Bounding and shaping the demand of generalized mixed-criticality sporadic task systems

Abstract

We generalize the commonly used mixed-criticality sporadic task model to let all task parameters (execution-time, deadline and period) change between criticality modes. In addition, new tasks may be added in higher criticality modes and the modes may be arranged using any directed acyclic graph, where the nodes represent the different criticality modes and the edges the possible mode switches. We formulate demand bound functions for mixed-criticality sporadic tasks and use these to determine EDF-schedulability. Tasks have different demand bound functions for each criticality mode. We show how to shift execution demand between different criticality modes by tuning the relative deadlines. This allows us to shape the demand characteristics of each task. We propose efficient algorithms for tuning all relative deadlines of a task set in order to shape the total demand to the available supply of the computing platform. Experiments indicate that this approach is successful in practice. This new approach has the added benefit of supporting hierarchical scheduling frameworks.

Appendix: Proof of Lemma 2

Before proving Lemma 2, we reformulate the function done _hi (τ _i ,ℓ) from Sect. 3.3 as

and show that it is an equivalent definition:

Lemma 4

$$ \operatorname{done^*\!}(\tau_{i}, \ell) = \mathrm{done}_{\textsc{hi}}( \tau_{i}, \ell). $$

Proof

We split the proof into three cases.

First case (D _i (hi)>ℓmodT _i ≥D _i (hi)−D _i (lo)): From ℓmodT _i ≥D _i (hi)−D _i (lo) we know that

(9)

With (9) we can rewrite $\operatorname{done^{*}\!}(\tau_{i}, \ell)$ as

Second case (D _i (hi)≤ℓmodT _i ): From D _i (hi)≤ℓmodT _i the equality (9) follows again. We can rewrite $\operatorname{done^{*}\!}(\tau_{i}, \ell)$:

We have C _i (lo)≤D _i (lo) and D _i (hi)≤ℓmodT _i . Therefore,

$$ \bigl(C_{i}({\textsc{lo}}) - D_{i}({\textsc{lo}})\bigr) + \bigl(D_{i}({\textsc {hi}}) - \ell\bmod T_{i}\bigr) \leq0 $$

and

$$ \operatorname{done^*\!}(\tau_{i}, \ell) = 0 = \mathrm{done}_{\textsc{hi}}(\tau _{i}, \ell). $$

Third case (ℓmodT _i <D _i (hi)−D _i (lo)): From ℓmodT _i <D _i (hi)−D _i (lo) and from C _i (lo)≤D _i (lo)≤D _i (hi)≤T _i we have⁹

$$\begin{aligned} \bigl(\ell- \bigl(D_{i}({\textsc{hi}}) - D_{i}({ \textsc{lo}})\bigr)\bigr) \bmod T_{i} =& T_{i} - \bigl(D_{i}({\textsc{hi}}) - D_{i}({\textsc{lo}})\bigr) + (\ell\bmod T_{i}) \\\geq& T_{i} - \bigl(D_{i}({\textsc{hi}}) - D_{i}({\textsc{lo}})\bigr) \\\geq& D_{i}({\textsc{lo}})\\\geq& C_{i}({\textsc{lo}}). \end{aligned}$$

Therefore,

$$ C_{i}({\textsc{lo}}) - \bigl(\bigl(\ell- \bigl(D_{i}({ \textsc{hi}}) - D_{i}({\textsc{lo}})\bigr)\bigr) \bmod T_{i} \bigr) \leq0 $$

and

$$ \operatorname{done^*\!}(\tau_{i}, \ell) = 0 = \mathrm{done}_{\textsc{hi}}(\tau _{i}, \ell). $$

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We can now prove Lemma 2.

Proof of Lemma 2

The lemma follows from straightforward substitutions, first:

To show the dbf _hi part, we consider full _hi and done _hi separately:

We use Lemma 4 for $\mathrm{done}_{\textsc{hi}}(\tau _{i}, \ell) = \operatorname{done^{*}\!}(\tau_{i}, \ell)$:

The dbf _hi part follows directly:

$$\begin{aligned} \mathrm{dbf}_{\textsc{hi}}(\tau_{i}, \ell) =& \mathrm {full}_{\textsc{hi}}(\tau_{i}, \ell) - \mathrm{done}_{\textsc {hi}}( \tau_{i}, \ell) \\=& \mathrm{full}_{\textsc{hi}}(\tau_{j}, \ell- \delta) - \mathrm {done}_{\textsc{hi}}(\tau_{j}, \ell- \delta) \\=& \mathrm{dbf}_{\textsc{hi}}(\tau_{j}, \ell- \delta). \end{aligned}$$

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