Eindhoven University of Technology BACHELOR. Price directed control of bike sharing systems. van der Schoot, Femke A.

Transcription

1 Eindhoven University of Technology BACHELOR Price directed control of bike sharing systems van der Schoot, Femke A. Award date: 2017 Link to publication Disclaimer This document contains a student thesis (bachelor's or master's), as authored by a student at Eindhoven University of Technology. Student theses are made available in the TU/e repository upon obtaining the required degree. The grade received is not published on the document as presented in the repository. The required complexity or quality of research of student theses may vary by program, and the required minimum study period may vary in duration. General rights Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights. Users may download and print one copy of any publication from the public portal for the purpose of private study or research. You may not further distribute the material or use it for any profit-making activity or commercial gain

2 Eindhoven University of Technology Bachelor Final Project Price Directed Control of Bike Sharing Systems Author: Femke van der Schoot Supervisor: Sem Borst July 7, 2017

3 Contents 1 Abstract 2 2 Introduction 3 3 Model Description Problem Description Goal Concepts Assumptions State Description Transition rates Queueing etwork Representation Translation Stationary Distribution Mean Value Analysis Procedure Mean Value Analysis Performance Analysis Price Analysis Price Allocation Price Allocation Price Allocation Price Allocation Influences of the Price Implementation umerical Experiments Goal Description Base Scenario Special Scenario Special Scenario Special Scenario Special Scenario Limitations of the Scenarios General Optimization Method Differences in umbers of Requests Differences in Routing Probabilities General Systems Conclusion 29 References 31 Appendices 32 A Mean Value Analysis with 4 Stations, Price Allocation 2 32 B Mean Value Analysis with 4 Stations, Price Allocation 3 34 /Department of Mathematics and Computer Science 1

4 1 Abstract Over the last few decades bike sharing systems have become more and more prevalent. They have emerged as a popular transport mode in urban trips. The main goal of implementing these systems is to jointly solve problems such as traffic congestion, traffic noise, et cetera, but they are hampered when requests arrive at empty stations. This report considers a bike sharing system that can be represented as a closed queueing network. We have used mean value analysis to gain insight in how the system behaves on average. The goal of the project is to optimize the total revenue for certain price allocations for bikes. A secondary goal is to see how the distribution of the bikes among the stations is influenced by the optimization of the average total revenue. In this project we have developed an optimization method to compute a price allocation for bikes that leads to an optimal total revenue. Further, we have observed numerically that optimizing the average total revenue typically leads to a more balanced bike distribution among stations, though the precise nature of this relation is hard to infer. /Department of Mathematics and Computer Science 2

5 2 Introduction A bike sharing system consists of multiple stations where people can rent bikes for a certain amount of time. In The etherlands for example, there is the OV-fiets. The systems act as a public transport mode for urban trips. Further, they are regarded as an effective way to jointly solve problems including traffic congestion, parking difficulties, traffic noise and air pollution. Over the last few decades, the amount of bike sharing systems has increased fast. In June 2014, these systems were used on five continents, including 712 cities, operating about 806, 000 bikes at 37, 500 stations [Shaheen et al., 2014, Li et al., 2016]. How do such bike sharing systems look like in reality? The OV-fiets works as follows: if one wants to rent a bike, one has to go to a train or bus station and rent a bike. One pays e3.85 euro per 24 hours. After a couple of days one returns the bike at the same station. In this system, the stations are far away from each other and one has to pay more when returning the bike at a different station than where it was picked up. Therefore, this bike sharing system is facing fewer problems in the balancing than other systems. For example, Paris has a bike sharing system implemented with more than 23, 600 bikes and 1, 800 stations over the city. When one rents a bike, the first 30 minutes are free of charge. Then the first additional half-hour costs e1, euro, the second additional half-hour costs e2, euro and from then on, each additional half-hour costs e4, euro. It means that the bike is mostly used for small trips through the city. It is also possible to buy a 1-day or 7-day ticket, which will mostly be used by tourists. Further, the system uses an application, where one can see at which station there are bikes or parking places available. Besides these systems, there are also systems that make use of rebalancing operations. They use trucks that transport bikes to rebalance them over the stations. The goal of this project is to maximize the average total revenue per time unit in a likewise model of a bike sharing system. A secondary goal will be to study the relation between the average total revenue and the distribution of the bikes among the stations; in other words, we study the balance of the system. A study similar to this project is done by Adelman [2007]. History, impacts, models of provision and future of bike sharing systems are reviewed by DeMaio [2009]. Fishman et al. [2013] provided a synthesis of the literature for bike sharing systems. Some balancing problems are solved or alleviated by rebalancing operations. A study on this subject is done by Schuijbroek et al. [2013]. More studies on bike sharing systems can be found in the references of the above-mentioned studies. While bike sharing systems solve or alleviate some problems, they also create new problems. For example, there are systems with over 50, 000 bikes. One can imagine that a station does not have 50, 000 parking places, which sometimes can lead to full stations, where temporarily people cannot return a bike. Also, some stations can be more popular than other stations, which can lead to an imbalance of bikes distributed over the stations. Requests at a station cannot be met if this station is empty, which is clearly undesirable. In this project we will take a look at such a bike sharing system and develop an effective method to partly solve or alleviate these kinds of problems. We want to do this by finding a way to adjust the price allocations for bikes such that the system will have a better balance and the problem of empty stations is reduced. Within this model, we do not take into account rebalancing operations. Further, we will look at systems, where the all prices for a bike are equal and determined per time unit. The OV-fiets is mostly picked up and returned at the same station. However, here we will consider a system that is designed for tourists, such as in Paris, where there are many stations not far from each other. Finally, we will assume an infinite capacity of the parking places at each station. This is far from reality, but it will make the model manageable to work with. ow, we can focus on average total revenues, empty stations and the related rejected requests. /Department of Mathematics and Computer Science 3

