LỊCH KHAI GIANG CAC LƠP TIENG ANH THANG 3

Ngày tải lên: 25/01/2013, 17:09

... task priority becomes variable. Ready tasks 4 8 11 15 18 22 26 30 Quantum = 4 C 1 = 20 t 0 τ 1 τ 1 τ 2 τ 3 C 2 = 7 τ 2 C 3 = 3 τ 3 Figure 1. 13 Example of Round-Robin scheduling 1.2 BASIC CONCEPTS ... CONCEPTS FOR REAL-TIME TASK SCHEDULING 19 Ready tasks 1 4 11 15 t Prio i C i 37 11 4 32 4 0 τ 4 τ 3 τ 2 τ 1 τ 1 τ 3 τ 4 τ 2 Figure 1.14 Example of priority scheduling (the l...

Ngày tải lên: 08/11/2013, 00:15

... schedule EDF schedule LLF schedule Task t 3 misses its deadline t 2 214567891 031 112 13 214567891 031 112 13 214567891 031 112 13 214567891 031 112 13 t t t t t 1 t 3 Task t 2 misses its deadline Figure ... response times of tasks τ 4 , τ 5 and τ 6 . 2 .3. 2 Answers Exercise 2.1: Task set schedulability Q1 U = 0 .33 + 0.25 + 0 .33 = 0.92 Major cycle = [0, LCM (3, 4, 6)] = [0, 12]...

Ngày tải lên: 08/11/2013, 00:15

... execution is shown in Figure 3. 23. τ 1 τ 2 τ 3 1 234 56789101112 131 41516171819202122 232 425 1 234 56789101112 131 41516171819202122 232 425 1 234 56789101112 131 41516171819202122 232 425 Priority inheritance Showing ... 75 τ 1 τ 2 τ 3 τ 4 t 1 2 3 4 5 6 7 8 9 101112 131 41516171819202122 232 4 t 1 2 3 4 5 6 7 8 9 101112 131 41516171819202122 232 4 t 1 2 3 4 5 6 7 8 9 101112 13...

Ngày tải lên: 15/12/2013, 11:15

... 20 0 t 1 t 2 t 3 t 3 p t 3 a 22 24 26 28 30 32 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 t t t Instance #1 Instance #2 Instance #3 Instance #4 Figure ... τ 2 (0, 6, 32 , 32 ), and task τ 3 with primary and 84 4 SCHEDULING SCHEMES FOR HANDLING OVERLOAD t 1 t 2 t 3 0 2 4 6 8 101214161820 0 22 24 26 28 30 32 0 2 4 6 8 10...

Ngày tải lên: 15/12/2013, 11:15

... SCHEDULING 1 234 56789101112 131 41516171819202122 232 425 1 234 56789101112 131 41516171819202122 232 425 1 2 3 1 234 56789101112 131 41516171819202122 232 425 t t t 1 234 56789101112 131 41516171819202122 232 425 t 1 234 56789101112 131 41516171819202122 232 425 t 4 5 : ... Figure 5.8. Since C + is equal to 6, C S /3 is equal to 4.66 (14 /3) and thus the t 1 t 2 t 3 t 4 Proc 1 Proc...

Ngày tải lên: 15/12/2013, 11:15

... ms M 2 M 1 M 1 M 1 M 1 M 1 M 1 M 1 M 1 M 1 M 1 M 1 M 1 M 3 M 4 M 5 M 6 M 6 5 M 1 10 M 2 M 2 15 M 3 M 3 20 M 2 M 4 M 5 M 5 25 M 1 30 M 2 M 3 M 6 M 6 35 M 1 40 M 2 M 4 M 5 M 5 45 M 3 M 3 50 M 2 M 2 55 M 1 Macrocycle =60 ... τ 1 and τ 3 . The preceding lemma is checked because we have: r ∗ 1 ≤ r ∗ 3 ,d ∗ 1 ≤ d ∗ 3 and C 1 + C 3 ≤ d ∗ 3 − r ∗ 1 (6.15) 6.6 EXERCISE 6.1: J...

Ngày tải lên: 15/12/2013, 11:15

... 1 AT 1 c,2 AT 1 c ,3 AT 1 c,4 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Switch 2 t AT 2 c,1 ExD 2 c,1 ExD 2 c,2 ExD 2 c ,3 ExD 1 c,4 AT 2 c,2 AT 2 c ,3 AT 2 c,4 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ... auxVC 3 s = 15. Thus packet P 1,1 is stamped with a virtual clock value equal to 4, and P 2 ,3 and P 3, 3 are stamped with a virtual clock value equal to 15. • At time...

Ngày tải lên: 15/12/2013, 11:15

... parallel modules signal Rail1On, Rail1Off, Rail1Free, Rail2On, Rail2Off, Rail2Free, Rail3On, Rail3Off, Rail3Free, Rail4On, Rail4Off, Rail4Free, Rail5On, Rail5Off, Rail5Free; % initialization module await ... Rail1Off; % switch off the power of rail 1 || await Rail3Free; % rail 3 must be free before entering % rail 2 ]; 182 8 SOFTWARE ENVIRONMENT 8.1 .3 RT-Linux The Linux operating system i...

Ngày tải lên: 15/12/2013, 11:15

... 50 30 000 5 modcomp 50 4 600 992 6 13 1 1 cond − activ 38 4 136 221 141 4 2 processing 36 4 92 496 106 4 2 storage 34 4 128 249 136 4 3 perturbo 33 4 112 218 120 4 3 demand 32 4 155 34 8 167 4 3 digigage ... (ms) Period (ms) Priority M 1 τ 1 τ 27 , τ 22 8 130 0.5078 10 12 M 2 τ 18 τ 11 , τ 5 , τ 23 3 82 0 .32 03 14 11 M 3 τ 2 τ 16 3 82 0 .32 03 20 10 M 4 τ 8 τ 4 2 73 0....

Ngày tải lên: 15/12/2013, 11:15