SIMC NUS High School of Mathematics and Science May 28, 2014

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1 SIMC 014 NUS High School of Mathematics and Science May 8, 014 Introduction Singapore is world-renowned for her world-class infrastructure, including a highly developed transport system. Relative to her size, Singapore is one of the most well-connected cities in the world, and enjoys rapid travel speeds between all corners of the island. However, due to an increasing population and limited land area, it is crucial for the transport system to continually adapt, in order to minimize wear and tear of public transport, keep up with the pace of Singapore s economy, and to ensure affordable and convenient travel for all residents. Here we consider several situations in which we attempt to maximize the utility of land transportation, which are simplified models of problems encountered by transport authorities around the world. 1 Part A : Tourist in a hurry 1.1 Question 1 a) Exhaustive Enumeration First, note that there are (n 1)! ways of arranging the order of the attractions to visit. Also, for a given arrangement of the (n 1) attractions, there are 3 ways to travel between any two consecutive attractions, giving a total of 3 n possible ways of traveling for each arrangement. Hence, a total of 3 n (n 1)! candidate solutions have to be examined. Since a linear scan of the path is required to determine the time taken and validity of each candidate solution, the time complexity of this brute force algorithm is Θ(3 n (n!)). Algorithm 1: Exhaustive Search Data: n, b, c i,j,k, t i,j,k, i, j {0, 1,..., n 1}, k {0, 1, } Result: Valid route with minimum traveling time begin P {1,,..., n 1} /* Order of attractions to visit */ T An array of n zeroes /* Means of transportation between locations */ optimalt ime optimalcost optimalp ath An array of n 1 zeroes optimalt ransport An array of n zeroes for all permutations of P and possible methods of transportation for all paths (T) do cost = c 0,P [0],T [i] + c P [n ],0,T [n 1] time = t 0,P [0],T [i] + t P [n ],0,T [n 1] for i {0, 1,,..., n 3} do cost = cost + c P [i],p [i+1],t [i] time = time + t P [i],p [i+1],t [i] if cost b and time < optimalt ime then optimalt ime time optimalcost cost optimalp ath P optimalt ransport T 1

2 b) More Effective Exact Solver Mixed Integer Linear Programming This problem can be formulated as a mixed integer program. Suppose the modes of transportation are A, B and C and the attractions are 1,,...,n. a ij, b ij, c ij At ij, Bt ij, Ct ij Ac ij, Bc ij, Cc ij Z δ in (G) δ out (G) For i j, 0 i, j n, 1 if transport mode A, B, C respectively is used to travel from attraction i to j, 0 otherwise. For i j, 0 i, j n, it is the time taken to travel from i to j using mode A, B and C respectively in minutes. For i j, 0 i, j n, it is the cost to travel from i to j using mode A, B and C respectively in cents. Budget of the tour. Total number of trips into any of the attractions in a connected subgraph G from an attraction not in the subgraph. Total number of trips from any of the attractions in a connected subgraph G to an attraction not in the subgraph. Using the variables described above, we can formulate a brief mixed integer linear programming model: Subject to 1 i,j n i j Minimize 1 i,j n i j (a ij Ac ij + b ij Bc ij + c ij Cc ij ) Z (a ij At ij + b ij Bt ij + c ij Ct ij ) n (a kj + b kj + c kj ) = 1 for all 1 k n j=1 n (a kj + b kj + c jk ) = 1 for all 1 k n j=1 a ij + b ij + c ij 1 for all 1 i, j n, i j δ in (G) = 1 for all connected subgraphs G. δ out (G) = 1 for all connected subgraphs G. This algorithm can be made to run very fast and does so in practice. This is indeed the case on average large number of attractions. Moreover, iterative improvement can approach the optimal solution at successive steps. Moreover, a bound of the answer can be made by cutting plane techniques. The exponential number of terms can be easily remedied by considering Lagrange multipliers. This a really fast algorithm and can easily solve instances of up to 1000 attractions. However, the time complexity of such an algorithm is still a open problem, with no known non-trivial bound. Dynamic Programming This problem can be modeled as a Dynamic Programming problem. This is the Travelling Salesman Problem, except with an extra constraint in the form of a budget. This budget will be an additional variable within the DP computation, represented by B in cents. The

