00:01
We're going to prove by induction here.
00:02
Base case, we have the for single task t1.
00:06
Utilization is c1 over t1, which is less than equal to 1.
00:11
And because there is only one task, so the processor is not overloaded, and the task can always meet its deadline as long as c1 is less than or equal to t1.
00:28
So the task set is scheduled.
00:32
Now, we assume that for any set of k -harmonic task, t1, t2, tk, with total utilization, ci over t -i, less than equal to 1, the task set is schedulable under fixed priority preemptive scheduling.
00:55
And we'll consider adding a new task tk plus 1.
01:01
And the period would be tk plus 1, which is an integer multiple.
01:09
Of tk.
01:10
So let's say m tk for m greater than or equal to 1.
01:15
And execution time would be ck plus 1...