Answer:
Check the explanation
Explanation:
Number of 4 TB hard drive need = 32/4 = 8 and cost = 200*8 = $1600
Number of 6 TB hard drive need = max(32/6) = 6 and cost = 250*6 = $1500
Number of 8 TB hard drive need = 32/8 = 4 and cost = 300*4 = $1200
Number of 10 TB hard drive need = 32/10 = 4 and cost = 350*4 = $1400
Hence using 4 hard drive of 8 TB will minimize the cost.
Your task is to break a number into its individual digits, for example, to turn 1729 into 1, 7, 2, and 9. It is easy to get the last digit of a number n as n % 10. But that gets the numbers in reverse order. Solve this problem with a stack. Your program should ask the user for an integer, then print its digits separated by spaces.
Answer:
stack = []
number = int(input("Enter a number: "))
while number > 0:
stack.append(number % 10)
number = int(number / 10)
while len(stack) > 0:
print(stack.pop(), end=" ")
Explanation:
Initialize an empty list to represent the stack
Get a number from the user
Create a while loop that iterates through the given number and append its last digit to stack
When the first loop is done, your stack is filled with the digits of the given number
Create another while loop that iterates through your stack and prints its elements, digits of the number, separated by spaces
Verify each computer in the LAN can ping the other computers. Also verify that each computer can ping the routerÕs gateway IP address [10.10.20.250]. If any of the assigned IP addresses fail to generate a reply, troubleshoot and correct the problem(s) until all devices can ping each other.
Answer:
The difficulty arises when your computer asks for an IP address and no one responds.
Whether due to a network problem — maybe not being on a network at all — or perhaps a lack of DHCP server to hand out IP addresses, the result is the same: the request for an IP address assignment goes unanswered.
Your machine waits for a while and then gives up.
But when it gives up, it invokes what’s called Automatic Private IP Addressing, or APIPA, and makes up its own IP address. And those “made up” IP addresses all begin with 169.254.
Limited connectivity
Situations resulting in a 169.254.x.x IP address are flagged as “limited connectivity” in Windows.
If you see that status associated with your network connectivity, then you’re facing the problem discussed here. Your machine almost certainly has a 169.254.x.x IP address.
Explanation:
Suppose a file system can have three disk allocation strategies, contiguous, linked, and indexed. We have just read the information for a file from its parent directory. For contiguous and linked allocation, this gives the address of the first block, and for indexed allocation this gives the address of the index block. Now we want to read the 10th data block into the memory. How many disk blocks (N) do we have to read for each of the allocation strategies?
Answer:
Since we are using three disk allocation strategies,continuous linked and indexed,therof to process and read allocation strategies we need 99830 disk blocks.
Write a program that does the following: • Alphabetizes a list of small (non-capital) letters. For consistency, make the string 12-15 random small letters. Label the sequence "string 1" in the data statements, followed by a ".space" directive to reserve 20-30 byte spaces (extra room in case you input a few extra characters, and so that you will still have a null-terminated string, since ".space" clears all byte spaces to 0). • Inputs the string from the keyboard using syscall 8 and places it in the space labeled "string 1." • Alphabetizes the string using a loop and outputs the alphabetized string to the simulated console. • Removes any characters that are not small letters (capitals, punctuation, numbers, etc.). • Outputs the ordered letters, preceded by a statement: "The alphabetized string is:
Answer:
Explanation:
find the attached below
Write a while loop that adds the first 100 numbers (from 1 to 100) together. To do so, use a while loop that continues while the condition of cur_num being less than or equal to stop_num is True. Inside the loop, it should add cur_num to the running total variable, and also add 1 to cur_num. Make sure to check at the end that the value of total reflects the total summation.
Answer:
I am writing JAVA and C++ program. Let me know if you want the program in some other programming language.
Explanation:
JAVA program:
public class SumOfNums{
public static void main(String[] args) { //start of main function
int cur_num = 1; // stores numbers
int stop_num= 100; // stops at this number which is 100
int total = 0; //stores the sum of 100 numbers
/* loop takes numbers one by one from cur_num variable and continues to add them until the stop_num that is 100 is reached*/
while (cur_num <= stop_num) {
total += cur_num; //keeps adding the numbers from 1 to 100
//increments value of cur_num by 1 at each iteration
cur_num = cur_num + 1;}
//displays the sum of first 100 numbers
System.out.println("The sum of first 100 numbers is: " + total); }}
Output:
The sum of first 100 numbers is: 5050
C++ program
#include <iostream> //for input output functions
using namespace std; //to identify objects like cin cout
int main(){ //start of main() function body
int cur_num = 1; // stores numbers
int stop_num= 100; // stops at this number which is 100
int total = 0;//stores the sum of 100 numbers
/* loop takes numbers one by one from cur_num variable and continues to add them until the stop_num that is 100 is reached*/
while (cur_num <= stop_num) {
//keeps adding the numbers from 1 to 100
//increments value of cur_num by 1 at each iteration
total += cur_num;
cur_num = cur_num + 1; }
//displays the sum of first 100 numbers
cout<<"The sum of first 100 numbers is: "<<total; }
The output is given in the attached screen shot.
