Department of Policy and Planning Sciences
Discussion Paper Series

No.1363

Spatial Analysis on Accuracy of Travelling
Distance on Networkby
ZHONG Dai, TAMURA Kazuki, OHSAWA Yoshiaki

November 2019

UNIVERSITY OF TSUKUBA
Tsukuba, Ibaraki 305-8573
JAPANSpatial Analysis on Accuracy of Travelling
Distance on Network
ZHONG Dai
Institute of Policy and Planning Sciences, University of Tsukuba
TAMURA Kazuki
Asian Growth Research InstituteOHSAWA Yoshiaki
Institute of Policy and Planning Sciences, University of Tsukuba

Abstract
It is usual that Traffic analysis deals with data aggregated at
representative points instead of dealing the whole existing data.
As a reason, there is an urgent need of expenditure and hightechnology to dealing with detailed data. Another reason is thatthe estimate of traffic of logistics or human is always linked withother socioeconomic analysis and the data, of which could be
more uncontrollable and changeable. Furthermore, in recent
years, from the viewpoint of protection of personal information,
the precision of the collected data may be dropped and cause theaggregation of the collected data. Therefore, in order to propose
a way to grasp the error between the two kinds of data, thiswork will build a model which represent the two kinds of datawith the continuous distance and discrete distance between two
points in the network to simulate the data before aggregated
and the aggregated data, and further discuss the gap of them,aiming at deriving the upper and lower limits of the gap.

Keyword：Continuous Distance，Discrete Distance，Traffic Data，Road
Network，Crofton's formula-2-

Contents

Cap.1.Introduction .............................................................................................- 4 Cap.2. Tow kinds of distance in the network......................................... - 4 Sec.2-1 Average distance between two points.................................. - 4 Sec.2-2 Antipodal and antipodal points ..................................................- 4 Cap.3.The gap of the two kinds of distance ..................................................- 5 Sec.3-1 the discrete average distance and continuous average distance 6Sec.3-2 Check of the deviation formula .....................................................- 8 Cap.4.The application of the general deviation formula ........................ - 13 Sec.4-1 tree .................................................................................................... - 13 Sec.4-2 lattice-network ............................................................................... - 14 Sec.4-3 the upper and lower limits of the general deviation formula- 15
Sec.4-4 Subdivision in the network ......................................................... - 15 Cap.5.The derive of gap by Crofton’s formula .......................................... - 17 Sec.5-1 tree .................................................................................................... - 18 Sec.5-2 lattice networks ............................................................................. - 19 Conclusion......................................................................................................... - 20 References ......................................................................................................... - 21 --3-

1.Introduction
As expounded in the Abstract, there are many reasons which could cause
gap between original data and analytical data. And they could be roughly
divided in two: the data could not be collected for some inevitable factors
such as time or economic cost; and the other one is that a part of data would
be discarded from collected data in order to keeping the data quality,
reducing the cost of processing or protecting the privacy for data acquisition
objects. Both of those reasons will have caused the aggregation of data. And
this work will build a model to study in the gap caused by aggregation.

2.Tow kinds of distance in the network
2-1

Average distance between two points

In this work, two kinds of methods are defined to measure the distance
between two points. The first one is to measure the average distance of two
points which is randomly distributed in the network. Dividing the networkin equal segments, and using midpoints to represent segments, the otherone is to measure the average distance between those midpoints which
represent those segments. And in all the figures of this work, all of therandomly distributed points are displayed in white points while those
midpoints are displayed as black points (fig.1).

Fig.1 continuous points and discrete points

As expounded in the introduction, in order to look into the error between
the data before aggregated and the aggregated data, this work transformthe error into a calculable mathematical problem. And research on thiserror could be able to be discussed by the analysis of the gap between thetwo kinds of methods to measure the distance between two points in the
network.

2-2

Antipodal and antipodal points

Before the introduction of the calculation, there is another conception
defined in this work should be clarified. Between two point distributed in-4-the network, there are always two pathways between them. If the twopathways have the same length, this pair of points are defined as antipodal
points. And if the segments in the network are completely consisted of
antipodal points, those pair of segments are defined as antipodal zones.
Further take consider of double count, this work defined 𝜆 as double of theamount of antipodal zones pairs.

