Scheduling Algorithm and Analysis Interrupts and non-RM Tasks - PowerPoint PPT Presentation

Scheduling Algorithm and Analysis Interrupts and non-RM Tasks (Module 31) Yann-Hang Lee Arizona State University yhlee@asu.edu (480) 727-7507 Summer 2014 Real-time Systems Lab, Computer Science and Engineering, ASU

Schedulability with Interrupts  Interrupt processing can be inconsistent with rate monotonic priority assignment.  interrupt handler executes with high priority despite its period  interrupt processing may delay execution of tasks with shorter periods  Effects of interrupt processing must be taken into account in schedulability model.  Question is: how to do that? 1 Real-time Systems Lab, Computer Science and Engineering, ASU

Determining Schedulability with Interrupts Task(i) Period(T) Execution Priority(P) Deadline Time(C) (D) τ 3 200 60 HW 200 τ 1 100 20 High 100 τ 2 150 40 Medium 150 τ 4 350 40 Low 350 τ 3 is an interrupt handler 2 Real-time Systems Lab, Computer Science and Engineering, ASU

UB Test with Any Fixed Priority  Test is applied to each task.  Determine effective utilization ( f i ) of each task i using when d i =p i e e 1 ∑ ∑ + + j i e k p p p f i = ∈ ∈ j H k H j i i n 1 Preemption from tasks Preemption form the Execution of a tasks that can hit more task under test That can hit only once than once (with period (with period greater than less that p i ) p i )  Compare effective utilization against bound, U(n). n = num(H n ) + 1 num(H n ) = the number of tasks in the set H n 3 Real-time Systems Lab, Computer Science and Engineering, ASU

UB Test with Any Fixed Priority  Test is applied to each task.  Determine effective utilization ( f i ) of each task i using when d i <p i (e i must be done within d i, not pi) e 1 ∑ e ∑ + + j i e k f i = p d d ∈ ∈ j H k H j i i n 1 Preemption from tasks Preemption form the Execution of a tasks that can hit more task under test That can hit only once than once (with period (with period greater than less that d i ) d i )  Compare effective utilization against bound, U(n). n = num(H n ) + 1 num(H n ) = the number of tasks in the set H n 4 Real-time Systems Lab, Computer Science and Engineering, ASU

UB Test with Interrupt Priority: τ 3  For τ 3 , no tasks have a higher priority: H = H n =H 1 = { }. C + + ≤ 3 f 3 = 0 0 U ( 1 ) T 3  Note that utilization bound is U(1): num( H n ) = 0. Plugging in numbers: C 60 = = < f 3 = 3 0 . 3 1 . 0 T 200 3 5 Real-time Systems Lab, Computer Science and Engineering, ASU

UB Test with Interrupt Priority: τ 1  For τ 1 , τ 3 has a higher priority: H = { τ 3 }; H n = {}; H 1 = { τ 3 }. C 1 ∑ + + k ≤ 1 f 1 = 0 C U ( 1 ) T T = k 3 1 1  Note that utilization bound is U(1): num( Hn ) = 0. Plugging in the numbers: C C 20 60 + T = + = < 1 3 f 1 = 0 . 800 1 . 0 100 100 T 1 1 6 Real-time Systems Lab, Computer Science and Engineering, ASU

UB Test with Interrupt Priority: τ 2  For τ 2 : H={ τ 1 , τ 3 }; H n ={ τ 1 }; H 1 ={ τ 3 }; C C 1 ∑ ∑ = + + ≤ j f 2 C U ( 2 ) 2 k T T T = = j 1 k 3 j 2 2  Note that utilization bound is U(2): num( H n ) = 1. Plugging in the numbers: 20 40 60 C C C = + + = + + = > f 1 2 3 0 . 867 0 . 828 2 T T T 100 150 150 1 2 3 7 Real-time Systems Lab, Computer Science and Engineering, ASU

UB Test with Interrupt Priority:  For τ 4 : H={ τ 1 , τ 2 , τ 3 }; H n ={ τ 1 , τ 2 , τ 3 }; H 1 ={}; C = ∑ C + + ≤ j 4 f 0 U ( 4 ) 4 T T = j 1 , 2 , 3 j 4  Note that utilization bound is U(4): num( H n ) = 3. Plugging in the numbers: C C C C = + + + 1 2 3 4 f 4 T T T T 1 2 3 4 20 40 60 60 = + + + = > 0 . 882 0 . 756 100 150 200 350 8 Real-time Systems Lab, Computer Science and Engineering, ASU

Supplementary Slides Real-time Systems Lab, Computer Science and Engineering, ASU