Fair lateness scheduling: reducing maximum lateness in G-EDF-like scheduling

Abstract

In prior work on soft real-time (SRT) multiprocessor scheduling, tardiness bounds have been derived for a variety of scheduling algorithms, most notably, the global earliest-deadline-first (G-EDF) algorithm. In this paper, we devise G-EDF-like (GEL) schedulers, which have identical implementations to G-EDF and therefore the same overheads, but that provide better tardiness bounds. We discuss how to analyze these schedulers and propose methods to determine scheduler parameters to meet several different tardiness bound criteria. We employ linear programs to adjust such parameters to optimize arbitrary tardiness criteria, and to analyze lateness bounds (lateness is related to tardiness). We also propose a particular scheduling algorithm, namely the global fair lateness (G-FL) algorithm, to minimize maximum absolute lateness bounds. Unlike the other schedulers described in this paper, G-FL only requires linear programming for analysis. We argue that our proposed schedulers, such as G-FL, should replace G-EDF for SRT applications.

Appendix: Additional proofs

Lemma 1

If ℓ≥Y _i , then

$$ \operatorname{DBF}(\tau_i, Y_i, \ell) \le U_i \ell+ C_i \biggl( 1 - \frac {Y_i}{T_i} \biggr). $$

Proof

As first demonstrated in Baruah et al. (1990), for arbitrary-deadline G-EDF,

$$ \operatorname{DBF}(\tau_i, \ell) = C_i \times\max \biggl \{ 0, \biggl\lfloor \frac{\ell- D_i}{T_i} \biggr\rfloor+ 1 \biggr\}. $$

Therefore, for arbitrary GEL schedulers, we instead define

$$ \operatorname{DBF}(\tau_i, Y_i, \ell) = C_i \times\max \biggl\{ 0, \biggl\lfloor \frac{\ell- Y_i}{T_i} \biggr\rfloor+ 1 \biggr\}. $$

(57)

If ℓ≥Y _i , then

$$\begin{aligned} U_i \ell+ C_i \biggl( 1 - \frac{Y_i}{T_i} \biggr) \ge{} & \mbox{\{By definition of the floor function\}} \\ {} & U_i \biggl( T_i \biggl\lfloor\frac{\ell- Y_i}{T_i} \biggr\rfloor+ Y_i \biggr) + C_i \biggl(1 - \frac{Y_i}{T_i} \biggr) \\ = {} & \mbox{\{Rearranging\}} \\ {} & C_i \biggl\lfloor\frac{\ell- Y_i}{T_i} \biggr\rfloor+ \frac{C_i Y_i}{T_i} + C_i - \frac{C_i Y_i}{T_i} \\ = {} & \mbox{\{Cancelling and rearranging\}} \\ {} & C_i \biggl( \biggl\lfloor\frac{\ell-Y_i}{T_i} \biggr\rfloor+ 1 \biggr) \\ = {} & \mbox{\{By (57)\}} \\ {} & \operatorname{DBF}(\tau_i, Y_i, \ell) \end{aligned}$$

 □

Lemma 2

$\operatorname{DBF}(\tau_{i}, Y_{i}, \ell) \le U_{i} \ell+ S_{i}(Y_{i})$.

Proof

We consider two cases.

We first consider ℓ∈[0,Y _i ). In that case, ℓ−Y _i <0, and $\lfloor\frac{\ell- Y_{i}}{T_{i}} \rfloor+ 1 \le0$. Therefore, by the definition of S _i (Y _i ) in (9), and (57), $\operatorname{DBF}(\tau_{i}, Y_{i}, \ell) = 0 \le U_{i} \ell+ S_{i}(Y_{i})$.

We then consider ℓ≥Y _i . By the definition of S _i (Y _i ) in (9),

$$ U_i \ell+ S_i(Y_i) \ge U_i \ell+ C_i \biggl( 1 - \frac{Y_i}{T_i} \biggr). $$

Therefore, by Lemma 1,

$$ \operatorname{DBF}(\tau_i, Y_i, \ell) \le U_i \ell+ S_i(Y_i). $$

 □

Lemma 5

If x is compliant and each job of τ _j with higher priority than J _i finishes with response time no greater than Y _j +x _j +C _j , then J _i finishes with response time no greater than Y _i +x _i +C _i .

