Traffic Flow Models 271 Traffic flow models and the evacuation problem Samuel W. malone Carl A. miller Daniel b. neill Duke University Durham, NC Advisor: David p Kraines Introduction We consider several models for traffic flow. A steady-state model employs model for car-following distance to derive the traffic-flow rate in terms of empirically estimated driving parameters. We go on to derive a formula for total evacuation time as a function of the number of cars to be evacuated The steady-state model does not take into account variance in speeds of vehicles. To address this problem, we develop a cellular automata model for traffic flow in one and two lanes and augment its results through simulation After presenting the steady-state model and the cellular automata models, we derive a space-speed curve that synthesizes results from both Ne address restricting vehicle types by anal rehicle speed variance To assess traffic merging, we investigate how congestion occurs We bring the collective theory of our assorted models to bear on five evac uation strategies Assumptions e Driver reaction time is approximately 1 sec Drivers tend to maintain a safe distance; tailgating is unusual All cars are approximately 10 ft long and 5 ft wide Almost all cars on the road are headed to the same destination The UIMAP Journal 22 (3)(2001)271-290. @Copyright 2001 by COMAP, Inc. All rights reserved Permission to make digital or hard copies of part or all of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice. Abstracting with credit is permitted, but copyrights for components of this work owned by others than COMAP must be honored. To copy otherwise, to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP
Traffic Flow Models 271 Traffic Flow Models and the Evacuation Problem Samuel W. Malone Carl A. Miller Daniel B. Neill Duke University Durham, NC Advisor: David P. Kraines Introduction We consider several models for traffic flow. A steady-state model employs a model for car-following distance to derive the traffic-flow rate in terms of empirically estimated driving parameters. We go on to derive a formula for total evacuation time as a function of the number of cars to be evacuated. The steady-state model does not take into account variance in speeds of vehicles. To address this problem, we develop a cellular automata model for traffic flow in one and two lanes and augment its results through simulation. After presenting the steady-state model and the cellular automata models, we derive a space-speed curve that synthesizes results from both. We address restricting vehicle types by analyzing vehicle speed variance. To assess traffic merging, we investigate how congestion occurs. We bring the collective theory of our assorted models to bear on five evacuation strategies. Assumptions • Driver reaction time is approximately 1 sec. • Drivers tend to maintain a safe distance; tailgating is unusual. • All cars are approximately 10 ft long and 5 ft wide. • Almost all cars on the road are headed to the same destination. The UMAP Journal 22 (3) (2001) 271ñ290. c Copyright 2001 by COMAP, Inc. All rights reserved. Permission to make digital or hard copies of part or all of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice. Abstracting with credit is permitted, but copyrights for components of this work owned by others than COMAP must be honored. To copy otherwise, to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP
272 The UMAP Journal 22.3 (2001) erns density d: the number of cars per unit distance Occupancy n: the proportion of the road covered by cars Flow g: the number of cars per time unit that pass a given point Separation distance s: the average distance between midpoints of successive cars Speed v: the average steady-state speed of cars Travel Time: how long a given car spends on the road during evacuation Total Travel Time: the time until the last car reaches safety The Steady-State model Develo pmei Car-following is described successfully by mathematical models; following Rothery [1992, 4-1], we model the average separation distance s as a function of common speed u s=a+ Bu+r where a, B, and y have the physical interpretations the effective vehicle length L B=the reaction time, and the reciprocal of twice the maximum average deceleration of a following vehicle This relationship allows us to obtain the optimal value of traffic density (and speed)that maximizes flow Theorem. For q= kv(the fundamental equation for traffic Hlow) and (1), traffic How g is maximized at q*=(+21/2L12)-1, (L/7)1/2 B(/L)1 32 Proof: Consider N identical vehicles, each of length L, traveling at a steady- state speed v with separation distance given by (1). If we take a freeze-frame picture of these vehicles spaced over a distance D, the relation D=NL+Ns