Fig.2 antipodal points: path1=path2

For example, in the 1x2 lattice-network, there are 5 pairs of antipodal zones
while 𝜆 will be 10 in this situation.

Fig.3 1x2 lattice-network

Furthermore, as the premise of the simulation model, all of the segments
have the same length, and the pathways between two points are uniformly
and independent in the network. Finally, all the antipodal zones are allcontained by segments in networks.

3.The gap of the two kinds of distanceIn this chapter, we will discuss about the calculation of the discrete average
distance and continuous average distance. Then based on that we will
derive the general formula of the gap of the two kinds of distance.
Furthermore, in order to check the correctness of the formula, a series ofinspection will be conducted later.

-5-3-1 the discrete average distance and continuous
average distanceIn this work, it is assumed that the trip occurs in networks are uniformly
and independently. In the situation of the discrete average distance, eachsegment is considered as one segment the point which represent thatsegment is placed at the middle of this segment. Let 𝑙𝑖𝑗 be the shortest
distance between segment 𝑖 and segment 𝑗, the amount of the segments be
𝑘 , the average distance between those midpoints 𝑑̄ (discrete average
distance) will be presented as followed:𝑘𝑘1𝑑̄ ＝ 2 ∑ ∑ 𝑙𝑖𝑗𝑘𝑖＝1 𝑗≠𝑖

When it comes to the continuous average distance, the method of calculation
will be derived based on the result of discrete average distance. The average
distance could always be divided as two parts in whatever discrete average
distance or continuous average distance: the average distance between thepair of points which distributed in the same segment in the network (innerdistance). And the average distance between the pair of points which comefrom different segments in the network (approximately distance).
Firstly, in the situation of the calculation of the average inner-distance, fordiscrete average distance the inner distance will be zero while thecontinuous average distance is obviously not equal to zero (fig.4).

Fig.4 inner-distance

Supposed that there are two points(𝑥1 ,𝑥2 ) which are randomly distributedin the same segment which have a length of a . and the shortest distance
between them is 𝑑 (fig.5). And this distance could be represent in thecoordinate system like fig.6. Causing that the average distance of all the1
random-distribution pairs of points will be 3 𝑎 according to the fig.7. Hence,compared to the discrete average distance which the inner-distance equals
1
to 0, the continuous average distance will be longer by 3 𝑎 in every segment.

-6-

Fig.5 arbitrarily distribute two points

Fig.6 distance in coordinate system

Fig.7 continuous average distanceAnd for the calculation of the approximately distance, the pathway between
two points in networks are always be two, causing the shortest pathway
between antipodal points is always shorter than that of the distance
between two midpoints (fig.8).

Fig.8 antipodal points and aantipodal points

When calculate the approximately distance of continuous average distancebased on that of discrete average distance, take the base unit in latticenetwork for example. In the base unit, discrete average distance will be
2𝑎 while the continuous average distance in the whole unit will be a(fig.9).

Fig.9 discrete and continuous average distances in base unit
-7-

Taking apart the situation of the continuous average distance, thecontinuous average distance could be divided into 3 parts A, B and C(fig.10). And based on that the continuous average distance in part.A and1
part.B are 3 𝑎 and 𝑎 (referring to former calculation), and the occurrence111

accuracy of part.A, part.B and part.C are 4, 2, 4. Let 𝑥 be the continuous
average distance of part.C, the equation could be established as followed:
1 1
1
1𝑎＝ × 𝑎 + × 𝑎 + × 𝑥
4 3
2
4
It is not complicated that the continuous average distance of part.C equals
5
to 3 𝑎, which means that the continuous average distance between the pair5of points which exist in antipodal segments is 3 𝑎. Hence, compare to thediscrete average distance which the approximately distance equals to 2𝑎,1
the continuous average distance will be shorter by 3 𝑎 in every antipodal
links pair.

