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P/g. e. if we decompose the interval [0, P ] into P/g intervals of length g, then, due to their periods, an instance of o1 and an instance of o2 will be executed into a same interval of length g. So g < E(o1 ) + E(o2 ) means that I1 and I2 overlap. This proves the sufficiency of (1). To prove the necessity of (1) we show that if g < GCD(T (o1 ), T (o2 )), tasks o1 and o2 cannot be scheduled. Let assume that g < GCD(T (o1 ), T (o2 )) and without loss of generality we assume also that S(o1 ) = 0.

Finally, due to the hard real-time nature of the considered systems we deal with non-preemptive tasks. Nonpreemptive scheduling is important for a variety of reasons [2]: Real-time systems are often designed using preemptive scheduling to guarantee the execution of high priority tasks. For multiple reasons there is a great interest in exploring non-preemptive scheduling in the case of hard real-time systems where missing deadline leads to catastrophic situations. This paper presents a necessary and sufficient schedulability condition for determining whether a task will satisfy its period and precedences constraints when some tasks have already been scheduled.

This condition is proposed by the following theorem. Theorem 2 Tasks of the set {i ∈ N, i ≤ n, (oi : E(oi ), T (oi ))} are schedulable if n E(oi ) ≤ GCD(∀i, T (oi )) (3) i=0 The question is: can we generalize the theorem 1 for more than two tasks? Only the condition of the theorem 1 is used in the schedulability analysis (start times are not known yet). But we can easily observe that this condition cannot be useful since it compares only two tasks at the same time. Thus, we need a more general condition which takes into account a set of tasks.

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