Which of the following is NOT a unit for stress?

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?

Which process is MOST likely to suffer starvation in priority scheduling? 

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?

Which process state indicates the process is waiting for some external event?

Consider the following processes scheduled using Shortest-Remaining-Time-First (SRTF) scheduling: Process Arrival Time Burst Time P1 0 7 P2 2 4 P3 4 1 P4 5 4 Which of the following is the CORRECT execution sequence (Gantt Chart)?

Consider the following RAG:  Which statement is CORRECT?

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

A process: Arrives at time = 2 ms Completes at time = 17 ms What is the turnaround time?