MaMaEuSch Management Mathematics for European Schools http://www.mathematik.unikl.de/ mamaeusch Connection and costs distribution problems Federico Perea Justo Puerto MaMaEuSch Management Mathematics for European Schools 94342 - CP - 1-2001 - DE - COMENIUS - C21 University of Seville This project has been carried out with the partial support of the European Union in the framework of the Sokrates programme. The content does not necessarily reflect the position of the European Union, nor does it involve any responsibility on the part of the European Union. 0
1 Previous concepts 1.1 What is a graph? In this work we describe some situations which can be represented by graphs. In order to investigate such situations we do not need to study graphs in detail. We start by introducing the idea of a graph through some examples. 1.1.1 Examples Route Maps The diagram in figure 1 is a map of what will be the Sevilla Underground. Like all maps, it does not represent every feature of the city in question, but only those of relevance to the people who use it. In the case of that map, the exact geographical locations of the stations are unimportant. What is important, however, is the way in which the various stations are interconnected, so that a passenger can plan a route from one station to another. The map is simply a diagrammatic way of indicating how the stations are interconnected. Architectural Floor Plans The plan of the lower floor of a house is represented in figure 2. For small plans like this, such a diagram is very convenient for showing which rooms have mutual access, but for large plans a less cumbersome representation is useful. One such representation is to draw the rooms as small solid circles as shown in figure 3. Such diagrams are known to architects as circulation diagrams, because of their use in analyzing the movements of people in large buildings. In particular, they have been used in the designing of airports, and in planning the layout of supermarkets. Such diagrams are useful in representing the connection between the various rooms, but they do not give us any information about the size or shape of the rooms. 1.1.2 The definition of a graph The common feature in all the preceding examples is that in each case we have a system of objects which are interrelated in some way. In the first example the objects are stations interconnected by rails and in the second example they are rooms with mutual access. In each case we can draw a diagram in which the objects are represented by points and the interconnections are represented by lines. Such a diagram is called a graph. The points representing the objects are called vertices (also called nodes or points), and the lines representing the 1
Figure 1: Future Sevilla Underground. interconnections are called edges (also called arcs or just lines). For example, the circulation diagram of the house is a graph with seven vertices (corresponding to the playroom, kitchen, hall, etc.) and ten edges (corresponding to the interconnections between these rooms). We can formalize these ideas as follows: Definition 1.1 A graph is a diagram consisting of points, called vertices, joined together by lines, called edges; each edge joins exactly two vertices. We also need to know two concepts: cycle and connection. We say that a graph G is connected if there is a path in G between any given pair of vertices. Both the example on the underground and the example on the lower floor of a house give us connected graphs. The graph shown in figure 4 is disconnected, that is, not connected. A path that connects a vertex to itself is called a cycle. In figure 3 the path 2
Figure 2: Lower floor. living room - sitting room - study - living room is a cycle, because it joins the living room to itself. Those concepts lead us to an entity we are using in this work: a tree network. A connected graph with contains no cycles is called a tree. Some examples are shown in figure 5. If G is a connected graph, then a spanning tree in G is a tree which includes every vertex of G. For example let us consider the graph shown in figure 6. The number of spanning trees in a graph can be very large (bounded above by 2 n 2, where n is the number of nodes that the network has). In figure 7 we present three possible spanning trees for the graph shown in figure 6. 1.2 Cooperative situations In many situations we find several agents that, in case they join their efforts, can either get a higher profit or have lower costs after their actions (for instance business) are taken. Those situations are called cooperative situations, because it is allowed for the agents to cooperate. Let us see one example. 1.2.1 Example Three friends, Justin, Paula and Malcom, are looking into the possibilities of starting a care center for elderly people from 60 to 80 years. After a thorough investigation of the local regulations and market, they come to the following assessment of the situation. They need 3