7.8.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new = (old + 1) * 2. Ex: If xVal = 3 and yVal = 4, then xValNew is 8 and yValNew is 10. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 #include using namespace std; /* Your solution goes here */ int main() { int xValNew; int yValNew; int xValUser; int yValUser; cin >> xValUser; cin >> yValUser; CoordTransform(xValUser, yValUser, xValNew, yValNew); cout << "(" << xValUser << ", " << yValUser << ") becomes (" << xValNew << ", " << yValNew << ")" << endl; return 0; } 1 test passed All tests passed Run Feedback? How was this section?
Answer:
Here is the CoordTransform() function:
void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){
xValNew = (xVal + 1) * 2;
yValNew = (yVal + 1) * 2; }
The above method has four parameters xVal and yVal that are used as input parameters and xValNew yValNew as output parameters. This function returns void and transforms its first two input parameters xVal and yVal into two output parameters xValNew and yValNew according to the formula: new = (old + 1) *
Here new variables are xValNew and yValNew and old is represented by xVal and yVal.
Here the variables xValNew and yValNew are passed by reference which means any change made to these variables will be reflected in main(). Whereas variables xVal and yVal are passed by value.
Explanation:
Here is the complete program:
#include <iostream>
using namespace std;
void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){
xValNew = (xVal + 1) * 2;
yValNew = (yVal + 1) * 2;}
int main()
{ int xValNew;
int yValNew;
int xValUser;
int yValUser;
cin >> xValUser;
cin >> yValUser;
CoordTransform(xValUser, yValUser, xValNew, yValNew);
cout << "(" << xValUser << ", " << yValUser << ") becomes (" << xValNew << ", " << yValNew << ")" << endl;
return 0; }
The output is given in the attached screenshot
Currently when an animal other than a cat or dog is entered (such as a hamster), the program asks if it is spayed or neutered before displaying the message that only cats and dogs need pet tags. Find a way to make the program only execute the spay/neuter prompt and input when the pet type is cat or dog.
Answer:
int main() {
string pet; //stores the pet string which is cat or dog
char spayed; // stores the choice from y or n
cout << "Enter the pet type (cat or dog): "; //prompts user to enter pet type
cin >> pet; //reads the input from user
if(pet=="cat"|| pet=="dog") { // if user enters cat or dog
cout << "Has the pet been spayed or neutered (y/n)? ";
//asks user about pet been spayed of neutered
cin >> spayed;} //reads y or n input by the user
else // else part will execute if user enters anything other than cat or dog
cout<<"only cats and dogs need pet tags";
// displays this message if input value of pet is other than cat or dog
Explanation:
The answer to the complete question is provided below:
#include <string>
using namespace std;
int main() {
string pet; // "cat" or "dog"
char spayed; // 'y' or 'n'
// Get pet type and spaying information
cout << "Enter the pet type (cat or dog): ";
cin >> pet;
if(pet=="cat"|| pet=="dog") {
cout << "Has the pet been spayed or neutered (y/n)? ";
cin >> spayed;}
else
cout<<"only cats and dogs need pet tags";
// Determine the pet tag fee
if (pet == "cat")
{ if (spayed == 'y' || spayed == 'Y') //lowercase or upper case y is accepted
cout << "Fee is $4.00 \n";
else
cout << "Fee is $8.00 \n";
}
else if (pet == "dog")
{ if (spayed == 'y' || spayed=='Y')
cout << "Fee is $6.00 \n";
else
cout << "Fee is $12.00 \n";
}
return 0; }
According to part a) of this question the OR operator is used in this statement if (spayed == 'y' || spayed=='Y') so that if the user enters capital Y or small y letter both are acceptable.
According to the part b) if(pet=="cat"|| pet=="dog") statement is used to make program only execute spay/neuter prompt and input when the pet type is cat or dog otherwise the program displays the message: only cats and dogs need pet tags.
Final answer:
The student's issue with the program asking about spay/neuter for non-applicable animals can be fixed by adding a conditional statement that checks for the pet type and only asks the question if the pet is a cat or dog.
Explanation:
To address the issue presented in the student's question, a conditional statement should be implemented in the program's code. This statement would check the type of pet entered by the user and only execute the spay/neuter prompt and input sequence if the pet is a cat or dog. Here is a simple pseudo-code example to illustrate:
if (petType == "cat" || petType == "dog") {
// Ask if the pet is spayed or neutered
} else {
// Display a message that only cats and dogs need pet tags
}
By using such a conditional structure, the program will prompt for spay/neuter information appropriately based on the pet type, improving its functionality and user experience.
