美赛2011B统稿(第12次)

Geography Segregation and Frequency Assignments: Modeling the Distribution

Schemes for Repeaters

Team # 10731 February 14, 2011

Summary

The use of repeaters is a cheap, effective method of realizing wider signals coverage. A natural question for today‟s communication engineers is “how to meet all the requirements in a given area using the minimum number of repeaters”. Nowadays, with the use of CTCSS, each repeater will be associated with a specific PL tone to help decision-making on signal selection, and thus to reduce interferences to the lowest level. Using mathematics, it is possible to distill the relationship between essential parameters and to find the optimal repeater distribution scheme for 1000-user case, 10000-user case and the mountainous area case.

We considered mainly two aspects to work out the distribution schemes: geography segregation and frequency assignments. To obtain the scheme based on the geography segregation, we assume that users are evenly distributed in the given circular area and signals transmitted by repeaters are strong enough to cover the whole area. Thus, the distribution of repeaters is determined by r0, the longest distance signals can reach from a user to a repeater. With this in mind, the goal of the geography model was simplified as full coverage of the 40-mile radius circular area by a minimum number of small circles with a radius of r0. Inspired by honeycombs and coverage calculation in satellite engineering, we proposed two algorithms and derived two corresponding distribution schemes. To get the scheme based on frequency assignments, we made an assumption that all the users have a good knowledge of the PL tones in all the sub-areas. Then, we divided the spectrum into several frequency pairs, calculated the maximum number of frequencies a repeater can respond simultaneously and worked out a corresponding distribution scheme. Finally, we made a comparison of these distribution schemes and chose the best one.

In the case of 1000 simultaneous users, the geography factor is decisive in determining the minimum number of repeaters. Our results show that there must be at least 43 repeaters. When it came to the 10000-user case, because of inadequate PL tones, the factor of frequency assignments plays a more important role. We proposed two modified models based on frequency assignments. Our results show that 46 or 54 repeaters are needed in the two modified models, respectively. As an extension of our model, we discussed the mountainous area case. The negative impacts of mountains on repeater distributions were analyzed and two solutions were proposed: putting repeaters on top of the mountains to overcome the shelter of signals, and putting some additional repeaters on slopes or bottoms of the mountains to expand the repeaters‟ effective areas.

The main advantage of our models is that they are very simple and allow us to analyze the distribution problem under the consideration of both geography segregation and frequency assignments. Moreover, our models can be generalized to cases with much more users. However, there are inevitable disadvantages in our models such as their inability of analyzing the cases in which users do not pre-know the specific PL tone in each sub-area.

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Table of Contents

1 Introduction ............................................................................................ 2 2 Assumptions............................................................................................ 2 3 Modeling the 1000-user case ................................................................. 3 3.1 Distribution scheme based on geography segregation ............................... 3 3.1.1 Calculation of small circle radius ...................................................... 3 3.1.2 Coverage of the circular area ........................................................... 4 3.1.2.1 The Hexagon Method ................................................................ 4 3.1.2.2 The Ring Method ...................................................................... 5 3.1.3 Comparison and conclusions ............................................................ 9 3.2 Distribution scheme based on frequency assignments ............................... 9 3.2.1 Additional assumptions .................................................................... 9 3.2.2 Descriptions of a typical repeater ...................................................... 9 3.2.3 Analysis of the spectrum ................................................................ 10 3.2.4 Distributing repeaters .................................................................... 10 3.3 Conclusions ........................................................................................ 12 4 Modeling the 10000-user case ............................................................. 12 4.1 Distribution scheme based on geography segregation ............................. 13 4.2 Distribution scheme based on frequency assignments ............................. 13 4.3 Model modifications ............................................................................ 14 4.3.1 The method of sharing frequency ................................................... 14 4.3.2 Reducing transmitting frequencies ................................................ 15 4.4 Conclusions ......................................................................................... 17 5 Extension: model for the mountainous-area case ............................. 17 5.1 Analysis of mountainous areas ........................................................... 18 5.2 Solutions to the problems ..................................................................... 18 5.3 Conclusions ......................................................................................... 19 6 Conclusions ........................................................................................... 19 6.1 Strengths ............................................................................................ 19 6.2 Weaknesses ......................................................................................... 20 7 References ............................................................................................. 21 8 Appendix ............................................................................................... 22

