gauss-seidel method · gauss–seidel method first we discuss the method and then we will discuss...

Gauss-Seidel methodDr. Motilal Panigrahi

Dr. Motilal Panigrahi, Nirma University

Solving system of linear equations

We discussed Gaussian elimination with partial pivoting

Gaussian elimination was an exact method or closed

method

Now we will discuss an open method or iteration method


Gauss–Seidel method

Gauss–Seidel method, also known as the Liebmann

method or the method of successive displacement

It is an iterative method

It is named after the German mathematicians Carl

Friedrich Gauss and Philipp Ludwig von Seidel, and is

similar to the Jacobi method.


https://en.wikipedia.org/wiki/Germany

https://en.wikipedia.org/wiki/Mathematician

https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss

https://en.wikipedia.org/wiki/Philipp_Ludwig_von_Seidel

https://en.wikipedia.org/wiki/Jacobi_method


First we discuss the method and then we will discuss when

the solution converges or diverges.

As noted in the previous slide, method of successive

displacement means first we assume with an initial

solution and then we displace the values successively.



We are given a system of equation

𝐴𝑥 = 𝑏

Say 𝑥 = 𝑥1, 𝑥2, … , 𝑥𝑛𝑇

First we make two assumptions,

(i) the diagonal of the matrix is nonzero, that is 𝑎𝑖𝑖 ≠ 0,

For 𝑖 = 1,2,… , 𝑛

(ii) The system has a solution.



We write the given system

𝐴𝑥 = 𝑏 (1)

in the following way

𝑥1 =𝑏1−𝑎12𝑥2−⋯−𝑎1𝑛𝑥𝑛

𝑎11… 2.1

𝑥2 =𝑏2−𝑎21𝑥1−𝑎23𝑥3…−𝑎2𝑛𝑥𝑛

𝑎22… 2.2

…

𝑥𝑛 =𝑏𝑛−𝑎𝑛1𝑥1−𝑎𝑛2𝑥2…−𝑎𝑛𝑛−1𝑥𝑛−1

𝑎𝑛𝑛… 2. 𝑛



We start with an initial values 𝑥1 = 0, 𝑥2 = 0,… , 𝑥𝑛 = 0

Then we use the equations (2.1) to (2.n) successively.

That in the first iteration

Eqn (2.1) will give the value of 𝑥1(1)

, we use this value of

𝑥1, and the other previous values of 𝑥𝑖′𝑠, in Eqn(2.2) to get

𝑥2(1). Then values 𝑥1

(1), 𝑥2

(1)and the other previous values

of 𝑥𝑖′𝑠, in Eqn(2.3) to get 𝑥3

(1)and so on till we get 𝑥𝑛

(1).


After first iteration

We have got 𝑥1(1), 𝑥2(1)

,…, 𝑥𝑛(1)

.

Then we start the second iteration. And so on.


But when should we stop?

Again as we were doing previously, calculate the

percentage relative error for each variable and find the

maximum among them which should be less than a

tolerance values.


Does the Method always converge?

No.

It depends whether the system is diagonally

dominant or not?

If diagonally dominant then it converges.


Diagonally Dominant

A system

𝑎11𝑥1 + 𝑎12𝑥2 +⋯+ 𝑎1𝑛𝑥𝑛 = 𝑏1𝑎21𝑥1 + 𝑎22𝑥2 +⋯+ 𝑎2𝑛𝑥𝑛 = 𝑏2

…

𝑎𝑛1𝑥1 + 𝑎𝑛2𝑥2 +⋯+ 𝑎𝑛𝑛𝑥𝑛 = 𝑏𝑛

is said to be diagonally dominant if it satisfy

𝑎𝑖𝑖 > 𝑎𝑖1 + 𝑎𝑖2 +⋯+ 𝑎𝑖,𝑖−1 + 𝑎𝑖,𝑖+1 +⋯+ 𝑎𝑖𝑛

For each 𝑖 = 1,2, … , 𝑛Dr. Motilal Panigrahi, Nirma University

That is

𝑎𝑖𝑖 > 𝑗=1,𝑗≠𝑖𝑛 𝑎𝑖𝑗

For 𝑖 = 1,2, … , 𝑛

Then we say that system is diagonally dominant.


Example 1

Solve using Gauss-Seidel method

8𝑥 − 3𝑦 + 2𝑧 = 20

4𝑥 + 11𝑦 − 𝑧 =33

6𝑥 + 3𝑦 + 12𝑧 = 35


Answer

The system in the given example is diagonally

dominant, as

8𝑥 − 3𝑦 + 2𝑧 = 20 8 > −3 + 2 = 5

4𝑥 + 11𝑦 − 𝑧 =33 11 > 4 + −1 = 5

6𝑥 + 3𝑦 + 12𝑧 = 35 12 > 6 + 3 = 9


Rewriting the equations, we get

8𝑥−3𝑦+2𝑧=20⇒ 𝑥 =20+3𝑦−2𝑧

8… (1.a)

4𝑥+11𝑦−𝑧=33⇒ 𝑦 =33−4𝑥+𝑧

11… (1.b)

6𝑥+3𝑦+12𝑧=35⇒ 𝑧 =35−6𝑥−3𝑦

12… (1.c)


Initial values assigned

𝑥(0) = 0, 𝑦(0) = 0, 𝑧(0) = 0

Now use equation (1.a), which gives


First iteration

𝑥(1) =20+3𝑦(0)−2𝑧(0)

8=20

8= 2.5

𝑦(1) =33−4𝑥 1 +𝑧(0)

11=33−4×2.5+0

11=

2.090909

𝑧(1) =35−6𝑥 1 −3𝑦(1)

12=35−6×2.5−3×2.090909

12

= 1.1439394Dr. Motilal Panigrahi, Nirma University

Second iteration

𝑥(2) =20+3×2.090909−2×1.1439394

8=2.998106

𝑦(2) =33−4×2.998106 +1.1439394

11= 2.013774

𝑧(2) =35−6×2.998106 −3×2.013774

12= 0.91417


i=0 i=1 i=2 i=3 i=4 i=5 i=6

𝑥(𝑖) = 0 2.5 2.998106 3.026623 3.016512 3.0166 3.016768

𝑦(𝑖) = 0 2.090909 2.013774 1.982516 1.985607 1.985964 1.985891

𝑧(𝑖) = 0 1.143939 0.91417 0.907726 0.912009 0.911875 0.91181


Percentage error calculation

i=2 i=3 i=4 i=5 i=6

%ea(x) 16.61402 0.942195 0.335179 0.00293 0.005546

%ea(y) 3.830369 1.576674 0.155661 0.017986 0.003675

%ea(z) 25.13419 0.709894 0.469584 0.014639 0.007174


Example 2

Solve using Gauss-Seidel method

10𝑥 − 2𝑦 + 𝑧 = 12

3𝑥 + 9𝑦 − 𝑧 =21

2𝑥 + 3𝑦 + 8𝑧 = 30


Answer

The system in the given example is diagonally

dominant, as

10𝑥 − 2𝑦 + 𝑧 = 12 10 > −2 + 1 = 3

3𝑥 + 9𝑦 − 𝑧 =21 9 > 3 + −1 = 4

2𝑥 + 3𝑦 + 8𝑧 = 30 8 > 2 + 3 = 5


Rewriting the equations, we get

10𝑥 − 2𝑦 + 𝑧 = 12 ⇒ 𝑥 =12+2𝑦−𝑧

10… (2.a)

3𝑥+9𝑦−𝑧=21⇒ 𝑦 =21−3𝑥+𝑧

9… (2.b)

2𝑥+3𝑦+8𝑧=30⇒ 𝑧 =30−2𝑥−3𝑦

8… (2.c)


Initial values assigned

𝑥(0) = 0, 𝑦(0) = 0, 𝑧(0) = 0

Now use equation (2.a), (2.b), (2.c),

which gives


i=0 i=1 i=2 i=3 i=4 i=5 i=6

𝑥(𝑖) = 0 1.2 1.314167 1.379892 1.372959 1.373259 1.373268

𝑦(𝑖) = 0 1.933333 2.198056 2.161946 2.163935 2.163945 2.163933

𝑧(𝑖) = 0 2.725 2.597188 2.594297 2.595284 2.595206 2.595208


Percentage error calculation

i=2 i=3 i=4 i=5 i=6

%ea(x) 8.687381 4.763103 0.504966 0.021792 0.000716

%ea(y) 12.04347 1.670253 0.091956 0.000458 0.000555

%ea(z) 4.921189 0.111406 0.038032 0.003026 7.87E-05


gauss-seidel method · gauss–seidel method first we discuss the method and then we will discuss...

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