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

numbers, there is obviously no utilisation factorthat can be equal to m(2 1 m − 1) , except when m = 1(one task) and in this case, the task (set) is triviallyschedulable. For a scheduling algorithm in general,we expect that v(m) is not an element of u(S 1 )(in other words, u(S 1 ) has no minimum element),which means that a set of tasks whose utilisationfactor is equal to v(m) is schedulable. Suppose thisis not true and assume an unschedulable task setΓ whose utilisation factor is v(m). We can reducethe execution time of one task of Γ by a very smallamount, and the new task set is still unschedulable.The utilisation factor of the new task set is less thanv(m), hence a contradiction. Note we do not considerthis argument as a proof, because it lacks therigor that is usually required in mathematics.3. Static SchedulingFor static scheduling, the priorities of tasks arefixed. Without loss of generality, we hereafter assumepriorities are in decreasing order, namely, fora set of tasks τ 1 , τ 2 , · · · , τ m , the priority order isτ 1 > τ 2 > · · · > τ m . In this section, we discussstatic scheduling and shall not state this all the time.A set of tasks is schedulable if the timing requirementof each instance of every task is satisfiedbefore the corresponding deadline. Let us take onetask, say τ k , to consider. We first study the conditionfor its first request to be satisfied. Suppose thetiming requirement is satisfied at time t (t ≤ T k ).Over the interval (0, t), the processor is occupiedby τ 1 , τ 2 , · · · , τ k , i.e., one of them is running andno tasks τ k+1 , τ k+2 , · · · , τ m can be executed, sinceτ k ’s request is outstanding until t. It follows that∑ ki=1⌈ tT i⌉C i ≤ t (where ⌈ tT i⌉ is the smallest integergreater than or equal to tT i), since tasks with higherpriorities need ∑ k−1i=1 ⌈ tT i⌉C i executing time. On theother hand, if there exists 0 < t ≤ T k , such that∑ ki=1⌈ tT i⌉C i ≤ t, then obviously there is enough timefor task τ k , along with the tasks with higher priorities,to complete the execution time. We next investigatea condition for an arbitrary request of atask.Lemma 2. The j-th request of task τ k is satisfied,if for any b ≤ ( j − 1)T k , there exists t such that( j − 1)T k < t ≤ jT k , ∑ ki=1⌈ t−bT i⌉C i ≤ (t − b).Proof: We search backwards from ( j − 1)T k forthe first time point b such that the processor isnot used by a task from τ 1 to τ k (either no taskis running or a task from τ k+1 to τ m is running)over interval (b − ɛ, b) where ɛ is arbitrarily small.3If such a point cannot be found, then let b = 0.It follows that the processor is totally occupiedby τ 1 , τ 2 , · · · , τ k over (b, ( j − 1)T k ) and there areno outstanding requests of τ 1 , τ 2 , · · · , τ k that startbefore b. The total amount of requested executiontime from τ 1 , τ 2 , · · · , τ k during (b, t) is at most∑ ki=1⌈ t−bT i⌉C i . The processor will continue to be occupiedby τ 1 , τ 2 , · · · , τ k from ( j − 1)T k until theirrequests are satisfied. Since ( j − 1)T k < t ≤ jT kand ∑ ki=1⌈ t−bT i⌉C i ≤ (t − b), there is enough time forτ k to complete execution by t.✷Liu and Layland [6] discovered a property,which became well known afterwards, about thestatic scheduling: the critical instant (the arrivingtime of a task that needs the maximal time to complete)of a task occurs when the task is requestedsimultaneously with requests of all higher prioritytasks. Apart from its application in the schedulabilitytheorem, it also provides the basis for the responsetime method studied by many researchers,for example, Joseph and Pandya [5]. However, theproof was informal in the form of illustrative diagramsand found to have flaws by Goossens [4].In [3], we formally proved a similar result: underthe static scheduler, if the timing requirement of atask’s first instance is satisfied, then all instancesof this task will be satisfied. The proof in [3] wasgiven in Duration Calculus [7], an interval temporallogic. For easy reference, we reformulate theproof using traditional mathematics and now theproof follows immediately from Lemma 2.Theorem 4. For any task, the timing requirementsof all its instances are satisfied under the staticscheduler if and only if its first instance is satisfied.Proof The only if part is by the definition. Wenext show the if part. Consider any task τ k . Its firstinstance is satisfied, and therefore there exists time0 < t ≤ T k such thatk∑⌈ t ⌉C i ≤ t (1)T ii=1For any instance j > 0, it follows from 0 < t ≤ T kthat for any b ≤ ( j − 1)T k there exists n > 0 suchthat ( j − 1)T k < b + nt ≤ jT k . Thus,∑ ki=1⌈ (b+nt)−bT i⌉C i ≤ ∑ ki=1n⌈ tT i⌉C i( by ⌈ ntT i⌉ ≤ n⌈ tT i⌉)= n ∑ ki=1⌈ tT i⌉C i≤ ntTherefore, by Lemma 2, the j-th instance of τ k issatisfied.✷