Laxity dynamics and LLF schedulability analysis on multiprocessor platforms

Abstract

LLF (Least Laxity First) scheduling, which assigns a higher priority to a task with a smaller laxity, has been known as an optimal preemptive scheduling algorithm on a single processor platform. However, little work has been made to illuminate its characteristics upon multiprocessor platforms. In this paper, we identify the dynamics of laxity from the system’s viewpoint and translate the dynamics into LLF multiprocessor schedulability analysis. More specifically, we first characterize laxity properties under LLF scheduling, focusing on laxity dynamics associated with a deadline miss. These laxity dynamics describe a lower bound, which leads to the deadline miss, on the number of tasks of certain laxity values at certain time instants. This lower bound is significant because it represents invariants for highly dynamic system parameters (laxity values). Since the laxity of a task is dependent of the amount of interference of higher-priority tasks, we can then derive a set of conditions to check whether a given task system can go into the laxity dynamics towards a deadline miss. This way, to the author’s best knowledge, we propose the first LLF multiprocessor schedulability test based on its own laxity properties. We also develop an improved schedulability test that exploits slack values. We mathematically prove that the proposed LLF tests dominate the state-of-the-art EDZL tests. We also present simulation results to evaluate schedulability performance of both the original and improved LLF tests in a quantitative manner.

Appendices

Appendix A: Proof of Theorem 1

The basic idea of the proof is to use mathematical induction.

(Basis) Each Inequality (10) for x=1 and x=2 holds by Observations 2 and 4, respectively.

(Inductive step) We will prove the following statements: for all x≥2, if Inequality (10) for x holds, then Inequality (10) for x+1 also holds. We consider two cases depending on the value of $\mathbf{1}_{t_{0}-x-1}$.

(Inductive step: case A) Assume $\mathbf{1}_{t_{0}-x-1}=1 \iff\sum_{j=0}^{x} N_{j}(t_{0}-x-1) \le m$.

All tasks in S _j (t ₀−x−1) for 0≤j<x are serviced in [t ₀−x−1,t ₀−x), and thus L _k (t ₀−x−1)=L _k (t ₀−x) for all tasks τ _k ∈S _j (t ₀−x−1) where 0≤j<x. So, the following condition holds:

(25)

Applying Eq. (25) to Inequality (10), we get the following condition:

(26)

Multiplying the above condition by $\frac{x+1}{x}$, we get the following condition:

(27)

We now look at the three terms of the LHS of Inequlaity (27). The first term can be upper-bounded as follows:

(28)

This is because we observe the following condition for x≥1, x≥j≥0:

(29)

If p=x+1, then $\mathbf{1}_{t_{0}-p}=1$ and $(1+\frac{\mathbf {1}_{t_{0}-p}}{p-1}) = \frac{x+1}{x}$. Therefore, the second term of the LHS of Inequlaity (27) can be rearranged if k=x+1 as follows:

(30)

And, the third term of the LHS of Inequality (27) can be also rearranged as follows:

(31)

Then, once we apply Eqs. (28), (30) and (31) to Inequality (27), then the LHS of Inequality (27) can be upper-bounded as follows:

(32)

We can now check that Inequality (10) holds for x+1 by adding the non-negative term for j=x to the first summation of the RHS of Inequality (32). Finally, we conclude that if Inequality (10) for x is true, then Inequality (10) for x+1 is also true when $\mathbf {1}_{t_{0}-x-1}=1$.

(Inductive step: case B) Assume $\mathbf{1}_{t_{0}-x-1}=0 \iff\sum_{j=0}^{x} N_{j}(t_{0}-x-1) > m$.

Since at most m tasks can be serviced in [t ₀−x−1,t ₀−x), there exists a minimum number y (≤x) such that at least one of the tasks in S _y (t ₀−x−1) is not serviced in [t ₀−x−1,t ₀−x). It holds that y≥1 because otherwise N ₀(t ₀−x−1)>m, which means t ₀−x is the first instant when there is a task with a negative laxity. Thus, all tasks in S _j (t ₀−x−1) for j=0,…,y−1 are serviced while all tasks in S _j (t ₀−x−1) for j=y+1,…,x are not serviced. Among tasks in S _y (t ₀−x−1), $\sum_{j=0}^{y}N_{j}(t_{0}-x-1)-m$ tasks are not serviced and $m-\sum_{j=0}^{y-1}N_{j}(t_{0}-x-1)$ tasks are serviced. Considering that serviced tasks keep their laxity values and non-serviced ones reduce their laxity values by one, we establish the following relationship between N _j (t ₀−x) and N _j (t ₀−x−1):