Figure 3: Graph. one nurse per four people from 71 to 80, and one nurse per 10 people from 60 to 70. For each person from 71 to 80 they are required to have 8 square meters indoors inside and 4 outdoors. For a person from 60 to 70 the requirements are 5 and 6 respectively. After calculating all costs they realize that they can make a net profit of 200 Euros per month per elderly man/woman from 71 to 80 and a net profit of 150 Euros per month per elderly man/woman from 60 to 70. Justin knows 9 people whom he can hire as nurses. He has the possibility of renting 260 square meters inside and 200 outside. Therefore he has to solve a linear programming problem to calculate the maximum profit that he can make on his own while complying with all regulations. If we call x 1 the number of elderly people from 71 to 80 and x 2 those that are from 60 to 70, the problem to solve in order to calculate the maximum profit that Justin can have is: 1 max 200x 1 + 150x 2 1 s.t. x 4 1 + 1 x 10 2 9 8x 1 + 5x 2 260 4x 1 + 6x 2 200 x 1 0, x 2 0 1 For further details on linear programming see: H.W.Hamacher, E.Korn, R.Korn, S.Schwarze Production Planning and Linear Optimization in MaMaEuSch web site http://www.mathematik.uni-kl.de/ mamaeusch/allgemein e/frames2 e.html (2004). 4
Figure 4: Disconnected Graph. After doing that he can get a profit of 7000 Euros on his own by having 20 elderly people of each group. Paula knows 5 people whom she can hire as nurses and Malcom knows 14. Paula has the possibility of renting 120 square meters indoors and 200 outdoors, and Malcom can rent 590 square meters inside and 400 outside. When solving the linear programming problems for Paula and Malcom, we have that Paula can have a maximum profit of 3600 Euros if acting on her own, and Malcom can earn 14000 Euros. They can also decide to combine forces. We assume that there is no overlap between the people that they can hire as nurses. If Justin and Paula decide to work together they will have a profit of 11000 Euros. If Justin and Malcom worked together they would get a profit of 22500 Euros. Paula and Malcom would have a profit of 19500. And, if all three of them worked together they would get a profit of 26500 Euros. That situation can be summarized in the following table: S v(s) S v(s) S v(s) S v(s) {1} 7000 {3} 14000 {1,3} 22500 {1,2,3} 26500 {2} 3600 {1,2} 11000 {2,3} 19500 0 where we denote Justin by number 1, Paula by number 2 and Malcom by number 3, and v(s) is the maximum profit that the group S can get by acting on its own. 5
Figure 5: Some trees. Their total profit is maximized if they combine forces. In this case they will have to figure out a way to divide the profit of 26500 Euros. 2 Connection and costs distribution problems In many daily situations we face up to optimization problems. It normally happens that, when several agents (people, companies,etc.) join their efforts in order to carry out an action in common which will serve all of them, the cost generated is lower than the cost that would be generated without acting allied. It is clear that they will join efforts, because they will spend less money acting together and they will get the same service. But then a new problem arises: how to allocate the costs generated by the common action. Let us imagine that several farmers are going to build a wire fence to isolate their farmlands. Since those fences delimit two land properties, they are used by more than one farmer. Therefore it seems logical to install the fences between them, and it seems illogical that they build the fences individually. But the time to pay approaches and, consequently, problems 6
Figure 6: This graph is not a tree. arises. In many similar situations it is agreed to pay the whole total cost in equal shares (from now on, this rule will be called proportional rule), and it is even thought to be the fairest way. Through some examples, we will see that the proportional rule can be unfair, taking into account the problems that the word fairness can bring on. 3 On cost allocation in connection problems The first situation we deal with belongs to the class of minimum spanning tree problems. In these problems we have several users (located in one node of a graph) who want to benefit by a product coming from a fixed source (could be water, electricity, petrol, etc.). We want to connect every user to the source. The network connection is not fixed, that is, the nodes can be connected in any possible way. Example 3.1 Let us consider a group of villages, each one of them needs to be connected to a reservoir directly or through other villages. Each possible connection has an associated cost and the problem is how to connect all the villages to the reservoir so that the total expense generated is as low as possible. The network with the minimum cost associated is called minimum spanning tree. Nevertheless, the construction of the minimum spanning tree is only one part of the problem. Besides minimizing the total expenses, a problem about the cost allocation must be considered, 7
