An optimal boundary fair scheduling

Abstract

Nowadays, many real-time operating systems discretize the time relying on a system time unit. To take this behavior into account, real-time scheduling algorithms must adopt a discrete-time model in which both timing requirements of tasks and their time allocations have to be integer multiples of the system time unit. That is, tasks cannot be executed for less than one time unit, which implies that they always have to achieve a minimum amount of work before they can be preempted. Assuming such a discrete-time model, the authors of Zhu et al. (Proceedings of the 24th IEEE international real-time systems symposium (RTSS 2003), 2003, J Parallel Distrib Comput 71(10):1411–1425, 2011) proposed an efficient “boundary fair” algorithm (named BF) and proved its optimality for the scheduling of periodic tasks while achieving full system utilization. However, BF cannot handle sporadic tasks due to their inherent irregular and unpredictable job release patterns. In this paper, we propose an optimal boundary-fair scheduling algorithm for sporadic tasks (named BF$^2$), which follows the same principle as BF by making scheduling decisions only at the job arrival times and (expected) task deadlines. This new algorithm was implemented in Linux and we show through experiments conducted upon a multicore machine that BF$^2$ outperforms the state-of-the-art discrete-time optimal scheduler (PD$^2$), benefiting from much less scheduling overheads. Furthermore, it appears from these experimental results that BF$^2$ is barely dependent on the length of the system time unit while PD$^2$—the only other existing solution for the scheduling of sporadic tasks in discrete-time systems—sees its number of preemptions, migrations and the time spent to take scheduling decisions increasing linearly when improving the time resolution of the system.

Appendix: Proof of optimality for PD\(^{2*}\)

In this appendix, we present the updated proof of the only lemma of the proof of optimality of PD$^2$ for the scheduling of sporadic tasks (Lemma $6$ in Srinivasan and Anderson 2002), which is impacted by the new definition of the group deadline proposed in Sect. 7.1. For the sake of clarity, every modification to the initial proof presented in Srinivasan and Anderson (2002) is made to bolditalics in the current document. Note that Lemma $6$ of Srinivasan and Anderson (2002) is identical to Lemma $12$ introduced in Srinivasan and Anderson (2005) which proves the optimality of PD$^2$ for the scheduling of dynamic task sets based on the generalized sporadic task model under some constraints. Hence, PD$^{2*}$ is also optimal for the scheduling of dynamic task sets based on the generalized sporadic tasks model under the same constraints (the other lemmas are not impacted in Srinivasan and Anderson 2005).

Remember that in Srinivasan and Anderson (2002), the proof of optimality of PD$^2$ is made by contradiction. Hence, they assume that there exists a task set $\tau$ respecting some properties noted (T1), (T2) and (T3) such that a deadline is missed during the schedule $S$ produced with PD$^2$. The first time-instant after the beginning of the schedule corresponding to a deadline miss is denoted by $t_d$.

Note that in the following proof, we use the notion of displacement defined in Srinivasan and Anderson (2002). A displacement is denoted by $\Delta _i = \left\langle X^{(i)}, t_i, X^{(i+1)}, t_{i+1} \right\rangle$ and means that the subtask $X^{(i)}$ which was initially scheduled at time $t_i$ is now replaced by the subtask $X^{(i+1)}$ which was scheduled at time $t_{i+1}$. A displacement $\Delta _i$ is therefore valid if and only if $X^{(i+1)}$ is eligible to be scheduled at time $t_i$ (which is denoted by $e(X^{(i+1)}) \le t_i$).

Lemma 10

Let $\tau _{i,j}$ be a subtask of a light task scheduled at $t' < pd(\tau _{i,j})$ in $S$. If the eligibility time of the successor of $\tau _{i,j}$ is at least $pd(\tau _{i,j}) + 1$, then there cannot be holes in both $pd(\tau _{i,j})$ and $pd(\tau _{i,j}) +1$.

