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vivado [14]
3 years ago
6

A disk drive has 2000 cylinders, numbered 0 to 1999. The drive is currently servicing a request at cylinder 783, with the previo

us request at cylinder 707. The queue of pending requests, in FIFO order, is 75, 678, 1623, 247, 248, 1529, 182. What distance, in tracks, must the disk read-write arm move in order to service the requests for each of the following disk scheduling algorithms?
a) FCFS
b) SSTF
c) SCAN
d) LOOK
e) C-SCAN
f) C-LOOK
Computers and Technology
1 answer:
Ket [755]3 years ago
5 0

Answer:

Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving a request at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is

Explanation:

You might be interested in
In the Sender Message Channel Receiver (SMCR) model, the ______ is the large, bureaucratic organization that produces messages.
mafiozo [28]

Answer:

Sender.

Explanation:

The Sender Message Channel Receiver (SMCR) model of communication was developed and created by David Berlo in 1960. SCMR was developed from the Shannon-Weaver model of communication of 1949.

The SCMR model of communication comprises of four (4) main components and these are;

1. Sender (S): this is typically the source of information (message) or the originator of a message that is being sent to a receiver. Thus, they are simply the producer of a message.

2. Channel (C): this is the medium used by the sender for the dissemination or transmission of the message to the recipient. For example, telephone, television, radio, newspapers, billboards etc.

3. Message (M): this is the information or data that is being sent to a recipient by a sender. It could be in the form of a video, audio, text message etc.

4. Receiver (R): this is typically the destination of information (message) or the recipient of a message that is being sent from a sender.

Hence, in the Sender Message Channel Receiver (SMCR) model, the sender is the large, bureaucratic organization that produces messages.

Additionally, encoding is the process of transmitting the message from the sender to the recipient while decoding is the process of interpreting the message that is being received from a sender.

7 0
3 years ago
Find the domain and range of Y=x^3-9/x-3​
dmitriy555 [2]
847nduak388jejdidmenrf838484858686
6 0
3 years ago
Which of the following is the fundamental unit of the virtualized client in an IaaS deployment? a)Workload b)Workspace c)Work un
forsale [732]

Answer:

a. Workload

Explanation:

The most fundamental unit of the virtualized client in an IaaS deployment is the workload because it simulates the potential of a certain server to perform a specific amount of work given.

6 0
3 years ago
Describe the effect of a pull up resistor
Vesna [10]
With a pull-up resistor, the input pin will read a high state when the button is not pressed. In other words, a small amount of current is flowing between VCC and the input pin (not to ground), thus the input pin reads close to VCC. When the button is pressed, it connects the input pin directly to ground.
3 0
4 years ago
C++ Code Outputs.
mojhsa [17]

Answer:

#include

#include

#include

#include

using namespace std;

struct courseInfo{

string name;

int unit;

char grade;

};

struct Student {

string fName;

string lName;

string idNumber;

courseInfo courses[2];

int unitCompleted;

double gpa;

};

Student s;

bool openFile(ifstream &in);

void Print_info_one(Student s);

void Read_info(Student &s);

float Find_points(char c) ;

bool openFile(ifstream &inFile){

string line;

int i=0,k=0;

string fName="", lname="", id="", name1="", name2="";

char grade1, grade2;

int unit1, unit2;

if (inFile.is_open())

{

while (getline(inFile, line))

{

while (line[i] != ',')

{

fName += line[i];

i++;

}

i++;

i++;

while (line[i] != ' ')

{

lname += line[i];

i++;

}

i++;i++;

while (line[i] != ' ')

{

id += line[i];

i++;

}

i++;

int count=0;

while (count <2)

{

name1 += line[i];

i++;

if(line[i] == ' ' ) count++;

}

i++;

grade1 = line[i];

i++;i++;

unit1 = line[i]-'0';

i++;i++;

count=0;

while (count <2)

{

name2 += line[i];

i++;

if(line[i] == ' ' ) count++;

}

i++;

grade2 = line[i];

i++;i++;

unit2 = line[i]-'0';

}

inFile.close();

s.fName = fName;

s.lName = lname;

s.idNumber = id;

s.courses[0].name = name1;

s.courses[0].grade = grade1;

s.courses[0].unit = unit1;

s.courses[1].name = name2;

s.courses[1].grade = grade2;

s.courses[1].unit = unit2;

s.unitCompleted = unit1 + unit2;

s.gpa = (unit1*Find_points(grade1) + unit2*Find_points(grade2))/(unit1+unit2);

}

else

{

cout << "Error reading file\n";

return false;

}

return true;

}

void Print_info_one(Student s){

cout << "Name: " << s.fName << ", " << s.lName << " ID Number: " << s.idNumber << " Course 1 Name: " << s.courses[0].name << " Grade: "

<< s.courses[0].grade << " Units: " << s.courses[0].unit << " Course 2 Name: " << s.courses[1].name << " Grade: "

<< s.courses[1].grade << " Units: " << s.courses[1].unit << " Unit completed: " << s.unitCompleted << " GPA:" << s.gpa << endl;

}

void Read_info(Student &s){

}

float Find_points(char grade){

switch (grade)

{

case 'A':

return 4.0;

break;

case 'B':

return 3.0;

break;

case 'C':

return 2.0;

break;

case 'D':

return 1.0;

break;

case 'F':

return 0;

break;

default:

break;

}

return 0;

}

int main() {

ifstream inFile;

std::fstream fs;

fs.open ("input.txt", std::fstream::in );

Print_info_one(s);

return 0;

Explanation:

5 0
3 years ago
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