3 subproblem in this DP problem is, given a bitmask which shows which locations that have been visited, the current position of the tourist and the amount of remaining budget, find the route that passes through all the remaining locations with the minimum traveling time while keeping within said budget. The state for this DP is {bitmask, position, budget}. Within the transition, the next path in the route will be from the tourist s position and one of the unvisited locations using one of the 3 modes of transportation if it fits within the budget. The answer to the subproblem would be the the minimum of the sum of the traveling time of a path and DP (bitmask next, next, budget cost), where next is the destination of the chosen path and cost is the cost of said path. This value will be memoized. When all the locations have been visited, indicated by all the values in the bitmask turned on, the final path in the route will be from the tourist s position and his hotel, where the 3 modes of transportation will be tested if they fit within the remaining budget, and will store and return the path with minimum traveling time. There are a total of N possible states for the bitmask, N possible values for position and B possible values for budget, meaning the number of states is of complexity Θ(N N B). Each subproblem explores all unexplored locations with 3 modes of transportation, meaning the transition is of complexity Θ(3N) = Θ(N). The overall complexity of this algorithm is Θ(N N B). c) Fast Approximate Algorithm We utilise a two-stage greedy strategy. The first step is to determine which order of the roads to adopt. This is determined by a simple traversal. We first start from the place of origin (in this case, Marina Bay Sands) and select the cheapest road, the one which takes the least time to walk to that has yet to be visited. This results in a cycle which is Θ(log N) of the optimal answer (by considering only walking). Searching for the next road from each location takes Θ(N) time, making this step Θ(N ). The second stage of the algorithm is to optimize the route. We look at the possibility of improving the path we have chosen by exploring more expensive options, such as taking public transport or a cab. For each path between two locations within the chosen route, we consider the effectiveness Time Decrease of switching to a more expensive method, calculated by and compare all of these Cost Increase values. Should the cost increase be equal to 0, the effectiveness will be taken to be infinite. The improvement with the highest effectiveness that fits within the budget will be implemented. After this improvement, the effectivenesses of further improvement of that path, for example, from Public Transport to Taxi, will also be calculated. This process of choosing the most effective improvement will continue until no further improvements can be implemented. This is merely a heuristic which chooses the most worthwile improvement. There are some disadvantages with this algorithm, namely the biggest one being the inability to find the correct ordering of the roads as it does not take into account improvements that could be made. However, cost in real life tends to be roughly proportional to distance travelled, it is not an unreasonable assumption to make that it gives a rather close solution. Overall this gives an Θ(N ) algorithm. In this case, the optimal solution takes 05 minutes and costs $ This is achieved by taking the taxi from Marina Bay Sands (MBS) to Singapore Flyer, then taking the public transport to the Zoo then similarly to Vivo City and then walk to Resort World Sentosa. Finally taking the taxi to Buddha Tooth Relic Temple and then to MBS. Comparably, our greedy solution takes 08 minutes and costs $ This is achieved by taking the taxi from MBS to Singapore Flyer, then similarly to Buddha Tooth Relic Temple and then to Vivo City, then taking the public transport to Resorts World Sentosa, then similarly to the Zoo and finally back to MBS. 3