Write an application that displays a series of at least five student ID numbers (that you have stored in an array) and asks the user to enter a numeric test score for the student. Create a ScoreException class, and throw a ScoreException for the class if the user does not enter a valid score (less than or equal to 100). Catch the ScoreException, display the message Score over 100, and then store a 0 for the student’s score. At the end of the application, display all the student IDs and scores. g
Answer:
See explaination for the program code
Explanation:
// ScoreException.java
class ScoreException extends Exception
{
ScoreException(String msg)
{
super(msg);
}
}
// TestScore.java
import java.util.Scanner;
public class TestScore
{
public static void main(String args[]) throws ScoreException
{
Scanner sc = new Scanner(System.in);
int studentId[] = { 1201, 1202, 1203, 1204, 1205 };
int scores[] = new int[5];
int score = 0;
for(int i = 0; i < studentId.length; i++)
{
try
{
System.out.print("Enter a numeric test score for the student"+(i+1)+" of Id "+studentId[i]+" : ");
score = sc.nextInt();
if(score < 0 || score > 100)
{
scores[i] = 0;
throw new ScoreException("Input should be between 1 and 100");
}
else
{
scores[i] = score;
}
}
catch(ScoreException ex)
{
System.out.println("\n"+ex.getMessage()+"\n");
}
}
//displaying student details
System.out.println("\n\nStudent Id \t Score ");
System.out.println("=============================");
for(int i = 0; i < studentId.length; i++)
{
System.out.println(studentId[i]+"\t\t"+scores[i]);
}
}
}
MSSQL
Create sp_Q2 as a stored procedure, which takes no input but it performs the tasks
in the order listed below:
(1) sp_Q2 first finds the country of suppliers who supply products with the top 2
highest unit price. It does not matter if these two products belong to the same
supplier or not. Also, it does not matter if the supplier of these two products
are in the same country or not.
(2) sp_Q2 does not return the countries, instead, it calls sp_Q1 (completed
in Q1) using the countries it finds.
Like in Q1, you should include 'go' to indicate the end of script that creates the
procedure and include a testing script to test sp_Q2.
The hint below is just a hint. You don't have to use it if you have a better approach.
Hint: save the countries in a table of any type (e.g., table variable, temporary table,
or using 'select...into') and retrieve one country a time into a variable.
Answer:
The code is given below
{
string cipher = "";
//Where cipher is an algorithm used.
for (int y = 0; iy< msg.length(); i++)
{
if(msg[i]!=' ')
/* applying encryption formula ( c x + d ) mod m
{here x is msg[i] and m is 26} and added 'A' to
bring it in range of ascii alphabet[ 65-90 | A-Z ] */
cipher = cipher +
(char) ((((a * (msg[i]-'A') ) + b) % 26) + 'A');
else
cipher += msg[i];
}
return cipher;
}
"Suppose that instead of always selecting the first activity to finish, we instead select the last activity to start that is compatible with all previously selected activities. De- scribe how this approach is a greedy algorithm, and prove that it yields an optimal solution."
The greedy algorithm described selects the last activity to start that is compatible with all previously selected activities, and it can be proved optimal using greedy-choice and optimal substructure properties. This algorithm's real-life application is similar to planning a timely route to an event, optimizing decisions step by step.
Explanation:The greedy algorithm approach to solving problems involves making the locally optimal choice at each stage with the hope of finding a global optimum. In the context of activity selection, the approach described in the question is indeed a greedy strategy, where instead of selecting the activity that finishes earliest, you select the activity that starts last while still being compatible with the previously selected activities. The method assumes that by maximizing the starting time of the current activity, we leave as much room as possible for other activities to take place before it.
To prove that this yields an optimal solution, we can use a greedy-choice property and an optimal substructure property. The greedy-choice property ensures that a locally optimal choice leads toward a globally optimal solution. By choosing the last activity to start, we reserve the maximum possible time for all other activities, thus leading to an optimal selection. The optimal substructure property suggests that if we have an optimal solution for a set of activities, then for any subproblem (such as a subset of these activities), a corresponding optimal solution can also be found. Hence, by demonstrating that each greedy step of choosing the latest-starting activity yields a solution as good as any other compatible activity, we prove that the overall selection is optimal.
An analogy for using a greedy algorithm in everyday life is when you're planning to attend an event with a specific time constraint, like a wedding. If you have to arrive by a certain time and consider traffic congestion, you work backwards from the event and plan your departure to optimize your route. This approach is a simple manifestation of a greedy algorithm, where you make the best immediate decision given the information you have.
A greedy algorithm for the activity selection problem involves selecting the last activity to start that is compatible with previously selected activities. This approach is optimal as it maximizes the number of non-overlapping activities by always making the best local choice.
