Chapter 7

Question 1

Device management involves four basic functions:

Answer 1

• Monitoring the status of each device, such as storage drives, printers, and other peripheral devices
• Enforcing preset policies to determine which process will get a device and for how long
• Allocating the devices
• Deallocating them at two levels—at the process (or task) level and then at the job level

Question 2

The system’s peripheral devices generally fall into one of three categories:

Answer 2

dedicated
shared
virtual

Question 3

are assigned to only one job at a time; they serve that job for the entire time it’s active or until it releases them.

Answer 3

dedicated devices

Question 4

type of device that can be assigned to several processes

Answer 4

Shared devices

Question 5

are a combination of the first two devices: They’re dedicated devices that have been transformed into shared devices.

Answer 5

Virtual devices

Question 6

The USB controller assigns bandwidth to each device depending on its priority

Answer 6

• Highest priority is assigned to real-time exchanges where no interruption in the data flow is allowed, such as video or sound data.
• Medium priority is assigned to devices that can allow occasional interrupts without jeopardizing the use of the device, such as a keyboard or joystick.
• Lowest priority is assigned to bulk transfers or exchanges that can accommodate slower data flow, such as printers or scanners.

Question 7

Regardless of the specific attributes of the device, the most important differences
among them are:

Answer 7

speed and degree of sharability

Question 8

Storage media are divided into two groups: ______________, which store records sequentially, one after the other; and _____________, which can store either sequential or direct access files. There are vast differences in their speed and sharability.

Answer 8

sequential access media
direct access storage devices (DASD)

Question 9

the density of the tape (measured in bpi), multiplied by the tape drive speed, called transport speed, which is measured in inches per second (ips)

Answer 9

transfer rate

Question 10

transfer rate formula

Answer 10

transfer rate (ips) = density * transport speed

Question 11

Blocking has two distinct advantages:

Answer 11

• Fewer I/O operations are needed because a single READ command can move an entire block, the physical record that includes several logical records, into main memory.
• Less tape is wasted because the size of the physical record exceeds the size of the gap.

Question 12

The two disadvantages of blocking seem mild by comparison:

Answer 12

• Overhead and software routines are needed for blocking, deblocking, and recordkeeping.
• Buffer space may be wasted if you need only one logical record but must read an entire block to get it.

Question 13

______________ are devices that can directly read or write to a specific place

Answer 13

Direct access storage devices (DASDs)

Question 14

DASDs can be grouped into three categories:

Answer 14

magnetic disks
optical discs
flash memory

Question 15

A ________________ looks like a large CD or DVD covered with magnetic film that has been formatted, usually on both sides, into concentric circles.

Answer 15

fixed-head magnetic disk

Question 16

advantage and disadvantage of fixed-head disk

Answer 16

advantage: its very fast, faster than the movable-head disks

disadvantage: high cost and its reduced storage space compared to a movable-head disk

Question 17

_____________________, such as computer hard drives, have one read/write head that floats over each surface of each disk. Disks can be a single platter, or part of a disk pack, which is a stack of magnetic platters.

Answer 17

Movable-head magnetic disks

Question 18

Two of the most important measures of optical disc drive performance are:

Answer 18

sustained data transfer rate and average access time

Question 19

There are several types of optical-disc systems, depending on the medium and the capacity of the discs:

Answer 19

CDs, DVDs, and Blu-ray

Question 20

A ___________ is the same physical size as a DVD or CD but the laser technology used to read and write data is quite different.

Answer 20

Blu-ray disc

Question 21

a type of electrically erasable programmable read-only memory

Answer 21

Flash memory

Question 22

Flash memory uses a phenomenon (known as ______________) to send electrons through a floating gate transistor where they remain even after power is
turned off

Answer 22

Fowler-Nordheim tunneling

Question 23

Depending on whether a disk has fixed or movable heads, there can be as many as
three factors that contribute to the time required to access a file:

Answer 23

seek time
search time
transfer time.

Question 24

formula of total access time of fixed-head drives

Answer 24

search time (rotational delay)
+ transfer time (data transfer)
= access time

Question 25

formula of total access time of movable-head drives

Answer 25

seek time (arm movement) search time (rotational delay) + transfer time (data transfer) = access time

Question 26

At the start of an I/O command, the information passed from the CPU to the channel is this:

Answer 26

* I/O command (READ, WRITE, REWIND, etc.) * Channel number * Address of the physical record to be transferred (from or to secondary storage) * Starting address of a memory buffer from which or into which the record is to be transferred

Question 27

The Device Manager relies on several auxiliary features to keep running efficiently under the demanding conditions of a busy computer system, and there are three problems that must be resolved:

Answer 27

* It needs to know which components are busy and which are free. * It must be able to accommodate the requests that come in during heavy I/O traffic. * It must accommodate the disparity of speeds between the CPU and the I/O devices.

Question 28

The traffic controller has three main tasks:

Answer 28

(1) it must determine if there’s at least one path available; (2) if there’s more than one path available, it must determine which to select; and (3) if the paths are all busy, it must determine when one will become available.

Question 29

A ___________ for the I/O device handler is the predetermined policy that the device handler uses to allocate access to the device among the many processes that may be waiting for it. It determines the order in which the processes get the device, and the goal is to keep seek time to a minimum.