6 In section 3, we will give a detailed model description of a bike sharing system, including the goal, concepts, assumptions and more. We then conclude that a bike sharing system behaves under suitable conditions like a Markov process. After that, in section 4, we give a representation of the model in terms of queues. From this queueing network representation, we can derive the stationary distribution and with that we can perform a mean value analysis, which is derived in section 5. We also carry out a price analysis in section 6. In this section, multiple ways of allocating prices to bikes are considered and formulas for average total revenues corresponding to these price allocations are derived. These formulas, together with the mean value analysis, will be the basis of the overall optimization method. Before developing the method, we will do numerical experiments in section 7 to gain insight in how the network behaves for various price allocations. Then, in section 8 a general optimization method will be developed based on the numerical experiments. Finally, the conclusion of this project is given in section 9. /Department of Mathematics and Computer Science 4

7 3 Model Description 3.1 Problem Description We will consider a system that consists of stations and bikes. At each of these stations people can rent one of these bikes immediately, if any available. Bikes cannot be reserved in advance. The renters can ride on the bike and finally bring it back to one of the stations, not necessarily the same one where the bike was rented. However, we want to avoid the situation that at a certain point in time most of the bikes are parked at a few stations while all the other stations are empty. It is certainly possible that some routes are used more often than other routes and therefore some stations are more used for renting the bikes and other stations for returning the bikes. This can lead to undesirable situations, since empty stations cannot meet requests anymore. We also have to take into account that each bike rental has a specific price, which can influence the number of requests or the station where one returns the bike. We will use these price influences to achieve a well-distributed network, which brings us to the goal of this project. 3.2 Goal The goal of this project is to achieve a well-distributed system by adjusting the prices of the bike rentals. Suppose we get a system in which every parameter is known, then our goal is to optimize the average total revenue per time unit. When performing this, the distribution of the bikes over the stations and routes is automatically taken into account, since this leads to a higher rate of circulation and therefore, more requests to be accepted. If the price is too low, the average total revenue will suffer from that. But on the other hand, if the price is too high, the number of requests will decrease. In this project, we will try to find an efficient method we can use to compute a price allocation that maximizes the average total revenue and still takes into account the distribution of the bikes and the rate of circulation. 3.3 Concepts Before we describe a model that fits this network, we first need to get a better insight in all the relevant concepts. Within these concepts, some useful notation will be introduced. Stations. The total number of stations is. The set of stations will be denoted by S, where station i will be denoted by S i. So S = {S i 1 i } with S =. Routes. Between each pair of stations there exist two routes, one in each direction. In reality, these routes may of course contain some overlap or multiple routes may exist. However, we call the route only on the stations where the bike is picked up and returned. We also take the routes into account from a station to itself. This results in a total number of 2 routes. The set of routes will be denoted by R and R ij denotes the route from station S i to S j, for each 1 i, j. ote that R ji does also exist, but is a different route, namely from station S j to S i. So R = {R ij 1 i, j } with R = 2. We will consider the routes as virtual stations in our network. A bike spends a certain amount of time on a route and it does not matter where it is exactly on that route. Therefore, as long as the bike has left station S i and has not arrived at station S j, it is at route R ij, which you can therefore consider as a virtual station. This will be useful in the translation from the model to a queueing network. So besides the real stations, we have 2 virtual stations. Hence, in total there will be + 2 stations, real and virtual. We will refer to both real and virtual stations as nodes in our network to avoid confusion. We denote the set of nodes as M = S R and conclude that M = + 2. Bikes. We denote the total number of bikes by K. These bikes circulate around the stations and the routes between them. At one of the stations, one comes to rent a bike. Then he or /Department of Mathematics and Computer Science 5

8 she will ride on this bike along one of the routes and finally, return the bike again at one of the stations. We assume that a bike cannot leave the network, see the next section Assumptions. If a bike is parked at station S i, we say that this bike has state S i. Say, for example, the bike is rented and the person wants to return the bike at station S j. Then while riding the bike, the bike has state R ij. Finally, once brought back at station S j, the bike has state S j. Requests. From outside the system requests arrive at one of the stations per time. We assume that for each station S i, with 1 i, the requests arrive according to a Poisson process with parameter λ i > 0. So on average there are 1/λ i units of time between two requests at station S i. A request can only be met if there is at least one bike at that station. If there is no bike at the station, the request is lost and will not return later. Otherwise one can rent a bike and ride on it. If a bike is rented at station S i, the probability of going to station S j will be denoted by the routing probabilities p ij > 0 for 1 i, j. It is possible to rent and return the bike at the same station, so p ii > 0. ote that j=1 p ij = 1 must hold for all 1 i. We can also describe the circulation of bikes in terms of nodes as follows. First, the bike is parked at node S i, for a certain 1 i. ext, a request comes in at this node and the bike is taken to another node, R ij with probability p ij, for 1 i, j. At this moment, j is not known. However, this does not matter for the model, since we know that there is some j for which the bike is at node R ij for a certain period of time. Finally, it arrives at node S j, where j is known and may be equal to i. Riding time. The riding time of a bike is the time between hiring and returning the bike. In other words, it is the time that the bike spends at a node R ij, for 1 i, j. We assume this time is an exponential random variable with parameter µ ij > 0, for 1 i, j, so the expected riding time is 1/µ ij. ext we will give a schematic model with 4 stations and 25 bikes, see Figure 1. /Department of Mathematics and Computer Science 6

