SURVEY ADJUSTMENT

Chapter 4 : Weight of Observation

Step 2: Create table

Instrument Distance AB, Weight = ̅ − .
EDM 15.231 m 3 0.0005 7.5 x10-7
0.0045
Disto 15.235 m 2 4.05 x10-5
meter 0.0105
1.1x10-4
Tape 15.220 m 1 0.00015
Total 6

Step 3: Calculate std. dev of weighted mean
σ 2

. ℎ , ҧ = √(σ )( − 1)
0.00015

ҧ = √ (6)(2) = 0.0035

Step 3: Calculate std. dev of weighted observation

. ℎ , = √ σ 2
( − 1)

Weight for each
observation

Std deviation of weighted observation distance using EDM
0.00015

= √ 3(2) = 0.005

NFZ-JKA PMM 43

Chapter 4 : Weight of Observation
Std deviation of weighted observation for distance using disto meter

0.00015
= √ 2(2) = 0.006

Std deviation of weighted observation for distance using tape
0.00015

= √ 1(2) = 0.009

Step 4: Calculate std. dev of weight unit

. ℎ , = √ σ 2
( − 1)

= √0.00015 = 0.009
2

NFZ-JKA PMM 44

Chapter 4 : Weight of Observation

Example 2
An angle is observed on three different days with the following results,
calculate: -

i.Weight mean
ii.Std. dev of weighted mean
iii.Std. dev of weighted observation
iv.Std. dev of weighted unit

Day Observation Weight
1 30° 10′20" 1
2 30° 10′30" 3
3 30° 10′50" 2
4 30° 10′45" 3
5 30° 10′50" 4

Step 1 : Calculate weighted mean
σ ( )

ℎ , ҧ = σ
ҧ = 392° 19′05" = 30° 10′41.92"

13

NFZ-JKA PMM 45

Chapter 4 : Weight of Observation

Step 2 : create table

day Observation, x Weight, w × = ̅ − .
1 30° 10′20" 21.92" 0.13"
2 30° 10′20" 1 90° 31′30" 11.92" 0.12"
3 60° 21′40" − 8.08" 0.04"
4 30° 10′30" 3 90° 32′15" −3.08" 0.01"
5 120° 43′20" −8.08" 0.07"
30° 10′50" 2 392° 19′05" 0.37"
total 30° 10′45" 3
30° 10′50" 4

13

Step 3: Calculate std. dev of weighted mean

. , ҧ = √(σ σ 2 1)
)(
−

0.37"
ҧ = √(13)(4) = 5.06"

Step 3: Calculate std. dev of weighted observation

. ℎ , = √ σ 2
( − 1)

Std deviation of weighted observation for 1st day
0.37"

1 = √ 1(4) = 18.25"

NFZ-JKA PMM 46

Chapter 4 : Weight of Observation
Std deviation of weighted observation for 2nd day

0.37"
2 = √ 3(4) = 10.54"

Std deviation of weighted observation for 3rd day
0.37"

3 = √ 2(4) = 12.9"

Std deviation of weighted observation for 4th day
0.37"

4 = √ 3(4) = 10.54"

Std deviation of weighted observation for 5th day
0.37"

5 = √ 4(4) = 9.12"

Step 4: Calculate std. dev of weight unit

. ℎ , = √ σ 2
( − 1)

= √0.37" = 18.25"
4

NFZ-JKA PMM 47

Chapter 4 : Weight of Observation

Example 3
Based on data below, calculate: -

1. Calculate weight mean
2. Std. dev of weighted mean
3. Std. dev of weighted observation
4. Std. dev of weighted unit

BIL Distance, . ,

1 30.467 ±0.020

2 30.453 ±0.014

3 30.448 ±0.020

4 30.457 ±0.010

5 30.462 ±0.010

Step 1 : Calculate weight for each observation

1
ℎ , = 2

1
1 = 0.0004 = 2500

1
2 = 0.0002 = 5000

1
3 = 0.0004 = 2500

1
4 = 0.0001 = 10000

1
5 = 0.0001 = 10000

NFZ-JKA PMM 48

Chapter 4 : Weight of Observation

Step 2: Calculate weight mean
σ ( )

ℎ , ҧ = σ
913742.5

ҧ = 30000 = 30.45808
Step 3 : create table

BIL Distance, . . .

±0.020 0.0004 2500 76167.5 -0.0089 0.19802
1 30.467 m

2 30.453 m ±0.014 0.0002 5000 152265 0.0051 0.13005
3 30.448 m ±0.020 76120 0.0101 0.25503
0.0004 2500

4 30.457 m ±0.010 0.0001 10000 304570 0.0011 0.0121

5 30.462 m ±0.010 0.0001 10000 304620 -0.0039 0.1521

TOTAL 30000 913742.5 0.0035 0.7473

Step 3: Calculate std. dev of weighted mean
σ 2

. ℎ , ҧ = √(σ )( − 1)
913742.5

ҧ = √(30000)(4) = 2.759

NFZ-JKA PMM 49

Chapter 4 : Weight of Observation
Step 3: Calculate std. dev of weighted observation

σ 2
. ℎ , = √
( − 1)