(33)

where

The detailed calculation of Q is as follows. For j=0,…,y−1, all tasks in S _j (t ₀−x−1) keep their laxity values at t ₀−x, and therefore the first two cases include N _j (t ₀−x−1) terms. Among tasks in S _y (t ₀−x−1), $(\sum_{k=0}^{y}N_{k}(t_{0}-x-1) -m)$ tasks do not perform their executions in [t ₀−x−1,t ₀−x) and then each of their laxity values is y−1 at t ₀−x. In turn, $(m-\sum_{k=0}^{y-1}N_{k}(t_{0}-x-1) )$ tasks perform its execution in [t ₀−x−1,t ₀−x) and then each of their laxity values is y at t ₀−x. Therefore, the second and third cases have the corresponding terms. For j=y,…,x−1, all tasks in S _j+1(t ₀−x+1) do not perform their executions in [t ₀−x−1,t ₀−x) and then each of their laxity values is j at t ₀−x. Thus, the third and fourth cases includes N _j+1(t ₀−x−1) terms.

By putting the above equation into Inequality (10), we derive the following condition:

(34)

We re-arrange terms as follows:

(35)

To re-arrange the first two lines of the above inequality, we use the following trivial equations.

• $\sum_{j=y}^{x-1} (x-j)\cdot N_{j+1}(t_{0}-x-1)$ is equal to $\sum_{j=y+1}^{x} (x-j+1)\cdot N_{j}(t_{0}-x-1)$.

• $(x-(y-1))\cdot\sum_{j=0}^{y} N_{j}(t_{0}-x-1) - (x-y)\cdot\sum_{j=0}^{y-1} N_{j}(t_{0}-x-1)$ is equal to $(x-(y-1))\cdot N_{y}(t_{0}-x-1)+\sum_{j=0}^{y-1} N_{j}(t_{0}-x-1)$

Using the above equations, we derive the following equation.

(36)

We now re-arrange the fourth and fifth lines of the Inequality (35). Using $\mathbf{1}_{t_{0}-x-1}=0$, we obtain $\prod_{p=x+1}^{x+1}(1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})=1$, and then the fourth line is equal to $\prod_{p=x+1}^{x+1}(1+\frac{\mathbf {1}_{t_{0}-p}}{p-1})\sum_{j=0}^{x-1} (x-j)\cdot (Z_{j}(t_{0}-x-1)-W_{j}(t_{0}-x-1) )$. Once we substitute k=x+1, the fourth line is equal to $\prod_{p=x+1}^{k}(1+\frac{\mathbf {1}_{t_{0}-k}}{k-1})\sum_{j=0}^{k-2} (k-j-1)\cdot (Z_{j}(t_{0}-k)-W_{j}(t_{0}-k) )$. Then,

(37)

We add m for both the LHS and RHS of the Inequality (35) to eliminate the third line, and replace the first two lines, and the third and fourth lines by Eqs. (36) and (37), respectively. We finally obtain the following inequality.

(38)

The last condition is identical to Inequality (10) for x+1. Finally, we conclude that if Inequality (10) for x is true, then Inequality (10) for x+1 is also true when $\mathbf{1}_{t_{0}-x-1}=0$.

By (Inductive step: case A) and (Inductive step: case B), the inductive step is correct.

Appendix B: Proof of Lemma 4

We show that the LHS of Inequality (19) is equal to or larger than that of Inequality (10) for x=1,2,3,…,∞. Then Inequality (19) is a necessary condition of Inequality (10), and then the lemma directly follows from Theorem 1, where Inequality (19) is as follows:

and Inequality (10) is as follows:

We investigate how much individual tasks contribute to the LHS of Inequality (10) and to that of Inequality (19) for given x. Then we prove that the contribution of task τ _k to the LHS of Inequality (19) is always equal to or larger than that to the LHS of Inequality (10). We let (A _x ) and (B _x ) denote the LHS of Inequality (19) and the LHS of Inequality (10) for given x, respectively. We consider two cases depending on whether task τ _k is active at t ₀−x or not.