4 Part B : Bus Stops.1 Question a) Consider a subsection on the street from point α to β and suppose it contains exactly one stop at γ as shown below. Also, suppose the stop at γ is the nearest stops to both points α and β. Then, the expected walking time to the bus stop, E(W, α, β) can be given by: E(W, α, β) = (γ α) + (β γ) W (β α) (β α) 4W (β α) = β α 4W (1) where the inequality is true by Cauchy-Schwarz inequality, and equality holds iff γ = α+β. Now, consider the whole stretch of road, with bus stops at 0 = x 1 < x < < x n 1 < x n = L. Let z i define the zone of the road that is dominated by the stop at x i, meaning that for any point in the zone, the stop at x i is the one closest to it. Note that we can define: z i = [0, x 0+x 1 ] i = 0 [ x i 1+x i, x i+x i+1 ] 1 i n 1, L] i = n [ x n 1+x n We shall define z i, z i, z i as the length, rightmost point and leftmost point of z i respectively. Now we shall formulate an expression for E(A) as shown (not used in this part but used in question 3 for further generalisation): E(A) = 1 B n j=0 n ( z i z j x i x j + i j L) + 1 W i=0 n i=0 z i ( z i x i ) + ( z i x i ) z i Lemma: We prove that given any configuration of bus stops, supposing there are 3 consecutive bus stops, A, B, C, it is optimal to place B at the midpoint of AC. () Without loss of generality, let the bus stops be at c q,..., c, c 1, 0, x, 1, c 1, c,..., c r in this order (after normalization and translation), and we aim to optimize the position of x. We claim that this is at x = 1 regardless of B, W and the values of the other c i. Consider two possible of the above configurations A 1, A with x = x 1, x respectively. Considering E(A 1 ) E(A ), we notice that only several factors have to be considered - i) expected duration of an arbitrary bus journey from [0, 1] to [c q, 0] or [1, c r ], ii) expected duration of a bus journey within [0, 1], and iii) the expected walking time within [0, 1]. 4

5 i) We claim that this value does not depend on the choice of x. This can be shown easily algebraically, and involves the idea that the size of the interval dominated by x is always 1. ii) It can be evaluated that the expeted duration is simply x 1 1 x (x 0)+ 1 (1 x)+ x 1 x (1 0). This is computed to be x x+1, which is also minimized at x = 1. 4 iii) From the earlier equation (1), we have shown that it is optimal for x to be 1. As x is optimal for two cases, and independent of the third, it is clear that x = 1 Hence, the lemma is true. is optimal. Since the lemma is true, it is clear that the given solution with equal distances between stops is the only local minima for waiting time. As the problem states that a global minimum exists, this must be the optimum. b) We first determined a poor upper bound on the optimal number of bus stops. For n = 1, the expected time can be found to be Walking Time + Traveling Time. E(Walking Time) = 10 1 =, while E(Traveling Time) = 1 0 ( ) = By linearity of expectation, the 5 0 expected traveling time for n = 1 is.55. In order to deterine an upper bound on the optimal n 0, given n uniformly distributed bus stops, we simply subdivide the street into 3 equal portions, and consider the extreme ends. The expected waiting time for commuters traveling from one end to another is at least ( n ). 3 Thus, the expected waiting time is at least 1 of the above value. We then find the minimal n 9 0 such that the above value exceeds the corresponding value for n = 1. As the above term is linear in n, we can always find such a bound in general. In this case, our bound is n = After obtaining the bound, we utilise a Monte Carlo algorithm in order to determine the number of bus stops such that the expected time of travel from a random point to another is minimised. For each iteration, we considered the starting and ending point of one passenger. This is done with a uniform distribution over the interval (0, 0). Subsequently, each passenger is simulated and the time taken is determined. We simply repeated this process 10 7 times, and averaged out the timings. By computing from 1 to our aforementioned lower bound, we have obtained the optimal number of intervals is 11 and the expected value is ± Question 3 The model suggested is far from realistic. Here we consider the possible improvements to the model that would make it more realistic as a real-life model. 1. We can consider waiting time for buses. We assume buses to leave a terminal uniformly every k hours. This generalisation is trivially solved. We realise that since that each trip that has to wait means that it has to travel for more than a stop. This means we only have to consider the cases where we do not need to take the bus for more than a stop as they have no reason to wait for a bus. In this cases, we realise that if we consider only the size of each zone z i, there would only be a 1 z i cost additional. By Cauchy-Schwarz inequality, this additional cost is minimal where all z are equal. Hence, the solution we have determined in Part (a) holds true here. Hence, the optimal solution occurs where it is uniform.. We should note that realistically, the commuters will only take a bus if necessary. We can instead assume that the commuters will always take the optimal path which requires the minimal travelling time. In this model, the uniform model described in Question (a) does not remain optimal. In fact, it is trivially shown that for three bus stops, placing stops at (0.5, 0.50, 0.75) 5