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1 Introduction

Someone compared the use of a repeater as a drinking spot for marathon runners: when a signal is send to a repeater, it picks up a cup of water, drink it and then run further. „A repeater is an electronic device that receives a signal and retransmits it at a higher level and/or higher power, or onto the other side of an obstruction, so that the signal can cover longer distances‟ [1]. By the help of repeaters low-power users can communicate with one another in situations where direct user to user contact would not be possible.

With the rapid growth of repeaters, communication engineers noticed that in case where too many repeaters were built in a limited area, serious problem may arise. Some repeaters were built too close while others may work on the same frequencies, both of which will cause conflicts and interference. Till now, this problem hasn‟t been completely solved, as can be proved by “the California repeater wars” that happened in Oct.10th, 2010[2]. Therefore, some engineers began to think of distributing repeaters in a properly organized form within a limited area. Thus, a natural question that today‟s communication engineers are faced with is that, “How do we meet all the requirements in a given area using the minimum number of repeaters?”

The purpose of our paper is to use mathematical ways to derive the best distribution schemes in which all the users‟ signals can be responded by minimal repeaters ,meanwhile ,the interference is reduced to the lowest level. This paper is mainly divided into four sections as follows:

Section One deals with the modeling of 1000-user case. We proposed a model in this section that considered the distribution problem from the aspects of geography segregation as well as frequency assignments. The result of 1000-user case is given at the end of this section. In Section Two, we analyzed the 10000-user case in depth and modified the model we build in Section One. Section Three discussed the mountain area case and gave a series of steps to distribute repeaters. Concluding remarks are presented in Section Four.

2 Assumptions

The following assumptions were used in our models:

Working within CTCSS: All the distribution schemes we got in our models were based on the assumption that, the objects we studied, which includes users and repeaters are all working within the “continuous tone-coded squelch system”. Thus, each user in the area has a specific PL tone; each repeater will also be associated with a fixed PL tone.

Strong transmitters: The power of a repeater‟s transmitter is strong enough for the transmitted signals to cover the whole area while that of users are much weaker. Therefore, what we are focused on was how lone the user‟s signal can travel. In other words, how large is a repeater‟s effective area is.

Working conditions: We didn‟t take into considerations the possibilities that the repeaters or user devices may break down. In our models, all the facilities are with

Team #10731 Page 3 of 22 excellent performances.

Distribution of users: For the purpose of simplifying our models, we assume that the users are distributed evenly in the given circular area.

Pre-known of the PL tones: Users always have a good knowledge of the PL tones in all the sub-areas that are divided by the effective area of repeaters. As the positions of users are not fixed, when they travelled from one to another sub-area in which their original PL tones will not be recognized, they will change them wisely by themselves.

3 Modeling the 1000-user case

To determine the minimum number of repeaters in the case of 1000 simultaneous users, we have to take into consideration both the geographical separation and frequency assignment. On the one hand, the distance between repeaters and users will greatly affect the number and position of repeaters in the given area. On the other hand, the maximum number of users a repeater can respond is also a decisive factor in determining the distribution schemes.

Thus, we firstly figured out a distribution scheme with minimum number of repeaters by considering only the geometrical separations. Then, we focused mainly on the frequency assignment and worked out another distribution scheme. Finally, we made a comparison between the above two types of schemes and chose an optimal one that meets the all the requirements.

3.1 Distribution scheme based on geography segregation

For the purpose of distributing repeaters economically in the circular flat area of 40 miles radius, the first step we have to take is to cover this circular area with minimum number of small circles. The center of each circle can be seen as a repeater position and the radius is determined by the longest distance a user‟s signal can reach.

Therefore, we obtained firstly the radius of small circles, and then adopted 2 algorithms to cover the circular area with these small circles. After this, we derived the number of repeaters by counting the minimum number of small circles that can fully cover the circular area.

