Selected solutions for. TDDD07 Real-time Systems

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1 Selected solutions for TDDD07 Real-time Systems Massimiliano Raciti Jordi ucurull Simin Nadjm-Tehrani Scheduling and resource sharing Real-Time Systems Laboratory Department of omputer and Information Science Linköping University, Sweden November 20 opyright 20 Simin Nadjm-Tehrani

2 Suggested Solutions. Scheduling Q.: The task set is schedulable using cyclic scheduling, but the tasks suffer jitter. Here is an example of scheduling, in which you can see that tasks are affected by jitter Major cycle: lcm(50,0,20) = 00ms Minor cycle: gcd(50,0,20)=0ms You can try to change the order of execution of the tasks, but the jitter still remains. One of the possible ways to schedule this task set without jitter is by reducing the periods, thus executing tasks more frequently than required (see lternative 2 in the slides of Lecture ). Task New eriod (ms) Major cycle: lcm(0,20,20) = 0ms Minor cycle: gcd(0,20,20)=20ms

3 Q.2: a) Task eriod WET (Trajectory follower) (Sensor & measurement) (Disk storage) D (Ground communication) Major cycle: lcm(500,500,000,500) = 000ms Minor cycle: gcd(500,500,000,500)=500ms The task set is not schedulable. This is due to the disk storage task, that cannot be fitted in any place inside the major cycle. The task set becomes schedulable if we assume that the disk storage task can be divided into 6 processes as following: =00ms,2=200ms,=200ms,=00ms,5=200ms, 6=200ms. a) The new task set is: Task eriod WET Trajectory follower Sensor & measurement 000 (incr. from 500) 00 Disk storage Ground communication (reduc. from 200) RMS: U % G for tasks is: n G n( 2 ) (2 ) 75% The task set is then schedulable using RMS. However, the jitter requirements are not met. Since the trajectory follower is not the process with higher priority, it can be preempted by higher priority processes, thus causing jitter.

4 b) This phenomenon is the priority inversion. When the sensor recording task is blocked waiting for the busy resource hold by the disk storage (which has the lowest priority), middle priority processes can preempt this one causing the sensor recording to wait for them as well. This problem can be solved using the priority inheritance protocol or the priority ceiling protocol, although the second one is preferable since it prevents deadlocks. Q.: a) ccording to the description we have the following task set: rocess eriod WET Stabiliser 50ms 5ms 2 Star follower 20ms 5ms Energy manager 0ms 0ms Minor cycle: gcd (50, 20, 0) = 0ms Major cycle: lcm (50, 20, 0) = 00 ms b) rocess eriod WET Shares ritical Section riority (π) Time Stabiliser 50ms 5ms R2 osition ms 2 Star follower 20ms 5ms R Memory 2ms 5 (Highest) Energy manager 0ms 0ms - - icture storage 50ms - R Memory 2ms ommunication 00ms - R2 osition ms (Lowest) eiling (R) = max (π2, π) = 5 eiling (R2) = max (π, π5) =

5 The blocking times are calculated. In order to simplify the process it is recommended to go through the lower to higher priority processes. Let lp {Z} denote the set of processes with lower priority than Z: 5 : There is no process with lower priority lp { 5} {}. No process can block 5 then 5 0 ms. : In this case lp{ } { 5}. We have to check if can be blocked by the lower priority process 5: (i) rocess 5 locks R2 and eiling( R2) 5 2. rocess 5 can block. rocess can be blocked by process 5. The blocking time is max( t 5) ms. R2, : In this case lp{ } {, 5}. We have to check if can be blocked by any of these two lower priority processes: (i) rocess 5 locks R2 and eiling( R2). rocess 5 can block. (ii) rocess locks R and eiling( R) 5. rocess can block. rocess can be blocked by processes 5 and. Then, the blocking time is max( tr2, 5, tr, ) 2ms. : In this case lp{ } { 5,, }. We have to check if can be blocked by any of these lower priority processes: (i) rocess 5 locks R2 and eiling( R2). rocess 5 cannot block. (ii) rocess locks R and eiling( R) 5. rocess can block. (iii) rocess locks R and eiling( R2). rocess cannot block. rocess can only be blocked by process. Then, the blocking time is max( t ) ms. R, 2 2 : In this case lp{ 2} { 5,,, }. We have to check if 2 can be blocked by any of these lower priority processes: (i) rocess 5 locks R2 and eiling( R2) 2 5. rocess 5 cannot block 2. (ii) rocess locks R and eiling( R) rocess can block 2. (iii) rocess locks R and eiling( R2) 2 5. rocess cannot block 2. (iv) does not use resources hence it cannot block 2.

