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A system in an unsafe state will ultimately deadlock.

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Last updated

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Date created

Mar 1, 2020

Cards (68)

Section 1

(50 cards)

A system in an unsafe state will ultimately deadlock.

Front

False

Back

Multiprogramming increases CPU utilization.

Front

True

Back

Every system call requires a change from user mode to kernel mode.

Front

True

Back

Each thread of a process has a separate stack and PC value.

Front

True

Back

Is busy waiting desirable or undesirable?

Front

Undesirable

Back

Each thread has its own register set and stack.

Front

True

Back

Multiprogramming cannot increase CPU utilization in a single processor system.

Front

False

Back

Global variables are shared between the threads of a process.

Front

True

Back

FCFS turn around time

Front

burst time + waiting time

Back

Issuing an I/O request may cause a process to move from running state to waiting state.

Front

True

Back

All processes in UNIX first translate to a zombie process upon termination.

Front

True

Back

A system call is triggered by hardware.

Front

False

Back

It is possible to create a thread library without any kernel-level support.

Front

True

Back

FIFO (first in, first out)

Front

The oldest or 1st entry in the queue is processed first

Back

If a resource allocation graph has a cycle, then the system must be in a deadlock state.

Front

False

Back

FCFS waiting time

Front

start time - arrival time

Back

The exec() system call creates a new process.

Front

False

Back

Every child process has a separate address space than its parent

Front

False

Back

I/O devices and the CPU can execute concurrently.

Front

True

Back

Application programmers typically use an API rather than directly invoking system calls.

Front

True

Back

Issuing an I/O request may cause a process to move from ready state to waiting state.

Front

False

Back

For a single-processor system, there will never be more than one process in the Running state.

Front

True

Back

Is bounded waiting desirable or undesirable?

Front

Desirable

Back

Round-robin (RR) scheduling degenerates to first-come-first-served (FCFS) scheduling if the time quantum is too long.

Front

True

Back

Semaphore is a synchronization tool that might cause busy waiting.

Front

True

Back

For a single-processor system, there will never be more than one process in the Ready state.

Front

False

Back

A trap is another name for software interrupt.

Front

True

Back

Incorrect use of semaphores might cause deadlocks.

Front

True

Back

Peterson's solution for mutual exclusion fulfills the bounded waiting property.

Front

True

Back

The operating system kernel consists of all system and application programs in a computer.

Front

False

Back

Each thread has its own global variables.

Front

False

Back

A trap is another name for hardware interrupt.

Front

False

Back

The value of a counting semaphore can range only between 0 and 1.

Front

False

Back

Each thread of a process has a separate address space

Front

False

Back

First-Come First-Served (FCFS) and Shortest Job First (SJF) algorithms are both preemptive CPU scheduling algorithms.

Front

Fasle

Back

Round Robin (RR) is a non-preemptive CPU scheduling algorithm.

Front

False

Back

Timer interrupt may cause the currently running process to move from running state to ready state.

Front

True

Back

System calls can be run in either user mode or kernel mode.

Front

False

Back

A batch system should have a preemptive scheduler.

Front

False

Back

System calls are part of the kernel code.

Front

False

Back

SJF scheduling provides the optimal average turnaround time if all processes arrive at the same time to a batch system.

Front

True

Back

Round Robin (RR) is a preemptive CPU scheduling algorithm.

Front

True

Back

Bounded Waiting

Front

There exists a bound, or limit, on the number of times other processes are allowed to enter their critical sections after a process has made request to enter its critical section and before that request is granted.

Back

The difference between a program and a process is that a program is an active entity while a process is a passive entity.

Front

False

Back

In shared memory, kernel is not involved during the communication process.

Front

True

Back

Completion of an I/O request may cause a process to move from waiting state to ready state.

Front

True

Back

Busy waiting

Front

Busy waiting is a technique in which a process repeatedly checks to see if a condition is true, such as whether keyboard input or a lock is available.

Back

How does Round Robin work?

Front

lets each process run for the quantum amount then it gets preempted and added to the back of the ready queue.

Back

In message queues, kernel is not involved during the communication process.

Front

False

Back

Interrupts may be triggered by either hardware or software.

Front

True

Back

Section 2

(18 cards)

Busy waiting is a good way to solve the mutual exclusion problem since it never allows CPU to be idle.

Front

False

Back

Need matrix is found by

Front

Max - Allocation

Back

It is less expensive to create a thread than a process.

Front

True

Back

how do you find work?

Front

current work + allocation

Back

The Round Robin (RR) CPU scheduling algorithm behaves pretty much like First-Come First-Served (FCFS) as quantum becomes very large.

Front

True

Back

Threads share a heap and a stack.

Front

False

Back

suppose P1 wants to request (1,0,2) what do you do?

Front

Add that to its allocation amount and subtract it from available.

Back

Interrupts can be triggered by both hardware and software.

Front

True

Back

If a resource allocation graph has no cycle, then the system might or might not be in a deadlock state.

Front

False

Back

process state transition diagram

Front

Back

If a graph contains no cycles then

Front

No deadlock

Back

If at least one of the conditions mentioned for a deadlock is eliminated, is it guaranteed that the system will never be in a deadlock state?

Front

yes

Back

Timer interrupt may cause the currently running process to move from running state to waiting state.

Front

False

Back

If the shared resources are numbered 1 through N and a process can only ask for resources that are numbered higher than that of any resource that it currently holds, then deadlock can never happen.

Front

True

Back

If a graph contains a cycle

Front

Deadlock if only one instance per resource type. If several instances per resource type possibility pf deadlock

Back

If not given available in bankers algo what do you do to find it?

Front

Add up everything in allocation matrix and subtract it from the amount of each given resource type.

Back

What are the four necessary conditions for a deadlock to occur in a system

Front

1. Mutual Exclusion 2.hold and wait 3.no preemption 4.circular wait

Back

If all the necessary conditions mentioned for a deadlock are satisfied in a system, is it guaranteed that the system is in a deadlock state?

Front

No

Back