Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm
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
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The most upstream process with issues would be a good location to start exploring the cause of the variance.
A manufacturing technique would be a specific procedure for generating a commodity.
Throughout manufacturing, a six sigma process has been utilized just to generate a product throughout which 99.99966 percent among all possibilities to produce certain aspects of a part seem to be likely toward being defect-free.
Thus the response above is correct.
Find out more information about chain processes here:
Answer: The furthest upstream process that has problems.
A process in manufacturing is a particular method used for producing a product.
A six sigma process is used in processing to produce a product that is 99.99966% of all opportunities to produce some feature of a part are statistically expected to be free of defects.
According to the rules of the six sigma process, when there's a defect, the best thing to do is investigate the furthest upstream process that has problems.
Answer:
volume = 53.747 m3 = 14198.138 gal
weight = 526652 N = 118396.08 lbf
Explanation:
We know that volume of water
where A' = 61% of A
=1898.015 ft^3
=526652 N
Answer:
The solution is given in the attachments.
Answer:
The answer is below
Explanation:
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.
Given that:
number of poles (p) = 4, frequency (f) = 60 Hz
1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:
2) The slip (s) = 0.05
The speed of the motor (n) is the speed of the rotor, it is given as:
3) s = 0.04
The rotor frequency is the product of the supply frequency and slip it is given as:
4) At standstill, the motor speed is zero hence the slip = 1:
The rotor frequency is the product of the supply frequency and slip it is given as: