Computer Aided Analysis Program Print

8/10/2019 Computer Aided Analysis Program Print

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B ALIUAG UNIVERSITYBaliwag, Bulacan

COLLEGE OF ENVIRONMENTAL DESIGN AND

ENGINEERING

CE 520

Computer Aided Analysis

COMPILATION OF PROGRAMSMARCH 17, 2014

Submitted by:Viola, Randy V.BSCE-V

Submitted to:Engr. Leonardo V. Surio, M. Eng.


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I. Preparation of Computer Program for :

A. Matrix AdditionB. Matrix MultiplicationC. Inverse of Matrix

D. Gauss Elimination to Solve System of Equations.

(Note: These programs of matrix are made using Microsoft Visual (Basic, C++)2010 express. It runs only if there are installed Microsoft Visual Basic 2010express in your computer).

A. Matrix Addition

In this program we compute addition of two matrices A and B which hassame order. First we enter order of matrix then enter numbers for matrix A and B.

After that gives resultant matrix.Suppose matrices have order 33.

Matrix A:

4 6 51 3 12 3 7

Matrix B:

9 2 38 0 64 1 2

Resultant Matrix is addition of matrices A and B:

13 8 89 3 7

6 4 9


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Program code for Matrix Addition:

#include#include using namespace std;

int main(){int a[10][10];int b[10][10];int x,y,i,j;

printf("\t\t\t MATRIX ADDITION\n");printf(" \n");

coutx>>y;

cout


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}cout


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OUTPUT:


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B. Matrix Multiplication

It is a Matrix multiplication in Microsoft Visual Basic using 2010 express. A programto multiply matrices (two dimensional array), this program multiplies two matrices which

will be entered by the user. Firstly user will enter the order of a matrix. If the enteredorders of two matrix is such that they can't be multiplied then an error message isdisplayed on the screen.

Suppose matrices have order 33.

Matrix A : :

1 2 00 1 1

2 0 1

Matrix B : :

1 1 22 1 11 2 1

Product of Matrix A & Matrix B is : :

5 3 43 3 23 4 5


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Program code for Matrix Multiplication:

Module Module1

Sub Main()

Dim matrix1 As Integer(,), matrix2 As Integer(,)

Dim row1 As Integer, col1 As Integer, row2 As Integer, col2 As Integer, i AsInteger, j As Integer, _

k As Integer, temp As Integer = 0

Console.WriteLine(" MATRIX MULTIPLICATION ")

Console.WriteLine(" ")

Console.WriteLine("Enter the number of rows in matrix A :: ")

row1 = Integer.Parse(Console.ReadLine())

Console.WriteLine("Enter the number of columns in matrix A :: ")

col1 = Integer.Parse(Console.ReadLine())

matrix1 = New Integer(row1 - 1, col1 - 1) {}

Console.WriteLine("Enter the number of rows in matrix B :: ")

row2 = Integer.Parse(Console.ReadLine())

Console.WriteLine("Enter the number of columns in matrix B :: ")

col2 = Integer.Parse(Console.ReadLine())

matrix2 = New Integer(row2 - 1, col2 - 1) {}

If col1 row2 Then

Console.WriteLine("Multiplication is not applicable!!!")

Console.WriteLine("Note : Number of columns of matrix A must be equal toNumber of rows of matrix B.")

Else


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Console.WriteLine("Enter Elements for matrix A")

For i = 0 To row1 - 1

For j = 0 To col1 - 1

matrix1(i, j) = Integer.Parse(Console.ReadLine())

Next

Next

Console.WriteLine("Enter Elements for matrix B")



matrix2(i, j) = Integer.Parse(Console.ReadLine())

Next

Next

Console.Clear()

Console.WriteLine("You have entered :: ")

Console.WriteLine("Matrix A ::")



Console.Write(matrix1(i, j))

Console.Write(" ")

Next

Console.WriteLine()

Next

Console.WriteLine("Matrix B ::")


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Console.Write(matrix2(i, j))

Console.Write(" ")

Next

Console.WriteLine()

Next

Console.WriteLine("Product of Matrix A & Matrix B is ::")



For k = 0 To row2 - 1

temp = temp + (matrix1(i, k) * matrix2(k, j))

Next

Console.Write(temp & " ")

temp = 0

Next

Console.WriteLine()

Next

End If

Console.Read()

End Sub

End Module


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INPUT:

OUTPUT:


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C. Inverse of Matrix

Given a matrix A, the inverse A -1 can be multiplied on either side of A to get theidentity. That is, AA 1 = A 1 A = I. Keeping in mind the rules for matrix multiplication, thissays that A must have the same number of rows and columns; that is, A must be

square. (Otherwise, the multiplication wouldn't work.) If the matrix isn't square, it cannothave a (properly two-sided) inverse. However, while all invertible matrices are square,not all square matrices are invertible.

Suppose matrices have order 33.

Matrix A :

2 1 30 0 2

4 1 1

The Inverse of Matrix (A -1) :

-0.5 0.5 0.52 -2.5 -10 0.5 0

Program Code for Inverse of Matrix:

#include#include using namespace std;int main(){

int i,j,k,n;float a[10][10]={0},d;void clrscr();

cout


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for(j=1;j1;i--){if(a[i-1][1]


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for(j=n+1;j


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D. Gauss Elimination to Solve System of Equations.

