Consider a computing system with 14 dedicated devices. All jobs currently runnin

Question

Consider a computing system with 14 dedicated devices. All jobs currently running on this system require a maximum of five devices to complete their execution. The jobs run for long periods of time with just 3 devices, requesting the remaining 2 only at the very end of the run. Assume that the job stream is endless and that your operating system's device allocation policy is a very conservative one: No job will be started unless all the required drives have been allocated to it for the entire duration of its run.

          1. What is the maximum number of jobs that can be active at once? Explain your answer.
          2. What are the minimum and maximum number of devices that may be idle as a                result of this policy? Under what circumstances would an additional job be started?

Explanation / Answer

1.Well, theoretically if you have 14 drives and each job requires 3 to run. If operating system supports a very conservative device allocation policy no job will be started unless all tapes required have been allocated to it for the duration of its run, then a conservative scheduling program as described will only schedule 2 jobs at a time. It because the maximum number of active jobs is 2.let me explain because the device policy requires allocating 5 device for each job and there are at most 14 drives available. Here 14 will be greater than or equal to 5x, where x is the number of active jobs.

Hence, the max integer x be 2.

2. What are the minimum and maximum number of tape drives that may be idle as a result of this policy?

The minimum number of tape drives that may be idle as a result is 4.this is because

When both running jobs are using their maximum usage of 5 drives each.

=> 14 - (2 * 5) = 4

The maximum number of tape drives that may be idle as a result of this policy is 8 this is because when both running jobs are using their minimum job usage of 3 drives each

=> 14 - (2 * 3) = 8.

If we follow Banker's algorithm, No process will be granted more resources than what the system has and No process will be allocated more resources than its maximum assigned limit. So there could only be 2 jobs running at the same time, since each job require a maximum of 8 drives.

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