Figure 7: Some trees. that is, it must be decided how much each village should pay for the network. In the following examples we will propose several allocation rules and we will discuss the idea of fair allocation. Formally, a minimum spanning tree problem is a 3-upla T = (N,, t), where N = {1,..., n} is the set of nodes (villages, in the example above), is the source (the reservoir) and t : E N R + is the nonnegative costs function that assigns a cost to each arc (how much it would cost building a pipe). E S is defined as the subset of all arcs between pairs of nodes of S N = N, so (S, E S ) is the complete graph over S. E S = {{i, j} i, j S, i j}. Since the connection costs are nonnegative, it is obvious that a graph at minimum cost connecting every node to the source is a tree itself, and that explains the name of the problem minimum spanning tree. Given a minimum spanning tree problem T = (N,, t), the Bird allocation β R (T ) is built by assigning to each point i N the cost of the first arc in the path from the node i to the source. The computation of this allocation can be taken from Prim s algorithm, which 8
builds a minimum spanning tree, starting from the fixed root, by adding the arcs at lowest price, without making loops (paths that connect a point to itself). Let us see an example. In the case shown in figure 8, we want to connect three nodes; A, B and C to a source (S) located in the red point at minimum cost. Figure 8: Example of Prim algorithm. We first join the source to its nearest node, in this example it is B. After that we find which is the nearest node to B or the source, and it results A. A is nearer to B than to the source, so we connect A with B. To finish we have to connect C to the graph. To get the minimum spanning tree we have to do it at minimum cost, that is to say, we connect C to its nearest node between those that are already in the graph. In the example, the nearest node to C is the source, so we build the arc joining C to the source. The result is the fourth picture of figure 8, the minimum spanning tree of the nodes shown in the first picture. Now we have to face the other problem: how to allocate the expenses generated after building the minimum spanning tree. To do that we follow Bird s Rule. Let us see how the rule works (Bird s Rule). Input: a minimum spanning tree problem (N,, t). Output: a set of arcs R E N and the corresponding Bird allocation β R (T ). 1. Choose the source as the root. 9
2. Initialize R =. 3. Find an arc at minimum cost e = {i, j} E N \ R incident in, or in any of the nodes of an arc of R, in such a way that the inclusion of e in R does not create any loop. 4. One of the nodes i, j, let us say j, was connected previously to the source and the other one i is a node that was not connected to the source yet. Let us assign the cost β R i (T ) = t(e) to the node i. 5. We join e to R. 6. If there are some nodes that have not been connected to the root yet in the graph (N, R), go back to the step number 3. The next example helps us to understand these rule. Example 3.2 Let us consider the minimum spanning tree problem T, with N = {1, 2, 3} as shown in the figure 9, where the numbers over the arcs represent the costs associated to each arc. Figure 9: Minimum spanning tree problem T. When we apply the algorithm to this problem, the first arc we join to R could be {, 1} or {, 3} (these are the arcs at minimum cost starting from the source). Let us suppose that we choose the first one, then the cost β R 1 (T ) = 10. Later, following the Bird s algorithm, we add {1, 2} to R, and we get β R 2 (T ) = 6. To finish, we add {2, 3} to R, and it results β R 3 (T ) = 5. 10
This leads us to the following cost allocation: (10, 6, 5). On the other hand, if we started from {, 3}, we would finally obtain the cost allocation β R (T ) = (6, 5, 10). Both of these minimum spanning trees are represented in figure 10. Figure 10: Two minimum spanning trees. That is to say, in the first solution we have that, according to the Bird s rule, the node 1 has to pay 10 units, the second node 6 and the node number 3 has to pay 5. Analogously, in the second solution the node have to pay 6, 5 and 10 units respectively. 4 Application Spain is divided into 17 different regions, and one of them is called Andalusia. It is divided into 8 provinces, and in each one of them there is a capital. The geographical map of this region, where the eight capitals are denoted by a black dot, is shown in figure 11. We present the following cases as application of minimum spanning tree: 4.1 Case 1: connection without source Let us suppose that a telecommunication company wants to connect the eight capitals by a optical fibre cable. The distances between the cities are given in the following table: 11
Figure 11: Andalusia. Almería Cádiz Córdoba Granada Huelva Jaén Málaga Sevilla Almería 0 484 332 166 516 228 219 422 Cádiz 484 0 263 335 219 367 265 125 Córdoba 332 263 0 166 232 104 187 138 Granada 166 335 166 0 350 99 129 256 Huelva 516 219 232 350 0 336 313 94 Jaén 228 367 104 99 336 0 209 242 Málaga 219 265 187 129 313 209 0 219 Sevilla 422 125 138 256 94 242 219 0 Build the minimum spanning tree for this case and allocate how much each capital has to pay for its construction. 12
