Using genetic algorithms for early schedulability analysis and stress testing in real-time systems

Abstract

Reactive real-time systems have to react to external events within time constraints: Triggered tasks must execute within deadlines. It is therefore important for the designers of such systems to analyze the schedulability of tasks during the design process, as well as to test the system's response time to events in an effective manner once it is implemented. This article explores the use of genetic algorithms to provide automated support for both tasks. Our main objective is then to automate, based on the system task architecture, the derivation of test cases that maximize the chances of critical deadline misses within the system; we refer to this testing activity as stress testing. A second objective is to enable an early but realistic analysis of tasks' schedulability at design time. We have developed a specific solution based on genetic algorithms and implemented it in a tool. Case studies were run and results show that the tool (1) is effective at identifying test cases that will likely stress the system to such an extent that some tasks may miss deadlines, (2) can identify situations that were deemed to be schedulable based on standard schedulability analysis but that, nevertheless, exhibit deadline misses.

Appendices

Appendix A: Schedulability analysis for Example 2

The Generalized Completion Time Theorem (GCTT) determines whether periodic tasks are schedulable, i.e., whether tasks can meet their deadlines, given preemption by higher priority tasks and blocking time by lower priority tasks. First note that this applies only to periodic tasks. However, for schedulability analysis, aperiodic tasks are considered equivalent to periodic tasks whose periods are equal to the minimum inter-arrival times [5]. The theorem assumes all tasks are ready for execution at the start of the task's period: i.e., at time zero, all tasks are awaiting execution [16]. It is important to note that the GCTT can only determine whether a group of tasks is schedulable. It cannot show whether they are not schedulable. (Hence, if the GCTT check fails, the group of tasks at hand may still be schedulable.) Additionally, it is argued [16] that “when all the tasks are initiated at the same time (the worst-case phasing), if a task completes its execution before the end of its first period, it will never miss a deadline.”

As discussed before, GCTT can be graphically illustrated by a timing diagram [5]. We rather focus here on its mathematical formulation.

A set of n periodic tasks can be scheduled by the Rate Monotonic Algorithm⁴ for all tasks phasing if the following condition is satisfied:

$$ \forall i,1 \le i \le n,\mathop {\min }\limits_{(k,l) \in R_i } \left( {\sum\limits_{j = 1}^{i - 1} {\frac{{C_j }}{{lT_k }}\left\lceil {\frac{{lT_k }}{{T_j }}} \right\rceil } + \frac{{C_j }}{{lT_k }} + \frac{{B_i }}{{lT_k }}} \right) \le 1 $$

where

$$ \,R_i = \left\{ {(k,l)\,|\,1 \le k \le i,l = 1, \ldots ,\left\lfloor {\frac{{T_i }}{{T_k }}} \right\rfloor } \right\}. $$

(C _i and T _j are the execution time and period of task i. B _i is the worst-case blocking time for task i)⁵

The theorem checks if each task can complete its execution before its first deadline, by checking the amount of time spent waiting for higher priority tasks (preemption) and lower dependent priority tasks (blocking). In the formula, i denotes the task to be checked and k denotes each of the tasks that affects the completion time of task i, i.e., task i and the higher priority tasks. For a given i and k, each value of l represents the number of times task k can execute during task i's period, which corresponds to the amount of time during which task i is preempted. Note that when determining B _i , each task on which i depends can block i only once because of priority differences.

Let us apply this theorem to the example in Section 4.2. The task architecture is recalled here to ease reading: Task t1 is periodic (C ₁ = 1, T ₁ = 3); Task t2 is aperiodic but transformed into a periodic task for schedulability analysis purposes (C ₂ = 3, T ₂ = 9); Task t3 is periodic (C ₃ = 2, T ₃ = 9); t1 and t3 are inter-dependent; t1 has the highest priority, t2 has the second highest priority and t3 has the lowest priority.

Let us apply the theorem on each of the three tasks in sequence.

1.1 Task t1

$$ R_1 = \left\{ {(k,l)|1 < k < 1,l = 1, \ldots ,\left\lfloor {\frac{{T_1 }}{{T_k }}} \right\rfloor } \right\} = \{ (1,1)\} . $$

Then,

$$\displaylines{\mathop {\min }\limits_{(k,1) \in R_1 } \Bigg(\sum\limits_{j = 1}^0 {\frac{{C_j }}{{lT_k }}\left\lceil {\frac{{lT_k }}{{T_j }}} \right\rceil } + \frac{{C_1 }}{{lT_k }} + \frac{{B_1 }}{{lT_k }}\Bigg)\cr = \mathop {\min }\limits_{(1,1)} \bigg(\frac{{C_1 }}{{lT_k }} + \frac{{B_1 }}{{lT_k }}\bigg) = \frac{{C_1 }}{{T_1 }} + \frac{{B_1 }}{{T_1 }} = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} < 1. } $$

Indeed, t1 can only be blocked by lower priority task t3 once during a period of 2 time units: B ₁ = 2. This means that t1 meets its first deadline.