This question is about using a greedy algorithm to solve the activity selection problem by choosing the last activity to start that is compatible with all previously selected activities. Here is a step-by-step explanation:
Initial Sorting: Sort the activities by their start times in decreasing order.Selection Process: Initialize an empty set to keep track of the selected activities. Go through the list of activities, and for each activity, check if it is compatible with the already selected activities (i.e., it does not overlap in time with them).Greedy Choice: Always select the current activity if it is compatible (non-overlapping) with the previously selected activities. Add this activity to the set of selected activities.This approach works because, like the classic greedy approach, it makes a locally optimal choice at each step with the hope of finding a global optimum. When you select the last activity to start that is compatible with the previously chosen activities, you are ensuring that the remaining time is utilized as best as possible, essentially maximizing the number of non-overlapping activities.
Proof of Optimality:
To demonstrate that this greedy algorithm yields an optimal solution, consider the following:
Inductive Hypothesis: Assume that the set of activities selected so far is optimal.Greedy Choice Property: Adding the next last-starting compatible activity preserves the optimality because selecting any earlier starting activity would reduce the number of non-overlapping activities that can be included.Optimal Substructure: By selecting the last compatible activity, we reduce the problem size and solve the smaller problem optimally until no more activities can be selected.Therefore, this approach ensures an optimal solution by maintaining the greedy choice property and optimal substructure, essential characteristics of a greedy algorithm.
You are troubleshooting an IEEE 802.11 wireless client device that cannot connect to the wireless network. You notice the client utility shows good signal strength but will not connect to the network. After opening a command prompt and typing the ipconfig command, you notice the IP address is 169.254.12.50. You suspect the IP address is not valid and could be causing the problem. What potential problem could cause this IP address?
Answer:
A potential problem that could cause this is if there's an issue with the DHCP.
Explanation:
Firstly, without a valid IP address your computer cannot use the network. Some of the problems can that can make an IP address invalid, are address conflicts with other computers and network configuration problems.
The Dynamic Host Configuration Protocol (DHCP) is the network service in routers that offers a convenient way to automatically assign IP addresses to computers joining a network. So when the IEEE 802.11 wireless client device connects to the wireless network, it should be assigned an IP address automatically. However, DHCP-generated addresses can sometimes cause problems. This happens when Windows assigns the IEEE 802.11 wireless client device an address before your Internet router does. The IP address may be invalid if it conflicts with the network's address range. Therefore, making the device not to connect to the wireless network.
rames of 1 Kb are sent over a 1 Mbps channel using a geostationary satellite whose propagation time from earth is 270 ms. Acknowledgments are always piggybacked onto data frames and headers are very short. Three - bit sequence numbers are used. What is the maximum achievable channel utilization for:
(a) Stop-and-wait
(b) Protocol 5
(c) Protocol 6
Answer:
a) The maximum channel utilization for stop and wait protocol is 0.18%
b) The maximum channel utilization for protocol 5 (Go-back) is 1.29%
c) The maximum channel utilization for protocol 6 (Selective repeat) is 0.74%
Explanation:
Final answer:
Explanation on calculating maximum channel utilization for different protocols over a 1 Mbps channel using a geostationary satellite.
Explanation:
Channel Utilization Calculation:
Stop-and-wait: Utilization = 1 / (1 + 2a), where a is the propagation delay per frame transmission time ratio.Protocol 5: Utilization = 1 / (1 + 2a) + a / (1 + 2a).Protocol 6: Utilization = 1 / (1 + a) (1 + 2a).Stop-and-wait protocol involves sending a frame and waiting for its acknowledgment before the next frame can be sent. This protocol is highly inefficient for long propagation times, as the sender spends most of the time waiting rather than transmitting.
Protocol 5 (Go-Back-N) and Protocol 6 (Selective Repeat) are sliding window protocols designed to improve efficiency by allowing multiple frames to be in transmission before requiring acknowledgment. However, the effectiveness of these protocols in this scenario would be significantly limited by the three-bit sequence numbering, which restricts the number of outstanding frames.
Calculating exact channel utilization for each protocol requires considering the round trip time (RTT), the size of the frames, the sequence number limitations, and the transmission rate. The RTT is particularly crucial, as it includes the propagation time to and from the satellite, drastically reducing the channel utilization for stop-and-wait protocol compared to the sliding window protocols.
P5. (5pts) Recall that we saw the Internet checksum being used in both transport-layer segment (in UDP and TCP headers) and in network-layer datagrams (IP header). Now consider a transport layer segment encapsulated in an IP datagram. Are the checksums in the segment header and datagram header computed over any common bytes in the IP datagram
Answer:
see explaination
Explanation:
No common bytes are used to compute the checksum in the IP datagram. Only the IP header is used to compute the checksum at the network Layer.
At TCP/UDP segment, the IP datagram may have different protocol stacks used.
Therefore, when a TCP datagram is entered in the IP datagram, it is not necessary to use common byte to compute checksum.
The checksums in the transport-layer (TCP/UDP) and network-layer (IP) headers are computed over different sets of bytes, ensuring the integrity of their respective areas independently.
The IP header checksum covers only the IP header, while the TCP/UDP checksum includes their headers and data, along with some IP header information.
