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a. Assume the scheduler in the OS will not start a job unless there are four tape drives available. When a job is started, fourdrives are assigned immediately and are not released until the job finishes. What is the maximum number of jobs that can be in progress at once? What is themaximum and minimum number of tape drives that may be left idle as a result of this policy?
If a conservative policy is used, at most 20/4 = 5 processes can be active simultaneously. Because one of the drives allocated toeach process can be idle most of the time, at most 5 drives will be idle at a time. In the best case, none of the drives will be idle.
b. Suggest an alternative policy to improve tape drive utilization and at the same time avoid system deadlock. What is themaximum number of jobs that can be in progress at once? What are the bounds on the number of idling tape drives?
To improve drive utilization, each process can be initially allocated with three tape drives. The fourth one will be allocated ondemand. In this policy, at most floor(20/3) = 6 processes can be active simultaneously. The minimum number of idle drives is 0 and the maximum number is2.
Process Name | Arrival Time | Processing Time |
A | 0 | 3 |
B | 1 | 5 |
C | 3 | 2 |
D | 9 | 5 |
E | 12 | 5 |
Perform the same analysis as depicted for this set.
0 5 10 15 20
| | | | | | | | | | | | | | | | | | | | |
FCFS P1 ------
P2 ----------
P3 ----
P4 ----------
P5 ----------
RR q=1 P1 -- -- --
P2 -- -- -- -- --
P3 -- --
P4 -- -- -- -- --
P5 -- -- -- ----
RR q=4 P1 ------
P2 -------- --
P3 ----
P4 -------- --
P5 -------- --
SPN P1 ------
P2 ----------
P3 ----
P4 ----------
P5 ----------
SRT P1 ------
P2 ----------
P3 ----
P4 ----------
P5 ----------
HRRN P1 ------
P2 ----------
P3 ----
P4 ----------
P5 ----------
FB q=1 P1 -- -- --
P2 -- -- ---- --
P3 -- --
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