Std deviation of weighted observation for 1st data
0.7473

1 = √2500(4) = 0.009

Std deviation of weighted observation for 2nd data
0.7473

2 = √5000(4) = 0.006

Std deviation of weighted observation for 3rd data
0.7473

3 = √2500(4) = 0.009

Std deviation of weighted observation for 4th data
0.7473

4 = √10000(4) = 0.004

Std deviation of weighted observation for 5th data
0.7473

5 = √10000(4) = 0.004

NFZ-JKA PMM 50

Chapter 4 : Weight of Observation

Step 4: Calculate std. dev of weight unit

. ℎ , = √ σ 2
( − 1)

= √0.7473 = 0.432
4

NFZ-JKA PMM 51

Chapter 4 : Weight of Observation

Tutorial
Question 1
An angle was measured at four different times with the following results. What is the
most probable value for the angle and the standard deviation in the mean.

day Observation, x Std. dev
1 120° 30′20" ± 6.2"
2 120° 30′30" ± 9.8"
3 120° 30′50" ± 5.2"
4 120° 30′45" ± 4.7"

Question 2

The distance of the routes and the observed differences in elevations are
show below, calculate: -

1. Calculate weight mean

2. Std. dev of weighted mean

3. Std. dev of weighted observation

4. Std. dev of weighted unit

route Different distance
elevation
1 120 m
2 15.321 98 m
3 100 m
15.350

15.334

NFZ-JKA PMM 52

Chapter 5 : Least Square Adjustment

Chapter 5: Least Square Adjustment

LEAST SQUARE ADJUSTMENT APPLICATIONS

▪ A least square adjustment (LSA) is method to estimate or adjust the
observations to obtain the most accurate value on these observations
using statistical analysis.

▪ LSE is a systematic & simple method to compute estimated value of
variables for unknown quantities from redundant measurements when
then number of measurements more than number of variables

▪ Least-squares adjustment minimizes the sum of the squares of the
residuals or weighted residuals.

▪ Condition of least squares adjustment: number of observations must
equal or more than number of variables.

▪ LSE is not required if no redundant measurements

Step by step to solve LSA problem

1. Model the observation equation
2. Create matrix A, X and L
3. Find matrix
4. Find Determinant for
5. Find minor matrix for
6. Adjoint matrix
7. Inverse matrix
8. Find
9. Solve = ( )−1 .

NFZ-JKA PMM 53

Chapter 5 : Least Square Adjustment

EXAMPLE 1 : Distance

A baseline consists of four stations on a straight-line A, B, C and D are measured
using Electronic Distance Measurement device. In order to determine the
distance between the stations,

• Determine the number of observation (n) and variables (u).

• By using the matrix method, calculate the adjusted variables for the
distance of AB, BC and CD.

Table 4.1: Observation data for baseline

Observation Distance, m Note :
AB 25.051
BC 25.047 1. Identify all the
CD 25.110 data given
AC 50.091
BD 50.150 2. Change the
AD 75.200 data into simple
figure
Convert

1st variables 2nd variables 3rd variables

Figure 4.1 : Convert data into simple figure
Number of observation, n = 6
Variables, U = 3 AB , BC & CD

NFZ-JKA PMM 54

Chapter 5 : Least Square Adjustment

STEP 1 : Model the observation equation

AB = 25.051 + V1 NOTE : Set matrix A base on variables value
for each equation

BC = 25.047 + V2 AB BC CD
CD = 25.110 + V3
MATRIX 100

AB + BC = 50.091 + V4 010

BC + CD = 50.150 + V5 A = 0 0 1
1 1 0

AB + BC + CD = 75.200 + V6 011
[1 1 1]

STEP 2 : Create matrix A, X and L AB + BC + CD = 75.200 + V6

100 25.051

0 1 0 25.047
0 1 ; X =[ ] ; L =
A = 0 1 0 25.110
1 1 1 50.091

0 50.150
[1 1 1] [75.200]

NOTE : Matrix AT

STEP 3 : Find matrix ( ) Change Row to column

100

100 010 1 0 01 0 1
= [0 1 0 1 1 1]
1 0 0 1 0 1 0 1 0 A = 0 0 1
= [0 1 0 1 1 1] . 0 0 1 1 1 0 0 0 10 1 1
0 1 0 1 1 1 1 0
0 0 1 1 011
[1 1 1]
[1 1 1]

1+0+0+1+0+1 0+0+0+1+0+1 0+0+0+0+0+1
= [0 + 0 + 0 + 1 + 0 + 1 0 + 1 + 0 + 1 + 1 + 1 0 + 0 + 0 + 0 + 1 + 1]

0+0+0+0+0+1 0+0+0+0+1+1 0+0+1+0+1+1

3 21
= [ 2 4 2]

1 23

NFZ-JKA PMM 55

Chapter 5 : Least Square Adjustment

STEP 4: Find Determinant for matrix ( )

321 321 321 NOTE: Follow this
DET ( ) = [2 4 2] [2 4 2] [2 4 2] rule
123
123 123 +−+
[− + −]
= 3 [42 32] − 2 [12 23] + 1 [21 42] +−+