4.2 Case 2: connection with source Let us assume that the eight capitals must be connected to each other and to an energy source located in the city of Málaga. Build the minimum spanning tree for this case and decide how much each city has to pay taking into account that the city of Málaga should not pay anything for the network. 4.3 Solutions In the first case we do not have any source, so the first arc will be one that joins the nearest two cities to one another, in this case Sevilla and Huelva. After that we will join to the network the nearest city to Sevilla or Huelva choosing that city between the six ones that are not in the network yet. In our case Cádiz is the nearest to one of those two, and it is closer to Sevilla than to Huelva. So, the following arc to be in the network is Sevilla-Cádiz. We proceed this way with all the other cities and we get the minimum spanning tree made up by: Sevilla, Huelva, 94 Sevilla, Cádiz, 125 Sevilla, Córdoba, 138 Jaén, Córdoba, 104 Jaén, Granada, 99 Málaga, Granada, 129 Granada, Almería, 166 total distance, 855 For the case when we have a fixed source, the solution is the same and the only thing that changes is the order in which the arcs appear: Málaga, Granada, 129 Jaén, Granada, 99 Jaén, Córdoba, 104 Sevilla, Córdoba, 138 Sevilla, Huelva, 94 Sevilla, Cádiz, 125 Granada, Almería, 166 total distance, 855 13
The connection at minimum cost is shown in the figure 12. Figure 12: Minimum spanning tree. 4.4 Costs allocation In the example 1, if we suppose that each Km of the network costs 10000 Euros and the overall cost of the network must be paid by the cities, how much should each city pay? In the second example, if we suppose that Málaga paid the construction of the source that will supply the other cities and, because of that, it should not pay anything for the network, how much should each one of the other cities pay? 4.5 Allocation after the connection without source In the first case, since we do not have any source, we can not use Bird s rule. Many rules to allocate the expenses can be proposed and we will apply the next ones: Proportional allocation: we divide the total expense(8550000 Euros) between the number of cities connected by the network (8). Each city pays the same, (1068750 Euros). 14
Each city pays half of the price of the arcs that finish or begin in it, for instance, the connection between Sevilla-Huelva is paid by both cities in equal shares. 4.6 Allocation after the connection with source In the second case we do have a fixed source, Málaga. That is, in this case we can apply Bird s rule. Before then let us see how the proportional allocation works and the problems that it causes, and finally we will propose a third way of allocating costs. Proportional allocation. The total expense of the network amounted to 8550000, and because of that each of the seven cities connected to Málaga has to pay 8550000 = 1221428.6 Euros. Is that the 7 fairest way? Let us see what happens if Granada and Jaén decide to make their own network on their own, that is, without joining the rest. Applying the algorithm to build a minimum spanning tree as before, we would conclude that the shortest connection between both cities and Málaga has a length of 129 + 99 Km, that is to say, a total expense of 2280000 Euros. If those two cities had to accept the proportional rule, they would pay between them 2442857.2 Euros. Logically, these two cities do not agree with sharing the total expense in equal shares, since they can build a network by themselves that gives them the same service and at a lower price. This is an example where the proportional allocation does not make everyone happy. If when making an allocation we find subgroups of agents (cities in our case) that can connect themselves and pay less money, that allocation would probably be defeated. In Game Theory, those allocation that hold that there is no subgroup of players (in our case the players are the cities) that can spend less money by acting on their own, are called Core allocations. Let us see now the Bird s rule. Each city pays: Granada 1290000 Euros. Jaén 990000 Euros. Córdoba 1040000 Euros. Sevilla 1380000 Euros. 15
Huelva 940000 Euros. Cádiz 1250000 Euros. Almería 1660000 Euros. (Each city pays for the arc that connects it to the network). One can check that after this allocation there is no subgroup of cities that can spend less money by acting on their own, that is, we would say in Game Theory that this allocation belongs to the core. It is well known that in every minimum spanning tree problem the Bird s rule gives us an allocation belonging to the core. As a conclusion we present another kind of allocation used in connection problems. An allocation in which each city pays the proportional part of the arcs that it uses (if an arc is used by several cities, they pay the cost of that arc between all of them in equal shares). Let us see how this allocation works: The arc Málaga-Granada is used by all the cities, so they divide its cost (1290000 Euros) in seven parts. The arc Granada-Almería is only used by Almería, therefore this city bears the cost of it (1660000 Euros). Jaén, Córdoba, Sevilla, Cádiz and Huelva use the arc Granada-Jaén, so each one of them pays 990000 5 Euros. The arc Jaén-Córdoba is used to connect to the source by Córdoba, Sevilla, Cádiz and Huelva. Therefore each of these cities should pay 1040000 4 Euros. The connection Córdoba-Sevilla is used by Sevilla, Cádiz and Huelva. Because of that they have to pay 1380000 / 3 Euros each one. The connection Sevilla-Cádiz is only used by Cádiz, so this city bears the cost of that arc (1250000 Euros). Finally, the arc Sevilla-Huelva is only used by Huelva and therefore it has to pay the 940000 Euros for its construction. In conclusion, each city has to pay: Granada: 184285.7 Almería: 184285.7 + 1660000 = 1844285.7 Jaén: 184285.7 + 198000 = 382285.7 16
