Operating Systems

Start Lecture #7

Remark: We are doing 6.6 before 6.5 since 6.6 is easier and I believe serves as a good warm-up.

6.6: Deadlock Prevention

Attack one of the Coffman/Havender conditions.

6.6.1 Attacking the Mutual Exclusion Condition

The idea is to use spooling instead of mutual exclusion. Not possible for many kinds of resources.

6.6.2 Attacking the Hold and Wait Condition

Require each processes to request all resources at the beginning of the run. This is often called One Shot.

6.6.3 Attacking the No Preemption Condition

Normally not possible. That is, some resources are inherently pre-emptable (e.g., memory). For those, deadlock is not an issue. Other resources are non-preemptable, such as a robot arm. It is often not possible to find a way to preempt one of these latter resources. One exception is if the resource (say a CD-ROM drive) can be virtualized (recall hypervisors).

6.6.4 Attacking the Circular Wait Condition

Establish a fixed ordering of the resources and require that they be requested in this order. So if a process holds resources #34 and #54, it can request only resources #55 and higher.

It is easy to see that a cycle is no longer possible.

Homework: Consider Figure 6-4. Suppose that in step (o) C requested S instead of requesting R. Would this lead to deadlock? Suppose that it requested both S and R.

6.5 Deadlock Avoidance

Let's see if we can tiptoe through the tulips and avoid deadlock states even though our system does permit all four of the necessary conditions for deadlock.

An optimistic resource manager is one that grants every request as soon as it can. To avoid deadlocks with all four conditions present, the manager must be smart not optimistic.

6.5.1 Resource Trajectories

We plot progress of each process along an axis. In the example we show, there are two processes, hence two axes, i.e., planar. This procedure assumes that we know the entire request and release pattern of the processes in advance so it is not a practical solution. I present it as it is some motivation for the practical solution that follows, the Banker's Algorithm.

Homework: All the trajectories in Figure 6-8 are horizontal or vertical. Is is possible for a trajectory to be a diagonal.

Homework: 11, 12.

6.5.2 Safe States

Avoiding deadlocks given some extra knowledge.

Definition: A state is safe if there is an ordering of the processes such that: if the processes are run in this order, they will all terminate (assuming none exceeds its claim and assuming each would terminate if all its requests are granted).

Recall the comparison made above between detecting deadlocks (with multi-unit resources) and the banker's algorithm (which stays in safe states).

In the definition of a safe state no assumption is made about the running processes. That is, for a state to be safe, termination must occur no matter what the processes do (providing each would terminate if run alone and each never exceeds its claims). Making no assumption on a process's behavior is the same as making the most pessimistic assumption.

Remark: When I say pessimistic I am speaking from the point of view of the resource manager. From the manager's viewpoint, the worst thing a process can do is request resources.

Give an example of each of the following four possibilities. A state that is

  1. Safe and deadlocked—not possible.

  2. Safe and not deadlocked—a trivial is example is a graph with no arcs.

  3. Not safe and deadlocked—easy (any deadlocked state).

  4. Not safe and not deadlocked—interesting.

Is the figure on the right safe or not?

A manager can determine if a state is safe.

The manager then follows the following procedure, which is part of Banker's Algorithms discovered by Dijkstra, to determine if the state is safe.

  1. If there are no processes remaining, the state is safe.

  2. Seek a process P whose maximum additional request for each resource type is less than what remains for that resource type.
  3. The banker now pretends that P has terminated (since the banker knows that it can guarantee this will happen). Hence the banker pretends that all of P's currently held resources are returned. This makes the banker richer and hence perhaps a process that was not eligible to be chosen as P previously, can now be chosen.

  4. Repeat these steps.

Example 1

Consider the example shown in the table on the right.
A safe state with 22 units of one resource
processinitial claimcurrent allocmax add'l

Example 2

This example is a continuation of example 1 in which Z requested 2 units and the manager (foolishly?) granted the request.
An unsafe state with 22 units of one resource
processinitial claimcurrent allocmax add'l

Remark: An unsafe state is not necessarily a deadlocked state. Indeed, for many unsafe states, if the manager gets lucky all processes will terminate successfully. Processes that are not currently blocked can terminate (instead of requesting more resources up to their initial claim, which is the worst case and is the case the manager prepares for). A safe state means that the manager can guarantee that no deadlock will occur (even in the worst case in which processes request as much as permitted by their initial claims.)

