Operating Systems

Start Lecture #11

Remark: Don't forget that the labs are required. Currently 4 students have not yet submitted lab 1!

Strict Alternation

Now let's try being polite and really take turns. None of this wanting stuff.

    Initially turn=1

    Code for P1                      Code for P2

    Loop forever {                   Loop forever {
       while (turn = 2) {}              while (turn = 1) {}
       critical-section                 critical-section
       turn <-- 2                       turn <-- 1
       non-critical-section }           non-critical-section }

This one forces alternation, so is not general enough. Specifically, it does not satisfy condition three, which requires that no process in its non-critical section can stop another process from entering its critical section. With alternation, if one process is in its non-critical section (NCS) then the other can enter the CS once but not again.

The first example violated rule 4 (the whole system blocked). The second example violated rule 1 (both in the critical section. The third example violated rule 3 (one process in the NCS stopped another from entering its CS).

In fact, it took years (way back when) to find a correct solution. Many earlier solutions were found and several were published, but all were wrong. The first correct solution was found by a mathematician named Dekker, who combined the ideas of turn and wants. The basic idea is that you take turns when there is contention, but when there is no contention, the requesting process can enter. It is very clever, but I am skipping it (I cover it when I teach distributed operating systems in V22.0480 or G22.2251). Subsequently, algorithms with better fairness properties were found (e.g., no task has to wait for another task to enter the CS twice).

What follows is Peterson's solution, which also combines wants and turn to force alternation only when there is contention. When Peterson's algorithm was published, it was a surprise to see such a simple solution. In fact Peterson gave a solution for any number of processes. A proof that the algorithm satisfies our properties (including a strong fairness condition) for any number of processes can be found in Operating Systems Review Jan 1990, pp. 18-22.

    Initially P1wants=P2wants=false  and  turn=1

    Code for P1                        Code for P2

    Loop forever {                     Loop forever {
       P1wants <-- true                   P2wants <-- true
       turn <-- 2                         turn <-- 1
       while (P2wants and turn=2) {}      while (P1wants and turn=1) {}
       critical-section                   critical-section
       P1wants <-- false                  P2wants <-- false
       non-critical-section }             non-critical-section }

The TSL Instruction (Hardware Assist test-and-set)

Tanenbaum calls this instruction test and set lock TSL.

I call it test and set (TAS) and define
TAS(b), where b is a binary variable,
to ATOMICALLY set b←true and return the OLD value of b.

Of course it would be silly to return the new value of b since we know the new value is true.

The word atomically means that the two actions performed by TAS(x), testing (i.e., returning the old value of x) and setting (i.e., assigning true to x) are inseparable. Specifically it is not possible for two concurrent TAS(x) operations to both return false (unless there is also another concurrent statement that sets x to false).

With TAS available implementing a critical section for any number of processes is trivial.

    loop forever {
        while (TAS(s)) {}  ENTRY
        s<--false          EXIT
        NCS }

2.3.4 Sleep and Wakeup

Remark: Tanenbaum presents both busy waiting (as above) and blocking (process switching) solutions. We present only do busy waiting solutions, which are easier and used in the blocking solutions. Sleep and Wakeup are the simplest blocking primitives. Sleep voluntarily blocks the process and wakeup unblocks a sleeping process. However, it is far from clear how sleep and wakeup are implemented. Indeed, deep inside, they typically use TAS or some similar primitive. We will not cover these solutions.

Homework: Explain the difference between busy waiting and blocking process synchronization.

2.3.5 Semaphores

Remark: Tannenbaum use the term semaphore only for blocking solutions. I will use the term for our busy waiting solutions (as well as for blocking solutions). Others call our solutions spin locks.

P and V and Semaphores

The entry code is often called P and the exit code V. Thus the critical section problem is to write P and V so that

    loop forever
  1. Mutual exclusion.
  2. No speed assumptions.
  3. No blocking by processes in NCS.
  4. Forward progress (my weakened version of Tanenbaum's last condition).

Note that I use indenting carefully and hence do not need (and sometimes omit) the braces {} used in languages like C or java.

A binary semaphore abstracts the TAS solution we gave for the critical section problem.

The above code is not real, i.e., it is not an implementation of P. It requires a sequence of two instructions to be atomic and that is, after all, what we are trying to implement in the first place. The above code is, instead, a definition of the effect P is to have.

To repeat: for any number of processes, the critical section problem can be solved by

    loop forever

The only solution we have seen for an arbitrary number of processes is the one just above with P(S) implemented via test and set.

Remark: Peterson's solution requires each process to know its process number; the TAS soluton does not. Moreover the definition of P and V does not permit use of the process number. Thus, strictly speaking Peterson did not provide an implementation of P and V. He did solve the critical section problem.

To solve other coordination problems we want to extend binary semaphores.

Both of the shortcomings can be overcome by not restricting ourselves to a binary variable, but instead define a generalized or counting semaphore.

Counting semaphores can solve what I call the semi-critical-section problem, where you premit up to k processes in the section. When k=1 we have the original critical-section problem.

    initially S=k

    loop forever
       SCS   -- semi-critical-section

Solving the Producer-Consumer Problem Using Semaphores

Note that my definition of semaphore is different from Tanenbaum's so it is not surprising that my solution is also different from his.

Unlike the previous problems of mutual exclusion, the producer-consumer has two classes of processes

What happens if the producer encounters a full buffer?
Answer: It waits for the buffer to become non-full.

What if the consumer encounters an empty buffer?
Answer: It waits for the buffer to become non-empty.

The producer-consumer problem is also called the bounded buffer problem, which is another example of active entities being replaced by a data structure when viewed at a lower level (Finkel's level principle).

  Initially e=k, f=0 (counting semaphores); b=open (binary semaphore)

  Producer                         Consumer

  loop forever                     loop forever
     produce-item                     P(f)
     P(e)                             P(b); take item from buf; V(b)
     P(b); add item to buf; V(b)      V(e)
     V(f)                             consume-item