The checksums in the transport-layer segment (like TCP and UDP) and the network-layer datagram (IP) headers are designed to ensure data integrity by detecting errors that may have occurred during transmission.
However, the checksums are not computed over any common bytes of the IP datagram. The IP header checksum is computed exclusively over the IP header, which includes the source and destination IP addresses, protocol, and other fields. Conversely, the checksum for the TCP or UDP segment is computed over the entire segment, which includes the payload (data) and a pseudo-header. The pseudo-header contains certain fields from the IP header, like source and destination addresses, protocol, and segment length, but the checksum itself does not overlap with that of the IP header. Example:
If an IP datagram carries a TCP segment, the IP checksum will only ensure the integrity of the IP header.The TCP checksum ensures the integrity of the TCP header and its data.pendant publishing edits multi-volume manuscripts for many authors, for each volume, they want a label that contains the author's name, the title of the work, and a volume number in the form volume 9 of 9. design an application that reads records that contain an author's name, the title of the work, and the number of volumes. the application must read the records until eof is encountered and produce enough labels for each work
Answer:
See explaination for the program code
Explanation:
The code
#include<fstream>
using namespace std;
//Read data from file and returns number of records
int readFile(string authorName[], string title[], int volumeNo[], int volumeNo1[])
{
//Creates an object of ifstream
ifstream readf;
//Opens the file volume.txt for reading
readf.open ("volume.txt");
//Counter for number of records
int c = 0;
//Loops till end of file
while(!readf.eof())
{
//Reads data and stores in respective array
readf>>authorName[c];
readf>>title[c];
readf>>volumeNo[c];
readf>>volumeNo1[c];
//Increase the record counter
c++;
}//End of while
//Close file
readf.close();
//Returns record counter
return c;
}//End of function
//To display records information
void Display(string authorName[], string title[], int volumeNo[], int volumeNo1[], int len)
{
int counter = 0;
//Displays the file contents
cout<<"\n The list of multi-volume manuscripts \n";
cout<<("____________________________________________________");
//Loops till end of the length
for(int x = 0; x < len; x++)
{
cout<<"\n Author Name: "<<authorName[x];
cout<<"\n Title: "<<title[x];
cout<<"\n Volume "<<volumeNo[x]<<" of "<<volumeNo1[x]<<endl;
}//End of for loop
//Displays total records available
cout<<"\n The Records: "<<len;
}//End of function
//Main function
int main()
{
//Creates arrays to store data from file
string authorName[100], title[100];
int volumeNo[100];
int volumeNo1[100];
//To store number of records
int len;
//Call function to read file
len = readFile(authorName, title, volumeNo, volumeNo1);
//Calls function to display
Display(authorName, title, volumeNo, volumeNo1, len);
return 0;
}//End of main
volume.txt file contents
Pyari CPorg 1 2
Mohan C++ 3 4
Sahu Java 6 7
Ram C# 4 2
Sample Run:
The list of multi-volume manuscripts
____________________________________________________
Author Name: Pyari
Title: CPorg
Volume 1 of 2
Author Name: Mohan
Title: C++
Volume 3 of 4
Author Name: Sahu
Title: Java
Volume 6 of 7
Author Name: Ram
Title: C#
Volume 4 of 2
The Records: 4
For each of the two questions below, decide whether the answer is (i) "Yes," (ii) "No," or (iii) "Unknown, because it would resolve the question of whether P = NP." Give a brief explanation of your answer. (a) Let’s define the decision version of the Interval Scheduling Prob- lem from Chapter 4 as follows: Given a collection of intervals on a time-line, and a bound k, does the collection contain a subset of nonoverlapping intervals of size at least k? Question: Is it the case that Interval Scheduling ≤P Vertex Cover? (b) Question: Is it the case that Independent Set ≤P Interval Scheduling?
Final answer:
The original student question pertains to computational complexity and the relationship between different problems within the NP complexity class, seeking to understand polynomial-time reductions between Interval Scheduling, Vertex Cover, and Independent Set.
Explanation:
The questions provided revolve around computational complexity, particularly the concepts surrounding the complexity classes P, NP, and the relations between different computational problems. Unfortunately, the rest of the information provided is not related to the P vs NP question raised and therefore not relevant for forming an answer to the question at hand. The original question seems to ask whether 'Interval Scheduling' can be polynomial-time reduced to 'Vertex Cover' (Interval Scheduling ≤P Vertex Cover), and whether 'Independent Set' can be polynomial-time reduced to 'Interval Scheduling' (Independent Set ≤P Interval Scheduling). These questions are seeking to understand the computational complexity relationship between different NP problems.
Given a collection of n nuts and a collection of n bolts, arranged in an increasing order of size, give an O(n) time algorithm to check if there is a nut and a bolt that have the same size. The sizes of the nuts and bolts are stored in the sorted arrays NUT S[1..n] and BOLT S[1..n], respectively. Your algorithm can stop as soon as it finds a single match (i.e, you do not need to report all matches).