= 3(12 − 4) − 2(6 − 2) + 1(4 − 4)
= 3(8) − 2(4) + 1(0)
= 24 − 8 + 0
= 16
STEP 5: Find minor matrix

( ) = 321 321 321
[2 4 2] [2 4 2] [2 4 2]
123 123 123

321 321 321
[2 4 2] [2 4 2] [2 4 2]
123 123 123

321 321 321
[2 4 2] [2 4 2] [2 4 2]
123 123 123

( ) = [24 23] [21 23] [12 42]
[22 13] [13 13] [31 22]
[24 12] [23 12] [32 24]

12 − 4 6 − 2 4 − 4 840

( ) = [ 6 − 2 9 − 1 6 − 2 ] = [4 8 4]

4 − 4 6 − 2 12 − 4 0 4 8

NFZ-JKA PMM 56

Chapter 5 : Least Square Adjustment

STEP 6 : Find Adjoint matrix ( ) Note :
Adjoint matrix
( ) = ( )
( ) = ( )
+−+ So, for symmetric matrix
( ) = ( ) [− + −]
( ) = ( )
+−+

= 8 −4 0 8 −4 0
[−4 8 −4] = [−4 8 −4]

0 −4 8 0 −4 8

STEP 7: Matrix inverse ( )

( )−1 = 1 × ( )
det( )

= 1 +8 −4 +0
× [−4 +8 −4]
16 +0 −4 +8

+8 −4 0

16 16 16
= −4 8 −4

16 16 16
0 −4 +8
[16 16 16 ]

0.5 −0.25 0
= [−0.25 0.5 −0.25]

0 −0.25 0.5

NFZ-JKA PMM 57

Chapter 5 : Least Square Adjustment

STEP 8: Find

25.051

1 0 0 1 0 1 25.047
= [0 1 0 1 1 1] ×
0 1 0 1 1 25.110
0 50.091

50.150
[75.200]

1(25.051) + 0 + 0 + 1(50.091) + 0 + 1(75.200)
= [0 + 1(25.047) + 0 + 1(50.091) + 1(50.150) + 1(75.200)]

0 + 0 + 1(25.110) + 0 + 1(50.150) + 1(75.200)

150.342
= [200.488]

150.460

STEP 9: Solve = ( )− .

= ( )−1 .

0.5 −0.25 0 150.342

[ ] = [−0.25 0.5 −0.25] × [200.488]

0 −0.25 0.5 150.460

(0.5)(150.342) + (−0.25)(200.488) + (0)(150.460) 25.049

[ ] = [(−0.25)(150.342) + (0.5)(200.488) + (−0.25)(150.460)] = [25.0435 ]

(0)(150.342) + (−0.25)(200.488) + (0.5)(150.460) 25.108

AB = 25.049 m
BC = 25.0435 m
CD = 25.108 m

NFZ-JKA PMM 58

Chapter 5 : Least Square Adjustment

Example 2
Between four points A, B, C and D situated on a straight line in pairs distances AB, BC,
CD, AC, AD and BD were measured. The six measurements show in table. Calculate
the distances of AB, BC and CD by means of linear least squares adjustment.

Table 4.2 : Observation data

line Distance, m
AB 30.17
BC 10.12
CD 20.25
AC 40.31
AD 60.51
BD 30.36

AB C D

30.17 10.12 20.25

40.31

60.51

30.36

Figure 4.2 : Convert data into simple figure

Step 1: Model the observation equation

= 30.17 + 1
= 10.12 + 2
= 20.25 + 3
+ = 40.31 + 4
+ + = 60.51 + 5
+ = 30.36 + 6

NFZ-JKA PMM 59

Chapter 5 : Least Square Adjustment

STEP 2: Create matrix A, X and L

100 30.17

0 1 0 10.12
0 1 ; X =[ ] ; L = 20.25
A = 0 1 0 40.31
1 1 1 60.51

1
[0 1 1]
[30.36]

STEP 3: Find metric

100

1 0 0 1 1 0 0 1 0 3 2 1
= [0 1 0 1 1 1] . 0 0 1 = [2 4 2]
0 1 0 1 1 1 1 0 2 3
0 1 1 1 1

[0 1 1]

STEP 4: Find Determinant for matrix ( )

3 2 1 = 3 [42 23] − 2 [12 32] + 1 [21 24] = 16
Det = [2 4 2]
2 3
1

STEP 5: Minor matrix for ( )

[42 32] [12 23] [21 24] = 8 4 0
Minor = [22 13] [31 13] [31 22] [4 8 4]
12] [32 21] [32 24]] 0 4 8
[[42

STEP 6: Adjoint matrix

( ) = ( )

+−+
( ) = ( ) [− + −]

+−+

NFZ-JKA PMM 60

Chapter 5 : Least Square Adjustment

8 −4 0 8 −4 0
( ) = [−4 8 −4] ( ) = [−4 8 −4]

0 −4 8 0 −4 8

STEP 7: Matrix inverse ( )

( )−1 = 1 ( ( ))
det( )

1 8 −4 0 0.5 −0.25 0

= 16 × [−4 8 −4] = [−0.25 0.5 −0.25]
0 −4 80 −0.25 0.5

STEP 8: Find

30.17

1 0 0 1 1 0 10.12 130.99
= [0 1 0 1 1 1] . 20.25 = [141.30]
0 1 0 1 1 40.31
0 60.51 111.12

[30.36]

STEP 9: Solve = ( )− .
= ( )−1 .