3.1.1 Calculation of small circle radius

In engineering practice [3], the transmission loss of signals (LW) determined by

LW=SG (1)

Where SG represents the system gain, which indicates the maximum reduction of signals when they reach the corresponding repeater.

SGUTUA(RACLRS) (2)

UT: Transmitting power of the user; UA: Antenna gain of the user; RA: Antenna gain of the repeater; CL: Cable loss;

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RS: Sensibility of the repeater;

We collected the necessary data [4] and made a list as follows.

Table 1: Necessary Data in Calculation of SG Denotation Value UT 5W UA 0dBi RA 9.8dBi CL 2dB RS -116dBm In a flat area with VHF spectrum band from 145MHz to 155MHz, LW can be calculated using the following formula [3].

Lw88.120log(F)20log(h1h2)40log(r0) (3)

In which Fdenotes the frequency of the signal, r0 represents the longest distance that signals can reach, h1 and h2 is the heights of repeater and user antennas. Usually we take h130mand h21.5m[4].

Combine Eq. 1, Eq. 2 and Eq. 3, we get

Lw88.120log(F)20log(h1h2)40log(r0)UTUA(RACLRS)SG (4) Substitute all the necessary data into the Equation above, we get

r011.8km

Thus we have figured out the radius of small circles. 3.1.2 Coverage of the circular area

Now that we have a circular area with a radius of 40 miles, in which 1000 simultaneous users are evenly distributed, what we have to do is to distribute repeaters in this area in an appropriate way so that all the users‟ signals can be responded using minimal number of repeaters.

To achieve our goal, we simplified this problem as a geometrical problem:

How to fill a large circle with minimal amount of small circles?

In this problem, the large circle can be seen as the circular area with radius of 40 miles (about .37km) and the small circles with radius of 11.8km represent the area in which a repeater can work effectively.

Solving this problem perfectly is rather a big challenge for us. Luckily, we obtained inspirations from the honeycomb and satellites and proposed two algorithms: the Hexagon Method and the Ring Method. 3.1.2.1 Hexagon Method

The famous mathematician, Pappus of Alexandria (c. 290 – c. 350), proposed a world-known conjecture [5]: honeycomb conjecture. He stated that a regular hexagonal grid or honeycomb is the best way to divide a surface into regions of equal area with the least total perimeter. This conjecture was later proved by mathematician Thomas C. Hales [6].

Team #10731 Page 5 of 22 Inspired by this conjecture, we come to an idea of using inscribed hexagons to represent the small circles and then to divide the large circle with them.

Our idea proved reasonable by classic mathematical theory [7]: “Like squares and equilateral triangles, regular hexagons fit together without any gaps to tile the plane.” If we fill a large circle with equilateral triangles, equilateral triangles or hexagons as shown in Fig.1, respectively, the public areas of circumscribed circles in the three cases are different. We made a calculation and listed the results in Table.2:

Figure 1: Filling a Circle by the Hexagon Method

Table 2: The Relative Public Area of Three Figures’ Circumscribed Circle Figure The public square measure of circumscribed circle equilateral triangles 39.1% square 18.16% hexagon 5.77% From table.2 we can see clearly that hexagon is the best choice as the substitution of its circumscribed circle to fill the large circle.

With the help of AutoCAD, we worked out the result and show them in Fig.2:

Figure 2: the Distribution Scheme Using the Hexagon Method

As can be seen in Fig.2, using the Hexagon Method, at least 43 repeaters should be distributed in the area to meet the demand of simultaneous 1000 users.

3.1.2.2 Ring Method

Team #10731 Page 6 of 22 In the Hexagon Method, we filled the large circle from its center point and then add hexagons one by one to realize coverage to the boundary. This will result in a waste of area just outside the boundary. As we can see in Fig.2, the hexagons along the boundary are in very awkward positions: we can‟t fill the large circle without them while the area they contributed is rather small.

To avoid this problem, we decided to cover the boundary area firstly. However, it‟s a rather hard work to determine the position of small circles on the boundary if we go on using the same thought as that used in the Hexagon Method. Thus, we adopted the experience of satellite engineers and proposed the Ring Method.