272 The UMAP Journal 22.3 (2001) Terms Density d: the number of cars per unit distance. Occupancy n: the proportion of the road covered by cars. Flow q: the number of cars per time unit that pass a given point. Separation distance s: the average distance between midpoints of successive cars. Speed v: the average steady-state speed of cars. Travel Time: how long a given car spends on the road during evacuation. Total Travel Time: the time until the last car reaches safety. The Steady-State Model Development Car-following is described successfully by mathematical models; following Rothery [1992, 4-1], we model the average separation distance s as a function of common speed v: s = α + βv + γv2, (1) where α, β, and γ have the physical interpretations: α = the effective vehicle length L, β = the reaction time, and γ = the reciprocal of twice the maximum average deceleration of a following vehicle. This relationship allows us to obtain the optimal value of traffic density (and speed) that maximizes flow. Theorem. For q = kV (the fundamental equation for traffic flow) and (1), traffic flow q is maximized at q∗ = (β + 2γ1/2L1/2) −1, v∗ = (L/γ) 1/2, k∗ = β(γ/L)1/2 − 2γ β2 − 4γL . Proof: Consider N identical vehicles, each of length L, traveling at a steadystate speed v with separation distance given by (1). If we take a freeze-frame picture of these vehicles spaced over a distance D, the relation D = NL + Ns
Traffic Flow Models 273 must hold, where s is the bumper-to-bumper separation. Since s=s-L,we obtain k=N/D=N/(NL+NS=1/(L+s=l/s We invoke(1)to get k +Bu+r This is a quadratic equation in v; taking the positive root yields (k)= Applying q= ku, we have 9(k)k 21/41+(2-4D) Differentiating with respect to k, setting the result equal to zero, and wading hrough algebra yields the optimal values given. Interpretation and Uses We can estimate q*,k*, and u* from assumptions regarding car length(L), reaction time(B), and the deceleration parameter(). If welet L= 10 ft, B= ls and? x023 s /ft(a typical value [Rothery 1992]),we obtain 0.510cars/s,*=20.85ft/s,k=0.024cars/ft A less conservative estimate for y is ?=2(af -a1 ) where af and aL are the average maximum decelerations of the following and lead vehicles Rothery [1992]. We assume that instead of being able to stop instantaneously(infinite deceleration capacity), the leading car has deceleration capacity twice that of the following car. Thus, instead of y= 1/ 2a=023 s/ft, we use the implied value for a to computer=i(a-1-2a)=37=0.0115 s2/ ft and get 9=0.596 cars/s, v*=29.5 ft/s N 20 mph, k*=.020 cars/ft Going 20 mph in high-density traffic with a bumper-to-bumper separation of 40 ft is not bad The 1999 evacuation was far from optimal. Taking 18 h for the 120-mi trip from Charleston to Columbia implies an average speed of 7 mph and a bumper-to-bumper separation of 7 ft Limitations of the steady-State Model The steady-state model does not take into account the variance of cars' speeds. Dense traffic is especially susceptible to overcompensating or under- compensating for the movements of other drivers A second weakness is that the value for maximum flow gives only a first order approximation of the minimum evacuation time Determining maximum flow is distinct from determining minimum evacuation time
