 4 4 3 instead of 2! Consumer 1 n c 1 2 c 2 Both processes are started at the same time and consumer uses some old value initially t 3

A Race Condition Because of the timing and which process starts first n There is a chance that different executions may end up with different results n That is, because of interleaving, we can not predict what will happen next when processes are executing n 4

Why Processes Need to Communicate? n To synchronize their executions n To exchange data and information 5

Critical Sections n Critical Section n n A section of code in which a process accesses and modifies shared variables Mutual Exclusion n A method of prevention to ensure that one (or a specified number) of process(es) is/are in a critical section 6

Mutual Exclusion Problem Program producerconsumer; Procedure producer; begin repeat produce; putinbuffer; critical section until false; End; Procedure consumer; Begin repeat getfrombuffer; consume; until false; End; Begin (* main program *) cobegin producer; consumer coend End. critical section 7

Critical Section A critical section in the procedure of a process is the section in which the process refers and changes common variables (which are shared by other processes) n There may be several critical sections operating on the same or different common variables in a process n 8

Rules to Form Critical Sections n n n No two processes may be simultaneously inside their CS -(Rule of Mutual Exclusion) No assumptions are made about relative process speeds or number of CPUs - (Race Conditions) No process can remain inside a CS indefiniely (CS must be small in code and it’s execution must take minimal time) - (Rule of Deadlock and Starvation) A process outside a CS should not block other processes - (Rule of Deadlock and Starvation) No process should wait forever before entering its CS (When a process leaves a CS, the process manager must activate a waiting process to enter its CS. Processes waiting to enter a CS usually wait in a queue maintained in a FIFO manner or by a priority rule) (Rule of Starvation) 9

Correctness Specification n Correctness specifications required for a solution are : Mutual exclusion : Statements from the critical sections of two or more processes must not be interleaved n Freedom from Deadlock : If some processes are trying to enter their critical sections, then one of them must eventually succeed n Freedom from Starvation : If any process tries to enter its critical section, then that process must eventually succeed 10 n

Synchronization n n A synchronization mechanism (synchronization primatives) must be provided to handle the critical section problem This synchronization mechanism consists of additional statements that are placed before and after a critical section (preprotocol and postprotocol) 11

First Attempt (Critical Section Problem for two processes) Program mutualexclusion; Var turn : integer; Procedure P; begin repeat non-critical section; repeat (* do nothing *) until turn = 1; (* busy wait while P 2 is in CS*) critical section; turn : = 2; until false; End; Procedure Q; Begin repeat non-critical section; repeat (* do nothing *) until turn = 2; (* busy wait while P 1 is in CS *) critical section; turn : = 1; until false; End; Begin (* main program *) turn : = 1; (* we start with P 1 *) cobegin P; Q coend End. 12

Comments n n Do (* nothing *) is implemented by timeslicing (process loops in the while instruction ) Solution satisfies mutual exclusion, since only one process can be in CS Deadlock is not possible. Both processes can never be stuck at the while statement Process starvation is possible, if a process spends a lot of time in its non-critical section. Here we assume that the other process is busy waiting to enter the critical section. If processes spend finite times in noncritical sections then there will be no starvation 13

Drawback The processes are not loosely connected. That is, the right to enter a CS is being explicitly passed from one process to another n Suppose process P is waiting to enter the CS. If something happens to process Q while inside the non-critical section, process P is hopelessly deadlocked after the next pass n 14

First Attempt Using Book’s Convention n Await is blocking wait not a busy wait loop 15

Proving Correctness with State Diagrams n n n In the first attempt, states are triples of the form (pi, qj, turn) Variables used in the non-critical and critical sections are assumed to be distinct from the variables used in the protocols so that they have no effect at the correctness of the solution. So, they are left out of the states The mutual exclusion correctness property holds if the set of all accessible states does not contain a state of the form (p 3, q 3, turn) for some value of turn, because p 3 and q 3 are the labels of the critical 16 sections

State Diagrams n How many states can be in a state diagram? Suppose the algorithm has N processes with ni statements in process i, and M variables where variable j has mj possible values. n The number of possible states is n n 1 x. . . x n. N x m 1 x. . . x m. M n For the first attempt, the number of states is 4 x 4 x 2=32 since we have 4 statements per process and turn can be 1 or 2 17

State Diagram of the First Attempt n n The initial state is (p 1, q 1, 1) Lefthand side is P executing, righthand side is Q executing The incremental construction terminates after 16 of the 32 possible states have been constructed We do not have states (p 3, q 3, 1) or (p 3, q 3, 2) so mutual exclusion property for correctness holds for the first attempt 18

Abbreviating the State Diagram n n Statements executed in the non-critical and critical sections are irrelevant to the correctness of the synchronization algorithm. So, they are eliminated In fact, you can think of p as n n n p 1: non-critical section; await turn = 1 p 2: critical section; turn <- 2 Now we have 4 states to consider 19

Correctness of the First Attempt Mutual exclusion property holds (see the state diagram) n Is the algorithm deadlock free? n Which means : If some processes are trying to enter their critical sections, then one of them must eventually succeed n 20