As we all know, the signals a satellite send periodically from sky can be received in a circular area on the ground. When the satellite is moving, the satellite coverage area becomes a series of circles, as is shown in Fig.3. To simplify the calculation of the area where the signals can reach on the ground, engineers usually take the rectangular part marked in shadowed with red color in Fig.3 as the satellite coverage area.

Figure 3: the Effective Area for Calculation

Inspired by this idea, we thought of bending the rectangle, making it end to end to form a ring and then putting it in the large circle, just as shown in Fig.4. In order to testify the effectiveness of this method, we need a deeper insight into it.

N rings situation

1 ring situation

2 rings situation

Figure 4: Filling a Circle by the Ring Method

At the first glance, the number of small circles is affected by 2 factors: the number of rings and the number of small circles on each ring. However, the 2 factors are not independent of each other: (Ⅰ) the number of rings depends on their width. The narrower the rings are, the more rings we need; (Ⅱ) the width of rings depends on the distance between the centers of two circles

Team #10731 Page 7 of 22 nearby. In the same ring, the nearer two small circles are, the lower the width is. (Ⅲ) the distance will affect the number of small circles in each ring.

Thus, what we should do next is to figure out the relations between those parameters and to build a mathematical model to decide the distribution of small circles.

r0 d

L

Figure 5: Necessary Variables for the Model

We denote N1 as the total number of small circles in the large circle, n1 as the

number of rings and R0 as the radius of the large circle. Assume that there is a small circle whose center is the same as that of the large circle and denote r1 as its radius. Then we build a model to explore the relationship between N1 and other parameters, and thus determine the minimum number of N1:

minN1d2L22(2)(2)r0rRR0dr00d1St:Rn10dn12kdN(r)10L2k1n10;d0;r10;L0 (5)

With the help of MATLAB (corresponding program in Appendix 8.1), we figured out

the relationship between the width of rings (d) and the total number of small circles

Team #10731 Page 8 of 22 (N1) , as shown in Figure :

350 300250200N1150100500 0510d152025

Figure 6: the Result of Eqs.5

In Fig.6, it is clearly shown that, as the width of rings (d) increases, the number of small circles (N1) will change following the U-shaped trend. At the red point whered16, N1 reaches its lowest value 43.

Table 3: the Value of the Variables at the Red Point in Fig.6 n1 r1 L d N1 4 0 17.35 16 43 With the result listed in Table.3, we got the repeater distribution, as shown in Fig.7. Fig.7 confirmed the result of our model that at least 43 repeaters should be distributed in the area.

Figure 7: the Distribution Scheme Using the Ring Method

Team #10731 Page 9 of 22 3.1.3 Comparison and conclusion

In3.1.2 we discussed two algorithms: the Hexagon Method and the Ring Method. From what we have discussed, we can come to the conclusion that both methods led to the same result: the minimum number of repeaters is 43.

But the results of these two algorithms differ on how to position each repeater in the same area this difference can be easily found if we compare Fig.2 with Fig.7. Taking a deeper insight of the 2 scheme, we hold the opinion that, on the one hand, the position scheme of the Hexagon Method has an advantage in reducing waste and interference, as the superposition area is much smaller than that of the Ring method. On the other hand, we can easily find in Fig.2 that the positions of 6 repeaters are not in the given area with a radius of 40 miles. This may lead to some political and economical troubles in management while this problem is avoided by adopting the Ring Method.

3.2 Distribution scheme based on frequency assignments

3.2.1 Additional assumptions

In order to formulate a tractable analytic model based on available frequency resources, the following assumptions were also made:

Fixed channel spacing: To avoid interference, different users in a limited area usually have to send signals on different frequencies. In case of narrow band spectrum, the issue of frequency interval can never be ignored. However, regulations about channel spacing differ in different regions around the world. The channel spacing ranges from 15 kHz to 20 kHz and is decided by the local coordination policies [8]. In our paper, we take the policies of Colorado and assume that the channel spacing between different users is fixed as 15 kHz: a user will not send signals using a frequency the same as another user nearby. In other words, the frequency interval between adjacent frequency allocations is 15 kHz.

