Operating Systems

3.4.2: Detecting Deadlocks with Multiple Unit Resources

This is more difficult.

• The figure on the right shows a resource allocation graph with multiple unit resources.
  
  
• Each unit is represented by a dot in the box.
  
  
• Request edges are drawn to the box since they represent a request for any dot in the box.
  
  
• Allocation edges are drawn from the dot to represent that this unit of the resource has been assigned (but all units of a resource are equivalent and the choice of which one to assign is arbitrary).
  
  
• Note that there is a directed cycle in red, but there is no deadlock. Indeed the middle process might finish, erasing the green arc and permitting the blue dot to satisfy the rightmost process.
  
  
• The book gives an algorithm for detecting deadlocks in this more general setting. The idea is as follows.
  1. look for a process that might be able to terminate (i.e., all its request arcs can be satisfied).
  2. If one is found pretend that it does terminate (erase all its arcs), and repeat step 1.
  3. If any processes remain, they are deadlocked.
  
  
• We will soon do in detail an algorithm (the Banker's algorithm) that has some of this flavor.
  
  
• The algorithm just given makes the most optimistic assumption about a running process: it will return all its resources and terminate normally. If we still find processes that remain blocked, they are deadlocked.
  
  
• In the bankers algorithm we make the most pessimistic assumption about a running process: it immediately asks for all the resources it can (details later on “can”). If, even with such demanding processes, the resource manager can assure that all process terminates, then we can assure that deadlock is avoided.

3.4.3: Recovery from deadlock

Preemption

Perhaps you can temporarily preempt a resource from a process. Not likely.

Rollback

Database (and other) systems take periodic checkpoints. If the system does take checkpoints, one can roll back to a checkpoint whenever a deadlock is detected. Somehow must guarantee forward progress.

Kill processes

Can always be done but might be painful. For example some processes have had effects that can't be simply undone. Print, launch a missile, etc.

Remark: We are doing 3.6 before 3.5 since 3.6 is easier.

3.6: Deadlock Prevention

Attack one of the coffman/havender conditions

3.6.1: Attacking Mutual Exclusion

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

3.6.2: Attacking Hold and Wait

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

3.6.3: Attacking No Preempt

Normally not possible. That is, some resources are inherently preemptable (e.g., memory). For those deadlock is not an issue. Other resources are non-preemptable, such as a robot arm. It is normally not possible to find a way to preempt one of these latter resources.

3.6.4: Attacking Circular Wait

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: 7.

3.5: 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.

3.5.1 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.

• We have two processes H (horizontal) and V.
• The origin represents them both starting.
• Their combined state is a point on the graph.
• The parts where the printer and plotter are needed by each process are indicated.
• The dark green is where both processes have the plotter and hence execution cannot reach this point.
• Light green represents both having the printer; also impossible.
• Pink is both having both printer and plotter; impossible.
• Gold is possible (H has plotter, V has printer), but the system can't get there.
• The upper right corner is the goal; both processes have finished.
• The red dot is ... (cymbals) deadlock. We don't want to go there.
• The cyan is safe. From anywhere in the cyan we have horizontal and vertical moves to the finish point (the upper right corner) without hitting any impossible area.
• The magenta interior is very interesting. It is
  • Possible: each processor has a different resource
  • Not deadlocked: each processor can move within the magenta
  • Deadly: deadlock is unavoidable. You will hit a magenta-green boundary and then will no choice but to turn and go to the red dot.
• The cyan-magenta border is the danger zone.
• The dashed line represents a possible execution pattern.
• With a uniprocessor no diagonals are possible. We either move to the right meaning H is executing or move up indicating V is executing.
• The trajectory shown represents.
  1. H excuting a little.
  2. V excuting a little.
  3. H executes; requests the printer; gets it; executes some more.
  4. V executes; requests the plotter.
• The crisis is at hand!
• If the resource manager gives V the plotter, the magenta has been entered and all is lost. “Abandon all hope ye who enter here” --Dante.
• The right thing to do is to deny the request, let H execute moving horizontally under the magenta and dark green. At the end of the dark green, no danger remains, both processes will complete successfully. Victory!
• This procedure is not practical for a general purpose OS since it requires knowing the programs in advance. That is, the resource manager, knows in advance what requests each process will make and in what order.

Homework: 10, 11, 12.

3.5.2: Safe States

Avoiding deadlocks given some extra knowledge.

• Not surprisingly, the resource manager knows how many units of each resource it had to begin with.
  
  
• Also it knows how many units of each resource it has given to each process.
  
  
• It would be great to see all the programs in advance and thus know all future requests, but that is asking for too much.
  
  
• Instead, when each process starts, it announces its maximum usage. That is each process, before making any resource requests, tells the resource manager the maximum number of units of each resource the process can possible need. This is called the claim of the process.
  • If the claim is greater than the total number of units in the system the resource manager kills the process when receiving the claim (or returns an error code so that the process can make a new claim).
  • If during the run the process asks for more than its claim, the process is aborted (or an error code is returned and no resources are allocated).
  • If a process claims more than it needs, the result is that the resource manager will be more conservative than need be and there will be more waiting.

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).

Recall the comparison made above between detecting deadlocks (with multi-unit resources) and the banker's algorithm

• The deadlock detection algorithm given makes the most optimistic assumption about a running process: it will return all its resources and terminate normally. If we still find processes that remain blocked, they are deadlocked.
  
  
• The banker's algorithm makes the most pessimistic assumption about a running process: it immediately asks for all the resources it can (details later on “can”). If, even with such demanding processes, the resource manager can assure that all process terminates, then we can assure that deadlock is avoided.

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 the all terminate and to not exceed their claims). Making no assumption is the same as making the most pessimistic assumption.

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

1. Safe and deadlocked--not possible.
   
   
2. Safe and not deadlocked--trivial (e.g., no arcs).
   
   
3. Not safe and deadlocked--easy (any deadlocked state).
   
   
4. Not safe and not deadlocked--interesting.