Question

Three processes arrive at time zero with CPU bursts of 16, 20, and 10 milliseconds. If the scheduler has prior knowledge about the length of the CPU bursts, the minimum achievable average waiting for these three processes in a non-preemptive scheduler (rounded to the nearest integer) is ______ milliseconds.

Answer 1

The scheduler has prior knowledge about the length of the CPU bursts and it is non-preemptive.

Therefore shortest job first algorithms while giving minimum achievable average waiting

Gantt chart:

P3

P1

P2

0            10           26            46

Process table:

Process Id	Arrival Time (AT in ms)	Burst Time (BT in ms)	Completion Time (CT)	Turn Around Time (TAT)	Waiting Time (WT)
P1	0	16	26	26	10
P2	0	20	46	46	26
P3	0	10	10	10	0

 

Average waiting time = \(\frac{0~+10+26}{3} = 12\; ms\)

Average waiting time = 12 ms

Important Points:

TAT = CT – AT

WT = TAT – BT