Part.A
Part.B
Part.C
Fig.10 continuous average distance divided in 3 parts

Based on the result of incremental portion and reduce portion which has
been calculated, the continuous average distance 𝑟̄ could be presented asfollowed:1
𝜆𝑟̄ ＝𝑑̄ +
𝑎− 2𝑎
3𝑘
3𝑘
Furthermore, the general formula of deviation could be stated as followed:𝑘−𝜆𝐺 ≡ 𝑟̄ − 𝑑̄ ＝𝑎
3𝑘 2

3-2 Check of the deviation formula
3-2-1 tree
Take the tree-network which has three segments A, B, C with the length of
𝑎 for example.
In fig.11(a), midpoints represent the segments, and the distance between-8-them are represented in Table.1(a). From this table, we can calculate the
6𝑎2
average of them is 32 ＝ 3 𝑎 . In the situation of the calculation of thecontinuous average distance in this tree, the distance of the trips happen in
every segments are represented as Table.1(b), and causing the average
1 1
7
equals to 32 (3 𝑎 × 3 + 6𝑎) ＝ 9 𝑎. By the way this work mark the increased
parts of continuous average distance compared with discrete average
distance as red while the reduced parts of continuous average distancecompared with discrete average distance as blue in the calculation table.
1
The gap between them came out to be 9 𝑎 , which accordant with that
𝑘−𝜆

3−0

1

calculated by general deviation formula: 𝐺＝ 3𝑘 2 𝑎＝ 3×32 𝑎＝ 9 𝑎the number of the pairs of antipodal zones is 0.

because(a)(b)
Fig.11 tree-network with 3 segments(a)(b)
Table.1 discrete and continuous average distances in tree-network with 3
segments

3-2-2 circle
By the same token, in the circle which is divided as two equal parts. While
𝑘＝2 and 𝜆＝2 the gap calculated by general deviation formula is 𝐺＝
𝑘−𝜆
3𝑘 2

2−2

𝑎＝ 3×22 𝑎＝0

-9-

(a)
(b)
Fig.12 circle divided in two segments(a)(b)
Table.2 continuous and discrete average distances in circle

And according to Table.2, the gap between two kinds of average distancecomes to be 0, witch is accordant to the result from general deviation
formula.

3-2-3 lattice-network
In the situation of 1 2 lattice-network of which every segment have alength of a .(fig.13)

Fig.13 1x2 lattice-network

According to the calculation in the Table.3, the gap between them is 𝐺＝
65

66

1

𝑘−𝜆

7−10

𝑟̄ − 𝑑̄ ＝ 49 𝑎 − 49 𝑎＝ − 49 𝑎. In the 1 × 2 lattice network, 𝑘＝7 , 𝜆＝10, andthe result of the calculation through the general deviation formula comes1

out to be 𝐺＝ 3𝑘 2 𝑎＝ 3×72 𝑎＝ − 49 𝑎.

- 10 -

(a)(b)

Table.3 discrete and continuous average distances in 1x2 lattice-network

3-2-4 general lattice-network
Furthermore, take the general lattice-network raised by Koshizuka[2] forconsideration(fig.14).

Fig.14 general lattice-network

According to the function graph of the distribution of the average distance
10𝑏∫𝑟𝑓(𝑟)𝑑𝑟on the network, continuous average distance come to be 𝑟̄ ＝ 0 (36𝑎)2
631＝

𝑎 . (𝑓 (𝑟) represent the function of the distribution of the continuous
average distance on networks(fig.15).
162

Continuous average distanceFig.15 continuous average distance derived by Koshizuka- 11 -

According to the symmetry of the lattice-network, discrete average distance1263
1
could be figure out and 𝑑̄ ＝𝑎, causing the gap of them 𝐺＝ −
𝑎. In
324

324

this lattice-network, 𝑘＝36, 𝜆＝48 and the result of the calculation by the
𝑘−𝜆

36−48

1

general deviation formula 𝐺＝ 3𝑘 2 𝑎＝ 3×362 𝑎＝ − 324 𝑎.

3-2-5 hexagon-network
In the hexagon-network (fig.16) the inspection will be conducted similarly.