User Selection based on PL tones: As is assumed in (1), under the CTCSS system, all the users have a specific tone [9]. Thus, PL tone plays an important role in the user selection. We assume that a repeater will only responds to received signals with a specific PL tone that is associated to it initially. That is to say, a repeater will select signals only basing on their PL tone, for example, “PL 1A”. Their working frequency will not affect the user-selection process. 3.2.2 Descriptions of a typical repeater

Currently there are many different working manners for repeaters around the world. For example, some repeaters may receive signals with any frequencies and transmit them with corresponding signals while others may transmit all the signals on the same frequency.

In this part of our paper, a typical repeater receive signals with special PL tones (they may have different frequencies) and transmit the received signals on frequencies 600 kHz below or above the original frequencies. A typical working process of a repeater is shown in Fig.8

Team #10731 Page 10 of 22 145MHz, PL=1A (145+0.6)MHz, 145.015MHz, PL=1A (145.015+0.6)MHz, 145.015MHz, PL=2A Repeater (PL code=1A)

Figure 8: the Typical Working Process of a Repeater

3.2.3 Analysis of the spectrum

Now that we have frequencies ranging from 145MHz to 148MHz at hand, we can figure out the exact number of frequencies available for users.

Repeater Repeater User frequency User frequency Waste frequency frequency frequency 146.8MHz 145.6MHz 145MHz 146.2MHz 147.4MHz 148MHz

Figure 9: Divisions of Frequencies Available

As can be seen in Fig.9, if users send signals on frequencies ranging from 145 to 145.6MHz, then they will not be allowed to take use of frequencies from 145.6 to 146.2 as these frequencies have been used by the transmitter of a repeater. So are the frequencies between 146.8 and 147.4MHz. For signals with frequencies between 147.4 and 148 MHz, as repeaters can transmit them neither in higher nor in lower frequencies, these frequencies are wasted. In a word, there are only a few frequencies available for users.

3.2.4 Distributing repeaters

For the case shown in Fig.9, we can figure out the maximum number of frequencies which can be associated to users with the same PL code. In other words, the maximum number of signals for a repeater to response is:

fmaxfmin(145.6145)103(146.8146.2)103n80 (6)

finterval15If all the repeaters are working at full capacities to accommodate as many as 1000

simultaneous users, then the minimum number of repeaters needed in frequency space can be calculated theoretically as follows:

N1000]13 (7) nmin[][n80We denote r0' as the maximum radius for a repeater to take effect and R as the radius of the circular area, and get

nminr0'2R2 (8)

Thus,

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r0'17.75

What we should emphasize here is that, practically 13 repeaters whose maximum radius is 17.75km can‟t fill the circular area at all as their coverage area are all circles. But the work we have done above does provide us with the maximum value of r0' with which we can figure out a more precise answer by the help of the Ring Method. By modifying Eq.5, we have

minN1d2L2'2()()r022'rr00Rn10 (9) dSt.rR0n1dr010d2r0n12kdN(r)012k1LL0What made our new model differ from Eq.5 is that in this case, there are 2 variables:

r0' and d in the new model while in Eq.9,N1 the total number of small circles in the large circle can be decided all by one variable.

With the help of MATLAB (The corresponding Program in Appendix 8.2), we figured out the relationship betweenN1, r0' andd, the result are shown in Fig.10 below.

Figure 10: the Result of Eqs.9

Team #10731 Page 12 of 22 A deeper insight into Fig.10 led to the following conclusions:

(Ⅰ) the value of N1 will decrease with the increase ofr0'. The larger r0' is, the smaller N1 is.

(Ⅱ) N1 will change in a “U” trend with the increase of variabled.

By analyzing the results we found that N1 will get its smallest value 23 at the point where d21.7 andr0'17.65.

The distribution scheme based on frequency assignments is shown in Fig.11

Figure 11: the Distribution Scheme Based on Frequency

3.3 Conclusions

In the distribution scheme based on geography segregation, the minimum number of repeaters is 43 while in the scheme based on frequency assignments, the number is only 23. Thus we can come to the conclusion that, geographical factors play the main role in affecting distributing schemes in case of 1000 simultaneous users. In the flat area with a radius of 40 miles, we have to build at least 43 repeaters.