Traffic Flow Models 273 must hold, where s is the bumper-to-bumper separation. Since s = s − L, we obtain k = N/D = N/(NL + Ns )=1/(L + s )=1/s. We invoke (1) to get k = 1 α + βv + γv2 . This is a quadratic equation in v; taking the positive root yields v(k) = 1 2γ 4 γ k + (β2 − 4γL) − β 2γ . Applying q = kv, we have q(k) = k 2γ 4 γ k + (β2 − 4γL) − kβ 2γ . Differentiating with respect to k, setting the result equal to zero, and wading through algebra yields the optimal values given. Interpretation and Uses We can estimate q∗, k∗, and v∗ from assumptions regarding car length (L), reaction time (β), and the deceleration parameter (γ). If we let L = 10 ft, β = 1 s, and γ ≈ .023 s2/ft (a typical value [Rothery 1992]), we obtain q∗ = 0.510 cars/s, v∗ = 20.85 ft/s, k∗ = 0.024 cars/ft. A less conservative estimate for γ is γ = 1 2 (a−1 f − a−1 l ), where af and al are the average maximum decelerations of the following and lead vehicles Rothery [1992]. We assume that instead of being able to stop instantaneously (infinite deceleration capacity), the leading car has deceleration capacity twice that of the following car. Thus, instead of γ = 1/2a = .023 s2/ft, we use the implied value for a to compute γ = 1 2 a−1 − 2a−1 = 1 2 γ = 0.0115 s2/ft and get q∗ = 0.596 cars/s, v∗ = 29.5 ft/s ≈ 20 mph, k∗ = .020 cars/ft. Going 20 mph in high-density traffic with a bumper-to-bumper separation of 40 ft is not bad. The 1999 evacuation was far from optimal. Taking 18 h for the 120-mi trip from Charleston to Columbia implies an average speed of 7 mph and a bumper-to-bumper separation of 7 ft. Limitations of the Steady-State Model The steady-state model does not take into account the variance of carsí speeds. Dense traffic is especially susceptible to overcompensating or undercompensating for the movements of other drivers. A second weakness is that the value for maximum flow gives only a firstorder approximation of the minimum evacuation time. Determining maximum flow is distinct from determining minimum evacuation time.
274 The UMAP Journal 22.3 (2001) Minimizing evacuation Time with the steady-State Model Initial Considerations The goal is to keep evacuation time to a minimum, but the evacuation route must be as safe as possible under the circumstances. How long on average it takes a driver to get to safety( Columbia)is related to minimizing total evacu ation time but is not equivalent A General Performance measure A metric M that takes into account both maximizing traffic flow and mir mizing individual transit time T is M=W+(1-W) where< w< I is a weight factor, D is the distance that to traverse, I is the number of lanes, and n is the number of cars to evacuate. This metric assumes that the interaction between lanes of traffic(passing)is negligible, so that total flow is that of an individual lane times the number of lanes. Giver w, minimizing M amounts to solving a one-variable optimization problem in either v or k. Setting W= l corresponds to maximizing flow, as in the preceding section Setting W=0 corresponds to maximizing speed, subject to the constraint u< cruise, the preferred cruising speed; this problem has solution M=D/cruise. The model does not apply when cars can travel at cruise Setting W=1/2 corresponds to minimizing the total evacuation time The evacuation time is the time D/v for the first car to travel distance d plus the time N/lq for the n cars to flow by the endpoint. To illustrate that maximizing traffic flow and maximizing speed are out of sync, we calculate the highest value of W for which minimizing M would result in an equilibrium speed of cruise. This requires a formula for the equilibrium value u* that solves the problem minimize M(u)=WW(L +Bu+u2) subject to0<v≤ Cruis
274 The UMAP Journal 22.3 (2001) Minimizing Evacuation Time with the Steady-State Model Initial Considerations The goal is to keep evacuation time to a minimum, but the evacuation route must be as safe as possible under the circumstances. How long on average it takes a driver to get to safety (Columbia) is related to minimizing total evacuation time but is not equivalent. A General Performance Measure A metric M that takes into account both maximizing traffic flow and minimizing individual transit time T is M = W N lq + (1 − W) D v , where 0 ≤ W ≤ 1 is a weight factor, D is the distance that to traverse, l is the number of lanes, and N is the number of cars to evacuate. This metric assumes that the