Write c code to implement round robin with q
=
5
Once each process finishes executing its burst time it go back to re
-
execute again after a certain come back time of each process. Simulate it until
2
0
0
time units

Question

Write c code to implement round robin with q
=
5
Once each process finishes executing its burst time it go back to re
-
execute again after a certain come back time of each process. Simulate it until 
2
0
0
 ﻿time units

Accepted Answer

```c #include #include #define MAX_PROCESSES 10 // Process structure to hold process details typedef struct { int pid; int burst_time; int comeback_time; int remaining_time; int last_executed_time; } Process; // Function to initialize the processes void initialize_processes(Process processes[], int n, int burst_time[], int comeback_time[]) { for (int i = 0; i < n; i++) { processes[i].pid = i + 1; processes[i].burst_time = burst_time[i]; processes[i].comeback_time = comeback_time[i]; processes[i].remaining_time = burst_time[i]; processes[i].last_executed_time = -1; // -1 indicates not yet executed } } // Function to find the next process to execute using round robin scheduling int find_next_process(Process processes[], int n, int current_time, int q) { int index = -1; for (int i = 0; i < n; i++) { if (processes[i].remaining_time > 0 && (processes[i].last_executed_time == -1 || current_time - processes[i].last_executed_time >= processes[i].comeback_time)) { index = i; break; } } return index; } // Function to simulate round robin scheduling void round_robin(Process processes[], int n, int q) { int current_time = 0; bool done; while (current_time < 200) { done = true; int index = find_next_process(processes, n, current_time, q); if (index != -1) { done = false; if (processes[index].remaining_time > q) { current_time += q; processes[index].remaining_time -= q; } else { current_time += processes[index].remaining_time; processes[index].remaining_time = 0; } processes[index].last_executed_time = current_time; printf("Process %d executed from time %d to %d ", processes[index].pid, processes[index].last_executed_time - q, current_time); } if (done) { break; } } } // Main function to test round robin scheduling int main() { int burst_time[] = {10, 5, 8}; int comeback_time[] = {20, 25, 15}; int n = sizeof(burst_time) / sizeof(burst_time[0]); Process processes[MAX_PROCESSES]; initialize_processes(processes, n, burst_time, comeback_time); round_robin(processes, n, 5); return 0; } ``` ∻Explanation∻ ⚹This C code simulates a round robin scheduling algorithm with a quantum time of 5 and a simulation time limit of 200 time units. Each process, once finished executing its burst time, will wait for its comeback time before it can be scheduled again. The code includes functions to initialize processes, find the next process to execute, and simulate the round robin scheduling. The `main` function initializes an array of processes with their respective burst and comeback times and then calls the round robin function to simulate the scheduling.⚹ ∻Step-by-step Instruction∻ ‖ ▪ ⋮ Define the `Process` structure to hold the details of each process. ‖ ‖ ▪ ⋮ Implement the `initialize_processes` function to initialize the processes with their burst and comeback times. ‖ ‖ ▪ ⋮ Implement the `find_next_process` function to find the next process to execute based on the round robin scheduling and comeback time. ‖ ‖ ▪ ⋮ Implement the `round_robin` function to simulate the round robin scheduling until 200 time units or until all processes are done. ‖ ‖ ▪ ⋮ Write the `main` function to set up the processes and call the round robin scheduling function. ‖ ∻Time Complexity∻ ⚹The time complexity of the round robin function is O(n) for each iteration, where n is the number of processes, because it iterates over the list of processes to find the next one to execute. However, since the simulation runs until 200 time units, the overall time complexity depends on the number of processes and their comeback times.⚹ ∻Space Complexity∻ ⚹The space complexity is O(n), where n is the number of processes, because the program stores an array of `Process` structures.⚹

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