9 Figure 1: Schematic Model 3.4 Assumptions ow, all concepts have been described. In the next chapter we will describe a queueing network representation that fits the problem description. But before that, we have to introduce a few assumptions to make the model somewhat easier and manageable to work with. We assume that at every station, there is enough parking space, so for at least K bikes. In other words, it is possible for all K bikes to be stored at one station at a certain point in time. aturally, we want to avoid this situation, but this assumption makes it more convenient to analyze the model. We assume that there is a route between each pair of stations, even from a station to itself. Since we consider bikes, it is imaginable that the renters are tourists who want to explore the city in which multiple of these stations are placed. Then it is possible to deliver the bike at any of the stations, just like the tourist wants. In this situation, it should of course be possible to return the bike just where one rented it. We assume that a bike cannot leave or enter the system. With that, we automatically assume that no bikes will be stolen or broken. Thus, all bikes will stay within the system and are always either at a station or on a route. We assume that if a request cannot be met, since there are, for example, no bikes at a station, the request immediately leaves the system and will not return later. In other words, if a station does not have a single bike, no requests can be accepted until someone returns a bike at that station. /Department of Mathematics and Computer Science 7

10 We assume that all prices for a bike rental are equal. We have already assumed that a riding time is exponential distributed. However, we also assume that there is a maximum riding time. In other words, one brings back the bike within a certain time. In real systems, this could for example be 24 hours. Later in this project, we will see that the distribution of the riding time does not matter, as long as the expectation is 1/µ ij for route R ij. The advantage of this assumption is the fact that the price does not depend on the riding time, but can be determined per time unit, for example these 24 hours. 3.5 State Description We define Q(t) to be a random vector of dimension + 2, representing the + 2 nodes in our system. The first entries denote the number of bikes at various stations at time t. The 2 entries that follow denote the number of bikes on the various routes at time t. So to visualize this: Q(t) = (Q 1 (t), Q 2 (t),..., Q (t), Q 1,1 (t), Q 1,2 (t),..., Q 1, (t), Q 2,1 (t),..., Q, (t)), (1) with Q i (t) the number of bikes at station S i at time t and Q ij (t) the number of bikes on route R ij at time t. ote that this random vector also depends on K, the total number of bikes. We will use this fact at a later stage. Furthermore, we denote a specific state to be a vector q, which clearly has also dimension + 2. q = (q 1, q 2,..., q, q 1,1, q 1,2,..., q 1,, q 2,1,..., q 1, q,1,..., q, ) (2) We use the notations q i for the number of bikes at station S i and q ij for the number of bikes on the route R ij in state q. The state space is Ω = {q + 2 q i + q ij = K} (3) with size Ω = ( + 2 ) K. We also define a i to be the unit vector with a 1 at place q i (and the other places zeros) and b ij to be the unit vector with a 1 at place q ij (and the other places zeros), both with dimension + 2. We will need these at a later stage. Since all arrivals of requests and riding times are exponential, we can conclude that {Q(t), t > 0} evolves as a Markov process. 3.6 Transition rates ext, we will have a look at the transition rates of the Markov process. We do this in the following way: suppose we are in state q and want to go to another state q by only one step. This can be done in two ways: a bike is leaving a station and goes en route or a bike is leaving a route and arrives at a station. Looking at the first option we can describe this as q = q a i + b ij for all i, j = 1,...,, where q i > 0. In this case, the transition rate from q to q is p ij λ i 1{q i > 0}. ote that if q i = 0, there are no bikes at station S i and therefore no bike can leave this station. Looking at the second option, we can describe this as q = q + a i b ij for all i, j = 1,...,, where q ij > 0 for the same reason as above. In this case the transition rate from q to q is µ ij q ij 1{q ij > 0}. We do not need the last indicator function, since q ij is already a term in the sum. However, for later ease we will include this indicator function. ow suppose we are in state q and want to describe the transition rate from another state q to this state q. We have to ask ourselves: how do we get into this state via another state q by only making one step. Again, we can get there when a bike that was not en route before, leaves a station and goes en route and we can get in state q when a bike, that was not at a station before, arrives at that station via some route. /Department of Mathematics and Computer Science 8