Córdoba: 184285.7 + 198000 + 260000 = 642285.7 Sevilla: 184285.7 + 198000 + 260000 + 460000 = 1102285.7 Cádiz: 184285.7 + 198000 + 260000 + 460000 + 1250000 = 2352285.7 Huelva: 184285.7 + 198000 + 260000 + 460000 + 940000 = 2042285.7 Although this allocation seems to be fair it does not belong to the core, since if Sevilla, Cádiz and Huelva decide to make their own network to connect them to Málaga, the costs would amount to 2190000 (Sevilla-Málaga) + 940000 (Sevilla-Huelva) +1250000 (Sevilla-Cádiz), what means a total expense of 4380000 Euros, a lower price than the 5496855 Euros that they have to pay after the allocation described above. That is, this allocation does not satisfy the proposed fairness condition and, therefore, it will never be taken. 5 How to pay a new elevator in a resident s association? Nowadays there still exist buildings which do not have any elevator. At a given point in time their neighbours think that this is the time to install it. In those cases the controversy of how to pay the elevator appears. The way of allocating the common costs motivate that some neighbours decide to act on their own without taking into account the remainder. This may happen when there is a group of neighbours that can spend less money if they build an elevator only for their own use than the money they would spend if they had to pay their share for the construction of the elevator for the whole community. Let us see the following case: In a five storey block, with one apartment in each floor, an elevator is going to be installed. The company in charge of building the elevator has a fixed price depending on the number of floors the elevator has. Due to technical problems the cost of each floor is increased with its height, that is, it is more expensive to build the part of the elevator between the sixth and the seventh floor than the one between the first and the second. The prices of the company are shown in the following table: Number of storeys Cost of the elevator 1 10000 2 21000 3 33000 4 46000 5 60000 17
The draft of the elevator is shown in figure 13. Figure 13: Elevator draft. In the resident s association it was proposed the following cost allocation: each one pays proportionally, that is, 12000 Euros. Is this allocation fair? How would you allocate the expenses among all the neighbours? 5.1 Sharing expenses Let us see the three rules to share costs that we saw before in our new situation: 1. Proportional allocation says that each neighbour has to pay 12000 Euros. But let us thing a bit, if the neighbour of the first floor wanted to build an elevator only for his own use, he would have to pay less (10000 Euros), so he will not want to be part of the joint construction of the elevator. 2. Bird s rule. When using that rule to share costs no subgroup of neighbours will be able to build an elevator for themselves at less cost. In this case, Bird s rule proposes the following cost allocation: The neighbour of the first floor pays 10000 Euros (the cost of building an elevator from the hall to the first floor). The neighbour of the second floor pays 11000 (the cost of building the elevator from the first floor to the second) and so on. With this allocation rule the neighbour of the third floor pays 12000 Euros, 13000 has to pay the one living 18
in the fourth floor and the neighbour of the fifth floor pays 14000. (Recall that those allocations, so that there is no subgroup of agents that can spend less money by acting by themselves, are known in Cooperative Game Theory as core allocations.) 3. Another rule that can be proposed is the next: each neighbour pays for the number of floors that he has to pass through in order to get to his floor, that is, the neighbour of the third floor has to pay for the first part of the elevator, the second and the third. In this case each neighbour pays: The first part costs 10000 Euros and it is paid among all the neighbours, because all of them use it ( 10000 5 = 2000 Euros). The part between the first and the second storey is paid by the neighbours from the second floor to the fifth ( 11000 4 = 2750 Euros). The third part costs 12000 and is paid by the neighbours of the third, fourth and fifth floor ( 12000 3 = 4000 Euros). The part between the third and the fourth storey is 13000 Euros and it has to be paid by those neighbours who use it, that is the ones living in the fourth and the fifth floor, ( 13000 2 = 6500 Euros). The last part, between the fourth and the fifth is paid only by the neighbour of the fifth floor, ( 14000 1 = 14000 Euros). So the payoff vector is: (2000, 4750, 8750, 15250, 29250), where the i th neighbour pays the i th component of the vector above. Although that is a stable allocation belonging the Core of the corresponding cooperative game, it seems to be excessive that the neighbour of the fifth floor pays 15 times what the neighbour of the first storey pays. What do you think about the concept of fairness in allocation of expenses? 19