Proof

Note that by part (c) of Lemma 2 (in Srinivasan and Anderson 2002), $pd(\tau _{i,j}) \le t_d$. Therefore, $t' < t_d$. Let $pd(\tau _{i,j}) = t$. If $t=t_d$, then $t$ satisfies the stated requirements because there is no holes in slot $t_d$ (by part (d) of Lemma 2 in Srinivasan and Anderson 2002). In the rest of the proof we assume that $t < t_d$, and hence $t+1 \le t_d$. Suppose that there are holes in both $t$ and $t+1$. Because there is a hole in slot $t$ and (from the statement of the lemma) the eligibility time of the successor of $\tau _{i,j}$ is at least $t+1$, by Lemma 5 (in Srinivasan and Anderson 2002), either $pd(\tau _{i,j}) < t$ or $pd(\tau _{i,j}) = t ~\wedge ~ {b}(\tau _{i,j}) = 1$. Because $pd(\tau _{i,j}) = t$, the latter in fact must hold, i.e., $pd(\tau _{i,j}) = t ~\wedge ~ {b}(\tau _{i,j}) = 1$. We now show that $\tau _{i,j}$ can be removed without causing the missed deadline to be met, contradicting (T2) from Srinivasan and Anderson 2002. In particular, we show that the sequence of left-shifts caused by removing $\tau _{i,j}$ does not extend beyond slot $t+1$.

Let the chain of displacements caused by removing $\tau _{i,j}$ be $\Delta _1$, $\Delta _2$, ..., $\Delta _k$, where $\Delta _i = \left\langle X^{(i)}, t_i, X^{(i+1)}, t_{i+1} \right\rangle$, $X^{(1)} = \tau _{i,j}$ and $t_1 = t'$. By Lemma 3 (in Srinivasan and Anderson 2002), we have $t_{i+1} > t_i$ for all $i \in [1,k]$. Also, the priority of $X^{(i)}$ is greater than the priority of $X^{(i+1)}$ at $t_i$, because $X^{(i)}$ was chosen over $X^{(i+1)}$ in $S$. Because $pd(\tau _{i,j}) = t ~\wedge ~ {b}(\tau _{i,j}) = 1$, this implies the following property (from Prioritization Rules 4 presented in Sect. 7.1):

(P) For all $i \in [1, k+1]$, (i) $pd(X^{(i)}) > t$ or (ii) ${\varvec{pd}}({\varvec{X}}^{\varvec{(i)}}) = \varvec{t}$ and $\varvec{b}({\varvec{X}^{\varvec{(i)}})} =\varvec{0}$ or (iii) $\varvec{pd}(\varvec{X}^{\varvec{(i)}}) = \varvec{t}$ and ${\varvec{GD}}^{\varvec{*}}(\varvec{X}^{\varvec{(i)}}) \le \varvec{GD}^{*}(\varvec{{\tau _{i,j}})}$

Suppose this chain of displacements extends beyond $t+1$, i.e., $t_{k+1} > t+1$. Let $h$ be the smallest $i \in [1, k+1]$ such that $t_i > t+1$. Then, $t_{h-1} \le t+1$.

If $t_{h-1} < t+1$, then $X^{(h)}$ is eligible to be scheduled in slot $t+1$ because $e({X^{(h)}}) \le t_{h-1}$ (by the validity of displacement $\Delta _{h-1}$). Because there is a hole in slot $t+1$ in $S$, $X^{(h)}$ should have been scheduled there in $S$ (which is a contradiction with the assumption that $t_h > t+1$). Therefore, $t_{h-1} = t+1$ and by Lemma 4 (in Srinivasan and Anderson 2002), $X^{(h)}$ must be a successor of $X^{(h-1)}$. By similar reasoning, because there is a hole in slot $t$, $t_{h-2} = t$ and $X^{(h-1)}$ must be the successor of $X^{(h-2)}$ (see Figure 6(b) in Srinivasan and Anderson 2002).