0.5 −0.25 0 130.99 30.17

[ ] = [−0.25 0.5 −0.25] × [141.30] = [10.1225]

0 −0.25 0.5 111.12 20.235

AB = 30.170 m
BC = 10.1225 m
CD = 20.235 m

NFZ-JKA PMM 61

Chapter 5 : Least Square Adjustment

Example 3
EDM instrument is placed at point A and reflector is placed successively at
point B, C and D. The observed value AB, AC, AD, BC, CD are show in table.
Calculate the unknown value AB, BC and CD

Table 4.3 : Observation data
line Distance, m
AB 10.231
AC 30.452
AD 52.223
BC 20.225
BD 41.995

A BC D
10.231 m
30.452 m
52.223 m

20.225 m
41.995 m

Figure 4.3 : Convert data into simple figure

Step 1: Model the observation equation

= 10.231 + 1
+ = 30.452 + 2
+ + = 52.223 + 3
= 20.225 + 4
+ = 41.995 + 5

NFZ-JKA PMM 62

Chapter 5 : Least Square Adjustment

STEP 2: Create matrix A, X and L

100 10.231

110 30.452

A = 1 1 1 ; X =[ ] ; L = 52.223
010
[0 1 1] 20.225
[41.995]

STEP 3: Find metric
100

1 1 10 0 1 1 0 3 2 1
= [0 1 1 1 1] . 1 1 1 = [2 4 2]

0 0 10 1 0 1 0 1 2 2
[0 1 1]

STEP 4: Find Determinant for matrix ( )

3 2 1 = 3 [42 22] − 2 [12 22] + 1 [12 24] = 8
Det = [2 4 2]
2 2
1

STEP 5: Minor matrix for ( )

[42 22] [21 22] [12 24] = 4 2 0
Minor = [22 12] [31 21] [13 22] [2 5 4]
21] [23 21] [23 42]] 0 4 8
[[24

STEP 6: Adjoint matrix

( ) = ( ) 4 −2 0
( ) = [−2 5 −4]
+−+
( ) = ( ) [− + −] 0 −4 8

+−+

4 −2 0
( ) = [−2 5 −4]

0 −4 8

NFZ-JKA PMM 63

Chapter 5 : Least Square Adjustment

STEP 7: Inverse matrix ( )

( )−1 = 1 ( ( ))
det( )

1 4 −2 0 0.5 −0.25 0

= 8 × [−2 5 −4] = [−0.25 0.625 −0.5]
0 −4 80 −0.5 1

STEP 8: Find matrix ( )

10.231
1 1 1 0 0 30.452 92.906
= [0 1 1 1 1] . 52.223 = [144.895]
0 0 1 0 1 20.225 94.218

[41.995]

STEP 9: Solve = ( )− .

= ( )−1 .

0.5 −0.25 0 92.906 10.2293

[ ] = [−0.25 0.625 −0.5] × [144.895] = [20.2239]

0 −0.5 1 94.218 21.7705

AB = 10.2293 m
BC = 20.2239 m
CD = 21.7705 m

NFZ-JKA PMM 64

Chapter 5 : Least Square Adjustment

Example 4 : levelling

Given the height of point TBM 1 is 100.500m. Calculate the adjusted height
variable for points B, C and D using Least Square Adjustment observation
equation method.

FROM TO DIFFERENT HEIGHT TBM 1 B
D
TBM 1 B 0.046 C
B D 0.265 V
TBM 1 D 0.312
TBM 1 C -0.024
C B 0.070
C D 0.336

Step 1: Model the observation equation

– = 0.046 + 1
– = 0.265 + 2
− = 0.312 + 3
− = −0.024 + 4
− = 0.070 + 5
− = 0.336 + 6

Insert known value New equation

– . = 0.046 + 1 B = 100.546 + V1
– = 0.265 + 2 – = 0.265 + 2
− . = 0.312 + 3 D = 100.812 + V3
− . = −0.024 + 4 C = 100.476+ V4
− = 0.070 + 5 − = 0.070 + 5
− = 0.336 + 6 − = 0.336 + 6

NFZ-JKA PMM 65

Chapter 5 : Least Square Adjustment

STEP 2: Create matrix A, X and L

10 0 100.546

−1 0 1 0.265
1 ; X =[ ] ; L = 100.812
A = 0 0 0 100.467
0 1 0
0.070
1 −1 1] [ 0.336 ]
[0 −1

STEP 3: Find metric

1 00

1 −1 0 0 1 0 −1 0 1 3 −1 −1
= [0 0 0 1 −1 −1] . 00 1 = [−1 3 −1]
1 1 0 0 1 01 0 −1 3
0 1 −1 0 −1

[0 −1 1]

STEP 4: Find Determinant for matrix ( )

3 −1 −1 3 [−31 −31] − (−1) [−−11 −31] + (−1) [−−11 −31] = 16
Det = [−1 3 −1]=
−1 3
−1

STEP 5: Minor matrix for ( )