11 The first option gives us: q = q + a i b ij, for all i, j = 1,..., where q ij > 0. If q ij = 0, there would be no bikes on route R ij, hence we cannot reach state q by transferring the bike from S i to R ij. The rate is now p ij λ i 1{q ij > 0}. The second option gives us q = q a i + b ji, for all i, j = 1,...,, where q i > 0 for the same reason as described above. Also note that it is b ji and not b ij, since we are going from state q to q and therefore we have to leave route R ji to arrive at station S i. This yields a transition rate of µ ji (q ji + 1) 1{q i > 0}. ote that +1 again comes from the fact that we are going from state q to state q. Therefore, in state q the service rate of route R ji was µ ji (q ji + 1). Knowing all the transition rates from and to a state q Ω, we will have a look at the steadystate distribution. Since all states are reachable and due to the fact that all parameters are defined to be greater than 0, the Markov process is irreducible and therefore the steady-state distribution exists. We define π(q) = P{Q = q} and denote the steady-state distribution by the vector π = {π(q) q Ω}. In other words, we want to know π(q) for all states q in the state space. For π to be a stationary distribution, it has to satisfy 0 π(q) 1 for all q Ω; q Ω π(q) = 1; q Ω π(q) g(q, q ) = q Ω π(q ) g(q, q) for all q Ω. (4) Here, g(q, q ) is the transition rate from state q to state q. This is the formula we have derived above. The left hand side of the third item represents the rate which state q is left. The right hand side represents the rate which state q is reached. We start with expanding the left hand side. By taking together the two different options which are described above we can rewrite the left hand side of the equation as: q Ω π(q) g(q, q ) = π(q) p ij λ i 1{q i > 0} + π(q) µ ij q ij 1{q ij > 0} We can simplify this first sum, since j=1 p ij = 1. So we get: q Ω π(q) g(q, q ) = π(q) λ i 1{q i > 0} + π(q) µ ij q ij 1{q ij > 0} In the same way, we can expand the right hand side of the equation by again taking the two different options together: q Ω π(q ) g(q, q) = π(q +a i b ij ) p ij λ i 1{q ij > 0}+ π(q a i +b ji ) µ ji (q ji +1) 1{q i > 0} ow, we have the left and the right hand sides of the equations. Together with the first and second properties of π, we should now be able to find the stationary distribution. However, the vector π has dimension ( + 2 ) K, which can get immense if and K become large. Even for a computer it would take very long to compute π(q) for all q Ω. Besides, we would like to have an expression for π for each and K. To achieve this, we will translate this model into a closed queueing network representation. Since for this equivalent model we know the stationary distribution, we can use this to solve the equations for the stationary distribution of the original problem. /Department of Mathematics and Computer Science 9

12 4 Queueing etwork Representation In the previous chapter, we have derived the equations for the stationary distribution of our model. Since these are very hard to solve, we will translate the model into a queueing network. For this network the stationary distribution is known and we can use that to solve the equations for the original model. 4.1 Translation In this new network the bikes will be the customers and the renters will be the servers. The stations turn into single-server queues and the routes turn into infinite-server queues. We will explain how this fits the previous model. Let us first look at a station S i. For convenience, we say that the bike, which is parked for the longest time, is leaving the station first. This is according to the first in, first out-principle. We know that requests arrive according to a Poisson process with parameter λ i. We can also say that bikes leave station S i with exponentially distributed intervals with parameter λ i as long as there are any bikes present. This is exactly what happens to customers at a single-server queue, which makes the bikes act like customers. Since each renter makes sure that a bike is leaving the station, the renter performs the job of the server in the queue. ow we look at a route R ij. We know that the riding time is exponential with µ ij. Furthermore, we know that there can be as many bikes (maximum K) as possible on this route at the same time. We can also describe this as bikes that are leaving route R ij at an exponential rate with parameter µ ij q ij, where q ij is the number of bikes at route R ij. Again this is exactly what happens to customers at an infinite-server queue. So also in this queue the bikes act like customers. Because the number of bikes being served at the same time can be K, there are at least K servers. Since there are only K bikes, the number of customers is at most K and therefore we can interpret this node as an infinite-server queue. The renters are the ones that are riding on all these bikes and make sure that a bike is leaving a route. Therefore, again the renters perform the job of the servers in the queue. The network is closed, since customers, or bikes, cannot leave the system. They are traveling from station to route, to station et cetera, or in other words from single-server queue to infiniteserver queue, to single-server queue et cetera. ote that there are + 2 queues. 4.2 Stationary Distribution As said before, for closed queueing networks with single-server or infinite-server queues and probabilistic routing we know the equations of the steady state and therefore the stationary distribution. First, we will introduce the equations of the throughput, also known as traffic equations. Every node has a throughput, denoted by γ i for station S i and γ ij for route R ij. The throughput for a station S i is determined by the bikes that arrive from a route R ji. For a route R ij the throughput is determined by the number of bikes that are sent onto that route. However, this is only possible via station S i, from which bikes are sent onto route R ij with probability p ij. Taking this all together, we get the following equations: We can substitute the second equation into the first one: γ i = γ ji and γ ij = γ i p ij (5) j=1 γ i = γ j p ji and γ ij = γ i p ij (6) j=1 /Department of Mathematics and Computer Science 10