Answer:
See explaination
Explanation:
Keep two iterators, i (for nuts array) and j (for bolts array).
while(i < n and j < n) {
if nuts[i] == bolts[j] {
We have a case where sizes match, output/return
}
else if nuts[i] < bolts[j] {
this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1
}
else {
this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1
}
}
Since we go to each index in both the array only once, the algorithm take O(n) time.
2. Consider the two-dimensional array A: int A[][] = new int[100][100]; where A[0][0] is at location 200 in a paged memory system with pages of size 200. Asmall process that manipulates the matrix resides in page 0 (locations 0 to 199). Thus, every instruction fetch will be from page 0. For three page frames, how many page faults are generated by the following array-initialization loops? Use LRU replacement, and assume that page frame 1 contains the process and the other two are initially empty. a. for (int j = 0; j < 100; j++) for (int i = 0; i < 100; i++) A[i][j] = 0; b. for (int i = 0; i < 100; i++) for (int j = 0; j < 100; j++) A[i][j] = 0;
Final answer:
The problem is about calculating the number of page faults for two different loop orderings in a paged memory system under LRU replacement. Loop (a) has a column-major ordering, likely causing a new page fault for each new column accessed, while loop (b) has a row-major ordering, causing fewer page faults due to contiguous memory access.
Explanation:
A student has asked about the number of page faults generated by array-initialization loops in a paged memory system, specific to a given scenario with LRU replacement policy. We have two different loop orderings to consider with three page frames available.
For the first loop (part a), which is column-major order:
Page fault will occur for each new column since each element of a column is on a different page (due to row-major storage in C-like languages).
There are 100 columns, so initially, we will have 100 page faults for the first 100 pages.
As pages are brought into memory and replaced using LRU, subsequent accesses to those pages within the loop may not cause additional faults if they remain in the frame.
For the second loop (part b), which is row-major order:
Since each row is contiguous in memory, fewer page faults will occur.
The loop accesses 200 int elements before moving to a new page, thus causing a page fault.
We can fill two page frames with array data since one frame is occupied by the process.
Server farms such as Google and Yahoo! provide enough compute capacity for the highest request rate of the day. Imagine that most of the time these servers operate at only 60% capacity. Assume further that the power does not scale linearly with the load; that is, when the servers are operating at 60% capacity, they consume 90% of maximum power. The servers could be turned off, but they would take too long to restart in response to more load. A new system has been proposed that allows for a quick restart but requires 20% of the maximum power while in this "barely alive" state. a. How much power savings would be achieved by turning off 60% of the servers? b. How much power savings would be achieved by placing 60% of the servers in the "barely alive" state? c. How much power savings would be achieved by reducing the voltage by 20% and frequency by 40%? d. How much power savings would be achieved by placing 30% of the servers in the "barely alive" state and 30% off?
Answer:
a) Power saving = 26.7%
b) power saving = 48%
c) Power saving = 61%
d) Power saving = 25.3%
Explanation:
Why might it be problematic for Tiktok’s parent company to be based in China, a country whose practices around free speech differ greatly from those of the United States?
Answer:
most likely tiktok will have similar, if not the same free speech restrictions as china's
Explanation:
The idea is to simulate random (x, y) points in a 2-D plane with domain as a square of side 1 unit. Imagine a circle inside the same domain with same diameter and inscribed into the square. We then calculate the ratio of number points that lied inside the circle and total number of generated points.
Answer:
For calculating Pi (π), the first and most obvious way is to measure the circumference and diameter.
π = Circumference / diameter
For the random points in the 2-D plane, we calculate the ratio of number points inside the circle and generate random pairs and then check
x² + y² < 1
If the given condition is true, then increment the points and used them. In randomized and simulation algorithms like Monte Carlo. In this case, if the number of the iteration is more then more accurate, the result is.
Explanation:
For simulating the random points in 2-D using Monte Carlo algorithms:
Monte Carlo
In this algorithm, we calculate the ratio of the number of points that lied in the circle. We do not need any graphics or simulations to generated points. In this, we generate random pairs and check x² + y² < 1.
π ≅ 4 × N inner / N total
The other way for calculating Pi (π) is to use Gregory-Leibniz Series.
Series which converges more quickly is Nilakangtha Series, which is developed in the 15th century. It is the other way to calculate Pi (π).
Similarly, the Windows server OS can run regular workstation applications such as MS Office or Adobe Photoshop. Why is this a bad idea? Can you find a situation where it might be appropriate?
Answer:
It affects the system speed
Explanation:
The normal windows server OS can run the regular workstation applications like the MS Office or the Adobe Photoshop. Photoshop is an editor that have one of its function as being an editor that helps to edit the picture itself by changing the pixels or edit anything in the image.
On the other hand, Ms Office would help to crop or resize or adjust the brightness etc.
When it comes to cropping or reducing/increasing the brightness of the images then Office is of agreat help as it is free where as for photo shop advanced editing to the images can be done.There is the situation where both needs to be used.
