The way a programmer describe a pre-emptive dialog by purely graphical means is; by producing a window that covers the entire screen to make it the currently selected window.
In a graphics - based interaction, it is supposed that the user can only interact with parts of the system that are visible. However, In a windowing system, the user can only direct input to a single window that was currently selected and the way to change that selected window is to indicate with some gesture within that window.
Finally, to create a pre-emptive dialog, the system would do so through the production of a window that covers the entire screen to make it the currently selected window. Thereafter, all user input would be directed to that window and the user would have no means of selecting any other window. Then the covering window will now pre-empt any other user action with the exception of that which it is defined to support.
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Answer:
In an illustrations based communication, it is expected that the client can just associate with parts of the framework that are obvious. In a windowing framework, for instance, the client can just direct contribution to a solitary, at present chosen window, and the main methods for changing the chose window would be by demonstrating with some signal inside that window. To make a preemptive exchange, the framework can create a window that covers the whole screen and make it the right now chosen window. All client information would then be coordinated to that window and the client would have no methods for choosing another window. The 'covering' window in this way preempts some other client activity with the exception of that which it is characterized to help
Answer:
b). False
Explanation:
Lumped body analysis :
Lumped body analysis states that some bodies during heat transfer process remains uniform at all times. The temperature of these bodies is a function of temperature only. Therefor the heat transfer analysis based on such idea is called lumped body analysis.
Biot number is a dimensionless number which governs the heat transfer rate for a lumped body. Biot number is defined as the ratio of the convection transfer at the surface of the body to the conduction inside the body. the temperature difference will be uniform only when the Biot number is nearly equal to zero.
The lumped body analysis assumes that there exists a uniform temperature distribution within the body. This means that the conduction heat resistance should be zero. Thus the lumped body analysis is exact when biot number is zero.
In general it is assume that for a lumped body analysis, Biot number 0.1
Therefore, the smaller the Biot number, the more exact is the lumped system analysis.
Answer:
The output will be (3, 4) becomes (8, 10)
Explanation:
#include <stdio.h>
//If you send a pointer to a int, you are allowing the contents of that int to change.
void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){
*xNew = (xVal+1)*2;
*yNew = (yVal+1)*2;
}
int main(void) {
int xValNew = 0;
int yValNew = 0;
CoordTransform(3, 4, &xValNew, &yValNew);
printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);
return 0;
}
Answer:
Minimum standard diameter for the PVC pipe = 14.26 inches
Minimum standard diameter for the steel pipe = 14.70 inches
Explanation:
Head loss = 6/1000...................................................(1)
Head loss = hf/l
............................(2)
Q = 2500 gal/min
a) Minimum standard diameter for PVC
C for PVC = 130
Equating (1) and (2) and putting C = 130
b) Minimum standard diameter for steel
C for steel = 120
Equating (1) and (2) and putting C = 120
The annual quantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.
To determine the break-even point between the manual arc welding cell and the robotic cell, we need to calculate the total costs for each method and then equate them.
For the manual arc welding cell:
Labor cost per hour = (welder's hourly rate x arc-on time) + (fitter's hourly rate x fitter's participation in the cycle) = ($30 x 0.25) + ($25 x 0.3) = $11.25
Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $11.25 x 15.4 / 60 = $2.89
Overhead cost per welded assembly = (labor cost per hour x (1 - arc-on time - fitter's participation in the cycle)) x cycle time per assembly / 60 = ($30 x 0.45) x 15.4 / 60 = $4.68
Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly = $2.89 + $4.68 = $7.57
Total cost per hour = total cost per welded assembly x production rate = $7.57 x 8 = $60.56
Total cost per year = total cost per hour x hours of operation per year = $60.56 x 2,000 = $121,120
For the robotic arc welding cell:
Labor cost per hour = fitter's hourly rate x fitter's participation in the cycle = $25 x 0.62 = $15.50
Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $15.50 x 15.4 / 60 = $3.97
Overhead cost per welded assembly = power and utility cost per hour + annual maintenance cost / production rate = $3.80 + $3,500 / (8 x 2,000) = $3.80 + $0.22 = $4.02
Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly + (installed cost / (production rate x service life)) = $3.97 + $4.02 + ($158,000 / (8 x 3)) = $3.97 + $4.02 + $6,208.33 = $14.19
Total cost per hour = total cost per welded assembly x production rate = $14.19 x 8 = $113.52
Total cost per year = total cost per hour x hours of operation per year = $113.52 x 2,000 = $227,040
To find the break-even point, we set the total cost of the manual arc welding cell equal to the total cost of the robotic arc welding cell and solve for the annualquantity of welded assemblies:
$121,120 + x($7.57) = $227,040 + x($14.19)
$7.57x - $14.19x = $227,040 - $121,120
$-6.62x = $105,920
x = $105,920 / $6.62
x = 15,982.7
Therefore, the annualquantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.
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The file KnapsackData1.txt and KnapsackData2.txt are sample input files
for the following Knapsack Problem that you will solve.
KnapsackData1.txt contains a list of four prospective projects for the upcoming year for a particular
company:
Project0 6 30
Project1 3 14
Project2 4 16
Project3 2 9
Each line in the file provides three pieces of information:
1) String: The name of the project;
2) Integer: The amount of employee labor that will be demanded by the project, measured in work weeks;
3) Integer: The net profit that the company can expect from engaging in the project, measured in thousands
of dollars.