6 is an improvement over (0.00, 0.50, 0.75) with B = 100, W = 5 and S = 0. However, we do not have an idea on how to approach this question. We conjecture that taking the optimal locations for n stops is the layout of n + stops, albeit without the endpoint. 3. The walking speeds of commuters are not fixed. For example, they can walking speed should be proportional to the distance they are walking. Alternatively, the walking speed of commuters could follow a normal distribution with mean W and standard deviation σ. A similar model could be used for buses, with their speed following a normal distribution. This generalisation is only possible to tackle considering the metric. The method depends on multiplying it into Equation (). After which, it may be possible to manipulate the inequalities to obtain a closed form, else we attempt tonumerically obtain the optimal values. 4. Commuters are unlikely to be spread evenly across the street. In occasional cases, there will be more people taking the bus at the middle. However, in most cases, people will be more congregated at the ends of a road, implying that the density at the end of a road is higher than at other parts. This generalisation can be solve via a normalisation of the cumulative frequency graph. From the graph above, we can see that we can take uniformly distributed points (1,, 3) to map to (1,, 3 ) new points on the length of the road. Thus, for this particular case, we would build 5 bus stops, two at the endpoints and 3 more at 1, and 3. Alternatively, the commuters might not alight uniformly across the road. The technique derived in the previous part does not simply generalise to this as both the starting and finishing locations obey a distinct distribution. The solution is actually to consider the probability density functions of the boarding and unboarding locations. After which, we convulate the two functions and consider the resulting probability density function. We transform this into a cumulative frequency graph and apply the normalisation step taken in the previous paragraph. This method works due to summation expression of the expected value. This method could be used in tandem with other generalisations. If we obtain the optimal value for some other case, we merely normalise it. 5. It is not necessary for there to be a bus stop at the two terminals. In particular, it could possibly be optimal to have no bus stops, or only 1 bus stop. (e.g. When B is very small, or S is very large). This generalisation can be solved in two parts. The first is where B is sufficiently large as compared to W and S isn t significant. In this case, we would like to minimise walking time, hence uniformly distributing the stations as done in Part (a) is optimal. The other case is where they are sufficiently large to play a part. This is where minimizing bus trips is more optimal. 6

7 This can be done by pushing all the bus stops very close to one of the endpoints of the road. If the two are in equal proportion, then any layouts of the stations would make no difference. 3 Part C : Traffic Junction 3.1 Question 4 a) We may assume that the cars are all points as we only need to consider the distance between the centers of the cars, and we may assume that car C is represented as a point on the y axis, and the cars on the horizontal road are represented as points on the x axis. Let y(t) represent the y-coordinate of car C at time t, and let x 1 (t) and x (t) be the x-coordinate of two cars, A and B, on the x axis. Without loss of generality, suppose the starting configuration is as shown in Figure 1, where x 1 (0) = 0, x (0) = d, y(0) = k. Since the speed of the cars are same, note that before car C reaches the x axis, y(t) + x 1 (t) = k Thus, it suffices to consider the minimum distance between cars A and C. In fact, we have the following formula for the distance between cars A and C with respect to time t:z(t) = [x 1 (t)] + [y(t)]. By Cauchy-Schwarz inequality, Hence, we have that: ([x 1 (t)] + [y(t)] )(1 + 1) (x 1 [t] + y[t]) = k (3) z(t) = [x 1 (t)] + [y(t)] k = k (4) Note that the equality holds in this case iff x 1 [t] = y[t] = k, and this occurs at t = k as shown v in Figure 3. Hence, suppose k = s. Then we have k = s. At t = k, the configuration shown v in Figure 4 occurs, after which, the relationship between cars C and B can modeled the as the reverse of what occurred between cars C and A. Hence, d k = s. As a road planner, we would check to see if the average gap between two adjacent cars on the horizontal road is greater than are equal to d. If it isn t, then the junction would need a traffic light. b) We may assume the car in the vertical lane, say C, waits at co-ordinate (0, s) as we have shown that this is the minimum distance of the car from the origin to ensure that it doesn t need to slow down. Suppose the first car on the horizontal road appears at time T. Let E(W ) refer to the expected waiting time. Let r = s v. E(W ) = E(W T r) + E(W T > r) = E(T + E(W ) T r) + 0 = E(T T r) + E(W )P (T r) Hence, we can express E(W ) by the following expression: E(W ) = E(T T r) 1 P (T r) (5) Since T is exponentially distributed with respect to λ, we have that E(T T r) = r 0 λte λt dt and 1 P (T r) = e λt. Therefore, we have that: r E(W ) = e λt λte λt dt = 1 λ (eλr λr 1) (6) 0 7