6 rocess 2 can only be blocked by process. Then, the blocking time is max( t ) ms. 2 R, 2 c) No, the schedulability test used for RMS cannot be applied because process 2 has a deadline D smaller than the period T. Q.: a) ccording to the description we have the following task set: rocess eriod WET Servo motor controller ms 0.5ms 2 Servo motor controller 2 ms 0.5ms Sensor analysis 0ms 2ms Decision maker 0ms ms 5 Logging? 2ms Since the scheduler used is EDF, we use the schedulability test formula to isolate the minimum possible period of the logging task: N i T i Ti 2 5 T 2 T T T T 5 T 5. ms The logging task could be run with a period of. ms or longer. ossible assumptions are:.system overheads are not considered 2.ontext switching is ignored. D T b) We calculate the major and minor cycles for the cyclic schedule: Minor cycle: gcd (,, 0, 0, 8) = 2ms Major cycle: lcm (,, 0, 0, 8) = 0ms

7 nd the resulting cycle schedule is: c) s the processes with the same period mixed this is the new task set: rocess eriod WET Locking riority (π) time M (shared memory) Servo motor controllers ms ms (Highest) 2 Sensor and decision maker 0ms ms 0.5ms (Lowest) Logging 8ms ms ms 2 To calculate the response time analysis we have to take into account the blocking times. Then, first we have to calculate the ceiling of the resource M: eiling (M) = max (π, π) = 2 The blocking times are calculated. In order to simplify the process it is recommended to go through the lower to higher priority processes. Let lp {Z} denote the set of processes with lower priority than Z: 2 : There is no process with lower priority lp { 2} {}. No process can block 2 then 2 0ms. : In this case lp{ } { 2}. We have to check if can be blocked by the lower priority process 2: (i) rocess 2 locks M and eiling( M ) 2 2. rocess 2 can block. rocess can be blocked by process 2. The blocking time is max( t 2) 0. ms. M, 5

8 : In this case lp{ } { 2, }. We have to check if can be blocked by any of these two lower priority processes: (i) rocess 2 locks M and eiling( M ) 2. rocess 2 cannot block. (ii) rocess locks M and eiling( M ) 2. rocess cannot block. rocess cannot be blocked by 2 neither. Then, the blocking time is 0ms. fter computing the blocking time, we have all we need to do the response time analysis. This time it is easier if we start with the highest priority task and we proceed down to the lowest. R : This is the process with highest priority, hence no other processes can interrupt it: R J 0 0 ms R : This process can be interrupted by. 0 w 0.5.5ms w T.5ms 2 w T 5.5ms w ms w w w T R w J ms R 2 : This process can be interrupted by and. 0 w ms 2 2 w 0 2 w 0 2 T T 0 8 7ms w 2 w 2 T T ms 2 2 w 2 2 w ms w 2 w 2 w 2 T R 2 w 2 J ms T

9 d) Sporadic task is a task that is not periodically scheduled, but for which we can set a minimum inter-arrival time. n example could be a task reacting to the external event produced by a user pressing the keys of a keyboard: here there is a minimum time we can consider before two consecutive events. Q.5: a) From the description we obtain the following set of tasks: rocess eriod WET ruise controller 0ms ms 2 Wheel pair regulators 5ms ms utomatic braking? ms In order to calculate the maximum possible period, we will create an equation with the formula that determines the U utilization of the task set and the maximum U utilisation allowed: N i T i Ti T 2 T T 0.7 T 5ms The period of must be equal or greater than 5ms. b) From the description we obtained the following set of tasks: rocess eriod WET ruise controller 0ms ms 2 Wheel pair regulators 5ms ms utomatic braking 5ms ms ower usage control 0ms 2ms Then we calculate the major and minor cycles: Minor cycle: gcd (0, 5, 5, 0) = 5ms Major cycle: lcm (0, 5, 5, 0) = 0ms

