Question

The traditional UNIX scheduler enforces an inverse relationship between priority numbers and priorities: the higher the numbe1~ the lower the priority. The scheduler recalculates process priorities once per second using the following function: 4. Priority = (Recent CPU usage / 2) + Base where base 40 and recent CPU usage refers to a value indicating how often a process has used the CPU since priorities were last recalculated Assume that recent CPU usage for process P is 20, process P2 is 16, and process P3 is 8. What will be the new priorities for these three processes when priorities are recalculated? Based on this information, does the traditional UNIX scheduler raise or lower the relative priority of a CPU-bound process?

Answer 1

Answer:

Based on various facts, the traditional UNIX scheduler reduces that relative priority of the CPU-bound operation.

Explanation:

$Priority = (\frac{current\:CPU\:usage}{2}) + Base$

$\therefore\;\;\;\;\;\;\;\;\;\;\;\; Base = 40$

$P = (\frac{20}{2} ) + 40 = 50\\ \\P2 = (\frac{16}{2}) + 40 = 48\\ \\P3 = (\frac{8}{2} ) + 40 = 44$

Based on various facts, the traditional UNIX scheduler reduces that relative priority of the CPU-bound operation.

Answer 2

According to the result of the calculated priority values, the scheduler lowers the relative priority of CPU bound processes.

Given the Parameters :

Recent usage :

• P = 20 ; P2 = 16 ; P3 = 8 ; Base = 40

Using the relation given :

• $priority = \frac{Recent \: CPU \: usage}{2} + Base$

Calculating new priorities :

At P = 20 :

• $priority = \frac{20}{2} + 40 = 50$

At P2 = 16

• $priority = \frac{16}{2}+ 40 = 48$

At P3 = 8 :

• $priority = \frac{8}{2} + 40 = 44$

Based on the result obtained, since the values of the processes declines, then the system lowers the relative priorities of CPU - bound processes.

Learn more : https://brainly.com/question/18120831