8 c) We can model this as a queue. We have a line of cars in a First In First Out model. Moreover, in this model, if a car starts moving, we can assume without loss of generality that it will exit the queue, as it only will move if there is a sufficient gap. We can make a few observations, the arrival rate of the queue is Markovian and memoryless. This is a Poisson process with rate λ = λ v. Moreover, the service rate is exponential (service is basically waiting for a sufficiently large gap). This has the rate µ = 1 λ (eλr λr 1) as we derived in part (b). There is only one server. This forms a M/M/1 queue. From this, we derive the utilization, ρ is λ. If ρ 1, it is clear that the expected number of cars queuing tends to. Moreover, if 0 ρ < 1, the expected number of cars queuing is ρ 1 ρ. 3. Question 5 Firstly, we can see that the two vertical lanes are independent, thus we can consider them separately. Due to this, note that we may flip the direction of one of the lanes due to symmetry, and we reflect it with respect to the center of one of the lanes perpendicular to it. An example of reflection is shown in Figure No further generalizations We can assume the following: The traffic in both directions is constant (λ and λ v ) apply to both directions Traffic lights can be placed at 4 possible locations (the corners of the junctions), to control traffic from the 4 directions independently. Traffic lights will only have red and green lights, and they switch between them instantaneously. 8 µ

9 No pedestrians will use the roads at any point in time. All cars will instantaneously brake to a speed of 0 as soon as they see a red light. The speed of cars in both directions is a constant, v. The cars moving in the horizontal direction will continue moving, and the vertical direction cars will try to make their way through the gaps. Similar to Question 4. To determine the methods to control the traffic lights, we suggest a few heuristics to minimize average waiting time and/or the number of cars queuing. These are: 1. We can determine various thresholds T horizontal and T vertical as the quotas for the number of cars we allow before switching the dominant direction of traffic. 1 This thresholds should be proportional to λ, the rate of the Poisson process. Unforunately, this is only achievable if we determine a method to evaluate the number of cars waiting.. Another possible approach is to switch the dominant direction after a fixed amount of time. The time between switches is also directly proportional to λ. 3. An implementation of the greedy approach is to always favour the side with longer queue. We alternate between the two sides, checking which side has a longer queue, and switching if the other side is longer. If the two sides are equal, we do not alter the dominant direction. 3.. Additional Considerations 1. The traffic in both directions might not be equal. We could thus model all 4 directions to have four different rate constants λ up, λ down, λ left, λ right. Our methods still extend to the aforementioned scenarios, however, due to different λ, we have to consider the sum of all λ. This is because in a Poisson process, we have to take the product to consider both of the cases. This would give us suitable ratios to apply the heuristics used in section Pedestrians could arrive in 4 possible directions at a Poisson rate as well. The pedestrians would be taken as points, and the travel at a speed of p 1, p, p 3, p 4. We treat this as an 8-laned equivalent of Question 4(c). 3. An orange traffic light could also be included, which would cause drivers to slow down linearly from v to 0 over time. 1 This is achieved by turning the currently dominant direction s corresponding traffic light to red, and then back to green 9

10 4. No priority would be given (i.e. Both sides try to give way to one another.) 5. Suppose the speeds of the cars are normally distributed with mean µ and standard deviation σ. For the last three generalizations, we may run a Monte Carlo algorithm to determine correct parameters of which to apply heuristics as mentioned in This is done by simulating the various traffic flows and parameters, then selecting the most efficient. 10