Before investigating each case, we first observe when task τ _k contributes to (B _x ) through Z _θ (t ₀−y), W _θ (t ₀−y) and N _θ (t ₀−y). Task τ _k contributes to (B _x ) through exactly one N _θ (t ₀−x)-term if it has an active job at t ₀−x, and there is no contribution through any N _θ (t ₀−x)-term if it has no active job at t ₀−x. Since Z _θ (t ₀−y) denotes the number of tasks whose jobs are released at t ₀−y+1 with a laxity of θ, task τ _k contributes to (B _x ) through Z _θ (t ₀−y)-terms only when its job is released at t ₀−y+1, and then θ is always D _k −C _k . Similarly, since W _θ (t ₀−y) denotes the number of tasks whose jobs are finished at t ₀−y+1 and have a laxity of θ at t ₀−y, task τ _k contributes to (B _x ) through W _θ (t ₀−y)-terms only when it finishes execution at t ₀−y+1. In addition, a job of τ _k can contribute to (B _x ) through at most one Z _θ (t ₀−y)- and at most one W _θ (t ₀−y)-terms.

We now derive upper bounds of contribution of some W-, Z- and N-terms in the following two lemmas.

Lemma 8

The sum of the contribution of the W-term of τ _k ’s qth job and the Z-term of τ _k ’s (q+1)th job to (B _x ) is upper-bounded by zero, if the finishing time of τ _k ’s qth job and the release time of τ _k ’s (q+1)th job are in [t ₀−x+1,t ₀).

Proof

Suppose that τ _k ’s qth job finishes its execution at t ₀−y+1 (1≤y≤x), and it has a laxity of θ at t ₀−y. And, suppose that τ _k ’s (q+1)th job released at t ₀−y′+1 (1≤y′≤y). Then, by definition, contribution of the W-term of τ _k ’s qth job and the Z-term of τ _k ’s (q+1)th job to (B _x ) is $-\prod_{p=y}^{x} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})\cdot (y-\theta-1)$ and $\prod_{p=y'}^{x} (1+\frac{\mathbf {1}_{t_{0}-p}}{p-1})\cdot(y'-(D_{k}-C_{k})-1)$, respectively. Then, the sum of contribution of the W-term of τ _k ’s qth job and the Z-term of τ _k ’s (q+1)th job to (B _x ) is calculated by

(39)

Once we apply θ≤D _k −C _k and $\prod_{p=y'}^{y-1} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})\ge1$, we derive that $( \theta -\prod_{p=y'}^{y-1} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})\cdot(D_{k}-C_{k}) )$ is upper-bounded by zero. Once we apply $\prod_{p=y'}^{y-1} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1}) \le\frac{y'}{y'-1}\cdot\frac {y'+1}{y'}\cdot\cdots \cdot\frac{y-1}{y-2}=\frac{y-1}{y'-1}$, we derive that $( -(y-1) + \prod_{p=y'}^{y-1} (1+\frac{\mathbf {1}_{t_{0}-p}}{p-1})\cdot(y'-1) )$ is also upper-bounded by zero. Therefore, the RHS of Eq. (39) is upper-bounded by zero. □

Lemma 9

The sum of the contribution of the N-term of τ _k ’s qth job, the W-term of τ _k ’s qth job, and the Z-term of τ _k ’s (q+1)th job to (B _x ) is upper-bounded by x−D _k +C _k , if τ _k ’s qth job is active at t ₀−x, the finishing time of τ _k ’s qth job is in [t ₀−x+1,t ₀), and the release time of τ _k ’s (q+1)th job is in [t ₀−x+1,t ₀).

Proof

Assume that τ _k ’s qth job has a laxity of θ′ and θ at t ₀−x and t ₀−y (1≤y≤x), respectively, and it finishes its execution at t ₀−y+1. And, assume that τ _k ’s (q+1)th job released at t ₀−y′+1 (1≤y′≤y). By definition, contribution of the N-term of τ _k ’s qth job amounts to x−θ′. Then, the sum of contribution of the N-term and W-term of τ _k ’s qth job and the Z-term of τ _k ’s (q+1)th job to (B _x ) amounts to x−θ′ plus the RHS of Eq. (39) as follows:

(40)

Once we apply θ≤θ′≤D _k −C _k to the RHS of Eq. (40), we can derive the following conditions:

(41)

Since $\prod_{p=y'}^{x} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})\ge\prod_{p=y}^{x} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})$, the RHS of Eq. (41) is upper-bounded by x−D _k +C _k , and this proves the lemma. □

We now prove Lemma 4 by proving (A _x ) is larger than or equal to (B _x ) for a given x. To do this, we use Lemmas 8 and 9, and the fact that the contribution of each W-term is no larger than zero.

Suppose that τ _k ’s qth job is released strictly before t ₀−x, and τ _k ’s (q+1)th job is released at t ₀−y (≥t ₀−x). And, τ _k ’s pth job is released strictly before t ₀, and τ _k ’s (p+1)th job is released at t ₀+v (≥t ₀). Then, among jobs invoked by τ _k , only the (q+1)th, (q+2)th, …, and pth jobs contribute to (B _x ) through their Z-terms. And the (q+1)th, (q+2)th, …, and (p−1)th jobs of τ _k definitely contribute to (B _x ) through their W-terms, and the qth and pth jobs of τ _k may or may not contribute to (B _x ) through their W-terms depending on their finishing times. We now consider two cases as follows.