interaction between lanes of traffic (passing) is negligible, so that total flow is that of an individual lane times the number of lanes. Given W, minimizing M amounts to solving a one-variable optimization problem in either v or k. Setting W = 1 corresponds to maximizing flow, as in the preceding section. Setting W = 0 corresponds to maximizing speed, subject to the constraint v ≤ vcruise, the preferred cruising speed; this problem has solution M = D/vcruise. The model does not apply when cars can travel at vcruise. Setting W = 1/2 corresponds to minimizing the total evacuation time N lq + D v . The evacuation time is the time D/v for the first car to travel distance D plus the time N/lq for the N cars to flow by the endpoint. To illustrate that maximizing traffic flow and maximizing speed are out of sync, we calculate the highest value of W for which minimizing M would result in an equilibrium speed of vcruise. This requires a formula for the equilibrium value v∗ that solves the problem minimize M(v) = W N(L + βv + γv2) lv + (1 − W) D v subject to 0 < v ≤ vcruise
Traffic Flow Models 275 The formula for M(u) comes from (1),q= ku, and k= 1/s. Differentiating with respect to u, setting the result equal to zero, and solving for speed yields Cruise, L For cruise to equal the square root, we need (uZruisey-L) Using N= 160,000 cars, D=633, 600 ft(120 mi), I=2 lanes, cruise =60 mph 88 ft/s, y=0115 s /ft, and L=10 ft, we obtain W N 1/11. Thus, minimizing evacuation time in situations involving heavy traffic flow is incompatible with allowing drivers to travel at cruise speed with a safe stopping distance Computing Minimum Evacuation Time rom the fact that T= 2M when W=1/2, we obtain "=ku V=IL+ DU/N B1/2L+DN]-1/2-27+乙+N DL/N] B2-4D-7+D/N1 The minimum evacuation time is lq Predictions of the Steady-State Model a For N large, evacuation time minimization is essentially equivalent to the ow maximization( Figure 1), and it can be shown analytically that Nim How(N) The predicted evacuation time of 40 h for N= 160,000 seems reasonable We can evaluate the impact of converting I-26 to four lanes by setting l=4 in the equation for minimum evacuation time, yielding T N 23 h. For the steady-state model, this prediction makes sense, since the model does not deal with the effect of the bottleneck that will occur when Columbia is swamped by evacuees. The bottleneck would be compounded by using four lanes instead of two. On balance, however, doubling the number of lanes would lead to a net decrease in evacuation time
Traffic Flow Models 275 The formula for M(v) comes from (1), q = kv, and k = 1/s. Differentiating with respect to v, setting the result equal to zero, and solving for speed yields v∗ = min vcruise, 1 γ L + (1 − W) W · Dl N . For vcruise to equal the square root, we need W = 1 + N Dl v2 cruiseγ − L −1 . Using N = 160,000 cars, D = 633,600 ft(120 mi), l = 2 lanes, vcruise = 60 mph = 88 ft/s, γ = .0115 s2/ft, and L = 10 ft, we obtain W ≈ 1/11. Thus, minimizing evacuation time in situations involving heavy traffic flow is incompatible with allowing drivers to travel at cruise speed with a safe stopping distance. Computing Minimum Evacuation Time From the fact that T = 2M when W = 1/2, we obtain q∗ = k∗v∗, v∗ = 1 γ [L + Dl/N], k∗ = βγ1/2[L + Dl/N] −1/2 − 2γ [L+ 1 2 Dl/N] [L+Dl/N] [β2 − 4γL] − γ [Dl/N]2 [L+Dl/N] . The minimum evacuation time is T ∗ = N lq∗ + D v∗ . Predictions of the Steady-State Model For N large, evacuation time minimization is essentially equivalent to the flow maximization (Figure 1), and it can be shown analytically that lim N→∞ Tflow(N) Tmin(N) = 1. The predicted evacuation time of 40 h for N = 160,000 seems reasonable. We can evaluate the impact of converting I-26 to four lanes by setting l = 4 in the equation for minimum evacuation time, yielding T ≈ 23 h. For the steady-state model, this prediction makes sense, since the model does not deal with the effect of the bottleneck that will occur when Columbia is swamped by evacuees. The bottleneck would be compounded by using four lanes instead of two. On balance, however, doubling the number of lanes would lead to a net decrease in evacuation time.