Fig.16 hexagon-network

In this situation, let’s confirm the𝜆firstly. In the surrounding segments of
the hexagon-network, there are six pairs of antipodal zones in eachsurrounding segments, 𝑎 − 𝑎1 , 𝑎 − 𝑎2 , 𝑎 − 𝑎3 and 𝑏 − 𝑏1 , 𝑏 − 𝑏2 , 𝑏 − 𝑏3 .
And in the inner segements, there are six pairs of antipodal zones in each
inner segments, 𝑐 − 𝑐1, 𝑐 − 𝑐2, 𝑐 − 𝑐3 and 𝑑 − 𝑑1, 𝑑 − 𝑑2 , 𝑑 − 𝑑3 (fig.17) .
and according to that, by the general deviation formula, the gap between
𝑘−𝜆

24−72

1

the two kinds of distance is 𝐺＝ 3𝑘 2 𝑎＝ 3×242 𝑎＝ − 36 𝑎.

Fig.17 antipodal zones in hexagon-network
187

When calculate the discrete average distance for inspection, 𝑑̄ ＝𝑎 could72
be derived by symmetry. And in the situation of continuous average distance,- 12 -

similarly take the base unit as explanation for that.

Fig.18 base unit of hexagon-network

In the triangular base unit of the hexagon-network(fig.18), the continuous
average distance is 𝑎 in the whole segments. And based on the result of thecontinuous average distance in the square, the continuous average distancein a single triangular unit 𝑟̄1 could be calculated as followed:
1 1
1
𝑎 𝑎
1
2 𝑎
23
𝑟̄1 ＝ × 𝑎 + × ( + ) + × (𝑎 + × ) ＝ 𝑎
4 2
2
2 2
4
3 2
24
1 1 1
,
,
are represent the probability of occurrence of the parts showed
4 2 4
in fig.19

Fig.19 continuous average distance divided in 3 parts

Based on that, 𝑟̅ in a hexagon-network could be calculated as

187
72

𝑎. and1

the gap between them will be 𝐺＝𝑟̄ − 𝑑̄ ＝ − 36 𝑎, which is exactly the samewith the result from the general deviation formula.
According to the check of the general deviation formula above, we could
reach to the conclusion that the general deviation formula could apply forthe tree, lattice, hexagon networks which have the same length as havebeen analyzed in examples.

4.The application of the general deviation formula
4-1 treeIn the shape of tree-networks, there will not have any pairs of antipodal
zones, which means that the general deviation formula for tree-networks
1
will be 𝐺＝ 3𝑘 𝑎.
- 13 -

From the general deviation formula, it can be tell that the deviation in treenetworks are only related with the numbers of segments. And the deviationis unrelated to the shapes of the tree-networks as long as they have thesame number of segments(fig.20). And the deviation in tree-networks go to
be underestimation.

Fig.20 tree-networks, 𝑘 = 9

4-2 lattice-network
In the general lattice-network, let the scale of the network be 𝑚 × 𝑛 (fig.21),
the pairs of antipodal zones:22
𝜆＝𝐶𝑛+1× 𝑚 × 2 + 𝐶𝑚+1
× 𝑛 × 2＝𝑚𝑛(𝑚 + 𝑛 + 2),

And the number of segments:
𝑘＝(𝑚 + 1)𝑛 + (𝑛 + 1)𝑚＝2𝑚𝑛 + 𝑚 + 𝑛
The general deviation formula for m n lattice-network will be(𝑚 + 𝑛)(1 − 𝑚𝑛)
𝑘−𝜆
𝐺＝
𝑎＝
𝑎
2
3𝑘
3(2𝑚𝑛 + 𝑚 + 𝑛)2
From the formula, it could be concluded that the deviation of latticenetworks go to be overestimation.

