How to Troubleshoot Apps for the Modern Connected Worker
Term paper
1. PAPER –
CAP643
Submitted To: -
Submitted By: Abhishek Kumar
Mrs. Kavisha
Mam
~1~
Roll No: - 1207B333
Reg No: - 11210885
Course: - MCA 3rd
Sem. Section: D1207
2. ACKNOWLEDGEMENT
The experience that we have gathered during this Design
Problem has been unique .For this we are pleased to express
our deepest sense of gratitude and regards to our respected
teacher Mrs. Kavisha Mam for their guidance, inspiration
and constructive suggestions that helps us in the preparation
of the design problem.
I am also thankful to my classmate
constant encouragement and support.
2
4. INTRODUCTION
Gauss Jacobi method is the first iterative method used to solve linear
system of equations. This project explains you how to solve the linear
equation using Gauss Jacobi iterative method
Abstract:This paper is concerned with the application of preconditioning
techniques to the well-known Jacobi iterative method for solving the
finite difference equations derived from the discretization of self-ad
joint elliptic partial differential equations. The convergence
properties of this one parameter preconditioned method are
analyzed and the value of the optimum preconditioning parameter
and the performance of the method determined for a variety of
standard problems.
Jacobi method involves rewriting equation 1 as follow:
X=D-1(L+U)x+D-1b
If we express it as an iterative method, we see it takes the form:
X(k+1)=Gx(k)+f
As a motivational example, we assume start with a linear system like
this one:
1
5
9
2
6
1
3
7
2
x1
x2
x3
*
4
=
4
8
3
5. Let us solve the ith equation for xi :
X1 = (4- 2x2-3x3)/1
X2 = (8- 5x1-7x3)/6
X3 = (3- 9x1-x2)/2
We can express this as an iterative method and rewrite it in a matrix
format.
x1
x2
x3
(k+1)
=
(k)
1/1
0
0 1/6
0
0
0
0
1/2
1/1
0
0 1/6
0
0
0
0
1/2
0
-5
-9
*
*
-2
0
-1
4
8
3
5
-3
-7
0
x1
* x2
x3
+
6. x1
x2
x3
(k+1)
=
0
-5/6
-9/2
-1/2 -1/3
0 -7/6
-1/2
0
*
x1
x2 +
x3
1/4
8/6
3/2
As you see, this takes the form x(k+1)=Gx(k)+f.
Let's implement Jacobi's method for this problem.
There are some steps which are followed by Gauss Jacobi
method to perform certain operation:
Step 1
• Find the value of x1 from the first equation by substituting the initial
values of other unknowns.
• Find the value of x2 from the first equation by substituting the initial
values of other unknowns.
• Find the value of x3 from the first equation by substituting the initial
values of other unknowns.
And so on till the value of xn is computed from the nth equation using
the initial values of x1, x2,…. xn-1.
6
7. Step 2
•
Find out the value of x1 from the first equation by substituting
the values of other unknowns got in the 1st iteration.
•
Find out the value of x2 from the first equation by substituting
the values of other unknowns got in the 1st iteration.
•
Find out the value of x3 from the first equation by substituting
the values of other unknowns got in the 1st iteration.
Step 3
•
Find out the value of x1 from the first equation by substituting
the values of other unknowns got in the 2nd iteration.
•
Find out the value of x2 from the first equation by substituting
the values of other unknowns got in the 2nd iteration.
•
Find out the value of x3 from the first equation by substituting
the values of other unknowns got in the 2nd iteration.
7
8. Example
x1+2x2+3x3 = 4
5x1+ 6x2+7x3 = 8
9x1+x2+2x3 = 3
Sol: As you see the system is diagonal system, therefore the
convergence is assured. Since we want the solution correct up to 4
significant digits, therefore the iterative process will terminate as soon
as we find the successive iteration do not produce any change at first
four significant positions.
We rewrite the given system of equations as
X1 = (4- 2x2-3x3)/1
X2 = (8- 5x1-7x3)/6
X3 = (3- 9x1-x2)/2
8
9. We start with initial approximation as
X1 = X2 = X3 = 0
Iteration 1: Substituting the initial values in the above equation, we
obtain
X1 =4.0000 X2 = 1.3333
X3 =1.5000
Iteration 2: Substituting the initial values in the above equation, we
obtain
X1 = -3.1667
X2 = -3.7500
X3 = -17.1667
Now compare Gauss Jacobi method from
other linear equation like Gauss Seidel
Method
Gauss Seidel Method
In Jacobi’s method, even though the new values of unknowns are
computed in each iteration, but the values of unknowns in previous
iterations are used in the subsequent iterations.
That is, although a new value of x1 is computed from the first
equation in a current iteration, but it is not used to compute the new
values of other unknowns in the current iteration.
9
10. Note that the new values of the unknown are better than the old
values, and should be used in preference to the poorer values
Step 1
• Find out the value of x1 from the first equation by substituting the
initial values of other unknowns.
• Find out the value of x2 from the second equation by substituting
current value of x1 and the initial values of other unknowns.
• Find out the value of x3 from the first equation by substituting the
current value of x1 and x2 initial values.
And so on till the value of xn is computed from the nth equation using
the initial values of x1, x2,…. xn-1.
10
11. Now we will solve same equation by Gauss Seidel method and
then we will find the basic difference between both the methods.
Our equation is:
x1+2x2+3x3 = 4
5x1+ 6x2+7x3 = 8
9x1+x2+2x3 = 3
Sol: As you see the system is diagonal system, therefore the
convergence is assured. Since we want the solution correct up to 4
significant digits, therefore the iterative process will terminate as soon
as we find the successive iteration do not produce any change at first
four significant positions.
We rewrite the given system of equations as:
X1 = (4- 2x2-3x3)/1
X2 = (8- 5x1-7x3)/6
X3 = (3- 9x1-x2)/2
We start with initial approximation as
X1 = X2 = X3 = 0
Iteration 1: Substituting X2 = X3 = 0 in the first equation, we Obtain
11
12. X1 =4.0000
Substituting X1 = 4.0000 X3 = 0 in the second equation, we Obtain
X2 = -2.0000
Substituting
X1 = 4.0000 X2 = -2.0000 in the third equation, we
Obtain
X3 =-15.5000
Thus, we obtain
X1 =4.0000
X2 = -2.0000
X3 =-15.5000
Iteration2: Now Substituting X2 =-2.0000 X3 = -15.5000 in the first
equation, we Obtain
X1 =54.5000
Substituting X1 = 54.5000 X3 = -15.5000 in the second equation, we
Obtain
X2 = -26.0000
Substituting X1 = 54.5000 X2 = -26.0000 in the third equation, we
Obtain
X3 =-230.7500
Thus, we obtain
X1 =54.5000
X2 = -26.0000
X3 =-230.7500
Note: - now you can easily compare with both the output. Then
you got different answer in both the iterative method.
12
14. float z;
z=(3-9*x-y)/2;
return z;
}
void main()
{
float x=0,y=0,z=0,tx,ty,tz;
int i,n;
clrscr();
printf("Enter the number of iteration:");
scanf("%d",&n);
for(i=0;i<n;i++)
{
tx=fx(y,z);
ty=fy(x,z);
tz=fz(x,y);
x=tx;
y=ty;
z=tz;
}
printf("X=%fn",x);
printf("Y=%fn",y);
printf("Z=%f",z);
getch();
}
14
15. Advantages and disadvantages
Advantages:
1. Iterative.
The Jacobi method first generates inexact results and subsequently
refines its results at each iteration, with the residuals converging at an
exponential rate. For many applications, this is highly desirable.
Disadvantages:
1. Inflexible.
The Jacobi method only works on matrices A for which ρ(A) < 1, or ||
A|| < 1holds. This makes it inapplicable to a large set of problems.
Furthermore, determining whether a matrix satisfies the previous
conditions is expensive to compute.
2. Large Set-Up Time.
The Jacobi method cannot immediately begin producing results.
Before it can begin its iteration, a matrix −D −1(L+U) must be
computed. For large input matrices, this may not be a trivial
operation, as it takes O(n2) time to perform this matrix multiplication.
The result is a significant lag before any results can be output.
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