276 The UMAP Journal 22.3 (2001) 40000000080001000001200001400000008000200000 N 8 gure 1. Comparison of minimum evacuation time(lower line)and maximum flow evacuation me(upper line One-Dimensional Cellular automata model Development In heavy traffic, cars make repeated stops and starts, with somewhat arbi each time state, cars move according to the following rules ella ness into trary timing a good model of heavy traffic should take this randomness into account but also be simple enough to give an explicit formula for speed We divide a single-lane road into cells of equal length. A ontains one car or no car. A car is blocked if the cell directly in front of it is occupied. At . a blocked car does not move If a car is not blocked, it advances to the next cell with probability p The decisions of drivers to move forward are made independently A traffic configuration can berepresented by a functionf: Z-10, 1, where f(k)= l if cell k contains a car and f(k)=0 if not. Probability distributions on the set of all such functions are called binary processes
276 The UMAP Journal 22.3 (2001) Figure 1. Comparison of minimum evacuation time (lower line) and maximum flow evacuation time (upper line). One-Dimensional Cellular Automata Model Development In heavy traffic, cars make repeated stops and starts, with somewhat arbitrary timing; a good model of heavy traffic should take this randomness into account but also be simple enough to give an explicit formula for speed. We divide a single-lane road into cells of equal length. A cell contains one car or no car. A car is blocked if the cell directly in front of it is occupied. At each time state, cars move according to the following rules: • A blocked car does not move. • If a car is not blocked, it advances to the next cell with probability p. The decisions of drivers to move forward are made independently. A traffic configuration can be represented by a functionf : Z → {0, 1}, where f(k)=1 if cell k contains a car and f(k)=0 if not. Probability distributions on the set of all such functions are called binary processes
Traffic Flow Models 277 Given a process X, define a process Ip(X) according to the following rule X(i),X(i+1)=(1,0 then (Ip (X)(), IP()(+1))=1(0, 1), with probability p; (1,0), with probability 1-p This rule is identical to the traffic flow rule given above: If X represents the traffic configuration at time t, Ip(X) gives the traffic configuration at timet+1 We are interested in what the traffic configuration looks like after several iterations of I. Let Ip(X)mean Ip applied n times to X. The formula for traffic speed in terms of density comes from the following theorem Theorem. Suppose that X is a binary process of density d. Let 1-1-4d(1-d)p 2pd and let Mp, d denote the Markov chain with transition probabilities prob. 1 1, w/prob. r The sequence of processes X, Ip(X), IF(X), IS(X),.. converges to Mp. d Here"density"means the frequency with which Is appear, analogous to the averagenumber of cars per cell. This theorem tells what the traffic configuration looks like after a long period of time Knowing the transition probabilities allows us to compute easily the average speed of the cars in Mp. d: the average speed is the likelihood that a randomly chosen car is not blocked and advances to the next cell at the next time state Pr[(Mp, d(i))=0p, (i)=1 Prp, d(i+1=0 Mp, d(i=1] Pr[I(Mp. d(i))=0Mp d(i)=l and Mp, d(i+1)=o 7=(1-1-01=)p=1-4=D.a The model does not accurately simulate high-speed traffic and does not take into account following distance, and the stop-and-start model of carmovement is not accurate when traffic is sparse. The model is best for slow traffic(under 15 mph)with frequent stops IThis theorem is the result of previous research [Miller 2001] by an author of this paper. The Appendix on binary processes(written during the contest period) gives a relatively complete prod of the result. [EDITOR'S NOTE: We unfortunately must omit the Appendix J