Also running both the application on a egular workstation would make the system slow and also both the applications would not be required at the same time at many times.
Write a program that will open a file named thisFile.txt and write every other line into the file thatFile.txt. Assume the input file (thisFile.txt) resides in the same directory as your code file, and that your output file (thatFile.txt) will reside in the same location. Do not attempt to read from or write to other locations.
Answer:
Check the explanation
Explanation:
The programmable code to solve the question above will be written in a python programming language (which is a high-level, interpreted, general-purpose programming language.)
f = open('thisFile.txt', 'r')
w = open('thatFile.txt', 'w')
count = 0
for line in f:
if count % 2 == 0:
w.write(line)
count += 1
w.close()
f.close()
5. Modify the file by allowing the owner to remove data from the file: a. Ask the owner to enter a description to remove b. If the description exists, remove the coffee name and the quantity c. If the description is not found, display the message: That item was not found in the file. 6. Modify the file by allowing the owner to delete data from the file: a. Ask the owner to enter a description to delete b. If the description exists, delete the coffee name and the quantity c. Replace the name and quantity of the coffee removed in step b by asking the user to enter a new coffee name and quantity d. If the description i
Answer:
Check the explanation
Explanation:
Code:
# 3------reading Coffe shop Inventory file
f = open('coffeeInventory.txt','r+')#opening file in reading mode
f1 = f.readlines()
numPounds = 0
for i in range(len(f1)):
print(f1[i].split('\n')[0])
if i!=0:
numPounds+=(int((f1[i].split('\n')[0]).split(';')[1]))
#Total pounds of coffee
print("Total pounds of Coffee:",numPounds)
f.close()#closing file
#4---------
f = open('coffeeInventory.txt','a+')#opening file in appending mode
#appending records
f.write('\nGuatemala Antigua;22')
f.write('\nHouse Blend;25')
f.write('\nDecaf House Blend;16')
f.close()#closing file
#5--------
#removing the record as per owner's description
Description = input("enter Description to remove:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
#storing the previous data
f2= open('coffeeInventory_pre.txt','w+')
f2.writelines(f1)
f2.close()
#removing the data if found
if found==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
#modifying file after removing
modify_file = open('coffeeInventory.txt','w+')
for x in data_list:
modify_file.write(x)
modify_file.close()
f.close()
#6--------------
#deleting the record as per owner's description
Description = input("enter Description to delete:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
# deleting in the pre_Coffee_inventory file
#finding record
f2 = open('coffeeInventory_pre.txt','r')
f3 =f2.readlines()
found = -1
for i in range(len(f3)):
line = f3[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f3.remove(f3[found])#deleting record
data_list1 = f3
f2.close()
#modifying pre file
f2 = open('coffeeInventory_pre.txt','w+')
f2.writelines(data_list1)
f2.close()
#opening file in writing mode
modify_file = open('coffeeInventory.txt','w+')
modify_file.writelines(data_list)
modify_file.close()
Answer:
See explaination
Explanation:
#5
#removing the record as per owner's description
Description = input("enter Description to remove:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
#storing the previous data
f2= open('coffeeInventory_pre.txt','w+')
f2.writelines(f1)
f2.close()
#removing the data if found
if found==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
#modifying file after removing
modify_file = open('coffeeInventory.txt','w+')
for x in data_list:
modify_file.write(x)
modify_file.close()
f.close()
#6
#deleting the record as per owner's description
Description = input("enter Description to delete:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
# deleting in the pre_Coffee_inventory file
#finding record
f2 = open('coffeeInventory_pre.txt','r')
f3 =f2.readlines()
found = -1
for i in range(len(f3)):
line = f3[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f3.remove(f3[found])#deleting record
data_list1 = f3
f2.close()
#modifying pre file
f2 = open('coffeeInventory_pre.txt','w+')
f2.writelines(data_list1)
f2.close()
#opening file in writing mode
modify_file = open('coffeeInventory.txt','w+')
modify_file.writelines(data_list)
modify_file.close()
The material of this section assumes that search keys are unique. However, only small modifications are needed to allow the techniques to work for search keys with duplicates. Describe the necessary changes to insertion, deletion, and lookup algorithms, and suggest the major problems that arise when there are duplicates in each of the following kinds of hash tables: (a) simple (b) linear (c) extensible
Answer:
The description of the given points can be described as follows:
Explanation:
A hash table is a specific array, which is used to contain items of key value. It uses a hash to calculate an index of an array, which inserts an item, and the given point can be described as follows:
In option a, It includes several duplicates, which is a large number of unavailable tanks, although there is a large overflow list for the used tanks. In option b, multiple copies are painful to scan because we have to read and understand rather than a single line. In option c, It can break frequently, if there are several exact copies.Select the correct statement(s) regarding CATV. a. CATV began as a broadcast capability over shared coaxial cable medium; as such, security issues are a concern b. CATV DOCSIS systems rely on QPSK and QAM modulation c. today’s CATV system uses a combination of fiber optic and coaxial cable d. all statements are correct
Answer:
d. all statements are correct
Explanation:
CATV is a very common method used in transmitting television signal to homes, this is achieved through the use of radio frequency signals which is transmitted via coaxial cable. In recent times, CATV as been transmitted through optical fiber, this optical fiber makes use of light to transmit signal.