1.2 Task t2

$$R_2 = \left\{ {(k,l)|1 \le k \le 2,l = 1, \ldots ,\left\lfloor {\frac{{T_2 }}{{T_k }}} \right\rfloor } \right\}.$$

For $k = 1,2, {\rm that\,\, is} , l= 1,2,3.\,\, {\rm For}\,\, k = 2, l=1, \ldots, \lfloor {\frac{{T_2 }}{{T_2 }}} \rfloor$ that is, l = 1. Thus R ₂ = {(1, 1), (1, 2), (1, 3), (2, 1)}. Note that since t2 is not dependent on the other two tasks, especially t3 (lowest priority task), t2 cannot be blocked by t3 and then B₂ = 0.

$$\displaylines{ {\rm For}\,(k,l) = (1,1),\,\sum\limits_{j = 1}^1 {\frac{{C_j }}{{T_1 }}} \left\lceil {\frac{{T_1 }}{{T_j }}} \right\rceil + \frac{{C_2 }}{{T_1 }} + \frac{{B_2 }}{{T_1 }} = \frac{{C_1 }}{{T_1 }}\left\lceil {\frac{{T_1 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{T_1 }}\cr = \frac{{C_1 }}{{T_1 }} + \frac{{C_2 }}{{T_1 }} = \frac{1}{3} + \frac{3}{3} > 1. }$$

$$\displaylines{ {\rm For}\,(k,l) = (1,2),\sum\limits_{j = 1}^1 {\frac{{C_j }}{{2T_1 }}} \left\lceil {\frac{{2T_1 }}{{T_j }}} \right\rceil + \frac{{C_2 }}{{2T_1 }} + \frac{{B_2 }}{{2T_1 }} = \frac{{C_1 }}{{2T_1 }}\left\lceil {\frac{{2T_1 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{2T_1 }}\cr = \frac{{2C_1 }}{{2T_1 }} + \frac{{C_2 }}{{2T_1 }} = \frac{2}{6} + \frac{3}{6} < 1. } $$

$$\displaylines{{\rm For}\,(k,l) = (1,3),\sum\limits_{j = 1}^1 {\frac{{C_j }}{{3T_1 }}} \left\lceil {\frac{{3T_1 }}{{T_j }}} \right\rceil + \frac{{C_2 }}{{3T_1 }} + \frac{{B_2 }}{{3T_1 }} = \frac{{C_1 }}{{3T_1 }}\left\lceil {\frac{{3T_1 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{3T_1 }}\cr = \frac{{3C_1 }}{{3T_1 }} + \frac{{C_2 }}{{3T_1 }} = \frac{3}{9} + \frac{3}{9} < 1. }$$

$$\displaylines{{\rm For}\,(k,l) = (2,1),\sum\limits_{j = 1}^1 {\frac{{C_J }}{{T_2 }}} \left\lceil {\frac{{T_2 }}{{T_j }}} \right\rceil + \frac{{C_2 }}{{T_2 }} + \frac{{B_2 }}{{T_2 }} = \frac{{C_1 }}{{T_2 }}\left\lceil {\frac{{T_2 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{T_2 }}\cr = \frac{{3C_1 }}{{T_2 }} + \frac{{C_2 }}{{T_2 }} = \frac{3}{8} + \frac{3}{8} \le 1.} $$

This means that t2 meets its first deadline.⁶

1.3 Task t3

$$R_3 = \left\{ {(k,l)|1 \le k \le 3,l = 1, \ldots ,\left\lfloor {\frac{{T_3 }}{{T_k }}} \right\rfloor } \right\}.$$

For k = 1, $l = 1, \ldots ,\lfloor {\frac{{T_3 }}{{T_1 }}} \rfloor ,$ that is, l = 1, 2, 3. For k = 2, $l = 1, \ldots ,\lfloor {\frac{{T_3 }}{{T_2 }}} \rfloor ,$ that is, l = 1. For k = 3, $l = 1, \ldots ,\lfloor {\frac{{T_3 }}{{T_3 }}} \rfloor ,$ that is l = 1. Since there is no lower priority task than t3, B ₃ = 0.