13 ote that the throughput for a station can be described in terms of the throughput of other stations, but the routes are not involved anymore. This is logical, since a route R ij is nothing more than a temporary state of the bike between stations S i and S j. We can also note that there is more than one solution for the throughput equations, since every multiple of a solution is again a solution. To avoid confusion, we add two equations which makes the sum of all γ i as well as γ ij equal to 1. We get the following system of equations that describe the relative throughputs of the queueing network: γ i = j=1 γ j p ji γ ij = γ i p ij (7) γ i = 1 j=1 γ ij = 1 These throughputs are relative and therefore independent of the number of bikes in the system. For a fixed number of bikes, K, the throughput is also fixed. We then denote it by γ i (K) and γ ij (K) for i, j = 1... These fixed throughputs are defined by γ i (K) = ˆσ(K) γ i (8) γ ij (K) = ˆσ(K) γ ij where ˆσ(K) is the scaling factor depending on the number of bikes, K. The next step is to give the stationary distribution for the single-server and infinite-server nodes and combine them. Since we are working with an irreducible Markov process with a finite state space, the steady state is uniquely determined. Therefore, we can check whether the found stationary distribution is correct. We first consider two different kinds of queueing networks: a network with only single-server queues and one with only infinite-server queues. For the first network of single-server queues with service rate λ i, the steady state distribution is known to be π 1 (q) = G 1 1 ( γ i ) q i λ i [Harchol-Balter, 2013, elson, 2013]. Remember that q i denotes the number of bikes at station S i. G 1 is a constant which makes sure that the second property of π is satisfied, q Ω π(q) = 1. For the other network consisting of 2 infinite-server queues with service rate µ ij q ij, the steady state distribution is known to be π 2 (q) = G 1 2 j=1 1 q ij! ( γ ij µ ij ) q ij [Harchol-Balter, 2013, elson, 2013]. Again, G 2 is a constant which makes sure that the second property of π is satisfied. ow, we will consider our queueing network, which is a mix of the two groups described above. Therefore, we claim that the following equation denotes steady state distribution: in which π(q) = G 1 ( ( γ i ) q i ) ( λ i G = [( q Ω ) qi ) ( λ i ( γ i 1 q ij! ( γ ij µ ij ) q ij ), (9) 1 q ij! ( γ ij µ ij ) qij )], (10) /Department of Mathematics and Computer Science 11

14 which makes sure that q Ω π(q) = 1. With this, the first property of π is satisfied as well, namely that 0 π(q) 1 for all q Ω. Remark: in this distribution the riding times do not need to be exponential distributed, since the routes correspond to infinite-server queues. The distribution can be anything as long as the expectation of the riding time on route R ij is 1/µ ij [Kelly, 2011, p. 79]. This does not hold for the arrivals of requests. These must arrive according to a Poisson process, since the stations correspond to single-server queues. We now repeat the equation we obtained from the previous chapter. property of π, see equation (4), it must hold that According to the third π(q) λ i 1{q i > 0} + π(q) µ ij q ij 1{q ij > 0} (11) = π(q + a i b ij ) p ij λ i 1{q ij > 0} + π(q a i + b ji ) µ ji (q ji + 1) 1{q i > 0} It is already proven in Harchol-Balter [2013] and elson [2013] that the above given stationary distribution fits this equation. However, because this is difficult to see, we will still prove it. Therefore, we will first expand π(q + a i b ij ) and π(q a i + b ji ) in terms of π(q). Remember that π(q + a i b ij ) is a state with one bike more at station S i and one bike less at route R ij in comparison with π(q). In the same way, π(q a i + b ji ) is a state with one bike more at route R ji and one bike less at station S i in comparison with π(q). We have: π(q + a i b ij ) = π(q) γ i λ i q ijµ ij γ ij = π(q) γ i λ i q ijµ ij γ i p ij Here we have used that γ ij = γ i p ij, recall equation (7). We also have: π(q a i + b ji ) = π(q) λ i γ ji γ i µ ji (q ji + 1) = π(q) q ijµ ij λ i p ij We now will substitute these expressions into the right hand side of the equation and show that it is equal to the left hand side. π(q + a i b ij ) p ij λ i 1{q ij > 0} + π(q a i + b ji ) µ ji (q ji + 1) 1{q i > 0} = π(q) q ij µ ij λ i p ij p ij λ i 1{q ij > 0} + π(q) λ i γ i = π(q) q ij µ ij 1{q ij > 0} + π(q) γ ji µ ji (q ji + 1) µ ji(q ji + 1) 1{q i > 0} λ i γ i γ ji 1{q i > 0} = π(q) q ij µ ij 1{q ij > 0} + π(q) λ i 1{q i > 0} γ ji γ i In the last step we used the fact that j=1 = 1, recall equation (5). ow one can easily see that this equals equation (11). Since the steady state distribution is unique, it is given by equation (9): π(q) = G 1 ( ( γ i ) q i 1 ) ( λ i q ij! ( γ ij ) q ij ) µ ij /Department of Mathematics and Computer Science 12