Fig.21 𝑚 × 𝑛 lattice-network- 14 -

4-3 the upper and lower limits of the general deviation
formulaFrom the general deviation formula, the upper and lower limits could bederived out as:
𝑘−2
1
−
≤𝐺≤
𝑎
3𝑘
3𝑘
And when remove the 𝑘 from the general deviation formula, the upperand lower limits go to be as followed:
1
1− 𝑎<𝐺≤ 𝑎3
3
In that, the upper limit will be acquired as the numbers of the segments
are 1 and the pairs of antipodal zones are 0, which showed up as a 1segment tree, referring to fig.22(a). And in the other hand, the lower limit
will be acquired as there are infinity segments between two points,
referring to fig.22(b).

(a)

(b)
Fig.22 upper and lower limits

4-4 Subdivision in the network
When subdividing the segments in networks(fig.23), If s represent the
number of the divided segments, the antipodal links pairs in the network
will turn to 𝑠 times compared with the number of antipodal links pairs
when 𝑠 = 1. So the general deviation formula will change to be as
followed:
1
𝐺𝑠 = 2 𝐺𝑠

Fig.23 s=1, s=2, s=3

From this formula, the conclusion could be draw that the deviation- 15 -

between the discrete average distance and continuous average distancecould decrease as the segments in the networks being subdivided.
So with the help of ArcMap10.6，the distance could be derived under the
network analysis tool box to confirm the calculations. So the simulation
results will be showed in next.
0.8

Average distance(a)

0.78
0.76
0.74

7

Asymptote 𝑟̅ = 9 𝑎

0.72
0.7

𝑑̅ =

0.68

2
𝑎
3
The times of subdivision

0.66
1

3

5

7

9
2

7

Fig.24 the average distance in tree(k=3), 𝑑̅ = 3 𝑎, 𝑟̅ = 9 𝑎

1.35

Average distance(a)

𝑑̅ =

1.345

66
𝑎
49

1.34
1.335

7

Asymptote 𝑟̅ = 9 𝑎
1.33

The times of subdivision

1.325
1

2

3

4

5

6

7

66
65
Fig.25 the average distance in lattice-networks(1x2), 𝑑̅ = 49 𝑎, 𝑟̅ = 49 𝑎

- 16 -

1.44

Average distance(a)

1.42
𝑑̅ =

1.4

17
𝑎
12

1.38
1.36
1.34

187

Asymptote 𝑟̅ = 144 𝑎

1.32
1.3
The times of subdivision

1.28
1

3

5

7

9

17
187
Fig.26 the average distance in hexagon-networks, 𝑑̅ = 12 𝑎, 𝑟̅ = 144 𝑎

5.The derive of gap by Crofton’s formula
In the chapter.3, the general deviation formula has been derived asfollowed:𝑘−𝜆𝐺 ≡ 𝑟̄ − 𝑑̄ ＝ 2 𝑎
3𝑘(1)

And in this chapter, this work will derive the general deviation formula by
the using of Crofton’s formula.
Increments will be joint in every segment in networks. Let the length ofincrements as 𝛥 in each segment，and A represent the original parts as B
represent the increments(referring to fig.27). 𝜇𝐴𝐴 , 𝜇𝐴𝐵 , 𝜇𝐵𝐵 respectively
represent the inner-distance in part.A, the approximately distance
between part.A and part.B, the inner-distance in part.B. Based on that, let
𝜇 (𝑥 ) represent the inner-distance in segment which have the length of 𝑥.
For 𝑟̄ = 𝜇(𝑎):1(𝑎2 𝜇𝐴𝐴 + 2𝑎𝛥𝜇𝐴𝐵 + 𝛥2 𝜇𝐵𝐵 )𝜇 (𝑎 + 𝛥 ) =(𝑎 + 𝛥 )2
could be derived.
Furthermore, if the cross-section traffic volume from part.A to part.B is 𝑏,
the inner-distance in part.A could be written as 𝜇𝐴𝐴 = 𝜇(𝑎) + 𝑏𝛥. and theequation could be changed into as followed:
𝜇 (𝑎 + 𝛥 ) − 𝜇 ( 𝑎 )1= 2 (2𝑎𝜇𝐴𝐵 + 𝛥𝜇𝐵𝐵 − (2𝑎 + 𝛥)𝜇(𝑎 + 𝛥)) + 𝑏
𝛥𝑎- 17 -