4 Modeling the 10000-user case

In this case, the circular area doesn‟t change while the number of users is 10 times larger than that of the 1000-user case. The boost in population may lead to some problems especially in assignments of frequencies.

To model the 10000-user case, we carried out our analysis based on two factors as in the previous section: geometrical analysis and frequency assignments. Then we discussed the impact caused by population change, proposed some solutions and modified our model to fit for this new case.

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4.1 Distribution scheme based on geography segregation

During the process of modeling the 1000-user case, we adopted the method of covering the circular area with small circles. If the radius of the small circles is determined, the minimum number of circles that can fully cover the area will also be determined. It is obvious that the radius of small circles is related only to the power of signals, so the radius of small circles will not change at all in the 10000-user case. Therefore, the distribution schemes we get in the 1000-user case based on geography segregation is still effective, there should be at least 43 repeaters available.

4.2 Distribution scheme based on frequency assignments

In order to have a clearer insight into the influence brought about by the additional 9000 users, we substituted 1000 by 10000 and used the same method we adopted in 3.2.

N10000nmin125 (10)

n80nminr0'2R2 (11)

Thus,

r0'5.72km

Substitute the data above into the model that we have built by the Ring Method, we get the relationship betweenN1, r0' andd, as shown in Fig.12:

Figure 12: the Relationship betweenN1, dandr0

From Fig.12 we found that at the point where d8.1 andr0'5.7, N1, the number of repeaters, will drop to its smallest value: 141.

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4.3 Model modifications

If we go on thinking in the same way as in the 1000-user case, we can easily come to the conclusion that 141 is the minimum number of repeaters that are needed for complete coverage. However, we should never ignore the fact that there are only a limited number of PL signals.

As is assumed in section4.2.1, a repeater will only respond to signals with a special PL code that is associated to it. Now that we have 54 PL tones at hand, the maximum number of repeaters should not be larger than 54. But according to the result of our model, there should be at least 141 repeaters available in the area. As a result, it becomes an inevitable problem that, if no effective measures are taken, some repeaters will have to share the same PL tone.

The situation that 2 or more repeaters share the same PL code is not allowed in a typical CTCSS system. To tackle this problem, we proposed two methods of modifying our model: the method of sharing frequency and the method of reducing transmitting frequencies.

4.3.1 The method of sharing frequency

Though repeaters are not allowed to share the same PL code, the restrictions in frequency assignments are relatively loose. Therefore, we are thinking of allowing 2 or more users to share the same frequencies.

To get clearer about the mathematical relations between the maximum number of users sharing the same frequency and the number of repeaters in need, we introduced a parameter, to represent the maximum number of users allowed to share the same frequency and modified the model in 4.2:

nmin8010000 (12)

nminr0'2R2 (13)

Thus we have

r0'125Rkm (14)

Again with the help of the Ring Method, we got Fig.13. From Fig.13 we can find thatN1, the number of repeaters, decreases with the increase of. If3, the number of repeaters need for the 10000-user case will fall below the level of 54.

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140120100N180604020012345lambda6710

Figure 13: the Relationship betweenN1and 

Theoretically, the larger  is, the fewer repeaters we need. But in practice, if a certain frequency is shared by too many users, the quality of the service provided for each user will fall dramatically to an unacceptable level. What we should pursue here is to reduce the number of repeaters while ensure the quality of the service provided by the repeaters. Therefore, we pick the red point in Fig.13 as the optimal point, where3 andN146.

4.3.2 Reducing transmitting frequencies

Besides the method of sharing frequencies, we can also adopt the method of reducing the repeaters‟ transmitting frequencies to provide more frequencies for users. In other words, a repeater will have to work on a new transmitting mechanism: when it receives signals with a special PL code, a repeater will transmit them in just one, two or more frequencies (if they are 600 kHz above or below the original frequencies) rather than sending them with corresponding frequencies 600 kHz higher or lower. A typical working process of this mechanism can be illustrated in Fig. 14

145MHz, PL=1A

148MHz

145.015MHz, PL=1A 148MHz

147MHz, PL=1A 148MHz

Repeater (PL code=1A)

Figure 14: the Typical Working Process of the New Mechanism

Team #10731 Page 16 of 22 In this mechanism, except several frequencies for repeaters to transmit signals, most frequencies in the 145 to 148MHz spectrum can be made full use of, just as shown in the following figure.