Gaussian elimination (also known as row reduction) is an algorithm forsolving systems of linear equations. It is usually understood as a sequence ofoperations performed on the associated matrix of coefficients. This method can also be

used to find the rank of a matrix, to calculate the determinant of a matrix, and tocalculate the inverse of an invertible square matrix. The method is named after CarlFriedrich Gauss.

Suppose matrices have 3 numbers of equations.

3x1 + 6x 2 + 1x 3 = 162x1 + 4x 2 + 3x 3 = 131x1 + 3x 2 + 2x 3 = 9

Matrix :

3 6 1 : 162 4 3 : 131 3 2 : 9

The result is :

x1 = 1.00x2 = 2.00x3 = 1.00

Program Code for Gauss Elimination to Solve System of Equations:

# include# include# include using namespace std;void main(){int i,j,k,n;float a[10][10],x[10];float s,p;printf("\t\t\t GAUSS ELIMINATION \n");printf(" \n");printf("Enter the number of equations : ");scanf("%d",&n) ;
http://en.wikipedia.org/wiki/Algorithmhttp://en.wikipedia.org/wiki/System_of_linear_equationshttp://en.wikipedia.org/wiki/Matrix_(mathematics)http://en.wikipedia.org/wiki/Rank_(linear_algebra)http://en.wikipedia.org/wiki/Determinanthttp://en.wikipedia.org/wiki/Invertible_matrixhttp://en.wikipedia.org/wiki/Carl_Friedrich_Gausshttp://en.wikipedia.org/wiki/Carl_Friedrich_Gausshttp://en.wikipedia.org/wiki/Carl_Friedrich_Gausshttp://en.wikipedia.org/wiki/Carl_Friedrich_Gausshttp://en.wikipedia.org/wiki/Invertible_matrixhttp://en.wikipedia.org/wiki/Determinanthttp://en.wikipedia.org/wiki/Rank_(linear_algebra)http://en.wikipedia.org/wiki/Matrix_(mathematics)http://en.wikipedia.org/wiki/System_of_linear_equationshttp://en.wikipedia.org/wiki/Algorithm


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printf("\nEnter the co-efficients of the equations :\n\n");for(i=0;i


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for(i=0;i


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II. Areas for Plane Figures:

A. SquareB. RectangleC. Trapezoid

D. CircleE. Triangle

Areas of Plane Figures:

(Note: This program name AREA CALCULATOR runs only if there are installedMicrosoft Visual Basic 2010 express in your computer.)

It is a calculator for finding areas of a plane figures only. There are square,Rectangle, Trapezoid, Circle and Triangle in the program. Just click the icon or buttonon the menu that you want to execute in finding the area of the given plane figures.

A. Square

The area of a square is given by the formula area = wi dth heig ht, Butsince the width and height are by definition the same, the formula is usuallywritten as area = s 2 where s is the length of one side.

For example:Length = 6m its width is also 6m.

Area = 36 m 2


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OUTPUT:

Program Code for Square:

Public Class square

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e AsSystem.EventArgs) Handles Button1.Click

TextBox3.Text = TextBox1.Text * TextBox2.Text

End Sub


TextBox1.Text = ""TextBox2.Text = ""

TextBox3.Text = ""End Sub

Private Sub Button3_Click(ByVal sender As System.Object, ByVal e AsSystem.EventArgs)

Form1.Show()End Sub


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Private Sub TextBox1_TextChanged(ByVal sender As System.Object, ByVal e AsSystem.EventArgs) Handles TextBox1.TextChanged

End SubEnd Class

B. Rectangle

To find the area of a rectangle, multiply the length by the width. Theformula is: , where is the area, is the length, is the width,and means multiply.

For example:Length = 6mWidth = 4m.

Answer is: Area = 24 m 2

OUTPUT:
http://popupwindow%28%27rectangle%27%29/http://popupwindow%28%27rectangle%27%29/


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Program Code for Rectangle:

Public Class rectangle

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As

System.EventArgs) Handles Button1.ClickTextBox3.Text = TextBox1.Text * TextBox2.Text

End Sub


TextBox1.Text = ""TextBox2.Text = ""TextBox3.Text = ""

End Sub

Private Sub Button3_Click(ByVal sender As System.Object, ByVal e AsSystem.EventArgs)

Form1.Show()End Sub


End SubEnd Class

C. Trapezoid

A trapezoid is a 4-sided figure with one pair of parallel sides. To find thearea of a trapezoid, take the sum of its bases, multiply the sum by the height ofthe trapezoid, and then divide the result by 2. The formula for the area of atrapezoid is:

Area = ( ) For example:

Top Base = 6mBottom Base = 8mHeight = 4m

Answer is:

Area = 28 m 2
http://popupwindow%28%27parallel%27%29/http://popupwindow%28%27parallel%27%29/http://popupwindow%28%27parallel%27%29/


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OUTPUT:

Program Code for Trapezoid:

Public Class trapezoid


End Sub


TextBox4.Text = ((((Val(TextBox1.Text) + Val(TextBox2.Text)) * TextBox3.Text)) *0.5)

End Sub


TextBox1.Text = ""TextBox2.Text = ""TextBox3.Text = ""TextBox4.Text = ""

End Sub

Private Sub Label3_Click(ByVal sender As System.Object, ByVal e AsSystem.EventArgs) Handles Label3.Click


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End Sub


End SubPrivate Sub TextBox4_TextChanged(ByVal sender As System.Object, ByVal e As

System.EventArgs) Handles TextBox4.TextChanged

End SubEnd Class

D. Circle

The area of a circle can be found by multiplying pi ( = 3.14) by the

square of the radius If a circle has a radius of 4, its area is 3.14*4*4=50.24 If youknow the diameter, the radius is 1/2 as large.

For example:

Radius: 6m

Answer is:

Area = 113.0973 m 2

OUTPUT:


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Program Code for Circle:

Public Class circlePrivate Sub TextBox2_TextChanged(ByVal sender As System.Object, ByVal e As

System.EventArgs)

End Sub


TextBox3.Text = 3.141592654 * TextBox1.Text * TextBox1.TextEnd Sub


TextBox1.Text = ""TextBox3.Text = ""

End Sub


End Sub


End SubEnd Class

E. Triangle

It is easy to find the area of a right-angled triangle, or any triangle wherewe are given the base and the height.

It is simply half of b times Area = bxh

For example:

Given:

Height (h) = 6m

Base (b) = 3m

Answer is: Area = 9m 2


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OUTPUT:

Program Code for Triangle:

Public Class triangle

Private Sub TextBox2_TextChanged(ByVal sender As System.Object, ByVal e AsSystem.EventArgs)

End Sub


TextBox3.Text = (TextBox1.Text * TextBox2.Text) / 2End Sub

Private Sub Button2_Click(ByVal sender As System.Object, ByVal e As

System.EventArgs) Handles Button2.ClickTextBox1.Text = ""TextBox2.Text = ""TextBox3.Text = ""

End Sub



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End Sub


End SubEnd Class

III. Solution of Quadratic Equation:

(Note: This program name Quadratic Equation Solver runs only if there are installedMicrosoft Visual Basic 2010 express in your computer.)

A quadratic equation is an equation of the second degree, meaningit contains at least one term that is squared. The standard form is ax + bx + c = 0with a, b, and c being constants, or numerical coefficients, and x is an unknownvariable. One absolute rule is that the first constant a cannot be a zero.
http://www.yourdictionary.com/quadratichttp://www.yourdictionary.com/quadratic


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For Example number 1:Find the roots of the given equation:x + -3x + -4 = 0

Answer is:x1 = 4 and x 2 =-1

*there two possible answer for the given equation.For Example 2:

Find the roots of the given equation:5x + -3x + 3 = 0

Answer is:x1 = Imaginary and x 2 = Imaginary

*the answer is both imaginary.

OUTPUT:


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Program Code for the Solution of Quadratic Equation:

Public Class Form1

Dim a, b, c, d, x, y, z, real, img, ePrivate Sub Label1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles

Label1.ClickEndEnd Sub

Private Sub Button2_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) HandlesButton2.Click

TextBox1.Text = ""TextBox2.Text = ""TextBox3.Text = ""TextBox4.Text = ""TextBox5.Text = ""

End Sub

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles

Button1.ClickIf TextBox1.Text = "" ThenExit Sub

Elsea = Textbox1.textb = Textbox2.textc = Textbox3.textd = (b ^ 2 - 4 * a * c)If d > 0 Then

y = ((-b + System.Math.Sqrt(d)) / (2 * a))z = ((-b - System.Math.Sqrt(d)) / (2 * a))TextBox4.Text = yTextBox5.Text = z

ElseIf d = 0 Then

x = -b / (2 * a)TextBox4.Text = xElseif d < 0 Then

TextBox4.Text = "IMAGINARY"TextBox5.Text = "IMAGINARY"

End IfEnd If

End SubPrivate Function MsgBox() As String

Throw New NotImplementedExceptionEnd Function

Private Sub PictureBox1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) HandlesPictureBox1.Click

End Sub

Private Sub TextBox4_TextChanged(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles

TextBox4.TextChanged

End Sub

End Class


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III. CHOLESKY METHOD

Suppose matrices have 3 numbers of equations.

5x1 +2 + 1x 3 = 3

4x1 + 1x 2 + -1x 3 = -3-2x 1 + 3x 2 + -3x 3 = 6

Matrix A:

5 2 14 1 -1-2 3 -3

Matrix b:

3 -3 6

The solution using Cholesky Method is:

X1 = -1.00X2 = 3.00X3 = 2.00


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OUTPUT:

Program Code for Cholesky Method:

#include#includeusing namespace std;int main(){

int n,i,k,j,p;float a[10][10],l[10][10]={0},u[10][10]={0},sum,b[10],z[10]={0},x[10]={0};void clrscr();

cout


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for(i=k;i

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