Proof

Let δ denote the sum of the amount of execution of J _i before y _i and the amount of time by which it finishes early, so that C _i −δ units remain at time y _i . By Lemma 4, the remaining work at time y _i for all jobs in H (including J _i itself) is at most G(x,Y)+S(Y). Therefore, there can be at most G(x,Y)+S(Y)−(C _i −δ) units of competing work from other tasks. (This expression actually bounds all competing work, but is sufficient as an upper bound for work from other tasks.) Because each Y _i ≥0, no job releases after y _i affect the scheduling of J _i . As depicted in Fig. 10, a CPU becomes available for τ _i at the earliest time that any CPU completes its competing work. Therefore, in the worst case, all CPUs execute competing work with maximum parallelism, and a CPU becomes available for τ _i at time

$$ t_c = y_i + \frac{G(\mathbf{X}, \mathbf{Y}) + S(\mathbf{Y}) - C_i + \delta}{m} $$

(58)

We consider two cases, depicted in Fig. 11:

Fig. 10

Possible competing work schedules after y _i

Fig. 11

Cases considered in Lemma 5

Case 1 (Fig.  11 (a)) If some predecessor of J _i is running at time t _c , we denote the immediate predecessor of J _i as $J_{i}'$. Because all competing work from other tasks has completed, J _i can begin execution immediately upon completion of $J_{i}'$. Therefore, because $J_{i}'$ has a response time no greater than the assumed response bound (11), and J _i must have been released at least T _i units after $J_{i}'$, $J_{i}'$ must complete no later than Y _i +x _i +C _i −T _i units after the release of J _i . Because we assumed C _i ≤T _i , J _i must complete within Y _i +x _i +C _i −T _i +C _i ≤Y _i +x _i +C _i units of its release, and the lemma is proven.

Case 2 (Fig.  11 (b)) If no predecessor of J _i is running at time t _c , J _i is eligible for execution at time t _c unless it has already completed. At most C _i −δ units of execution remain, so it completes by time

$$\begin{aligned} y_i + \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_i + \delta}{m} + C_i - \delta\le{} & \mbox{\{Because $\delta> \frac{\delta}{m}$\}} \\ {} & y_i + \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_i}{m} + C_i \\ \le{} & \mbox{\{By Definition~{1}\}} \\ {} & y_i + x_i + C_i, \end{aligned}$$

and the lemma is proven. □

Lemma 6

x is a minimum near-compliant vector iff

$$ \forall i,\quad x_i = \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_i}{m}. $$

(27)

Proof

We consider two cases, depending on whether (27) holds.

( 27 ) Holds If (27) holds, then by Definition 1, x is near-compliant. Let δ>0 and x′ be such that, for some j,

$$ x'_j = x_j - \delta, $$

(59)

and for i≠j, $x'_{i} = x_{i}$. Then, by the definition of G(x,Y) in (13),

$$ G\bigl(\mathbf{x}', \mathbf{Y}\bigr) \ge G( \mathbf{x}, \mathbf{Y}) - U_j \delta. $$

(60)

Therefore,

$$\begin{aligned} \frac{G(\mathbf{x}', \mathbf{Y}) + S(\mathbf{Y}) - C_j}{m} \ge{} & \mbox{\{By (60)\}} \\ {} & \frac{G(\mathbf{x}, \mathbf{Y}) - U_j \delta+ S(\mathbf{Y}) - C_j}{m} \\ = {} & \mbox{\{Rearranging\}} \\ {} & \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_j}{m} - \frac{U_j \delta}{m} \\ = {} & \mbox{\{By (27)\}} \\ {} & x_j - \frac{U_j \delta}{m} \\ > {} & \mbox{\{Because $U_{j} > 0$, $m > 1$, and $\delta> 0$, and by (59)\}} \\ & x_j' \end{aligned}$$

Therefore, by Definition 1, x′ is not near-compliant. Since j and δ were arbitrary, x is a minimum near-compliant vector.