145MHz

Repeater

frequency

145.6MHz

146.2MHz

Repeater frequency 146.8MHz

147.4MHz

148MHz

Figure 15: Divisions of Frequencies Available under New Mechanism

Thus, we get the total number of frequencies available

nminfmaxfmin(148145)103200 (15)

finterval15We denote a as the number of user frequencies and b as the number of transmitting

frequencies. Then we have

abnmin200 (16)

fa (17) bWhere f indicates the number of signals that is transmitted by the repeater on the same frequency. Generally speaking, the larger f is, the lower the transmitting quality is.

To obtain the optimal distribution scheme under the new transmitting mechanism, we have to reduce the number of repeaters to the lowest level while at the same time ensure a as-high-as-possible transmitting quality. Thus, our first job is to get the relationship betweenN1, fandb. We modified the model in 4.2 and get

nmin(200b)10000 (18)

nminr0'2R2 (19)

Thus we have

r0'R(200b) (20)

10000

Substituting the values into the Eq.5 in the Ring Method, we get Fig.16

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120 200100150N180100605040 02040b60800100

Figure 16: the Relationship betweenN1, band the Number of PL tones

In Fig.16, the blue line represents the number of repeaters; the red line represents the maximum number of PL codes; the line with green color represents the changing trend of fasb, the number of transmitting frequencies increases.

For the purpose of reducing the number of repeaters to the lowest level, we prefer the value of b to be as low as possible. But as bdecreases, f will increase rapidly to an unacceptable level. Taking the two factors into consideration, we decide the point where the blue line and the red line intersect.

We marked that point on Fig.16 and come to the conclusion that whenb9,N154.

4.4 Conclusions

In the 10000-user case, the distribution scheme based on geography segregation provided with us an answer of 43 repeaters, which is the same as in 1000-user case. The distribution scheme based on frequency assignments we adopted in the 1000-user case does not fit for the new case as there are 9000 more users. Thus we proposed 2 methods to tackle with the problem and made some modifications on the model. According to the result of first method, there should be at least 46 repeaters, and the second method 54 repeaters.

From what we have got above, we can easily find that in the 10000-user case, the factor of frequency assignments plays a more important role in deciding the distribution scheme, which is different from the 1000-user case, in which geography segregation is the decisive factor.

5 Extension: model for the mountainous-area case

As we have discussed the influence caused by the boost of population in the circular area, there still exists another factor that may affect our distribution scheme: the topography.

Team #10731 Page 18 of 22 5.1 Analysis of mountainous areas

In case the circular area is not flat but mountainous, it is harder for signals to be received and transmitted because of 2 problems [10]:

(1) The mountains will hinder the transmission of signals in certain directions;

(2) The height of mountains will cause longer distance for signals to travel from users to repeaters compared with the situation in flat area;

5.2 Solutions to the problems

Currently, the most common way to solve the 1st problem is to build repeaters on top of the mountains to minimize the influence caused by the blockage of mountains. So what we focused on in our paper is the solution to the 2nd problem.

We simplified the mountains as a conical shape, thus we can simplify the case in mountainous area, as illustrated in Fig.17:

r0

h

r1  r Figure 17: the Simplification of Mountainous Area Cases

In Fig.17, r0represents the longest distance in which a repeater can take effect in a flat area (In 4.1.1 we have figured out that r011.8km). However, if we take an Aerial view of the mountain, we can find that the actual radius of a repeater‟s area it can affect isr1. By analyzing the geometric relationship between r1 and r0 we get

r1r0cosr0 (21)

Obviously, the effective area of a repeater is reduced when the repeater is placed on top of the mountain. As a result, some users in mountainous areas cannot send their signals to even the nearest repeater. The easiest way to solve this problem is to add some repeaters on hillsides or bottoms of the mountains, just as is illustrated in Fig.18

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Figure 18: the Distribution Scheme in Mountainous Areas

5.3 Conclusions

Modeling distribution schemes in mountainous areas is a bit more complex than in flat areas. Due to the lack of necessary detailed data about this kind of area, we cannot provide a perfect distribution schemes for this case.