[−31 −31] [−−11 −31] [−−11 −31] = 8 −4 4
Minor = [−−11 −31] [−31 −31] [−31 −−11] [−4 8 −4]
−−11] [−31 −−11] [−31 −31]] −4 8
[[−31 4

STEP 6: Adjoint matrix

( ) = ( )

+−+
( ) = ( ) [− + −]

+−+

NFZ-JKA PMM 66

Chapter 5 : Least Square Adjustment

844 844
( ) = [4 8 4] ( ) = [4 8 4]

448 448

STEP 7: Matrix inverse ( )

( )−1 = 1 ( ( ))
det( )

1 844 0.5 0.25 0.25

= 16 × [4 8 4] = [0.25 0.5 0.25]
4 4 8 0.25 0.25 0.5

STEP 8: Find

100.546

1 −1 0 0 1 0 0.265 100.351
= [0 0 0 1 −1 −1] . 100.812 = [100.061]
1 1 0 0 1 100.467
0 101.413
0.070
[ 0.336 ]

STEP 9: Solve = ( )− .
= ( )−1 .

0.5 0.25 0.25 100.351 100.544

[ ] = [0.25 0.5 0.25] × [100.061] = [100.4715]

0.25 0.25 0.5 101.413 100.8095

B = 100.544 m
C = 100.472 m
D = 100.810 m

NFZ-JKA PMM 67

Chapter 5 : Least Square Adjustment

EXAMPLE 5
Calculate the variables for the elevations of A, B and C. Use the least square
adjustment method. Given the elevation of BM 1 is 12.142m and BM2=10.523m

FROM TO DIFFERENT HEIGHT
A BM1 -4.425
2.210
BM 1 C -2.455
C B -3.827
C BM2 1.375
B 6.040
BM2 A 4.664
BM2 A

B

Step 1: Model the observation equation NOTE :

1 − = −4.425 + 1 ▪ Different height = fore sight - back sight
− 1 = 2.210 + 2 ▪ Insert know value into the equation
− = −2.455 + 3
2 − = −3.827 + 4
− 2 = 1.375 + 5
− 2 = 6.040 + 6
− = 4.664 + 7

New equation after insert know value = 16.567 + 1
= 14.352 + 2
. − = −4.425 + 1 − = −2.455 + 3
− . = 2.210 + 2 = 14.350 + 4
− = −2.455 + 3 = 11.898 + 5
. − = −3.827 + 4 = 16.563 + 6
− . = 1.375 + 5 − = 4.664 + 7
− . = 6.040 + 6
− = 4.664 + 7 NFZ-JKA PMM 68

Chapter 5 : Least Square Adjustment

STEP 2: Create matrix A, X and L

100 16.576

001 14.352

0 1 −1 −2.455

A = 0 0 1 X =[ ] ; L = 14.350

010 11.898

100 16.563
[1 −1 0] [ 4.664 ]

STEP 3: Find metric

100

001

1 0 0 0 0 1 1 0 1 −1 3 −1 0
= [0 0 1 0 1 0 −1] . 0 0 1 = [−1 3 −1]

0 1 −1 1 0 0 0 0 1 0 0 −1 3

100
[1 −1 0]

STEP 4: Find Determinant for matrix ( )

3 −1 0 = 3 [−31 −31] − (−1) [−01 −31] + 0 [−01 −31] = 21
Det = [−1 3 −1]
−1 3
0

STEP 5: Minor matrix for ( )

[−31 −31] [−01 −31] [−01 −31] = 8 −3 1
Minor = [−−11 03] [30 30] [03 −−11] [−3 9 −3]
−01] [−31 −31]] −3 8
[[−31 −01] [−31 1

STEP 6: Adjoint matrix

( ) = ( )

+−+
( ) = ( ) [− + −]

+−+

NFZ-JKA PMM 69

Chapter 5 : Least Square Adjustment

831 831
( ) = [3 9 3] ( ) = [3 9 3]

138 138

STEP 7: Inverse matrix ( )

( )−1 = 1 ( ( ))
det( )

1 831 831
= 21 × [3 9 3]
1 3 8 21 21 21
=3 9 3

21 21 21
138
[21 21 21]

STEP 8: Find matrix ( )

16.576
14.352
1 0 0 0 0 1 1 −2.455 37.803
= [0 0 1 0 1 0 −1] . 14.350 = [ 4.779 ]
0 1 −1 1 0 0 0 11.898 31.157
16.563
[ 4.664 ]

STEP 9: Solve = ( )− .

831

21 21 21 37.803 16.568
3 9 3 × [ 4.779 ] = [11.900]
[ ] =
21 21 21 31.157 14.352
1 3 8
[21 21 21]

A = 16.568 m

B = 11.900 m

C = 14.352 m

NFZ-JKA PMM 70

Chapter 5 : Least Square Adjustment

Example 6:
Calculate angle BAC, CAD and DAE using Least Square Adjustment
observation equation method.