( 27 ) Does Not Hold Suppose (27) does not hold. If, for some j, $x_{j} < \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_{j}}{m}$, then by Definition 1, x is not near-compliant. Thus, if x is near-compliant, then for some j,

$$ x_j > \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_j}{m}. $$

(61)

If there is more than one such j, we choose one arbitrarily. We define x′ such that

$$ x_j' = \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_j}{m}, $$

(62)

and for i≠j,

$$ x_i' = x_i. $$

(63)

By the definition of G(x,Y) in (13) and (61)–(62),

$$ G\bigl(\mathbf{x}', \mathbf{Y}\bigr) \le G( \mathbf{x}, \mathbf{Y}). $$

(64)

Therefore,

$$\begin{aligned} x_j' = {} & \mbox{\{By (62)\}} \\ {} & \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_i}{m} \\ \ge{} & \mbox{\{By (64)\}} \\ {} & \frac{G(\mathbf{x}', \mathbf{Y}) + S(\mathbf{Y}) - C_i}{m}, \end{aligned}$$

and, for i≠j,

$$\begin{aligned} x_i' = {} & \mbox{\{By (63)\}} \\ {} & x_i \\ \ge{} & \mbox{\{By Definition~{1}\}} \\ {} & \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) - C_i}{m} \\ \ge{} & \mbox{\{By (64)\}} \\ {} & \frac{G(\mathbf{x}', \mathbf{Y}) + S(\mathbf{Y}) - C_i}{m}. \end{aligned}$$

Thus, x′ is near-compliant, so x is not a minimum near-compliant vector. □

Lemma 7

If the minimum near-compliant vector for the task system exists, it is unique.

Proof

We use proof by contradiction. Assume that two distinct minimum near-compliant vectors exist for the task system, x with

$$ \forall i,\quad x_i = \frac{s-C_i}{m}, $$

(65)

and x′ with

$$ \forall i,\quad x_i' = \frac{s'-C_i}{m}. $$

(66)

Without loss of generality, assume s′<s.

We show

$$\begin{aligned} s = {} & \mbox{\{By Corollary~{1}\}} \\ & G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) \\ = {} & \mbox{\{By the definition of $G(\mathbf{x}, \mathbf{Y})$ in (13)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \bigl(x_i U_i + C_i - S_i(Y_i)\bigr) + S(\mathbf{Y}) \\ = {} & \mbox{\{By (65)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{s-C_i}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + S(\mathbf{Y}) \\ = {} & \mbox{\{Rearranging\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{s' - C_i}{m} \biggr) U_i + \biggl( \frac{s - s'}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + S(\mathbf{Y}) \\ < {} & \mbox{\{Because $U^{+}\le m$ and $U_{i} \le1$ holds for each $i$\}} \\ & s-s' + \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{s' - C_i}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + S(\mathbf{Y}) \\ = {} & \mbox{\{By (66)\}} \\ & s - s' + \sum_{U^{+}-1~\mathrm{largest}} \bigl( x_i U_i + C_i - S_i(Y_i) \bigr) + S(Y) \\ = {} & \mbox{\{By the definition of $G(\mathbf{x}, \mathbf{Y})$ in (13)\}} \\ & s - s' + G\bigl(\mathbf{x}', \mathbf{Y}\bigr) + S( \tau) \\ = {} & \mbox{\{By (66)\}} \\ & s - s' + s' \\ = {} & \mbox{\{Cancelling\}} \\ & s. \end{aligned}$$

(67)

(67) is a contradiction, so the lemma is proven. □

Lemma 9

For any assignment of variables satisfying Constraint Sets 1–4,

$$ G+ S_{\mathrm{sum}}\ge G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}). $$

(68)