10 nd the resulting cycle schedule is: c) The task set table that we can extract from the exercise description is the following: rocess riority (π) Locking time R Locking time S 2 0.5ms 2 ms 2ms ms The ceiling of the two resources are calculated: eiling (R) = max (π, π2) = eiling (S) = max (π2, π) = The blocking times are calculated. In order to simplify the process it is recommended to go through the lower to higher priority processes. Let lp {Z} denote the set of processes with lower priority than Z: : There is no process with lower priority lp { } {}. No process can block then 0ms. : In this case lp{ } { }. We have to check if can be blocked by the lower priority process : (i) rocess locks S and eiling( S) 2. rocess can block. rocess can be blocked by process. The blocking time is max( t ) ms. S, 2 : In this case lp{ 2} {, }. We have to check if 2 can be blocked by any of these two lower priority processes: (i) rocess locks S and eiling( S) 2. rocess can block 2. (ii) rocess locks R and eiling( R) 2. rocess can block 2. rocess 2 can be blocked by processes and. Then, the blocking time is 2 max( ts,, tr, ) ms.

11 : In this case lp{ } {,, 2}. We have to check if can be blocked by any of these lower priority processes: Then (i) rocess locks S and eiling( S). rocess cannot block. (ii) rocess locks R and eiling( R). rocess cannot block. (iii) rocess 2 locks R and S, and eiling( R) and eiling( S). Then process 2 cannot block. 0ms. d) The statement is true because a task that is not periodic neither sporadic is aperiodic. n aperiodic task does not allow an execution time analysis because it does not present a minimum inter-arrival time. Q.6: a) Q.7: E b b a a E E a E b E 2 E a a b b E The priorities of the tasks, according to RMS, are: rocess riority (π) 2 D (a) The ceiling of a semaphore is the maximum priority of all processes that may lock the semaphore. Denoting X the ceiling of the resource X, we have that S max(, ) S2 max(, D) (b) The blocking time for each process can be calculated with the following algorithm. Let lp {Z} denote the set of processes with lower priority than Z. It is convenient to start from the lowest priority process to the highest.

12 D : D is the lowest priority process, thus no other lower priority processes can block it. lp { D} {} thus 0. D : lp{ } { D}. rocess D uses S2 and S 2 2 then D can block process since it can get a higher priority when blocking S2. The blocking time is. t S 2, D : lp{ } { D, }. Let s examine which process can block. S2. rocess D can block. (i) rocess D uses S2 and (ii) rocess uses S and S. rocess can block. Thus, process is blocked at most once by the processes D and. This means that max( ts 2, D, ts, ). : lp{ } { D,, }. (i) rocess D uses S2 and S 2. rocess D cannot block. (ii) rocess uses S and S. rocess can block. (iii) rocess uses S2 and S 2. rocess cannot block. Only process can block process, then. t S, Q.8: The priorities of the tasks, according to RMS, are: rocess riority (π) 2 Denoting X the ceiling of the resource X, we have that: S max(,, ) S2 max(, ) S max( ) s before, let lp {Z} denote the set of processes with lower priority than Z. : is the lowest priority of the set, thus lp { } {}. cannot be blocked by any lower priority process, 0. : lp{ } { }. rocess uses S and S 2 meaning that process can block process. The blocking time is. : lp{ } {, }. t S, (ii) rocess uses S and (iii) rocess uses S and semaphore S. S. rocess can block. S. rocess can block when using the

13 (iv) rocess uses S2 and 2 S. rocess can block when using the semaphore S2. Thus, process is blocked at most once by the processes D and. This means that max( t S,, t S,, t S 2, ). We can now calculate the response time by solving the equation: w w i i i i T j jhp(i ) j R i w i J i gain, it is convenient to start from the highest priority process down to the lowest. Response time R : hp()={} so the response time is not affected by interference from higher priority processes, then 2 R Response time R : Only process has higher priority than, hp()={}. Interference is caused by and the response time is: w 0 w w 0 6 T w 2 w 6 T R w 2 J Response time R : rocesses and have higher priority then, hp()={,}. Interference has the contribution from and and the response time is: w 0 w w 0 w 0 9 T T w 2 w w T T w w 2 w 2 T T w w w T R w J 0 T