Question

Consider a uniprocessor system executing three tasks T₁, T₂ and T₃, each of which is composed of an infinite sequence of jobs (or instances) which arrive periodically at intervals of 3, 7 and 20 milliseconds, respectively. The priority of each task is the inverse of its period, and the available tasks are scheduled in order of priority, with the highest priority task scheduled first. Each instance of T₁, T₂ and T₃ requires an execution time of 1, 2 and 4 milliseconds, respectively. Given that all tasks initially arrive at the beginning of the 1^st millisecond and task pre-emption’s are allowed, the first instance of T₃ completes its execution at the end of _____________ milliseconds.

Answer 1

First instance of all three process arrive at t = 0ms

Their arrival times are like:

T₁ = 0, 3, 6, 9, 12, ……….

T₂ = 0, 7, 14, 21, ………

T₃ = 0, 20, 40, 60, …………

Because they arrive periodically at intervals of 3, 7 and 20 milliseconds.

Also, their burst time are given as 1, 2 and 4 ms respectively.

Priorities are in the inverse of period.

Process	Arrival time	Burst time	Priority
T1	0, 3, 6, 9, ….	1	1/3 (Highest)
T2	0, 7, 14, 21, ….	2	1/7
T3	0, 20, 40, 60 ……	4	1/20

 

As, they all first arrive at time 0, so Gantt chart will be like:

T1

T2

T1

T3

T1

T2

T1

T3

……..

0          1          3           4        6          7          9          10       12

So, in this way first instance of T₃ completes its execution at the end of 12 ms.