Proof

$$\begin{aligned} &G+ S_{\mathrm{sum}} \\ &\quad{}\ge \mbox{\{By (32) and Constraint~Set~{4}\}} \\ &\qquad\ \,{} \sum_{U^{+}-1~\mathrm{largest}} (x_i U_i + C_i - S_{i}) + \sum _{\tau_i \in\tau} S_{i} \\ &\quad{}\ge \mbox{\{By (31); observe that each $S_{i}$ appearing in the first summation} \\ &\qquad\ \,{} \mbox{also appears in the second\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \bigl(x_i U_i + C_i - S_i(Y_i)\bigr) + \sum _{\tau_i \in\tau} S_i(Y_i) \\ &\quad{}= \mbox{\{By the definition of $S(\mathbf{Y})$ in (10), and the definition of $G(\mathbf{x}, \mathbf{Y})$ in (13)\}} \\ &\qquad\ \,{} G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}). \end{aligned}$$

 □

Lemma 10

For any assignment of variables satisfying Constraint Sets 1–5, x is a near-compliant vector.

Proof

Consider an arbitrary assignment of variables satisfying Constraint Sets 1–5. For arbitrary i,

$$\begin{aligned} x_i = {} & \mbox{\{By Constraint~Set~{1}\}} \\ & \frac{s - C_i}{m} \\ \ge{} & \mbox{\{By Constraint~Set~{5}\}} \\ & \frac{G+ S_{\mathrm{sum}}- C_i}{m} \\ \ge{} & \mbox{\{By Lemma~{9}\}} \\ & \frac{G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{y}) - C_i}{m}. \end{aligned}$$

(69)

By (69) and Definition 1, x is a near-compliant vector. □

Lemma 11

If s<0, then Constraint Sets 1–5 are infeasible.

Proof

We use proof by contradiction. Assume Constraint Sets 1–5 are satisfied by some assignment of variables with

$$\begin{aligned} s <{}& 0. \end{aligned}$$

(70)

$$\begin{aligned} s \ge{} & \mbox{\{By Constraint~Set~{5}\}} \\ & G+ S_{\mathrm{sum}} \\ \ge{} & \mbox{\{By Lemma~{9}\}} \\ & G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) \\ = {} & \mbox{\{By (10) and (13)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \bigl(x_i U_i + C_i - S_i(Y_i)\bigr) + \sum _{\tau_i \in\tau} S_i(Y_i) \\ = {} & \mbox{\{By Constraint~Set~{1}\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{s-C_i}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + \sum_{\tau_i \in\tau} S_i(Y_i) \\ \ge{} & \mbox{\{By (70)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{-C_i}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + \sum_{\tau_i \in\tau} S_i(Y_i) \\ = {} & \mbox{\{Rearranging\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{C_i(m - U_i)}{m} \biggr) - S_i(Y_i) \biggr) + \sum _{\tau_i \in\tau} S_i(Y_i) \\ \ge{} & \mbox{\{Because each $S_{i}(Y_{i}) \ge0$ by the definition of $S_{i}(Y_{i})$ in (9); observe that each} \\ & \mbox{$S_{i}(Y_{i})$ in the first summation also appears in the second\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \frac{C_i(m-U_i)}{m} \\ \ge{} & \mbox{\{Because $U_{i} \le1 < m$\}} \\ & 0. \end{aligned}$$

(71)

(70) contradicts (71), so the lemma is proven. □

Lemma 12

Constraint Sets 1–5 are feasible.

Proof

We present an assignment of variables that satisfies Constraint Sets 1–5. Let C _max denote the largest value of C _i in the system. Let

$$\begin{aligned} &s = m^2 C_{\max} + m S(\mathbf{Y}), \end{aligned}$$

(72)

$$\begin{aligned} &\forall i,\quad x_i = \frac{s - C_i}{m}, \end{aligned}$$

(73)

$$\begin{aligned} &\forall i,\quad S_{i} = S_i(Y_i), \end{aligned}$$

(74)

$$\begin{aligned} &G= G(\mathbf{x}, \mathbf{Y}) , \end{aligned}$$

(75)

$$\begin{aligned} &b= \mbox{$\bigl(U^{+}-1\bigr)$th largest value of $x_{i} U_{i} + C_{i} - S_{i}$}, \end{aligned}$$

(76)

$$\begin{aligned} &\forall i,\quad z_i = \max\{0, x_i U_i + C_i - S_{i} - b\}, \end{aligned}$$

(77)

$$\begin{aligned} &S_{\mathrm{sum}}= S(\mathbf{Y}) . \end{aligned}$$

(78)

Constraint Set 1 holds by (73).