15 5 Mean Value Analysis We now want to know how the system behaves in order to gain insight into certain problems that could occur. In the previous chapter we have derived the stationary distribution of our closed queueing network. However, even for a computer it would take very long to compute the normalization constant G in equation (10), since the state space Ω can get immense for large values of K and. Therefore, we would rather avoid the computation of this constant. As it turns out, this can be done through mean value analysis if we are only interested in mean values, such as the mean number of bikes per node and the mean sojourn time of a bike per node. Mean value analysis is based on multiple properties of closed queueing networks, which together form a recursive procedure for calculating these mean values. We will now introduce these properties and relationships that underly this procedure. First, mean value analysis is mainly based on the following, specific property of queueing networks. There are K customers in the network. When an arrival takes place, the joint queue length distribution of the network at this arrival instant is the same as the joint queue length distribution of the network with K 1 customers in the steady state. In other words, at the exact moment when a customer arrives at a queue in a system where there are K customers, the system state is as if there have been K 1 customers forever. Remember that we have defined Q(t) as the random vector of dimension + 2, with Q i (t) the number of bikes at station S i at time t and Q ij (t) the number of bikes on route R ij at time t. As said before, this vector depends on the number of bikes, K. Since we are only going to look at mean values, in steady state Q will not depend on t anymore, but it will depend on K. Therefore, from now on we will write Q(K) as well as Q i (K) and Q ij (K) to reflect this dependence. Relating the above mentioned property to our network, we can write this down as P[Q a (K) = q] = P[Q(K 1) = q], (12) where Q a (K) denotes the random vector Q(K) at an arbitrary arrival instant, excluding the arriving customer. This means that the following holds for all stations S i and routes R ij : E[Q a i (K)] = E[Q i(k 1)] E[Q a ij (K)] = E[Q ij(k 1)] (13) Second, we will give an expression for the mean sojourn time of a customer at a certain queue. For a single-server queue, it depends on the number of customers in the queue at an arrival instant. In other words, when a bike arrives at a station, it has to wait a certain amount of time, depending on the number of bikes at that station, and then it has to be served for a certain amount of time. These times are E[Q a i (K)]/λ i and 1/λ i respectively. Furthermore, for an infinite-server queue, the mean sojourn time of a customer only consists of the serving time 1/µ ij since there cannot be a queue. Therefore, denoting the sojourn time of a bike at a station by W i (K) and a bike on a route by W ij (K), the following expressions hold for all i, j = 1...: E[W i (K)] = (E[Q a i (K)] + 1)/λ i = (E[Q i (K 1)] + 1)/λ i (14) E[W ij (K)] = 1/µ ij In the second step we have used equation (13). Third, we will apply Little s Law and get the following two equations: E[Q i (K)] = γ i (K) E[W i (K)] = ˆσ(K) γ i E[W i (K)] E[Q ij (K)] = γ ij (K) E[W ij (K)] = ˆσ(K) γ ij E[W ij (K)] (15) for all i, j = 1... In the second step we have used equation (8), where ˆσ(K) denoted the scaling factor depending on the number of bikes, K. /Department of Mathematics and Computer Science 13

16 Finally, a trivial but very useful relation we need is E[Q i (K)] + E[Q ij (K)] = K (16) ow we combine equations (15) and (16) to find that K ˆσ(K) = γ i E[W i (K)] + j=1 γ ij E[W ij (K)] (17) In the next and final step, we will present the recursive procedure to calculate the mean sojourn times per station and per route, E[W i (K)], E[W ij (K)] and the mean number of bikes per station and per route, E[Q i (K)], E[Q ij (K)], by combining all equations derived above. 5.1 Procedure Mean Value Analysis Assume that, p ij, µ ij and λ i are known. First of all, we have E[Q i (0)] = E[Q ij (0)] = 0. Step 0: Calculate γ i and γ ij from equation (7). Set K=1 for step 1. Step 1: Given E[Q i (K 1)] and E[Q ij (K 1)], calculate E[W i (K)] and E[W ij (K)] from equation (14) for all i, j = 1... Step 2: Calculate ˆσ(K) from equation (17). Step 3: Calculate E[Q i (K)] and E[Q ij (K)] from equation (15) for all i, j = 1... ow we can repeat steps 1, 2 and 3 by increasing K by 1 until the desired value of K has been reached. This procedure is derived from Harchol-Balter [2013] and [elson, 2013, chapter ]. 5.2 Performance Analysis From this mean value analysis we obtain multiple quantities, like ˆσ(K), E[Q i (K)] and E[Q ij (K)] which we can use for further computations. For example, we are able to calculate the rate of circulation, σ(k), in a system with K bikes by: σ(k) = γ i (K) + γ ij (K) = ˆσ(K)γ i + ˆσ(K)γ ij = ˆσ(K)( γ i + γ ij ) = 2ˆσ(K) (18) We also have the ability to calculate the number of accepted and rejected requests per time unit as follows. Define λ = λ i, the average total number of requests per time unit. Define κ as the average total number of accepted requests and ν as the average total number of rejected requests, both per time unit. It follows directly that κ + ν = λ. Furthermore, we define θ as the average number of bikes that goes en route per time unit. Since κ = γ i (K) and θ = γ ij (K), it follows according to equation (18) that κ + θ = σ(k) = 2ˆσ(K) We also know that κ = θ, since each accepted request ensures that a bike goes en route. We can show this in terms of γ i (K) and γ ij (K): κ = γ i (K) = ˆσ γ i = ˆσ 1 = ˆσ γ ij = γ ij (K) = θ, /Department of Mathematics and Computer Science 14

