Method of Least Squares Notes

SRI KRISHNA COLLEGE OF ENGINEERING AND TECHNOLOGY
(AN AUTONOMOUS INSTITUTION| AFFILIATED TO ANNA UNIVERSITY|ACCREDITED BY NAAC WITH “A” GRADE)
ACCREDITED BY NBA (CSE, ECE, IT, MECH ,EEE, CIVIL& MCT)
KUNIAMUTHUR, COIMBATORE, TAMILNADU, INDIA
PRINCIPLE OF LEAST SQUARE METHOD
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PROBLEM
By the method of principle of least squares find the best fitting straight line to the
data given below:
x 5 10 15 20 25
y 15 19 23 26 30
SOLUTION:
Let the straight line be 𝑦 = 𝑎𝑥 + 𝑏
The normal equations are
𝑎 ෍ 𝑥 + 5𝑏 = ෍ 𝑦 … … … … … … … … . . (1)
𝑎 ෍ 𝑥 2 + 𝑏 ෍ 𝑥 = ෍ 𝑥𝑦 … … … … … … … . (2)
To calculate σ 𝑥 , σ 𝑥 2 , σ 𝑦 , σ 𝑥𝑦 we form below the table.
𝑥 𝑦 𝑥2 𝑥𝑦
5 16 25 80
10 19 100 190
15 23 225 345
20 26 400 520
25 30 625 750
75 114 1375 1885
TOTAL
The normal equations are 75𝑎 + 5𝑏 = 114 … … … … … . (3)
1375𝑎 + 75𝑏 = 1885 … … … … . (4)
Eliminate b; Multiply (3) by 15,
1125𝑎 + 75𝑏 = 1710 … … … … . . (5)
(4)-(5) gives, 250𝑎 = 175, 𝑎 = 0.7; 𝑏 = 12.3
The best fitting line is 𝑦 = 0.7𝑥 + 12.3
PROBLEM
Fit a straight line to the data given below. Also estimate the value of 𝑦 𝑎𝑡 𝑥 = 2.5
x 0 1 2 3 4
y 1 1.8 3.3 4.5 6.3
SOLUTION:
Let the best fit be 𝑦 = 𝑎𝑥 + 𝑏 … … … … … … . . (1)
The normal equations are
𝑎 ෍ 𝑥 + 5𝑏 = ෍ 𝑦 … … … … … … … … … … … . . (2)
𝑎 ෍ 𝑥 2 + 𝑏 ෍ 𝑥 = ෍ 𝑥𝑦 … … … … … … … … … . (3)
To calculate σ 𝑥 , σ 𝑥 2 , σ 𝑦 , σ 𝑥𝑦 we form below the table.
𝑥 𝑦 𝑥2 𝑥𝑦
0 1.0 0 0
1 1.8 1 1.8
2 3.3 4 6.6
3 4.5 9 13.5
4 6.3 16 25.2
TOTAL 10 16.9 30 47.1
Substituting in (2) and (3), We get,
10𝑎 + 5𝑏 = 16.9
30𝑎 + 10𝑏 = 47.1
Solving, we get 𝑎 = 1.33 , 𝑏 = 0.72
The equation is 𝑦 = 1.33𝑥 + 0.72
𝑦 𝑎𝑡 𝑥 = 2.5 = 1.33 2.5 + 0.72 = 4.045.
PROBLEM
Fit a parabola, by the method of least squares, to the following data; also estimate
𝑦 𝑎𝑡 𝑥 = 6.
x 1 2 3 4 5
y 5 12 26 60 97
SOLUTION:
Let 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 be the best fit.
The normal equations are,
𝑎 ෍ 𝑥 2 + 𝑏 ෍ 𝑥 + 5𝑐 = ෍ 𝑦 … … … … … … . … … (1)
𝑎 ෍ 𝑥 3 + 𝑏 ෍ 𝑥 2 + 𝑐 ෍ 𝑥 = ෍ 𝑥𝑦 … … … … . . … (2)
𝑎 ෍ 𝑥 4 + 𝑏 ෍ 𝑥 3 + 𝑐 ෍ 𝑥 2 = ෍ 𝑥 2 𝑦 … … … … … (3)
To calculate σ 𝑥 , σ 𝑥 2 , σ 𝑦 , σ 𝑥𝑦 and so on, we form below the table.
𝑥 𝑦 𝑥2 𝑥3 𝑥4 𝑥𝑦 𝑥 2𝑦
1 5 1 1 1 5 5
2 12 4 8 16 24 48
3 26 9 27 81 78 234
4 60 16 64 256 240 960
5 97 25 125 625 485 2425
15 200 55 225 979 832 3672
Total
Hence the equations (1), (2), (3) become,

55𝑎 + 15𝑏 + 5𝑐 = 200 … … … … … … … … … … . (4)
225𝑎 + 55𝑏 + 15𝑐 = 832 … … … … … … … … … (5)
979𝑎 + 225𝑏 + 55𝑐 = 3672 … … … … … … … … (6)
Solving we get,𝑎 = 5.7143, 𝑏 = −11.0858, 𝑐 = 10.4001
The parabola is 𝑦 = 5.7143𝑥 2 − 11.0858𝑥 + 10.4001
𝑦 𝑥 = 6 = 149.6001
PROBLEM
From the table given below, find the best values of ‘a’ and ‘b’ in the law 𝑦 =
𝑎𝑒 𝑏𝑥 by the method of least squares.
𝑥 0 5 8 12 20
𝑦 3.0 1.5 1.0 0.55 0.18
SOLUTION:
𝑦 = 𝑎𝑒 𝑏𝑥
log10 𝑦 = log10 𝑎 + 𝑏𝑥 log10 𝑒
𝑌 = 𝐴 + 𝐵𝑥 … … … … … … … … … … … … … … … . (1)
The Normal equations are,
𝐵 ෍ 𝑥 + 5𝐴 = ෍ 𝑌 … … … … … … … … … … … … . (2)
𝐵 ෍ 𝑥2 + 𝐴 ෍ 𝑥 = ෍ 𝑥 𝑌 … … … … … … … … … … 3
𝑥 𝑦 𝑌 𝑥2 𝑥𝑌
0 3.0 0.4771 0 0
5 1.5 0.1761 25 0.8805
8 1.0 0 64 0
12 0.55 -0.2596 144 -3.1152
20 0.18 -0.7447 400 -14.894
Total 45 -0.3511 633 -17.1287
Using (2) and (3),
5𝐴 + 45𝐵 = −0.3511…………………………………..(4)
45𝐴 + 633𝐵 = −17.1287……………………………...(5)
Solving (4) and (5),
𝐴 = 0.4815, 𝐵 = −0.0613
𝑎 = 10𝐴 = 3.0304,
𝑏 log10 𝑒 = 𝐵 = −0.0613, 𝑏 = −0.1411
The curve is 𝑦 = 3.0304𝑒 −0.1411𝑥

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SRI KRISHNA COLLEGE OF ENGINEERING AND TECHNOLOGY

PRINCIPLE OF LEAST SQUARE METHOD

Hence the equations (1), (2), (3) become,

The parabola is 𝑦 = 5.7143𝑥 2 − 11.0858𝑥 + 10.4001

45𝐴 + 633𝐵 = −17.1287……………………………...(5)

Solving (4) and (5),

𝑏 log10 𝑒 = 𝐵 = −0.0613, 𝑏 = −0.1411

The curve is 𝑦 = 3.0304𝑒 −0.1411𝑥

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