Priority Scheduling with Round Robin Consider the following set of processes: Process Burst Time Priority P1 5 2 P2 3 1 P3 6 2 P4 4 3 P5 2 1 Assume the time quantum = 2. Which of the following is the correct execution sequence (Gantt order)?

Which topic from this course surprised you the most or changed the way you think about how computers work internally? Explain briefly.

Why can preemptive scheduling lead to race conditions?

Considering the code: #include #include #include pthread_mutex_t lock1; pthread_mutex_t lock2; void* thread1(void* arg) { pthread_mutex_lock(&lock1); printf(“Thread 1 acquired lock1\n”); sleep(1); pthread_mutex_lock(&lock2); printf(“Thread 1 acquired lock2\n”); pthread_mutex_unlock(&lock2); pthread_mutex_unlock(&lock1); return NULL; } void* thread2(void* arg) { pthread_mutex_lock(&lock2); printf(“Thread 2 acquired lock2\n”); sleep(1); pthread_mutex_lock(&lock1); printf(“Thread 2 acquired lock1\n”); pthread_mutex_unlock(&lock1); pthread_mutex_unlock(&lock2); return NULL; } int main() { pthread_t t1, t2; pthread_mutex_init(&lock1, NULL); pthread_mutex_init(&lock2, NULL); pthread_create(&t1, NULL, thread1, NULL); pthread_create(&t2, NULL, thread2, NULL); pthread_join(t1, NULL); pthread_join(t2, NULL); pthread_mutex_destroy(&lock1); pthread_mutex_destroy(&lock2); return 0; } What is the MOST likely output?

A system is considered to be in a safe state if:

In a client-server socket model, which sequence correctly represents the communication flow?

A process has: Arrival Time = 1 ms First CPU Allocation = 4 ms Completion Time = 15 ms Burst Time = 9 ms What is the waiting time?

What is the value of semaphore S after this sequence? Initial Value: S = 3 Execution: wait(S); wait(S); signal(S);

A system satisfies bounded waiting but allows two processes simultaneously inside the critical section. Which requirement of solution to critical-section problem fails? 

On a system with two CPUs, Process P1 disables interrupts on CPU1 and enters the critical section. What is the main problem?