Rate Monotonic Scheduling Re-analysed - Faculty of Science and ...

This indicates that a set of tasks {(C i , T i ) | 1 ≤ i ≤m} is schedulable if and only if for any 1 ≤ k ≤ mthere exists 0 < t ≤ T k such that (1) holds. In otherwords, a set of tasks is unschedulable if and only ifthere exists 1 ≤ k ≤ m such that for any 0 < t ≤ T k ,∑ ki=1⌈ tT i⌉C i > t.4. Rate Monotonic SchedulerThe RMS is a static scheduler where tasks withshorter periods have higher priorities. Recall wehave the convention that priorities are in decreasingorder, so for a set of tasks {(C i , T i ) | 1 ≤ i ≤ m}this implies T 1 ≤ T 2 ≤ · · · ≤ T m . In this section,we discuss the RMS and shall not mention this anymore.We now come to establish the greatest lowerbound of utilisation factors of all unschedulabletask sets under the RMS.Theorem 5. For the RMS, v(m) = m(2 1/m − 1).Although our starting point, and subsequentlysome proofs, are different from that of Liu and Layland[6], we use many ideas from their proofs. Inparticular, Liu and Layland suggested first considerthe case that one period is at least half of any otherperiod. We can express this formally by a predicateH(T 1 , . . . , T m ) = ∀1 ≤ i, j ≤ m. 0 < T i /T j < 2. Inthe definitions below, assume Γ = {(C i , T i ) | 1 ≤ i ≤m}. LetS 2 = { Γ | Unschedulable(Γ) ∧H(T 1 , . . . , T m ) }and we shall prove that the bound over it has thesame value as over the following set of tasksS 3 = { Γ | ( ∧ m−1i=1 C i = T i+1 − T i ) ∧C m = 2T 1 − T m ∧ H(T 1 , . . . , T m ) }To fit in our framework, we introduce another setof tasksS 4 = { Γ | ( ∧ m−1i=1 C i = T i+1 − T i ) ∧C m > 2T 1 − T m ∧ H(T 1 , . . . , T m ) }and we shall proveglb(u(S 1 )) = glb(u(S 2 )) = glb(u(S 4 ))= glb(u(S 3 )) = m(2 1/m − 1)Lemma 3. glb(u(S 2 )) ≤ glb(u(S 4 ))4Proof: We prove this by showing S 4 ⊆ S 2 , namely,any task set {(C i , T i ) | 1 ≤ i ≤ m} ∈ S 4 is unschedulable.We shall prove that the m-th task will missits deadline. Consider any 0 < t ≤ T m . For notationalconvenience, let T 0 = 0, then there exists0 ≤ k < m, T k < t ≤ T k+1 . It follows thatm∑⌈ t ⌉C iT ii=1= 2C 1 + · · · + 2C k + C k+1 + · · · +C m−1 + C m= 2(T 2 − T 1 ) + · · · + 2(T k+1 − T k ) +T k+2 − T k+1 + · · · + T m − T m−1 + C m= T k+1 + T m − 2T 1 + C m> T k+1≥ tThis completes the proof of the lemma.Lemma 4. glb(u(S 2 )) ≥ glb(u(S 4 ))Proof: For any task set {(C i , T i ) | 1 ≤ i ≤ m} ∈ S 2 ,we shall find another set of task {(C ′ i , T ′ i ) | 1 ≤ i ≤m} ∈ S 4 , ∑ mi=1C ′ iT ′ i✷≤ ∑ mi=1C iT i. Suppose the k-th taskmisses its deadline. For t = T 1 , . . . , T k , we have∑ ki=1⌈t/T i ⌉C i > t. Since T k > T i and T k < 2T i forall 1 ≤ i < k, it follows immediatelyC 1 + C 2 + · · · +C k−1 + C k > T 12C 1 + C 2 + · · · +C k−1 + C k > T 2.2C 1 + 2C 2 + · · · +2C k−1 + C k > T kif T 1 , . . . , T k are mutually distinct. If the periods arenot mutually distinct, these still hold. For example,if T i = T i+1 are the first two equal periods, then wehavetherefore2C 1 + · · · + 2C i−1 + C i + · · · + C k > T i2C 1 + · · · + 2C i + C i+1 · · · + C k > T i = T i+1It follows that there exist α i > 1, i = 1 . . . k suchthatC 1 + C 2 + · · · +C k−1 + C k = α 1 T 12C 1 + C 2 + · · · +C k−1 + C k = α 2 T 2..2C 1 + 2C 2 + · · · +2C k−1 + C k = α k T k