6.5.3 The Banker's Algorithm (Dijkstra) for a Single Resource

The algorithm is simple: Stay in safe states. For now, we assume that, before execution begins, all processes are present and all initial claims have been given. We will relax these assumptions very soon.

In a little more detail the banker's algorithm is as follows.

Homework: 16.

6.5.4 The Banker's Algorithm for Multiple Resources

At a high level the algorithm is identical to the one for a single resource type: Stay in safe states.

But what is a safe state in this new setting?

The same definition (if processes are run in a certain order they will all terminate).

Checking for safety is the same idea as above. The difference is that to tell if there are enough free resources for a processes to terminate, the manager must check that for all resource types, the number of free units is at least equal to the max additional need of the process.

Homework: Consider a system containing a total of 12 units of resource R and 24 units of resource S managed by the banker's algorithm. There are three processes P1, P2, and P3. P1's claim is 0 units of R and 12 units of S, written (0,12). P2's claim is (8,15). P3's claim is (8,20). Currently P1 has 4 units of S, P2 has 8 units of R, P3 has 8 units of S, and there are no outstanding requests.

  1. What is the largest number of units of S that P1 can request at this point that the banker will grant?
  2. If P2 instead of P1 makes the request, what is the largest number of units of S that the banker will grant?
  3. If P3 instead of P1 or P2 makes the request, what is the largest number of units of S that the banker will grant?

Remark: Remind me to go over this one next time.

Limitations of the Banker's Algorithm

Homework: 22, 25, and 30. There is an interesting typo in 22. A has claimed 3 units of resource 5, but there are only 2 units in the entire system. Change the problem by having B both claim and be allocated 1 unit of resource 5.

6.7 Other Issues

6.7.1 Two-phase locking

This is covered (MUCH better) in a database text. We will skip it.

6.7.2 Communication Deadlocks

We have mostly considered actually hardware resources such as printers, but have also considered more abstract resources such as semaphores.

There are other possibilities. For example a server often waits for a client to make a request. But if the request msg is lost the server is still waiting for the client and the client is waiting for the server to respond to the (lost) last request. Each will wait for the other forever, a deadlock.

A solution to this communication deadlock would be to use a timeout so that the client eventually determines that the msg was lost and sends another.

But it is not nearly that simple: The msg might have been greatly delayed and now the server will get two requests, which could be bad, and is likely to send two replies, which also might be bad.

This gives rise to the serious subject of communication protocols.

6.7.3 Livelock

Instead of blocking when a resource is not available, a process may (wait and then) try again to obtain it. Now assume process A has the printer, and B the CD-ROM, and each process wants the other resource as well. A will repeatedly request the CD-ROM and B will repeatedly request the printer. Neither can ever succeed since the other process holds the desired resource. Since no process is blocked, this is not technically deadlock, but a related concept called livelock.

6.7.4 Starvation

As usual FCFS is a good cure. Often this is done by priority aging and picking the highest priority process to get the resource. Also can periodically stop accepting new processes until all old ones get their resources.

6.8 Research on Deadlocks

6.9 Summary


Chapter 3 Memory Management

Also called storage management or space management.

The memory manager must deal with the storage hierarchy present in modern machines.

We will see in the next few weeks that there are three independent decision:

  1. Should we have segmentation.
  2. Should we have paging.
  3. Should we employ fetch on demand.

Memory management implements address translation.

Homework: What is the difference between a physical address and a virtual address?

When is address translation performed?

  1. At compile time
  2. At link-edit time (the linker lab)
  3. At load time
  4. At execution time


  1. Dynamic Loading
  2. Dynamic Linking.
Note: I will place ** before each memory management scheme.

3.1 No Memory Management

The entire process remains in memory from start to finish and does not move.

The sum of the memory requirements of all jobs in the system cannot exceed the size of physical memory.


The good old days when everything was easy.