By (P), either $pd({X^{(h-2)}}) > t$ or ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-2}})}) = {\varvec{t}} ~{\varvec{\wedge }}~ \varvec{b}({{\varvec{X}}^{({\varvec{h-2}})}}) = {\varvec{0}}$ or ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-2}})}) = {\varvec{t}} ~{\varvec{\wedge }}~ {\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}}) \le {\varvec{GD}}^{*}({{\varvec{\tau }}_{{\varvec{i,j}}}})$. In either case, ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) > {\varvec{t+1}}$ or ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) = {\varvec{t+1}} ~{\varvec{\wedge }}~ \varvec{b}({{\varvec{X}}^{({\varvec{h-1}})}}) = {\varvec{0}}$. To see this, note that if $pd(X^{(h-2)}) > t$, then because $X^{(h-1)}$ is the successor of $X^{(h-2)}$, by (6) in Srinivasan and Anderson (2002), $pd(X^{(h-1)}) > t+1$. If ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}}) = {\varvec{t}}$ and $\varvec{b}({{\varvec{X}}^{({\varvec{h-2}})}}) = {\varvec{0}}$ , then using (6) (in Srinivasan and Anderson 2002 ), it holds that ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) > {\varvec{t+1}}$ . Moreover, because $\varvec{b}({{\varvec{\tau }}_{{\varvec{i,j}}}}) = {\varvec{1}}$ , Definition 14 yields ${\varvec{GD}}^{*}({{\varvec{\tau }}_{{\varvec{i,j}}}}) = {\varvec{pd}}({\varvec{\tau }}_{{\varvec{i,j}}}) + {\varvec{1}} = {\varvec{t+1}}$ . Hence, if ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-2}})}) = {\varvec{t}}$ then ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}}) \le {\varvec{t+1}}$ . By Definition 14 , ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}}) \ge {\varvec{pd}}({\varvec{X}}^{({\varvec{h-2}})}) = {\varvec{t}}$ implying that ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}}) = {\varvec{t}}$ or ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}})={\varvec{t+1}}$ . If ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}}) = {\varvec{t}}$ then Definition 14 imposes that $\varvec{b}({{\varvec{X}}^{({\varvec{h-2}})}}) = {\varvec{0}}$ . Therefore, using (6) (in Srinivasan and Anderson 2002 ), it holds that ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) > {\varvec{t+1}}$ . On the other hand, if ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}})={\varvec{t+1}}$ then, according to Definition 14 , either ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}})= {\varvec{pd}}({\varvec{X}}^{({\varvec{h-2}})})+{\varvec{1}} ~{\varvec{\wedge }} {\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) - {\varvec{pd}}({\varvec{X}}^{({\varvec{h-2}})}) \ge {\varvec{2}}$, or ${\varvec{GD}}^{{\varvec{*}}}({{\varvec{X}}^{({\varvec{h-2}})}})={\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) ~{\varvec{\wedge }}~ \varvec{b}({{\varvec{X}}^{({\varvec{h-1}})}}) = {\varvec{0}}$ . Hence, in all cases, ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) > {\varvec{t+1}}$ or ${\varvec{pd}}({\varvec{X}}^{({\varvec{h-1}})}) = {\varvec{t+1}} ~{\varvec{\wedge }}~ \varvec{b}({{\varvec{X}}^{({\varvec{h-1}})}}) = {\varvec{0}}$ .

Now, because $X^{(h-1)}$ is scheduled at $t+1$, by part (b) of Lemma 2 (in Srinivasan and Anderson 2002), the successor of $X^{(h-1)}$ is not eligible before $t+2$, i.e., $e({X^{(h)}}) \ge t+2$. This implies that the displacement $\Delta _{h-1}$ is not valid. Thus, the chain of displacements cannot extend beyond $t+1$ and because $t+1 \le t_d$, removing $\tau _{i,j}$ cannot cause a missed deadline at $t_d$ to be met. This contradicts (T2) (in Srinivasan and Anderson 2002). Therefore, there cannot be holes in both $t$ and $t+1$. $\square$