Repeat problem #1 above using 4 P-threads for the computations(This is the question)

Problem #1: Write a C program that uses 4 P-threads to speed up an algorithm of your choice, such as finding the Shubert function minimum. In your report, explain the algorithm you chose and whether you are using public sample code. Each of n threads should perform 1/nth of the work. Note that you may wish to pass an argument to a thread function, and the textbooks sample code for P-threads demonstrates this. Print the local result for each thread, the overall (global) result, and the run time of the parent/server. Configure the settings such that the overall run time is at least 1 second. You may divide the work between the parent thread and 1 child thread or between 2 child threads. Include the output in your report. Below is an example of output for finding the Shubert minimum: Total time was 10.65 seconds. Local min = -62.12412 Local min = -186.73064 Global min = -186.73064

shubert.c

# include # include # include # include

// Prototype for computation double shubert(double x1, double x2);

int main () { double x1, x2, y, min = 0; struct timeval start_time, stop_time, elapsed_time; // timers gettimeofday(&start_time,NULL); // Unix timer // Loops for (x1 = -2; x1 <= 2; x1 += 0.5) { // x1 parameter of Shubert function for (x2 = -2; x2 <= 2; x2 += 0.5) { // x2 parameter of Shubert function y = shubert(x1, x2); // Evaluate Shubert function defined below printf("%.2f ", y); // Print value for demonstration purposes if (y < min) // Check for global minimum min = y; } printf(" "); // Print on next row } gettimeofday(&stop_time,NULL); timersub(&stop_time, &start_time, &elapsed_time); // Unix time subtract routine printf("Total time was %0.2f seconds. ", elapsed_time.tv_sec+elapsed_time.tv_usec/1000000.0); printf("minimum = %0.2f ", min); }

// Compute Shubert value for x1,x2 combination double shubert(double x1, double x2) { double sum1 = 0; double sum2 = 0; int i; for (i = 1; i <= 5; i++) { sum1 += i * cos((i + 1) * x1 + i); sum2 += i * cos((i + 1) * x2 + i); } return sum1 * sum2; }