$$\displaylines{ {\rm For}\,(k,l) = (1,1),\sum\limits_{j = 1}^2 {\frac{{C_j }}{{T_1 }}} \left\lceil {\frac{{T_1 }}{{T_j }}} \right\rceil + \frac{{C_3 }}{{T_1 }} + \frac{{B_3 }}{{T_1 }} = \frac{{C_1 }}{{T_1 }}\left\lceil {\frac{{T_1 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{T_1 }}\left\lceil {\frac{{T_1 }}{{T_2 }}} \right\rceil\cr \qquad + \frac{{C_3 }}{{T_1 }} + \frac{{B_3 }}{{T_1 }} = \frac{{C_1 }}{{T_1 }} + \frac{{C_2 }}{{T_1 }} + \frac{{C_3 }}{{T_1 }} + \frac{{B_3 }}{{T_1 }} = \frac{6}{3} > 1. } $$

$$\displaylines{ {\rm For}\,(k,l) = (1,2),\sum\limits_{j = 1}^2 {\frac{{C_j }}{{2T_1 }}} \left\lceil {\frac{{2T_1 }}{{T_j }}} \right\rceil + \frac{{C_3 }}{{2T_1 }} + \frac{{B_3 }}{{2T_1 }}\cr = \frac{{C_1 }}{{2T_1 }}\left\lceil {\frac{{2T_1 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{2T_1 }}\left\lceil {\frac{{2T_1 }}{{T_2 }}} \right\rceil + \frac{{C_3 }}{{2T_1 }} + \frac{{B_3 }}{{2T_1 }}\cr = \frac{{2C_1 }}{{2T_1 }} + \frac{{C_2 }}{{2T_1 }} + \frac{{C_3 }}{{2T_1 }} + \frac{{B_3 }}{{2T_1 }} = \frac{7}{6} > 1. }$$

$$\displaylines{ {\rm For}\,(k,l) = (1,3),\sum\limits_{j = 1}^2 {\frac{{C_j }}{{3T_1 }}} \left\lceil {\frac{{3T_1 }}{{T_j }}} \right\rceil + \frac{{C_3 }}{{3T_1 }} + \frac{{B_3 }}{{3T_1 }}\cr = \frac{{C_1 }}{{3T_1 }}\left\lceil {\frac{{3T_1 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{3T_1 }}\left\lceil {\frac{{3T_1 }}{{T_2 }}} \right\rceil + \frac{{C_3 }}{{3T_1 }} + \frac{{B_3 }}{{3T_1 }} \cr = \frac{{3C_1 }}{{3T_1 }} + \frac{{2C_2 }}{{3T_1 }} + \frac{{C_3 }}{{3T_1 }} + \frac{{B_3 }}{{3T_1 }} = \frac{{11}}{9} > 1. }$$

$$\displaylines{{\rm For}\,(k,l) = (2,1),\sum\limits_{j = 1}^2 {\frac{{C_J }}{{T_2 }}} \left\lceil {\frac{{T_2 }}{{T_j }}} \right\rceil + \frac{{C_3 }}{{T_2 }} + \frac{{B_3 }}{{T_2 }} = \frac{{C_1 }}{{T_2 }}\left\lceil {\frac{{T_2 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{T_2 }}\left\lceil {\frac{{T_2 }}{{T_2 }}} \right\rceil + \frac{{C_3 }}{{T_2 }} + \frac{{B_3 }}{{T_2 }}\cr = \frac{{3C_1 }}{{T_2 }} + \frac{{C_2 }}{{T_2 }} + \frac{{C_3 }}{{T_2 }} + \frac{{B_3 }}{{T_2 }} = \frac{8}{9} < 1. }$$

$$\displaylines{{\rm For}\,(k,l) = (3,1),\sum\limits_{j = 1}^2 {\frac{{C_j }}{{T_3 }}} \left\lceil {\frac{{T_3 }}{{T_j }}} \right\rceil + \frac{{C_3 }}{{T_3 }} + \frac{{B_3 }}{{T_3 }} = \frac{{C_1 }}{{T_3 }}\left\lceil {\frac{{T_3 }}{{T_1 }}} \right\rceil + \frac{{C_2 }}{{T_3 }}\left\lceil {\frac{{T_3 }}{{T_2 }}} \right\rceil + \frac{{C_3 }}{{T_3 }} + \frac{{B_3 }}{{T_3 }}\cr = \frac{{3C_1 }}{{T_3 }} + \frac{{2C_2 }}{{T_3 }} + \frac{{C_3 }}{{T_3 }} + \frac{{B_3 }}{{T_3 }} = \frac{{11}}{9} > 1. }$$

This means that t3 meets its first deadline.⁶

1.4 Conclusion

As a conclusion, all three tasks meet their first deadlines, which demonstrates their schedulability.

Table 6. Executing RTTT once for each task in the avionics application (complete results)

Table 7 Executing RTTT 10 times for each task in the avionics application (complete results)

Table 7 Continued

Appendix B:Additional results for the industrial case study

Tables 6 and 7 below show the complete results of executing RTTT for each task of the Avionics case study. Results in Table 6 are partially presented in Table 4 whereas results in Table 7 are partially presented in Table 5.