EE472 Final Exam SOLUTION KEY. Prof. Blake Hannaford Department of Electrical Engineering The University of Washington 9-DEC-2008

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1 EE472 Final Exam SOLUTION KEY Prof. Blake Hannaford Department of Electrical Engineering The University of Washington 9-DEC-2008 Problem 1 / 15 Problem 2 / 25 Problem 3 / 25 Problem 4 / 15 Problem 5 / 10 Total / 90 Rules Individual Work Closed Book, Closed Notes Questions from previous exams my be modified. Two-page (two sides total) crib sheet allowed. No copying from prior exam solutions. 60 minutes 100 points The last page is the group effort survey. You may turn it in with the exam or remove it and turn it in separately to Prof. Hannaford. NAME 1

2 1 Critical Sections (15 pts) A real-time embedded system (operating system: 472OS) has a preemptive priority based scheduler. There are five tasks, one data buffer, one 8-bit digital I/O port, and two serial ports. The five tasks are: (each task s pseudo code is inside a while(1) loop which is not shown). Task 1 1. Initialize digital I/O port for input 2. Read digital input 3. Count rising edges on each input 4. Sleep(5) Task 2 1. Read Data from serial port 1 Task 3 Task 4 Task 5 2. Check CRC byte for errors in serial data 3. if (data buffer is empty) {write data payload to buffer 1. Initialize digital I/O port to output 2. write data to digital outputs 3. toggle bit 0 of digital I/O port on and off (i.e. make a brief pulse on bit 0). 4. Sleep(10) 1. if (data buffer is full) {take data from buffer 2. Write data to serial port 2 1. Count number of bytes in data buffer 2. display byte count on LCD 3. Sleep(250) 1.1 (5 pts) Which resource or resources are shared by at least two tasks? Fill in the table with a group of tasks on the left and one or more resources on the right. Tasks Resource(s) shared by tasks 1,3 Digital I/O Port 2, 4, 5 Buffer 1.2 (5 pts) Identify the specific lines of pseudocode above which are critical sections. { , {2.3, { , { , { (5 pts) The following calls are given in 472OS: Sem Init() Establishes a semaphore Sem Pend(S) Pends on a semaphore Sem Post(S) Posts a semaphore How would you use these to protect the critical section(s) you identified above? 2

3 At initialization time we need to call 472_Sem_Init() twice to generate a semaphore for each shared resource. For example: (begin C code example) Typedef Semaphore {{something here Semphore Sdig, Sbuf; Sdig = 472Sem_init(); // Semaphore for digital I/O port Sbuf = 472Sem_init(); // " " data buffer (end C code example) Then call 472Sem_Pend(Sdig) at the beginning of any critical section involving the Digital I/O port, and call 472Sem_Post(Sdig) at the end. Similarly, use Sbuf with Pend/Post to protect critical sections which involve the data buffer. 3

4 2 C Programming (25 pts) Write a C function to solve the problem or state why it is not possible. 2.1 (4 pts) Write a function to return TRUE if a 32 bit int has any 4 bits set in a row. // Bit mask containing 1111 #define MASK 15 #define MAXSHIFT 32-4 int prob2_1(int x) { int i,j,k,m,n; for(j=0; j<maxshift; j++){ k = x & (MASK << j); n=0; for(m=j; m<(j+4); m++) { if(k & (1<<m)) n++; if(n==4) return(true); return(false); 2.2 (4 pts) Write a function to accept a void pointer and return the size (in bytes) of the thing it points to. int prob1_2(void *ptr){ printf ("This is impossible! A void can point to anything so the size information is lost.\n"); 2.3 (9 pts) Write a function to identify a 32-bit mirror word. A mirror word is one where the MSB == LSB, bit31== bit1, bit30 == bit2, etc. For example, an 8-bit mirror word is: Your function should return TRUE if its argument is a mirror word and false otherwise. int prob2_3(unsigned int x){ unsigned int mask1, mask2, i; unsigned int f1, f2; 4

5 mask1 = 1<<31; mask2 = 1; // MSB // LSB for(i=0;i<16;i++) { f1 = f2 = 0; if(mask1 & x) { f1 = 1; if(mask2 & x) { f2 = 1; if(f1!= f2) return(false); mask1 = mask1 >> 1; mask2 = mask2 << 1; return(true); 2.4 (8 pts) What will be printed by this code? void prob2_4(){ int a=5, b=7, c=-4, d[3] = {4,6,12; int *p1, *p2; int x,y,z; p1 = &d[0]; *(p1++); (*p1)++; p2 = &c; x = a + d[2]; y = *p1 + x; z = *p2 + b; *p2 += d[0]; printf("%d %d %d %d \n", y, z, c, d[1]); Answer:

6 3 Scheduler Choice and Application (25 pts) You are designing real-time embedded software to operate a robotic toy pet dog. The dog has four moveable legs, a video camera, several sensors, pre-programmed amusing behaviors, and a rechargable battery. Among the things it can do is recognize human faces. It can walk toward or away from human faces it recognizes depending on whether or not it likes that person. If it falls down, on-board accelerometers can detect its orientation with respect to gravity and it can execute a pre-programmed series of motions to raise itself to a standing position. The tasks are described as follows: Task Description C P D CPU Util (%) DL Margin (ms) A Control leg motion 200µs 2ms 300µs B C D E Generation motion paths for joints. Plan amusing behaviors Look for human faces in camera signal Control balance and detect falls 100µs 2ms 1ms ms 1sec 500ms ms 1sec 2sec ms 10ms 20ms F Get up from fall 500µs 100ms 50ms G Check battery 5µs 10sec 5sec H Check proximity sensors 100µs 20ms 10ms I Navigate and plan body motion 50ms 500ms 1sec pts Compute the CPU utilization and the Deadline Margin for each task. CPU Utilization is the fraction of CPU time used by a task. Deadline Margin is how much flexible time is available before the deadline (D-C). Express utilization as a percentage and margin in milliseconds. Enter your answers in the last two columns of the table above pts Assign Priorites to these tasks (bigger number == higher priority) using the following two rules: Rule 1 The highest priority should be assigned to the tasks with the lowest deadline margin. (use the pre-defined priorities and write in the task letters below). Priority Task G D I C F E H B A (5 = highest, 45 = lowest) Now try again using Rule 2: Rule 2 The highest priority should be assigned to the tasks with the Highest CPU Utilization. 6

7 Priority Task G H* F* C B E* I* A* D (* = tie) 3.3 5pts Assign Priorites to these tasks using... Rule 3 The priority should be assigned to the tasks According to Rate Monotonic Scheduling. Priority Task G C* D* I F H E B* A* 3.4 5pts Comment on which scheduler would be best for the toy robot. Consider Which tasks are critical to the robot working and which are less important if they fail to meet their deadlines? With each scheduler system, what happens if one of the lowest priorty tasks is delayed? In your opinion, are the deadline specifications correct for a robot you would like to buy? Does either one of the schedulers give a low priority to a critical task or a high priority to one which is not so critical? Based on their functional description, A, B, E, F are most critical because they involve physical safety of robot. H could possibly be in that category as well. H may protect the robot from dangerous collisions or from falling off the edge of a table. C, D and F will be important for making the toy fun to play with. We might have quite a bit of time flexibility for G. Rule 2 puts F and H in rather low positions. Delaying these could cause problems. Rule 1 gives all the critical task above average" priority. Rule 3 does as well. No! D, E, and I have deadlines after their period. This is meaningless. Beyond this, the subjective, fun", effect of tasks C and D might generate better human reactions if their deadlines were tighter (Imagine if another human took two full seconds (one - onethousand... two - onethousand ) to recognize your face. Task I seems to score a bit too high in Rule 2. It would be nice if F scored higher in all the systems. Perhaps we should change its period. On the other hand, while getting up from a fall (F), we could shut down several of the other tasks. None of the three choices are horrible. Rule 2 is a bit worrysome because task A could rank as low as 4. Also, Rule 2 puts task H next to last. I would rule out Rule 2. Rule 1 and Rule 3 result in pretty similar ordering. Rule 3 (RMS) has the advantage that there is a mathematical proof that the tasks could meet their deadlines. On that basis I would go with RMS (Rule 3) as a tie breaker. 7

8 4 Interrupts (15 points) pts. Explain what is meant by task context or process context (equivalent terms) and what it has to do with interrrupts. The context is all information need to resume a task. This includes the value of program counter (where we are in the task) the stack pointer (what is our function calling status), and the CPU registers (holding intermediate results and likely to be overwritten by another task or ISR). An interrupt occurs at a random time. The current process context must be saved so that the ISR does not destroy it. If this happend, we could not resume the current task after the ISR pts. Describe the process of how an embedded, real-time system responds to an interrupt. You do not need to go into minute detail, but explain it in 3-5 major steps. 1. Save task context (on the stack) 2. Start the ISR code 3. Restore task context (from the stack) 4. Restart the task pts. List two or three pros and cons of using interrupts in an embedded real-time software system. PROS: Low Response latency (i.e. short time between interrupt and start of ISR code. Determinism (in many cases). Determinism is the consistency of the response time. (we don t have to wait for the scheduler to consider various alternatives. CONS: Difficult to debug. Fraction of CPU time is out of the schedulers control Concurrrency issues now arise due to pre-emption. Need to use semphores or similar measures. 8

9 5 Pointers (10 pts) (Similar to questions in a Microsoft job interview!) For the following declarations and statements, // declarations int i,j,k[10]={12,14,16,18,20,22,24,26,28,30; int *ip; char a,b,c[10]={3,6,9,12,15,18,21,24,27,30; char *cp; void *vp; typedef struct stst { int i; char *p; st; // assignment statements st x; ip = &k[4]; vp = ip; cp = &c[0]; x.p = cp+3; x.i = 27; Evaluate the following expressions (if the expression is an error, indicate that with X ): Solution *(ip+2) + c[3] = 36 (*cp +1) = 4 *(ip) + 6 = 26 vp = &x ; ((st*)vp)->p; = X *(ip-2) = 16 9

EE472 Mid Term SOLUTION KEY. Rules. Prof. Blake Hannaford Department of Electrical Engineering The University of Washington 31-Oct-2005 NAME

EE472 Mid Term SOLUTION KEY. Rules. Prof. Blake Hannaford Department of Electrical Engineering The University of Washington 31-Oct-2005 NAME EE472 Mid Term SOLUTION KEY Prof. Blake Hannaford Department of Electrical Engineering The University of Washington 31-Oct-2005 NAME Rules Individual Work Closed Book, Closed Notes, No Calculators 1-page