Is it Deadlock Free? n n n A process is trying to enter its CS if it is executing p 1 or q 1 Consider the initial state. Turn is 1, which means p 1 passes through and q 1 loops on Next state is the lower right. Here p 2 executes CS and eventually sets turn to 2. While it is in the CS, Q is executing q 1 Next state is upper left in which Q enters its CS. Turn value decides which process should enter the CS Both processes eventually enter their CS in a mutually exclusive manner. Hence, the property of freedom from deadlock is satisfied 21

Is it Starvation Free? n n n Consider a section of the original state diagram. NCS stands for a non-critical section The state on the lower left is the one in which p 2 is waiting to enter its CS and q 1 is in the NCS If process Q decides to stay in its NCS (or broken down), P can not enter its CS so starve Hence, freedom from starvation property does not hold for this solution Second attempt tries to solve it 22

Second Attempt n n n Each process has its own flag which is true when process is in CS (p 3, q 3 and p 5, q 5) The other process waits until this process leaves the CS (p 2, q 2) If a process halts in its NCS then the other proceeds independently. 23

Second Attempt (Abbreviated) 24

State Diagram for Abbreviated Algorithm n n The last state (p 3, q 3, true) means that both processes enter CS at the same time Mutual exclusion property does not hold 25

A Scenario Showing that Mutual Exclusion Does Not Hold 26

Second Attempt (for j. BACI) Program secondattempt; Var c 1, c 2 : integer; Procedure p; Begin repeat while c 2 = 0 do; c 1 : = 0; CS; c 1 : = 1; NCS; until false; End; Procedure q; Begin repeat while c 1 = 0 do; c 2 : = 0; CS; c 2 : = 1; NCS; until false; End; Begin (* main program *) c 1: = 1; c 2: = 1; (* Both processes are not in their CS *) cobegin p; q coend End. loop on while p 2 is in CS process is entering the CS process leaves the CS if p 2 breaks down here, p 1 can still execute since it has its own flag c 1 27

Comments & Correctness n n n Each process periodically checks the other’s common variable cx. If the other process is not in CS (ie. , cx = 1); the other process enters its CS, sets its flag c to 0 to indicate that it will enter its CS This program may not work correctly if events are as shown in the table We have no mutual exclusion so deadlock! c 1 c 2 initially 1 1 P 1 checks c 2 1 1 P 2 checks c 1 1 1 P 1 sets c 1 0 1 P 2 sets c 2 0 0 P 1 enters CS 0 0 P 2 enters CS 0 0 Both processes are in CS 0 0 28

Third Attempt n n n This algorithm is a modification of the second attempt Here, “await” statements become a part of the CS Algorithm satisfies mutual exclusion (prove it yourself!) 29

Scenario for a Deadlock n n Both processes are locked at p 3 and q 3 The situation is a deadlock case but we may call it a livelock since processes are actively executing statements but nothing useful is done 30

Third Attempt (In j. BACI) Program thirdattempt; Var c 1, c 2 : integer; Procedure p 1; Begin repeat c 1 : = 0; while c 2 = 0 do; CS; c 1 : = 1; NCS; until false; End; Procedure p 2; Begin repeat c 2 : = 0; while c 1 = 0 do; CS; c 2 : = 1; NCS; until false; End; Begin (* main program *) c 1: = 1; c 2: = 1; (* Both processes are not in their CS *) cobegin p 1; p 2, coend End. loop on while p 2 is in CS process is entering its CS process leaves the CS 31

Comments & Correctness n n The third solution is a modification of the 2’nd attempt. Here, the variable cx (request to enter CS) is set before checking the other process whether it is in CS or not The solution is not correct as shown in the events of the table c 1 c 2 initially 1 1 P 1 sets c 1 0 1 P 2 sets c 2 0 0 P 1 enters CS 0 0 P 2 enters CS 0 0 Both processes are in CS 0 0 32

Dekker’s Algorithm 33

Dekker’s Algorithm (in j. BACI) Program Dekker; Var turn : integer; wantp, wantq : boolean; Procedure p; Begin repeat wantp : = true; while wantq do if turn = 2 then begin wantp: = false; repeat until turn = 1; wantp: =true end; CS; turn : = 2; wantp: = false; NCS; until false; End; Procedure q; Begin repeat wantq 2 : = true; while wantp do if turn = 1 then begin wantq: = false; repeat until turn = 2; wantq: =false end; CS; turn : = 1; wantq : = false; NCS; until false; End; Begin (* main program *) turn: = 1; wantp: = false; wantq: = false; cobegin p; q coend End. 34

Explanation of the Algorithm Procedure p; Begin repeat wantp : = true; (* 1 *) while wantq do if turn = 2 then begin wantp: = false; repeat until turn = 1; wantp: =true end; (* 2 *) CS; turn : = 2; (* 3 *) wantp : = false; (* 4 *) NCS; until false; End; 1. 2. 3. 4. Process makes a request to enter CS If the other process had made a request to enter CS before and if it is its turn then n Take back the request to enter CS by setting the “want” variable to false” n Wait for the other process to exit CS and change the turn variable n Make a new request to enter CS and then enter CS Flip the turn variable so that now the other process can enter CS Set “want” variable to false to indicate that the process is now out of CS 35

Comments Explicit control of transfer by a turn variable. Hence; turn = 1 means P’s turn, turn = 2 Q’s turn n If a process is blocked, the other process can still go n Dekker’s algorithm is correct (prove it!) n It satisfies the mutual exclusion property n It is free from deadlock and starvation n 36