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Then, for every subset Ωj we compute the number Γ(Ωj ) which is the number of intervals of length E(ocdt ) where ocdt cannot be scheduled because of conflicts with tasks of Ωj . The following theorem introduce a way to compute Γ for a given subset. E(oi ) Proof P GCD(g,T (ocdt ))− n i=0 E(oi ) repAs in the previous proof E(ocdt ) resents the number of intervals of length E(ocdt ) which are able to contain task ocdt in one interval of length T (ocdt ) GCD(g, T (ocdt )) and GCD(g,T (ocdt )) represents the number of sub-intervals of length GCD(g, T (ocdt )) in one interval of length equal to T (ocdt ) Example 2 Let (a : E(a) = 1, T (a) = 8), (b : E(b) = 1, T (b) = 12) and (c : E(c) = 1, T (c) = 16) be three tasks already proved schedulable.

S(on−1 ) + xn−1 T (on−1 ) + E(on−1 ) − 1]∩ [S(on ) + xn T (on ), S(on ) + xn T (on ) + E(on ) − 1]) = ∅ 28 have a condition which deals with all tasks whatever their periods are. We choose to group, according to theorem 3, schedulable tasks and to look for a new condition which takes into account the candidate task and a tasks group that has been proved schedulable. , [S(on−1 ) + xn−1 T (on−1 ), S(on−1 ) + xn−1 T (on−1 ) + E(on−1 ) − 1]} do not overlap. ∪ Figure 2. Scheduling Time Intervals ([S(on−1 ) + ln−1 g, S(on−1 ) + ln−1 g + E(on−1 ) − 1]∩ [S(on ), S(on ) + E(on ) − 1]) = ∅ Clearly, this must be the case since free intervals between the intervals ([S(o0 ) + l0 g, S(o0 ) + l0 g + E(o0 ) − 1] ∪ ...

Rn , we construct a linear program (Figure 4) on the following (2n + 1) variables: 4. Data-transmission is sequential, which means that data-transmission to the (i + 1)’th processor may commence (at time-instant si+1 ) only after datatransmission to the i’th processor has completed (at time-instant (si + αi σCm ). This is represented by the n Inequality (4)’s of the LP in Figure 4. 5. , (si + αi σCm + αi σCp )) is, by definition, no larger than the completion time ξ of the entire schedule. This is represented by the n Inequality (5)’s of the LP in Figure 4.

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