17 where we have used that γ i = j=1 γ ij = 1, see equation (7). ow we have that κ = θ = 1 σ(k) = ˆσ(K) 2 ν = λ κ = λ ˆσ(K) (19) ext, we define u i as the probability that station S i has 0 bikes to rent: u i = P(0 bikes at station S i ) = q Ω q i=0 However, we have already seen that it will be difficult to calculate such a probability, since π(q) contains a normalization constant G, which is very hard to compute. What we do know: We can also express this as follows: or equivalently π(q) (1 u i )λ i = γ i (K) (20) (1 u i )λ i = κ u i λ i = ν The next step is to look at different networks, compute with help of mean value analysis the optimal average total revenues per time unit and see how this influences the rate of circulation and the distribution of the bikes over stations and routes. First, we need formulas to compute the average total revenues per time unit. Therefore, in the next chapter we will do a price analysis by calculating the average total revenues per time unit with different pricing methods. /Department of Mathematics and Computer Science 15

18 6 Price Analysis In this chapter, we will do a price analysis for the initial bike rental model based on the equivalent network. From now on, it costs a certain amount of euros to rent a bike. We will distinguish four kinds of price allocations. The first one is the simplest one, where every bike costs the same amount of euros to rent, regardless of which station the bike is rented or where it is returned. In the second price allocation, the price depends on the station where one rents the bike, but it does not depend on where the bike is returned. The third price allocation takes into account the station where the bike is returned, but the price is independent of the station where the bike is rented. The last price allocation depends on both the stations where the bike is rented and returned. First we will describe these four scenarios, then we will go deeper into the relation between the number of requests and the price of a bike rental and the relation between the routing probabilities and the price of a bike rental. 6.1 Price Allocation 1 In this scenario each bike rental has the same price, say t euros. Then the average total revenue T per time unit can be determined as follows: T = λ i (1 u i ) t = t γ i (K) = t ˆσ(K) γ i = t ˆσ(K) = t κ, (21) where equations (7), (8), (19) and (20) are used. This is very logical, since the average total revenue per time unit can be described as the number of accepted requests per time unit times the price of a bike, when each bike costs the same amount of euros. 6.2 Price Allocation 2 In the second scenario each bike rental at station S i has price t i. So prices can differ between the stations, but they do not depend on where one returns the bike. ow the average total revenue T per time unit can be computed as follows: T = λ i (1 u i ) t i = γ i (K)t i = ˆσ(K) γ i t i, (22) where we have used equations (8) and (20). 6.3 Price Allocation 3 In this scenario, the prices of a bike depend on where the bike is returned, but not on where the bike is picked up. We therefore say that a bike that is brought to station S j has price t j. The average total revenue T per time unit is computed as follows: T = λ i (1 u i ) p ij t j = γ i (K) p ij t j = ˆσ(K) γ i p ij t j = ˆσ(K) γ j t j, (23) where equations (7), (8) and (20) are used. 6.4 Price Allocation 4 In the last scenario, we combine scenarios 2 and 3. amely, the price of a bike depends on both the stations where the bike is picked up and where the bike is returned. We denote the price of a bike that goes on route R ij by ˆt ij. ow the average total revenue T per time unit is determined as follows: T = λ i (1 u i ) p ij ˆt ij = γ i (K) p ij ˆt ij = ˆσ(K) γ i p ij ˆt ij = ˆσ γ ij ˆt ij, (24) j=1 /Department of Mathematics and Computer Science 16

19 where equations (7), (8) and (20) are used. This price allocation is difficult to do computations with, since influences of the prices on the numbers of requests and routing probability are hard to find. Also it is not realistic to implement in a real system, since it can confuse customers with so many different prices. Therefore, we will only look at price allocations 1, 2 and 3. For these price allocations, we need to find the influences of the prices on the numbers of requests and routing probabilities. This will be done in the next section. 6.5 Influences of the Price Price Allocations 1 and 2 In price allocations 1 and 2, the price of a bike does not depend on the station where the bike is returned, but it can depend on where the bike is picked up. Hence, we should take into account that each λ i depends on the price of the bikes, t i. Therefore, we want to find the relation between the number of requests and the price of the bike rental. We assume that λ i only depends on t i for fixed i. In other words, the number of requests at a station only depends on the price at that station itself and does not depend on prices at other stations. We choose this relation to be linear, so λ i = a i t i + b i with a i, b i R constants. This ensures that at a certain price, the number of requests will be zero. Therefore, it limits the range of price allocations. The constants a i and b i will be determined as follows: for a network with known parameters λ i and t i, we have two points that are known. The first one is the relation between λ i and t i in the current network, which gives a point ( t i, λ i ). The second one is the limit price t l, which is independent of i. For this price, we assume that no one wants to rent a bike and therefore we have a point (t l, 0). ow we can draw a line between those two points and find a i and b i : a i = λ i t l t i b i = a i t l = λ i t l (25) t l t i Remark: we could also have chosen for the relation to be hyperbolic. However, that would imply that the number of requests will always be positive, no matter how expensive the bike rental is. ext, for optimization of the total revenue, it would be advantageous to increase the price as much as possible. Since this is not realistic and has a bad influence on the rate of circulation, which will become very low, we have chosen not to further examine this relationship. Since the prices are not dependent on the station where the bike is returned, the routing probabilities p ij are not influenced by the price in price allocations 1 and Price Allocation 3 In price allocation 3, the price of a bike rental depends on the station where the bike is returned. This influences the routing probabilities, p ij. Likewise, we choose a linear relation: p ij = c j t j + d j with c j, d j R constants. However, we must take into account that j=1 p ij = 1. Therefore, we will first calculate the constants to find ˆp ij. Finally, we scale all ˆp ij such that j=1 p ij = 1 by ˆp p ij = ij. We compute the constants as follows: again, for a network with known parameters k=1 ˆp ik p ij and t j, we can find two points that are known. The first one is the relation between p ij and t j in the current network, which gives a point ( t j, p ij ). The second one is again the point (t l, 0), which implies that when a bike has the limit price when returning to station S j, no one will rent /Department of Mathematics and Computer Science 17