Constraint Set 2 holds by (74).

Constraint Set 3 holds by (75)–(77).

Constraint Set 4 holds by (74) and (78).

To see that Constraint Set 5 holds, note that

$$\begin{aligned} G+ S_{\mathrm{sum}}= {} & \mbox{\{By (75) and ( 78)\}} \\ & G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) \\ = {} & \mbox{\{By (13)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \bigl(x_i U_i + C_i - S_i(Y_i)\bigr) + S(\mathbf{Y}) \\ = {} & \mbox{\{By (72)--(73)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{m^2 C_{\max} + m S(\mathbf{Y}) - C_i}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + S(\mathbf{Y}) \\ < {} & \mbox{\{Because $U^{+}\le m$, $\forall i, C_{i} \le C_{\max} \wedge U_{i} \le1$\}} \\ & \sum_{m-1~\mathrm{largest}} \bigl(m C_{\max} + S( \mathbf{Y}) + C_{\max}\bigr) + S(\mathbf{Y}) \\ = {} & \mbox{\{Rearranging\}} \\ & \bigl(m^2 - 1\bigr) C_{\max} + m S(\mathbf{Y}) \\ < {} & \mbox{\{By (72)\}} \\ & s. \end{aligned}$$

 □

Lemma 13

Let C _max denote the largest value of C _i in the system. If U ⁺>1 and s<C _max, then Constraint Sets 1–5 are infeasible.

Proof

We use proof by contradiction. Assume U ⁺>1 and Constraint Sets 1–5 are satisfied by some assignment of variables with

$$\begin{aligned} s <{}& C_{\max}. \end{aligned}$$

(79)

$$\begin{aligned} s \ge{} & \mbox{\{By Constraint~Set~{5}\}} \\ & G+ S_{\mathrm{sum}} \\ \ge{} & \mbox{\{By Lemma~{9}\}} \\ & G(\mathbf{x}, \mathbf{Y}) + S(\mathbf{Y}) \\ = {} & \mbox{\{By the definition of $S(\mathbf{Y})$ in (10), and the definition of $G(\mathbf{x}, \mathbf{Y})$ in (13)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \bigl(x_i U_i + C_i - S_i(Y_i)\bigr) + \sum _{\tau_i \in\tau} S_i(Y_i) \\ = {} & \mbox{\{By Constraint~Set~{1}\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{s-C_i}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + \sum_{\tau_i \in\tau} S_i(Y_i) \\ \ge{} & \mbox{\{By (79)\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \biggl( \biggl( \frac{C_{\max}-C_i}{m} \biggr) U_i + C_i - S_i(Y_i) \biggr) + \sum_{\tau_i \in\tau} S_i(Y_i) \\ \ge{} & \mbox{\{Because $C_{\max} \ge C_{i}$ for all $i$\}} \\ & \sum_{U^{+}-1~\mathrm{largest}} \bigl(C_i - S_i(Y_i)\bigr) + \sum_{\tau_i \in\tau} S_i(Y_i) \\ \ge{} & \mbox{\{Because each $S_{i}(Y_{i}) \ge0$ by the definition of $S_{i}(Y_{i})$ in (9); observe that each} \\ & \mbox{$S_{i}(Y_{i})$ in the first summation also appears in the second.\}} \\ & \sum_{U^{+}-1~\max} C_i \\ = {} & \mbox{\{By the definition of $C_{\max}$, because $U^{+}> 1$\}} \\ & C_{\max}. \end{aligned}$$

(80)

(79) contradicts (80), so the lemma is proven. □