OS Review Questions - Ch6

Students working at individual PCs in a computer laboratory send their files to be printed by a server that spools the files on its hard disk. Under what conditions may a deadlock occur if the disk space for the print spool is limited? How may the deadlock be avoided?

Disk space on the spooling partition is a finite resource. Every block that comes in de facto claims a resource and every new one arriving wants more resources. If the spooling space is, say, 10 MB and the first half of ten 2-MB jobs arrive, the disk will be full and no more blocks can be stored so we have a deadlock. The deadlock can be avoided by allowing a job to start printing before it is fully spooled and reserving the space thus released for the rest of that job. In this way, one job will actually print to completion, then the next one can do the same thing. If jobs cannot start printing until they are fully spooled, deadlock is possible.

Consider the following state of a system with four processes, P1, P2, P3, and P4, and five types of resources, RS1, RS2, RS3, RS4,and RS5: Using the deadlock detection algorithm described in Section 6.4.2, show that there is a deadlock in the system. Identify the processes that are deadlocked. (Refer to image in the book)

First, the set of unmarked processes, P = (P1 P2 P3 P4) R1 is not less than or equal to A R2 is less than A; Mark P2; A = (02031); P = (P1 P3 P4) R1 is not less than or equal to A R3 is equal to A; Mark P3; A = (02032); P = (P1 P4) R1 is not less than or equal to A R4 is not less than or equal to A

Consider Fig. 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.

Neither change leads to deadlock. There is no circular wait in either case.

Explain the differences between deadlock, livelock, and starvation.

A deadlock occurs when a set of processes are blocked waiting for an event that only some other process in the set can cause. On the other hand, processes in a livelock are not blocked. Instead, they continue to execute checking for a condition to become true that will never become true. Thus, in addition to the resources they are holding, processes in livelock continue to consume precious CPU time. Finally, starvation of a process occurs because of the presence of other processes as well as a stream of new incoming processes that end up with higher priority that the process being starved. Unlike deadlock or livelock, starvation can terminate on its own, e.g. when existing processes with higher priority terminate and no new processes with higher priority arrive.

All the trajectories in Fig. 6-8 are horizontal or vertical. Can you envision any circumstances in which diagonal trajectories are also possible?

If the system had two or more CPUs, two or more processes could run in parallel, leading to diagonal trajectories.

The four conditions (mutual exclusion, hold and wait, no preemption and circular wait) are necessary for a resource deadlock to occur. Give an example to show that these conditions are not sufficient for a resource deadlock to occur. When are these conditions sufficient for a resource deadock to occur?

Suppose that there are three processes, A, B and C, and two resource types, R and S. Further assume that there are one instance of R and two instance of S. Consider the following execution scenario: A requests R and gets it; B requests S and gets; C requests S and gets it (there are two instances of S); B requests R and is blocked; A requests S and is blocked. At this stage all four conditions hold. However, there is no deadlock. When C finishes, one instance of S is released that is allocated to A. Now A can complete its execution and release R that can be allocated to B, which can then complete its execution. These four conditions are enough if there is one resource of each type.

A system has four processes and fiv e allocatable resources. The current allocation and maximum needs are as follows: What is the smallest value of x for which this is a safe state? (Refer to image in the book)

The needs matrix is as follows: 01002 02100 10300 00111 If x is 0, we have a deadlock immediately. If x is 1, process D can run to completion. When it is finished, the available vector is 11221. Unfortunately we are now deadlocked. If x is 2, after D runs, the available vector is 11321 and C can run. After it finishes and returns its resources the available vector is 22331, which will allow B to run and complete, and then A to run and complete. Therefore, the smallest value of x that avoids a deadlock is 2.

In Fig. 6-1 the resources are returned in the reverse order of their acquisition. Would giving them back in the other order be just as good?

Yes. It does not make any difference whatsoever.