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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, which is 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 1st millisecond and task preemptions are allowed, the first instance of T3 completes its execution at the end of_____________________milliseconds.

I understand how to answer this question but I have one doubt please help,

The priority of each task is the inverse of its period, and the available tasks are scheduled in order of priority.

I'm confused with this line, Do they mean Burst time as a period or the time which Task arrive periodically.

BTW for this question it doesn't matter what we take In the end T1 will always has higher priority but can you help me here

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  • $\begingroup$ I interpret the question as saying that T1 has priority (1/3), T2 has priority (1/7) and so on. $\endgroup$ – user130558 Sep 19 '20 at 2:34

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