In spite of this, we made an analysis of the influence caused by topographical factors and proposed a solution to repeater distribution in the mountainous area:

Step 1: Installing repeaters on top of the mountains;

Step 2: Measuring the slope of each mountain and using Eq.21 to decide whether or not the repeaters‟ effective areas is large enough to cover the whole mountain; Step 3: Installing repeaters on slopes or bottoms of the mountains according to the result of step 2;

Step 4: Arranging repeaters on the remained flat areas.

6 Conclusions

6.1 Strengths

• Creativity: Inspired by the shape honeycombs and the calculation methods adopted originally in satellites engineering, we innovatively proposed two new algorithms in our model to solve the coverage problems in 3.1.2.

• Generality: The generality of our models had been proved in Section Two. We made some small modifications to our model built for 1000-user case and then successfully obtained the distribution scheme in 10000-user case. In a word, our models can be generalized to cases with much more users.

Team #10731 Page 20 of 22 • Simplicity: We solve the problem of distributing repeaters by decomposing it creatively into two sub-problems that are much easier to deal with. By modeling the two sub-problems ,respectively, and then by comparing the results of these models, we can get the best distribution scheme in a simple way.

6.2 Weaknesses

• Dependency: Our method of making distribution schemes depends strongly on the assumption that users are distributed in the area evenly. Thus, we may encounter some unpredictable obstacles if users are distributed randomly in the given area. • Incompleteness: As we focused our attention mainly on finding the minimum number and optimal positions of repeaters, we ignored the discussion on the detailed mechanism of signal transmission process from repeaters to repeaters.

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7 References

[1] http://www.dictionary30.com/meaning/Repeater

[2]http://shortwaveamerica.blogspot.com/2010/10/history-of-california-repeater-wars.html

[3] http://www.walktalk.cn/question6.htm

[4] http://wenku.baidu.com/view/4c2b6204cc175527072208d0.html [5] http://en.wikipedia.org/wiki/Honeycomb_conjecture [6] http://mathworld.wolfram.com/HoneycombConjecture.html

[7] http://zh.wikipedia.org/zh/%E5%85%AD%E8%BE%B9%E5%BD%A2 [8] http://www.narcc.org/

[9] NARCC Policy and Procedure for Coordination and Sanction. Northern Amateur Relay Council of California Inc. Repeater and Auxiliary Coordination for Northern California USA.

[10] Peter L. Stenberg, Sania Rahman, M. Bree Perrin, and Erica Johnson. Rural Areas in the New Telecommunications Era

[11] Al Duncan. Amateur Radio FM Repeater Basics . October 2007.

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8 Appendix

8.1.A program solving problem of 3.1.2

d=1:0.1:23.6;L=[]; n=[];r1=[];n1=[]; N=length(d); for i=1:N

L(i)=sqrt(11.8^2-(d(i)/2)^2)*2; n1(i)=fix(/d(i)); r1(i)=-n1(i)*d(i); n(i)=0; for j=1:n1(i)

n(i)=n(i)+fix(2*pi*(r1(i)+(j+1)*d(i)/2)/L(i))+1; end

N1(i)=n(i)+1; end

8.2.A program solving problem of 3.2.4 function N0=zui(r) d=1:0.1:(2*r); L=[]; n=[]; r1=[]; n1=[];

N=length(d); for i=1:N

L(i)=sqrt(r^2-(d(i)/2)^2)*2; n1(i)=fix(/d(i)); r1(i)=-n1(i)*d(i); n(i)=0;

for j=1:n1(i)

n(i)=n(i)+fix(2*pi*(r1(i)+(j+1)*d(i)/2)/L(i))+1; end

N1(i)=n(i)+1; end

N0=min(N1);

A=199:-1:100; a=length(A); for i=1:a

N=fix(10000/A(i)); r(i)=/sqrt(N);

N0(i)=zui(r(i)); end