Let 𝜇1 be the average distance from the increased junction 𝑥 to the endof this original part, and the approximately distance between part.A andpart.B will be written as 𝜇𝐴𝐵 = 𝜇1 + 𝑂(𝛥), and the following formula could
be derived:
𝑑𝜇(𝑎)
𝑑𝑎

2

(2)

= 𝑎 (𝜇1 − 𝜇 (𝑎)) + 𝑏

For the coefficient b, if the increments are always be inserted in the end of
the segment, the cross-section traffic volume from part.A to part.B will be
0. thus the equation could change into as followed:
2

𝑏

(3)
In segment which has length of a referring to fig.24, there is increment11
2
in the junction 𝑥, thus 𝜇1 = (𝑥 2 − 𝑎𝑥 + 𝑎2 ), 𝑏 = 2 (𝑎𝑥 − 𝑥 2 ), based on
𝜇 (𝑎) = 3 (𝜇1 + 2 𝑎)

1

𝑎

2

equation.(3), 𝜇(𝑎) = 3 𝑎 could be derived.

𝑎

𝑎

1

1

Furthermore, in the circle likes fig.28, 𝜇1 = 4, 𝑏 = 4, thus 𝜇(𝑎) = 4 𝑎based on equation.(3).

Fig.27 increment in line

Fig.28 increment in circle

5-1 treeIn tree-network which has k segments, the average distance from the𝑎midpoint to the rest parts comes to be 4 , and in the whole network, all of
the average distance from the midpoints to the rest part of the treenetwork go to be as followed:𝑎
𝜇1 = 𝑑̄ + 4𝑘
(4)

Fig.29 increment in tree-network- 18 -

Referring to fig.29, there are increments in the middle of every segment.
By the same token, let the original part be part.A and the increased partas part.B, the cross-section traffic volume from part.A to
1part.B(represented as b) will be 2𝑘 when the movement is in the same𝑑2

segments, and will be 𝑎 when the movement taking place in different
segments. Based on the average distance between the midpoints 𝑑̄ , thecross-section traffic volume from part.A to part.B:
1𝑎𝑏 = 𝑎 (𝑑̄ + 2𝑘)(5)
And based on equation.(3), (4), (5), 𝜇 (𝑎)could be derived as followed:𝑎𝜇 (𝑎) = 𝑑̄ +3𝑘Which is the same with the result of the general deviation formula.

5-2 lattice networks
By the same token as fig.29, increments are inserted in the middle ofevery segment, and the original part as part.A while the increased part as
part.B(fig.30).

Fig.30 increment in general network

When consider the average distance from every midpoint to the rest of the
network, since the exist of antipodal links pair(fig.31), the average𝑎distance will decrease by 4 .

Fig.31 reduced part in average distance- 19 -

And based on equation.(4) and the average distance between midpoints,
the average distance from every midpoint to the rest of the network 𝜇1
could be derived as followed:
𝜇1 = 𝑑̄ +(𝑘−𝜆)𝑎4𝑘 2(6)1

And the the cross-section traffic volume b in the part.A will decrease by
2(fig.32), thus b will change into as followed:(𝑘−𝜆)𝑎1
𝑏 = 𝑎 (𝑑̄ + 2𝑘 2 )(7)
Based on equation.(5).

Fig.32 reduced part in cross-section traffic volume

And finally cased on equation.(3), (6), (7),𝑘−𝜆𝜇(𝑎) = 𝑑̄ +𝑎
3𝑘 2
could be derived and the result is coincident with the equation.(1)Conclusion
In this work, the general deviation formula has been derived and it proved
the relationship between the discrete average distance and continuous
average distance in networks. By analysis, it can be concluded that by
calculating the discrete average distances in the network, the continuous
average distance can also be derived. And in chapter.5 the generaldeviation formula is also derived by the using of Crofton’s formula. thus
the Crofton’s formula is applicable in situations in continuous spaces suchas line segments, circles, circular rings, ellipses, etc. [8] [9] [10]. And
further discussion such as derivation of other moments about distance andprobability density function will be conducted in next stage.
- 20 -

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