Position Angle
BAC 30° 38′56"
CAD 54° 25′20"
DAE 25° 18′40"
BAD 85° 04′24"
CAE 79° 43′55"
BAE 110° 22′50"

Step 1: Model the observation equation

= 30° 38′56" + 1
= 54° 25′ 20" + 2
= 25° 18′40" + 3
+ CAD = 85° 04′24" + 4
+ = 79° 43′ 55" + 5
+ CAD + DAE = 110° 22′50" + 6

STEP 2: Create matrix A, X and L

1 0 0 30° 38′56"
1 0 54° 25′ 20"
0 0 1 25° 18′40"
1 0 =[ ] 85° 04′24"
A = 0 1 1 ; X ; L = 79° 43′ 55"
1 1 1] [110° 22′50"]

0
[1

NFZ-JKA PMM 71

Chapter 5 : Least Square Adjustment

STEP 3: Find metric

100

1 0 0 1 0 1 0 1 0 3 2 1
= [0 1 0 1 1 1] . 0 0 1 = [2 4 2]
0 1 0 1 1 1 1 0 2 3
0 0 1 1 1

[1 1 1]

STEP 4: Find Determinant for matrix ( )

3 2 1 = 3 [24 32] − 2 [21 23] + 1 [21 24] = 16
Det = [2 4 2]
2 3
1

STEP 5: Minor matrix for ( )

[24 32] [12 32] [21 42] = 8 4 0
= [22 13] [31 13] [13 22] [4 8 4]
12] [23 12] [23 42]] 0 4 8
[[24

STEP 6: Adjoint matrix

( ) = ( )

+−+
( ) = ( ) [− + −]

+−+

8 −4 0 8 −4 0
( ) = [−4 8 −4] ( ) = [−4 8 −4]

0 −4 8 0 −4 8

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Chapter 5 : Least Square Adjustment

STEP 7: Inverse matrix ( )

( )−1 = 1 ( ( ))
det( )

1 8 −4 0 0.5 −0.25 0

= 16 × [−4 8 −4] = [−0.25 0.5 −0.25]
0 −4 80 −0.25 0.5

STEP 8: Find matrix ( )

30° 38′56"

1 0 0 1 0 1 54° 25′ 20" 226° 06′10"
= [0 1 0 1 1 1] . 25° 18′40" = [329° 36′29"]
0 1 0 1 1 85° 04′24"
0 79° 43′ 55" 215° 25′25"

[110° 22′50"]

STEP 9: Solve = ( )− .
= ( )−1 .

0.5 −0.25 0 226° 06′10" 30° 38′57.75"
× [329° 36′29"] = [54° 25′20.75"]
[ ] = [−0.25 0.5 −0.25]
215° 25′25" 25° 18′35.25"
0 −0.25 0.5

BAC = 30° 38′57.75"
CAD = 54° 25′20.75"
DAE = 25° 18′35.25"

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Chapter 5 : Least Square Adjustment

Example7: Condition adjustment
The three observations are related to their adjusted values and their residuals.
Calculate adjusted angle for A and B using Least Square Adjustment observation
equation method.

Point Angle
A 150° 20′30"
B 80°17′35"
C 129° 21′30"

Step 1: Model the observation equation

= 150° 20′30" + 1
= 80°17′35" + 2
= 129° 21′30" + 3

Condition equation
+ B + C = 360° 00′00"
C = 360° 00′00" − −

Substitute to 1st equation NEW EQUATION

° ′ " − − = 129° 21′30" + 3 = 150° 20′30" + 1
+ = 360° − 129° 21′30" + 3 = 80°17′35" + 2
+ = 230° 38′30" + 3 + = 230° 38′30" + 3

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Chapter 5 : Least Square Adjustment

STEP 2: Create matrix A, X and L

1 0 ; X =[ ] ; L = 150° 20′30"
A =[0 1] [ 80°17′35" ]
1 230° 38′30"
1

STEP 3: Find metric

= [01 0 11] . 1 0 = [12 12]
1 [0 1]
1 1

STEP 4: Find Determinant for matrix ( )

Det = [21 12]= (2 × 2) − (1 × 1) = 3

STEP 5: Minor matrix r for ( )

Cof = [21 12]

STEP 6: Adjoint matrix

( ) = ( )
+−+

( ) = ( ) [− + −]
+−+

( ) = [−21 −21] ( ) = [−21 −21]

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Chapter 5 : Least Square Adjustment

STEP 7: Inverse matrix ( )

( )−1 = 1 ( ( ))
det( )

2 −1
−21] = [−31
= 1 × [−21 3 ]
3 3 2

3

STEP 8: Find matrix ( )

= [10 0 11] . 150° 20′30" [380° 5596′′0050""]
1 [ 80°17′35" ]= 310°
230° 38′30"

STEP 9: Solve = ( )− .
= ( )−1 .

2 −1 5569′′0050""] [18500°°1270′4′338.3.333""]

[ ] = [−31 3 ] × [380° =
2 310°
3
3

A = 150° 20′38.33"
B = 80° 17′43.33"
C = 360° − ( 150° 20′38.33" + 80° 17′43.33" )

= 129° 21′38.34"

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Chapter 5 : Least Square Adjustment

SOLVE LEAST SQUARE ADJUSTMENT WITH WEIGHTS

Step by step to solve LSA problem with weights
10. Model the observation equation
11. Create matrix A, X, W and L
12. Find matrix
13. Find Determinant for
14. Find minor matrix for
15. Adjoint matrix
16. Inverse matrix
17. Find
18. Solve = ( )−1 .