Your task is to write a program that:
1) Prompts the user for the number of work weeks available (integer);
2) Prompts the user for the name of the input file (string);
3) Prompts the user for the name of the output file (string);
4) Reads the available projects from the input file;
5) Dolves the corresponding knapsack problem, without repetition of items; and
6) Writes to the output file a summary of the results, including the expected profit and a list of the best
projects for the company to undertake.
Here is a sample session with the program:
Enter the number of available employee work weeks: 10
Enter the name of input file: KnapsackData1.txt
Enter the name of output file: Output1.txt
Number of projects = 4
Done
For the above example, here is the output that should be written to Output1.txt:
Number of projects available: 4
Available employee work weeks: 10
Number of projects chosen: 2
Number of projectsTotal profit: 46
Project0 6 30
Project2 4 16
The file KnapsackData2.txt, contains one thousand prospective projects. Your program should also be able to handle this larger problem as well. The corresponding output file,
WardOutput2.txt, is below.
With a thousand prospective projects to consider, it will be impossible for your program to finish in a
reasonable amount of time if it uses a "brute-force search" that explicitly considers every possible
combination of projects. You are required to use a dynamic programming approach to this problem.
WardOutput2.txt:
Number of projects available: 1000
Available employee work weeks: 100
Number of projects chosen: 66
Total profit: 16096
Project15 2 236
Project73 3 397
Project90 2 302
Project114 1 139
Project117 1 158
Project153 3 354
Project161 2 344
Project181 1 140
Project211 1 191
Project213 2 268
Project214 2 386
Project254 1 170
Project257 4 427
Project274 1 148
Project275 1 212
Project281 2 414
Project290 1 215
Project306 2 455
Project334 3 339
Project346 2 215
Project356 3 337
Project363 1 159
Project377 1 105
Project389 1 142
Project397 1 321
Project399 1 351
Project407 3 340
Project414 1 266
Project431 1 114
Project435 3 382
Project446 1 139
Project452 1 127
Project456 1 229
Project461 1 319
Project478 1 158
Project482 2 273
Project492 1 142
Project525 1 144
Project531 1 382
Project574 1 170
Project594 1 125
Project636 2 345
Project644 1 169
Project668 1 191
Project676 1 117
Project684 1 143
Project689 1 108
Project690 1 216
Project713 1 367
Project724 1 127
Project729 2 239
Project738 1 252
Project779 1 115
Project791 1 110
Project818 2 434
Project820 1 222
Project830 1 179
Project888 3 381
Project934 3 461
Project939 3 358
Project951 1 165
Project959 2 351
Project962 1 316
Project967 1 191
Project984 1 117
Project997 1 187
Answer:
Explanation:
Code:
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
import java.util.Scanner;
public class Knapsack {
public static void knapsack(int wk[], int pr[], int W, String ofile) throws IOException
{
int i, w;
int[][] Ksack = new int[wk.length + 1][W + 1];
for (i = 0; i <= wk.length; i++) {
for (w = 0; w <= W; w++) {
if (i == 0 || w == 0)
Ksack[i][w] = 0;
else if (wk[i - 1] <= w)
Ksack[i][w] = Math.max(pr[i - 1] + Ksack[i - 1][w - wk[i - 1]], Ksack[i - 1][w]);
else
Ksack[i][w] = Ksack[i - 1][w];
}
}
int maxProfit = Ksack[wk.length][W];
int tempProfit = maxProfit;
int count = 0;
w = W;
int[] projectIncluded = new int[1000];
for (i = wk.length; i > 0 && tempProfit > 0; i--) {
if (tempProfit == Ksack[i - 1][w])
continue;
else {
projectIncluded[count++] = i-1;
tempProfit = tempProfit - pr[i - 1];
w = w - wk[i - 1];
}
FileWriter f =new FileWriter("C:\\Users\\gshubhita\\Desktop\\"+ ofile);
f.write("Number of projects available: "+ wk.length+ "\r\n");
f.write("Available employee work weeks: "+ W + "\r\n");
f.write("Number of projects chosen: "+ count + "\r\n");
f.write("Total profit: "+ maxProfit + "\r\n");
for (int j = 0; j < count; j++)
f.write("\nProject"+ projectIncluded[j] +" " +wk[projectIncluded[j]]+ " "+ pr[projectIncluded[j]] + "\r\n");
f.close();
}
}
public static void main(String[] args) throws Exception
{
Scanner sc = new Scanner(System.in);
System.out.print("Enter the number of available employee work weeks: ");
int avbWeeks = sc.nextInt();
System.out.print("Enter the name of input file: ");
String inputFile = sc.next();
System.out.print("Enter the name of output file: ");
String outputFile = sc.next();
System.out.print("Number of projects = ");
int projects = sc.nextInt();
int[] workWeeks = new int[projects];
int[] profit = new int[projects];
File file = new File("C:\\Users\\gshubhita\\Desktop\\" + inputFile);
Scanner fl = new Scanner(file);
int count = 0;
while (fl.hasNextLine()){
String line = fl.nextLine();
String[] x = line.split(" ");
workWeeks[count] = Integer.parseInt(x[1]);
profit[count] = Integer.parseInt(x[2]);
count++;
}
knapsack(workWeeks, profit, avbWeeks, outputFile);
}
}
Console Output:
Enter the number of available employee work weeks: 10
Enter the name of input file: input.txt
Enter the name of output file: output.txt
Number of projects = 4
Output.txt:
Number of projects available: 4
Available employee work weeks: 10
Number of projects chosen: 2
Total profit: 46
Project2 4 16
Project0 6 30
Answer:
The solution is given in the attachments.