Answer 29

seek strategy

Question 30

Every scheduling algorithm should do the following: (goal of a seek strategy)

Answer 30

* Minimize arm movement * Minimize mean response time * Minimize the variance in response time

Question 31

____________ is the simplest device-scheduling algorithm; it is easy to program and essentially fair to users.

Answer 31

First-come, first-served (FCFS)

Question 32

_______________ uses the same underlying philosophy as Shortest Job Next (described in Chapter 4), where the shortest jobs are processed first and longer jobs are made to wait. With SSTF, the request with the track closest to the one being served (that is, the one with the shortest distance to travel) is the next to be satisfied, thus minimizing overall seek time.

Answer 32

Shortest seek time first (SSTF)

Question 33

________ uses a directional bit to indicate whether the arm is moving toward the center of the disk or away from it. The algorithm moves the arm methodically from the outer to the inner track, servicing every request in its path. When it reaches the innermost track, it reverses direction and moves toward the outer tracks, again servicing every request in its path.

Answer 33

SCAN

Question 34

The most common variation of SCAN is __________, sometimes known as the elevator algorithm, in which the arm doesn’t necessarily go all the way to either edge unless there are requests there

Answer 34

LOOK

Question 35

_________ holds all new requests until the arm starts on its way back. Any requests that arrive while the arm is in motion are grouped for the arm’s next sweep.

Answer 35

N-Step SCAN

Question 36

With ___________ the arm picks up requests on its path during the inward sweep.

Answer 36

C-SCAN or Circular Scan

Question 37

________ is an optimization of C-SCAN, just as LOOK is an optimization of SCAN. In this algorithm, the sweep inward stops at the last high-numbered track request, so the arm doesn’t move all the way to the last track unless it’s required to do so.

Answer 37

C-LOOK

Question 38

Which strategy is best? It’s up to the system designer to select the best algorithm for each environment. It’s a job that’s complicated because the day-to-day performance of any scheduling algorithm depends on the load it must handle; but some broad generalizations can be made based on simulation studies:

Answer 38

* FCFS works well with light loads; but as soon as the load grows, service time becomes unacceptably long. * SSTF is quite popular and intuitively appealing. It works well with moderate loads but has the problem of localization under heavy loads. * SCAN works well with light to moderate loads and eliminates the problem of indefinite postponement. SCAN is similar to SSTF in throughput and mean service times. * C-SCAN works well with moderate to heavy loads and has a very small variance in service times.

Question 39

a way to optimize search times by ordering the requests once the read/write heads have been positioned.

Answer 39

rotational ordering

Question 40

an algorithm that adds extra, redundant bits to the data and is therefore able to correct single-bit errors and detect double-bit errors.

Answer 40

Hamming code

Question 41

________ is a set of physical disk drives that is viewed as a single logical unit by the oper ating system.

Answer 41

RAID

Question 42

____________uses data striping without parity, without error correction. It is the only level that does not provide error correction, or redundancy, and so it is not considered a true form of RAID because it cannot recover from hardware failure. simple words (chatgpt): Splits data into parts to make it faster — but no backup.

Answer 42

RAID Level 0

Question 43

___________ also uses striping, and is called a mirrored configuration because it provides redundancy by having a duplicate set of all data in a mirror array of disks, which acts as a backup system in the event of hardware failure. simple words (chatgpt): Makes an exact copy (mirror) of everything.

Answer 43

RAID Level 1

Question 44

___________ uses very small strips (often the size of a word or a byte) and uses a Hamming code to provide error detection and correction, or redundancy simple words (chatgpt): Uses math (Hamming code) to fix small mistakes.

Answer 44

RAID Level 2

Question 45

_________ is a modification of Level 2 that requires only one disk for redundancy. Only one parity bit is computed for each strip, and it is stored in the designated redundant disk. simple words (chatgpt): Uses one disk just to check and fix mistakes.

Answer 45

RAID Level 3

Question 46

____________ uses the same strip scheme found in Levels 0 and 1, but it computes a parity for each strip, and then it stores these parities in the corresponding strip in the designated parity disk. simple words (chatgpt): Has one “parity disk” that saves extra info to fix problems

Answer 46

RAID Level 4

Question 47

__________ is a modification of Level 4. Instead of designating one disk for storing parities, it distributes the parity strips across the disks, which avoids the bottleneck created in Level 4 simple words (chatgpt): Spreads those backup clues across all disks.

Answer 47

RAID Level 5

Question 48

was introduced by the Berkeley research team in a paper that followed its original outline of RAID levels. This system provides an extra degree of error detection and correction because it requires two different parity calculations. simple words (chatgpt): Has two sets of backup clues — even safer!

Answer 48

RAID Level 6

Question 49

There are several types of optical-disc systems, depending on the medium and the capacity of the discs:

Answer 49

CDs, DVDs, and Blu-ray

Question 50

_________ is a type of electrically erasable programmable read-only memory

Answer 50

Flash memory

Question 51

Time required to position the read/write head on the proper track

Answer 51

Seek time

Question 52

the time it takes to rotate the disk until the requested record is moved under the read/write head

Answer 52

Search time

Question 53

Other term for search time

Answer 53

Rotational delay

Question 54

the fastest of the three; that’s when the data is actually transferred from secondary storage to main memory.

Answer 54

Transfer time

Question 55

Describe how primary storage differs from secondary storage.

Answer 55

Primary storage (like RAM) is fast, temporary memory used by the CPU to store data and instructions currently in use. Secondary storage (like hard drives) is slower, permanent memory used to store data and programs long-term.

Question 56

Explain the differences between buffering and blocking.

Answer 56

Buffering - data is written to a temporary area while it waits to go to an I/O device, and are often used to allow a fast resource to work with a resource, or vice versa Blocking - a storage and I/O saving technique that groups individual records into a block that’s stored and retrieved as a unit