(Case A: task τ _k ’s qth job is not active at t ₀−x)

Since there is no active job of τ _k at t ₀−x, jobs of τ _k cannot contribute to (B _x ) through any N-term. By Lemma 8, the contribution of a job of task τ _k through its W-term and the next job of task τ _k through its Z-term to (B _x ) is upper-bounded by zero. Therefore, the contribution of the (q+1)th, …, and (p−1)th jobs of τ _k through their W-terms and the (q+2)th, …, and pth jobs of τ _k through their Z-terms is also upper-bounded by zero. And, the qth and pth jobs of τ _k may contribute to (B _x ) through their W-terms, but any contribution though W-terms is non-positive. What remains is to calculate the contribution of (q+1)th jobs to (B _x ) through its Z-term.

The contribution of task τ _k ’s (q+1)th job to (B _x ) through its Z-term is $\prod_{p=y}^{x} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})\cdot (y-(D_{k}-C_{k})-1)$. Since $\prod_{p=y}^{x} (1+\frac{\mathbf {1}_{t_{0}-p}}{p-1})$ is upper-bounded by $\frac{y}{y-1}\cdot\frac {y+1}{y}\cdot\cdots \cdot\frac{x}{x-1}=\frac{x}{y-1}$, $\prod_{p=y}^{x} (1+\frac{\mathbf{1}_{t_{0}-p}}{p-1})\cdot(y-(D_{k}-C_{k})-1)$ is also upper-bounded by $(y-(D_{k}-C_{k})-1)\cdot\frac{x}{y-1}= x - (D_{k}-C_{k})\frac {x}{y-1}\le x-D_{k}+C_{k}$. Therefore, the total contribution of jobs of task τ _k to (B _x ) is upper-bounded by x−D _k +C _k .

By definition of $\delta_{k}^{*}(\theta,x)$, it holds that θ≤D _k −C _k , and there exists only one θ that results in $\delta_{k}^{*}(\theta,x)=1$ for a given x. Thus, the contribution of jobs of τ _k to (A _x ) is at least x−D _k +C _k .

Therefore, the total contribution of jobs of τ _k to (A _x ) is larger than or equal to that to (B _x ).

(Case B: task τ _k ’s qth job is active at t ₀−x (i.e., an N-term contributes to (B _x ).))

This case implies τ _k ’s qth job is finished after t ₀−x, which means it contributes to (B _x ) through its W-term. We consider two sub-cases: (Case B-1) the execution of τ _k ’s qth job is finished strictly before t ₀; and (Case B-2) at or after t ₀.

(Case B-1) By Lemma 9, the contribution of the qth job through its N- and W-terms and the (q+1)th job through its Z- term is upper-bounded by x−D _k +C _k . By Lemma 8, the contribution of the (q+1)th, …, and (p−1)th jobs of τ _k through their W-terms and the (q+2)th, …, and pth jobs of τ _k through their Z-terms is also upper-bounded by zero. Here the pth job of τ _k may contributes to (B _x ) through its W-term, but the quantity is no larger than zero. Therefore, the total contribution of jobs of τ _k to (B _x ) is upper-bounded by x−D _k +C _k , and as shown in (Case 1) this means the total contribution of jobs of τ _k to (A _x ) is larger than or equal to that to (B _x ).

(Case B-2) In this case, the N-term of the qth job of τ _k is the only term that contributes (B _x ) by jobs of τ _k ; any other jobs of τ _k cannot contribute to (B _x ) through any other terms. Suppose that the qth job has a laxity of θ at t ₀−x, then the contribution through its N-term is x−θ. Since the finishing time of the qth job is at or after t ₀, the release time of the job, t ₀−z, is no earlier than t ₀−D _k (i.e., t ₀−z≥t ₀−D _k ⇔z≤D _k ). Then, t ₀−x is at x+D _k −z time units ahead of its deadline.

On the other hand, if δ _k (θ′,x)=1, then θ′≤θ because x≤x+(D _k −z). Therefore, the contribution of the job to (A _x ), x−θ′, is no smaller than x−θ. Therefore, the total contribution of task τ _k to (A _x ) is larger than or equal to that to (B _x ).

By (Case A) and (Case B), the contribution of jobs of τ _k to (A _x ) is larger than or equal to that to (B _x ).