CATV can be traced back to 1924 in which cable broadcasting was done through the use cable in Europe.
Implement a superclass Appointment and subclasses Onetime, Daily, and Monthly. An appointment has the date of an appointment, the month of an appointment, the year of an appointment, and the description of an appointment (e.g., "see the dentist"). In other words, four major instant variables in these classes. Then, write a method occcursOn(year, month, day) that checks whether the appointment occurs on that date. Ensure to have a test program to test your classes. In your test program, you should create a list of appointments using the concept of polymorphism as your appointment pool. In other words, creating multiple objects using your subclasses of Onetime, Daily, and Monthly. Once you have those appointments available, allow the user to input day, month, and year to check whether the day of the month matches and print out the information of that matched appointment. For example, for a monthly appointment, you must check whether the day of the month matches. Have the user enter a date and print out all appointments that occur on that date.
Final answer:
The question involves creating an appointment system in object-oriented programming, where a superclass 'Appointment' and its subclasses 'Onetime', 'Daily', and 'Monthly' are defined, each with a method to check if an appointment occurs on a specified date.
Explanation:
Implementing Appointment Superclass and Subclasses
To model an appointment system, we implement a superclass named Appointment with subclasses called Onetime, Daily, and Monthly. Each class contains instance variables for the date, month, year, and a description of the appointment. The key method, occursOn(year, month, day), determines if an appointment is scheduled for a specific date by checking the year, month, and day against the appointment's details.
Appointment Superclass
The superclass includes the four main instance variables and provides the foundational behavior for its subclasses.
Onetime, Daily, and Monthly Subclasses
The subclasses override the occursOn() method to provide specific checking logic for one-time, daily, and monthly appointments.
A test program can be created to instantiate different types of appointments and check if they occur on a user-input date. By utilizing polymorphism, we can maintain a list of appointments of various subclasses. The test program will iterate through this list, applying the occursOn() method to find and print out matching appointments for the given date.
a user's given name followed by the user's age from standard input. Then use an ofstream object named outdata (which you must declare) to write this information separated by a space into a file called outdata. Assume that this is the extent of the output that this program will do. Declare any variables that you need.
Answer:
See explaination for program code
Explanation:
#include <iostream>
#include <string>
#include <fstream>
using namespace std;
int main() {
string name;
int age;
cout << "Enter name of user: ";
getline(cin, name);
cout << "Enter age of user: ";
cin >> age;
ofstream outdata("outdata");
outdata << name << " " << age << endl;
outdata.close();
return 0;
}
A network administrator identifies sensitive files being transferred from a workstation in the LAN to an unauthorized outside IP address in a foreign country. An investigation determines that the firewall has not been altered, and antivirus is up-to-date on the workstation. Which of the following is the MOST likely reason for the incident?
A. MAC Spoofing
B. Session Hijacking
C. Impersonation
D. Zero-day
Answer:
D. Zero-day
Explanation:
It is clearly stated that the antivirus is still up-to-date on the workstation, and there has been no alterations to the firewall. Hence, this means the software is functional and up-to-date with all known viruses. This shows that this attack is unknown to the manufacturer of the antivirus software and hence no virus definition or patch fixing this problem has been developed yet. In this light, it is a zero-day.
A zero-day is a type of vulnerability in a software that is not yet known to the vendor or manufacturer, hence this security hole makes the software vulnerable to attackers or hacker before it is been fixed.
Yolanda lost her left foot during her military service, but she has been issued a prosthetic that enables her to walk normally, and if you didn't see this device, you wouldn't know that she was using it. This is an example of the way technology is a(n) __________ means of extending human abilities.
Answer:
The correct answer to the following question will be "Artificial and superficial".
Explanation:
Prosthetic is just an automated means of replacing or restoring the missing component, where Yolanda has missing her foot and sometimes a prosthetic leg has indeed been released that helps her to walk or stand normally and superficially, but once it is thoroughly investigated you wouldn't have seen Yolanda using this tool.And as such the specified situation seems to be an example of how innovation would be artificial as well as a superficial medium of human capacity expansion.Some have argued that Unix/Linux systems reuse a small number of security features in many contexts across the system, while Windows systems provide a much larger number of more specifically targeted security features used in the appropriate contexts. This may be seen as a trade-off between simplicity and lack of flexibility in the Unix/Linux approach, against a better targeted but more complex and harder to correctly configure approach in Windows. Discuss this trade-off as it impacts on the security of these respective systems, and the load placed on administrators in managing their security
Answer:
see my discussion as explained bellow
Explanation: