principles of least squares. introduction in surveying, we often have geometric constraints for our...
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Principles of Least Squares
Introduction
• In surveying, we often have geometric constraints for our measurements– Differential leveling loop closure = 0– Sum of interior angles of a polygon = (n-2)180°– Closed traverse: Σlats = Σdeps = 0
• Because of measurement errors, these constraints are generally not met exactly, so an adjustment should be performed
Random Error Adjustment
• We assume (hope?) that all systematic errors have been removed so only random error remains
• Random error conforms to the laws of probability
• Should adjust the measurements accordingly
• Why?
Definition of a Residual
If M represents the most probable value of a measured quantity, and zi represents the ith measurement, then the ith residual, vi is:
vi = M – zi
Fundamental Principle of Least Squares
minimum223
22
21
2 nvvvvv
In order to obtain most probable values (MPVs), the sum of squares of the residuals must be minimized. (See book for derivation.) In the weighted case, the weighted squares of the residuals must be minimized.
minimum2233
222
211
2 nnvwvwvwvwwv
Technically the weighted form shown assumes that the measurements are independent, but we can handle the general case involving covariance.
Stochastic Model• The covariances (including variances) and hence
the weights as well, form the stochastic model
• Even an “unweighted” adjustment assumes that all observations have equal weight which is also a stochastic model
• The stochastic model is different from the mathematical model
• Stochastic models may be determined through sample statistics and error propagation, but are often a priori estimates.
Mathematical Model• The mathematical model is a set of one or more
equations that define an adjustment condition• Examples are the constraints mentioned earlier• Models also include collinearity equations in
photogrammetry and the equation of a line in linear regression
• It is important that the model properly represents reality – for example the angles of a plane triangle should total 180°, but if the triangle is large, spherical excess cause a systematic error so a more elaborate model is needed.
Types of ModelsConditional and Parametric
• A conditional model enforces geometric conditions on the measurements and their residuals
• A parametric model expresses equations in terms of unknowns that were not directly measured, but relate to the measurements (e.g. a distance expressed by coordinate inverse)
• Parametric models are more commonly used because it can be difficult to express all of the conditions in a complicated measurement network
Observation Equations
• Observation equations are written for the parametric model
• One equation is written for each observation• The equation is generally expressed as a
function of unknown variables (such as coordinates) equals a measurement plus a residual
• We want more measurements than unknowns which gives a redundant adjustment
Elementary ExampleConsider the following three equations involving two unknowns. If Equations (1) and (2) are solved, x = 1.5 and y = 1.5. However, if Equations (2) and (3) are solved, x = 1.3 and y = 1.1 and if Equations (1) and (3) are solved, x = 1.6 and y = 1.4.
(1) x + y = 3.0
(2) 2x – y = 1.5
(3) x – y = 0.2
If we consider the right side terms to be measurements, they have errors and residual terms must be included for consistency.
Example - Continued
x + y – 3.0 = v1
2x – y – 1.5 = v2
x – y – 0.2 = v3
To find the MPVs for x and y we use a least squares solution by minimizing the sum of squares of residuals.
2222 )2.0()5.12()0.3(),( yxyxyxvyxf
Example - ContinuedTo minimize, we take partial derivatives with respect to each of the variables and set them equal to zero. Then solve the two equations.
0)1)(2.0(2)1)(5.12(2)0.3(2
0)2.0(2)2)(5.12(2)0.3(2
yxyxyxyf
yxyxyxxf
These equations simplify to the following normal equations.
6x – 2y = 6.2
-2x + 3y = 1.3
Example - Continued
Solve by matrix methods.
443.1
514.1
3.1
2.6
62
23
141
3.1
2.6
32
26
y
x
y
x
We should also compute residuals:
v1 = 1.514 + 1.443 – 3.0 = -0.044
v2 = 2(1.514) – 1.443 – 1.5 = 0.086
v3 = 1.514 – 1.443 – 0.2 = -0.128
Systematic Formation of Normal Equations
Resultant EquationsFollowing derivation in the book results in:
Example – Systematic ApproachNow let’s try the systematic approach to the example.
(1) x + y = 3.0 + v1
(2) 2x – y = 1.5 + v2
(3) x – y = 0.2 + v3
Create a table:a b l a2 ab b2 al bl
1 1 3.0 1 1 1 3.0 3.0
2 -1 1.5 4 -2 1 3.0 -1.5
1 -1 0.2 1 -1 1 0.2 -0.2
Σ=6 Σ=-2 Σ=3 Σ=6.2 Σ=1.3Note that this yields the same normal equations.
Matrix Method
mnmm
n
n
aaa
aaa
aaa
A
21
22221
11211
ml
l
l
L2
1
nx
x
x
X2
1
Matrix form for linear observation equations:
AX = L + V
Where:
mv
v
v
V2
1
Note: m is the number of observations and n is the number of unknowns. For a redundant solution, m > n .
Least Squares Solution
Applying the condition of minimizing the sum of squared residuals:
ATAX = ATL
or
NX = ATL
Solution is:
X = (ATA)-1ATL = N -1ATL
and residuals are computed from:
V = AX – L
Example – Matrix Approach
3.1
2.6
2.0
5.1
0.3
111
121
32
26
11
12
11
111
121
2.0
5.1
0.3
11
12
11
3
2
1
LA
y
x
y
xAXA
VL
v
v
v
y
xAX
T
T
Matrix Form With Weights
Weighted linear observation equations:
WAX = WL + WV
Normal equations:
ATWAX = NX = ATWL
Matrix Form – Nonlinear System
We use a Taylor series approximation. We will need the Jacobian matrix and a set of initial approximations.
The observation equations are:
JX = K + V
Where: J is the Jacobian matrix (partial derivatives)
X contains corrections for the approximations
K has observed minus computed values
V has the residuals
The least squares solution is: X = (JTJ)-1JTK = N-1JTK
Weighted Form – Nonlinear System
The observation equations are:
WJX = WK + WV
The least squares solution is: X = (JTWJ)-1JTWK = N-1JTWK
Example 10.2
Determine the least squares solution for the following:
F(x,y) = x + y – 2y2 = -4
G(x,y) = x2 + y2 = 8
H(x,y) = 3x2 – y2 = 7.7
Use x0 = 2, and y0 = 2 for initial approximations.
Example - Continued
yyFxF
41
1
yyG
xxG
2
2
yyH
xxH
2
6
Take partial derivatives and form the Jacobian matrix.
412
44
71
26
22
411
00
00
0
yx
yx
y
J
Example - Continued
3.0
0
0
87.7
88
)4(4
),(7.7
),(8
),(4
00
00
00
yxH
yxG
yxF
K
Form K matrix and set up least squares solution.
2.1
6.3
3.0
0
0
447
1241
8139
39161
412
44
71
447
1241
KJ
JJ
T
T
Example - Continued
00458.0
2125.0
2.1
6.3
8139
391611
X
01004.0
00168.0
75219.0
12393.0
40354.8175082.38
75082.3861806.1571
X
Add the corrections to get new approximations and repeat.
x0 = 2.00 – 0.02125 = 1.97875 y0 = 2.00 + 0.00458 = 2.00458
Add the new corrections to get better approximations.
x0 = 1.97875 + 0.00168 = 1.98043 y0 = 2.00458 + 0.01004 = 2.01462
Further iterations give negligible corrections so the final solution is:
x = 1.98 y = 2.01
Linear Regression
Fitting x,y data points to a straight line: y = mx + b
Observation Equations
bmxvy
bmxvy
bmxvy
bmxvy
DyD
CyC
ByB
AyA
D
C
B
A
D
C
B
A
y
y
y
y
D
C
B
A
D
C
B
A
v
v
v
v
y
y
y
y
b
m
x
x
x
x
1
1
1
1
In matrix form: AX = L + V
Example 10.3
point x y
A 3.00 4.50
B 4.25 4.25
C 5.50 5.50
D 8.00 5.50
Fit a straight line to the points in the table. Compute m and b by least squares.
In matrix form:
D
C
B
A
v
v
v
v
b
m
50.5
50.5
25.4
50.4
100.8
150.5
125.4
100.3
Example - Continued
663.3
246.0
7500.19
8125.105
0000.47500.20
7500.203125.121)()(
1
1 LAAAb
mX TT
13.0
48.0
46.0
10.0
50.5
50.5
25.4
50.4
663.3
246.0
100.8
150.5
125.4
100.3
LAXV
Standard Deviation of Unit Weight
48.024
47.02
0
nm
vS
Where: m is the number of observations and
n is the number of unknowns
Question: What about x-values? Are they observations?
Fitting a Parabola to a Set of Points
Equation: Ax2 + Bx + C = y
This is still a linear problem in terms of the unknowns A, B, and C.
Need more than 3 points for a redundant solution.
Example - Parabola
Parabola Fit Solution - 1
155
144
133
122
111
100
2
2
2
2
2
2
A
41.93
43.98
21.102
77.104
43.105
84.103
L
Set up matrices for observation equations
Parabola Fit Solution - 2
046.104
902.1
813.0
09.608
37.1482
53.5354
61555
1555225
55225979
)(
1
1 LAAAx TT
Solve by unweighted least squares solution
Compute residuals
180.0
216.0
225.0
172.0
295.0
206.0
LAXV
Condition Equations
• Establish all independent, redundant conditions
• Residual terms are treated as unknowns in the problem
• Method is suitable for “simple” problems where there is only one condition (e.g. interior angles of a polygon, horizon closure)
Condition Equation Example
Condition Example - Continued
Condition Example - Continued
Condition Example - Continued
Note that the angle with the smallest standard deviation has the smallest residual and the largest SD has the largest residual
Example Using Observation Equations
Observation Example - Continued
11
10
01
A
2
2
2
3.41
00
09.91
0
007.61
W
360"14'03142
"35'1783
"56'38134
L
06429.005408.0
05408.007636.0WAAT
6370848.12
7867721.14WLAT
Observation Example - Continued
"1.44'1783
"2.00'39134)()( 1
WLAWAAX TT
"7.15'03142"1.44'1783"2.00'391343603 a
Note that the answer is the same as that obtained with condition equations.
Simple Method for Angular Closure
Given a set of angles and associated variances and a misclosure, C, residuals can be computed by the following:
n
ii
ii
Cv
1
2
2
Angular Closure – Simple Method
39.1613.49.97.6 2223
1
2 i
i
"7.139.161
)3.4("15
"1.939.161
)9.9("15
"2.439.161
)7.6("15
2
3
2
2
2
1
v
v
v