20 this bike. From this it follows that for the unscaled routing probabilities ˆp ij Then p ij is computed by scaling p ij = c j = p ij t l t j d j = c i t l = p ij t l (26) t l t j ˆp ij k=1 ˆp ik. The number of requests in price allocation 3 also depends indirectly on the prices. We take the same relation as in price allocations 1 and 2, but now λ i depends on the average price at a 1 station: j=1 t j. Therefore, we get: λ i = a i 1 j=1 t j + b i (27) In the formulas of the average total revenues, the parameters λ i and p ij are fixed after the choice of t, t i and t j. In the next chapter we will do numerical experiments in order to reach the goal of this project. We will choose specific scenarios and will use mean value analysis to get a closer look into how the systems work. Before performing this analysis, we will extend the goal description in more specific terms. 6.6 Implementation When one chooses for a specific price allocation, these prices must be implemented in the real system. This can be done, for example, by signs at the stations with all prices or an application on the smart phone, where one can check the prices. We have assumed in the previous section that the number of requests at a station does only depend on the prices at that specific station and not on the prices at the other stations. This might be not be completely correct, since with an application one can choose on forehand at which station to rent or return the bike depending on the current prices. As said before, this implementation in a real system is difficult when each price depends on a specific route, since this would give 2 different price allocations. Customers can get confused by all these options, which is clearly not desirable. Price allocation 1, 2 and 3 are more easy to implement in a real system by signs or applications. /Department of Mathematics and Computer Science 18

21 7 umerical Experiments 7.1 Goal Description ow we have all the tools to compute average total revenues, rate of circulations and the distribution of the bikes. With this in hand we want to achieve our goal as follows: we consider a system in which all parameters λ i, µ ij, p ij are known for a specific price allocation. With this knowledge, we compute the constants in the formulas for λ i and p ij for each of the stations. Then we will maximize the average total revenue per time unit by varying the price allocation over different stations to find the best price allocation. We will start by considering a base scenario in which all bikes are well distributed. For this base scenario, the highest average total revenue with its price allocation will be computed, along with the rate of circulation. ext, we will consider more extreme scenarios in which the distribution of the bikes is unbalanced or where the rate of circulation is low. For these scenarios we will try to find the price allocation which gives the highest average total revenue per time unit and check whether this price allocation ensures a better distribution and a nice rate of circulation. We will also compare these results to the base scenario. Since the base scenario has a perfect balance in as well the distribution of the bikes as the routing probabilities, there is a good motivation to see the optimized base scenario as the best possible network. Therefore, we compare the special scenarios to the base scenario and see whether the average total revenues per time unit of the special scenarios come close the average total revenue per time unit in the optimized base scenario. For all the scenarios we have chosen the following parameters to be fixed: umber of stations: = 4. umber of bikes: K = 100. Average total number of requests: λ = λ i = 100. Average riding time for route R ij : µ ij = µ = 1 for i, j = 1, 2, 3, 4. Price for a bike: t ij = t = 5 euros for i, j = 1, 2, 3, 4. Price limit for a bike: t l = 10. These are the parameters of the initial networks. The number of requests and the routing probabilities, λ i and p ij, are different per scenario. ote that when maximizing the average total revenue per time unit, the values of λ, λ i, p ij and t ij may change. Since 5 euros is going to be the initial price and 10 euros is the price limit, we have chosen to vary the prices for each station or route between 3 and 8 euros to find the maximum average revenue per time unit. A cheaper or more expensive solution would not be reasonable. In all numerical experiments in this chapter, we will only look at integer prices to find a solution. Later, when we give a general heuristic method to find the optimal total revenue, we will explain how to find more accurate prices. All mean value analysis calculations and other computations are done in Mathematica. The complete program for price allocations 2 and 3 can be found in Appendix A and Appendix B, respectively. ote that price allocation 1 can be easily implemented in one of the two Mathematica programs by keeping all prices the same. otation mark: all values for total revenues, rates of circulation and numbers of requests at stations are presented in the next sections as average values per time unit. The values for the routing probabilities and the numbers of bikes at stations and en route are not per time unit, but they are average values. Furthermore, if all numbers of requests or all routing probabilities are equal, we call them uniform. In other words, we then have λ i = 25 and p ij = 1 for i, j = 1, 2, 3, 4. 4 /Department of Mathematics and Computer Science 19