Example 1
Calculate the variables for the elevations of A, B and C. Use the least square
adjustment method. Given the elevation of BM 1 is 15.384m and BM2=16.245m

From To Different Height weight
A BM1 2
-5.663 2
BM 1 C -2.929 2
C B 5.174 4
C BM2 3.790 1
B 1.378 2
BM2 A 4.802 4
BM2 A 3.420

B

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Step 1: Model the observation equation Chapter 5 : Least Square Adjustment

1 − = −5.663 + 1 NOTE :
− 1 = −2.929 + 2 ▪ Different height = fore sight - back sight
− = 5.174 + 3 ▪ Insert know value into the equation
2 − = 3.790 + 4
− 2 = 1.378 + 5
− 2 = 4.802 + 6
− = 3.420 + 7

New equation after insert know value = 21.047 + 1
= 12.455 + 2
. − = −5.663 + 1 − = 5.174 + 3
− . = −2.929 + 2 = 12.455 + 4
− = 5.174 + 3 = 17.623 + 5
. − = 3.790 + 4 = 21.047 + 6
− . = 1.378 + 5 − = 3.420 + 7
− . = 4.802 + 6
− = 3.420 + 7

STEP 2: Create matrix A, X, W and L

100 21.047 2000000
001 X =[ ] 12.455 0200000
0 1 −1 5.174 0020000
A= 0 0 1 L = 12.455 = 0 0 0 4 0 0 0
010 17.623 0000100
100 21.047 0000020
[1 −1 0] [ 3.420 ] [0 0 0 0 0 0 4]

Base on weighted data

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Chapter 5 : Least Square Adjustment

STEP 3: Find metric

2000000 100

0200000 001

10 0 001 1 0020000 0 1 −1
= [0 0 1 0 1 0 −1] × 0 0 0 4 0 0 0 × 0 0 1

0 1 −1 1 0 0 0 0 0 0 0 1 0 0 010
0000020
100
[ [0 0 0 0 0 0 4] ] [1 −1 0]

100

001

20 0 002 4 0 1 −1 8 −4 0
= [0 0 2 0 1 0 −4] × 0 0 1 = [−4 7 −2]

0 2 −2 4 0 0 0 010 0 −2 8

100
[1 −1 0]

STEP 4: Find Determinant for matrix ( )

8 −4 0 = 8 [−72 −82] − (−4) [−04 −82] + 0 [−04 −72] = 288
Det = [−4 7 −2]
−2 8
0

STEP 5: Minor matrix for ( )

[−72 −82] [−04 −82] [−04 −72] = 52 −32 8
Minor = [−−42 80] [80 08] [80 −−24] [−32 64 −16]
−02] [−84 −74]] −16 40
[[−74 −02] [−84 8

STEP 6: Adjoint matrix 52 32 8
( ) = [32 64 16]
52 32 8
( ) = [32 64 16] 8 16 40

8 16 40

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Chapter 5 : Least Square Adjustment

STEP 7: Inverse matrix ( )

( )−1 = 1 ( ( ))
det( )

1 52 32 8 13 1 1
= 288 × [32 64 16]
8 16 40 72 9 36
=1 2 1

9 9 18
115
[36 18 36]

STEP 8: Find matrix ( )

2000000 21.047

0200000 12.455

10 0 001 1 0020000 5.174
= [0 0 1 0 1 0 −1] × 0 0 0 4 0 0 0 × 12.455

0 1 −1 1 0 0 0 0 0 0 0 1 0 0 17.623
0000020
[0 0 0 0 0 0 4] 21.047
[ 3.420 ]

21.047

12.455

20 0 002 4 5.174 97.868
= [0 0 2 0 1 0 −4] × 12.455 = [14.291]

0 2 −2 4 0 0 0 17.623 64.382

21.047
[ 3.420 ]

STEP 9: Solve = ( )− .

13 1 1

72 9 36 97.868 21.103
[ ] = 1 2 1 × [14.291] = [17.641]

9 9 18
1 1 5 64.382 12.490
[36 18 36]

A = 21.103 m
B = 17.641 m
C = 12.490 m

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Chapter 5 : Least Square Adjustment

Example 2:
Calculate angle BAC, CAD and DAF using Least Square Adjustment observation
equation method.

Position Angle Std. dev
BAC 45° 38′56" 5”
CAD 48° 25′20" 2”
DAF 85° 55′45" 2”
BAD 94° 04′20" 5”
CAF 134° 21′05" 10”

Step 1: Model the observation equation New Equation

= 45° 38′56" + 1 = 45° 38′56" + 1
= 48° 25′ 20" + 2 = 48° 25′ 20" + 2
= 85° 55′45" + 3 + = 94° 04′15" + 3
+ = 94° 04′20" + 4 + = 94° 04′20" + 4
CAD + DAF = 134° 21′ 05" + 5 = 45° 38′ 55" + 5

Condition equation
BAC + CAD + DAF = 180°
DAF = 180° − (BAC + CAD)

Substitute into observation equation

= 45° 38′56" + 1
= 48° 25′ 20" + 2
° − − = 85° 55′45" + 3
+ = 94° 04′20" + 4
CAD + ( ° − − ) = 134° 21′ 05" + 5

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Chapter 5 : Least Square Adjustment

STEP 2: Create matrix A, X ,W and L

10 45° 38′56" 1000 0
48° 25′ 20" 52 1
0 1 ; X =[ ] ; 0 0 0
A= 1 1 L = 94° 04′15" 0 22 0 0
1 94° 04′20" W= 0 1 0
1 0 22 1
[45° 38′ 55" ] 1
[1 0] 0 0 0 52
[ 0 0 0 0 102]

STEP 3: Find metric

1 0 0 0 0 10
0 0 01
52 1 0 0 × 11
1 11
= [01 0 1 1 10] × 0 22 1 0 [1 0]
1 1 1 0 22
0 52 0
0 0
[0 0 0 0 1
0
102]

10

= [0.004 0 0.25 0.04 0.001] × 01 = [00..2394 00..5249]
0.25 0.25 0.04 11
11
[1 0]

STEP 4: Find Determinant for matrix ( )

Det = [00..2349 00..2549]= 0.0995

STEP 5: Minor matrix r for ( )
minor = [00..5294 00..3249]

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Chapter 5 : Least Square Adjustment

STEP 6: Adjoint matrix

( ) = ( )

+−+
( ) = ( ) [− + −]

+−+

( ) = [00..5294 00..3294] ( ) = [−00.5.249 −00.3.249]

STEP 7: Inverse matrix ( )

( )−1 = 1 ( ( ))
det( )

1080 −580
−00.3.249] = [−159890
= 1 × [−00.5.249 199 ]
0.0995 199 680

199

STEP 8: Find matrix ( )

1 0 0 0 0 45° 38′56"
1 0 0 0 48° 25′ 20"
[01 0 1 1 10] 52 1 0 0 94° 04′15"
1 1 1 0 22 1 0 94° 04′20"
= . 0 0 22 1 ∙ [45° 38′ 55" ]
0 0 52
0 0 0 0 102]
[0

45° 38′56"

= [0.004 0 0.25 0.04 0.001] ∙ 48° 25′ 20" = [2399°°3233′′1407""]
0.25 0.25 0.04 94° 04′15"
94° 04′20"

[45° 38′ 55" ]

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Chapter 5 : Least Square Adjustment

STEP 9: Solve = ( )− .
= ( )−1 .

1080 −580 × [2399°°3233′′1407""] = [4458°°3285′5′189.4.38""]

[ ] = [−159890 199 ]
680
199
199

BAC = 45° 38′58.48"
CAD = 48° 25′19.3"
DAF = 180° − ( 45° 38′58.48" + 48° 25′19.3" )

= 85° 55′42.22"

“You don’t have to be great to start,
but you have to start to be great”
-Zig Ziglar-

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Chapter 5 : Least Square Adjustment

Tutorial
Question 1
Calculate the adjustment length AD and its estimated error given Figure 3 and the
observation data below

line Distance, m

AB 3.17
BC 1.12
CD 2.25
AC 4.31
AD 6.51
BD 3.36

Question 2
The use of least square adjustment principle is to solve the redundant
equations. From the equations below:

2 + = 21 + 1
24 − 6 = 11 + 2
4 − 2 = 20 + 3
i. State the number of variables and observation
ii. Calculate the variable by using the principles of least square
adjustment.

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Chapter 5 : Least Square Adjustment

Question 3
Using the conditional equation method, what are the most probable values
for the three interior angles of a triangle that were measured as.

station angle Std. dev
A 58° 14′ 56" 5.2”
B 65° 03′ 34" 5.2”
C 56° 40′ 20" 5.2”

Question 4

Calculate the variables for the elevations of B, C and D. Use the least square
adjustment method. Given the elevation of BM 1 is 40.213m

From To elevation Std. dev

A B 10.509 0.006

BC 5.360 0.004

CD -8.523 0.005

DA -7.348 0.003

B D -3.167 0.004

A C 15.881 0.012

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Chapter 5 : Least Square Adjustment
Reference
Abdul Wahid Idris dan Halim Setan. (2001). Pelarasan Ukur. Kuala Lumpur: Percetakan
Dewan Bahasa Dan Pustaka.
Azman Mohd Suldi dan Kamaluddin Hj Talib. (1994). Monograf: Penghitungan
penyelarasan. Shah Alam: UiTM.
D.Ghilani, c. (2010). Adjustment Computations: Spatial Data Analysis fifth Edition. New
Jersy: John Wiley & Sons, INC.
Khalid, H. F. (2003). Nota Program KPSL JUPEM. Ipoh: PUO.
Mikhail, Edward M. (1981). Analysis and Adjustment of Survey Measurements. Van
Nostrand Reinhold

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SURVEY ADJUSTMENT provides the students with knowledge on adjustment.
The book emphasizes the calculation of adjustment using the least square
adjustment method through observation and condition equations in solving
surveyed data. Besides, it is also provides students with knowledge and
practical skills to calculate and adjust surveyed data.

True value of measurement is unknown
Actual size of error is unknown

Errors exist in measurement data & computed
results