Traffic Flow Models 277 Given a process X, define a process Ip(X) according to the following rule: If X(i), X(i + 1) = (1, 0) then Ip(X)(i), Ip(X)(i + 1) = (0, 1), with probability p; (1, 0), with probability 1 − p. This rule is identical to the traffic flow rule given above: If X represents the traffic configuration at time t, Ip(X) gives the traffic configuration at time t+ 1. We are interested in what the traffic configuration looks like after several iterations of I. Let In p (X) mean Ip applied n times to X. The formula for traffic speed in terms of density comes from the following theorem1: Theorem. Suppose that X is a binary process of density d. Let r = 1 − 1 − 4d(1 − d)p 2pd and let Mp,d denote the Markov chain with transition probabilities 0 −→ 0, w/ prob. 1 − r; 1, w/ prob. 1 − r; 1 −→ 0, w/ prob. r; 1, w/ prob. r. The sequence of processes X, Ip(X), I2 p (X), I3 p (X),... converges to Mp,d. Here ìdensityî means the frequency with which 1s appear, analogous to the average number of cars per cell. This theorem tells what the traffic configuration looks like after a long period of time. Knowing the transition probabilities allows us to compute easily the average speed of the cars in Mp,d: the average speed is the likelihood that a randomly chosen car is not blocked and advances to the next cell at the next time state: v = P r [I(Mp,d(i))=0 |Mp,d(i) = 1] = P r [Mp,d(i + 1)=0 |Mp,d(i) = 1] · P r [I(Mp,d(i))=0 |Mp,d(i)=1 and Mp,d(i + 1) = 0] = rp = 1 − 1 − 4d(1 − d)p 2pd p = 1 − 1 − 4d(1 − d)p 2d . (2) The model does not accurately simulate high-speed traffic and does not take into account following distance, and the stop-and-start model of car movement is not accurate when traffic is sparse. The model is best for slow traffic (under 15 mph) with frequent stops. 1This theorem is the result of previous research [Miller 2001] by an author of this paper. The Appendix on binary processes (written during the contest period) gives a relatively complete proof of the result. [EDITORíS NOTE: We unfortunately must omit the Appendix.]
278 The UMAP Journal 22.3 (2001) Low Speeds We must set three parameters Ar= the size of one cell, the space taken up by a car in a tight traffic jam; we set it to 15 ft, slightly longer than most cars At=the length of one time interval, the shortest time by a driver to move into the space in front; we take this to be 0.5 s p=the movement probability, representing the proportion of drivers whomove at close to the overall traffic speed; we let p=0.85 In(2)we insert the factor(. z/At)to convert from cells/time-state to ft/s 1-√1-4d(1-d)P Density d isin cars per cell, related to occupancy n by d=(15 it/10 ft)xn=3n/2 Table 1 gives v for various values of n and k Table 1 for various values of n and k 0.600.0600.90 0.550.055083 0.500.0500.75 0.400.0400.6010 0.350.0350.5312 0.300.03004514 0.250. 0.200.0200.30 6 Ine lane To explore the one-dimensional cellular automata model, we wrote a simple simulation in C++. The simulation consists of a 5,000-element long(circular) array of bits, with a l representing a car and a O representing a(car-sized)empty space. The array is initialized randomly based on a value for the occupan An element is initialized to 1 with probability n, or to O with probability I-n The array is iterated over 5,000 time cycles: On each cycle, a car moves forward with probability p if the square in front of it is empty. The flow q is calculated as the number of cars N passing the end of the array divided by the number of time cycles(i. e, q=N/5000), and thus the average speed of an individual car in cells per time cycle is v=q/n=N/5000n
278 The UMAP Journal 22.3 (2001) Low Speeds We must set three parameters: ∆x = the size of one cell, the space taken up by a car in a tight traffic jam; we set it to 15 ft, slightly longer than most cars. ∆t = the length of one time interval, the shortest time by a driver to move into the space in front; we take this to be 0.5 s. p = the movement probability, representing the proportion of drivers who move at close to the overall traffic speed; we let p = 0.85. In (2) we insert the factor (∆x/∆t) to convert from cells/time-state to ft/s: v = ∆x ∆t 1 − 1 − 4d(1 − d)p 2d . Density d is in cars per cell, related to occupancy n by d = (15 ft/10 ft)×n = 3n/2. Table 1 gives v for various values of n and k. Table 1. v for various values of n and k. nK d v ft−1 (mph) 0.60 0.060 0.90 2 0.55 0.055 0.83 4 0.50 0.050 0.75 5 0.45 0.045 0.68 8 0.40 0.040 0.60 10 0.35 0.035 0.53 12 0.30 0.030 0.45 14 0.25 0.025 0.38 15 0.20 0.020 0.30 16 One Lane To explore the one-dimensional cellular automata model, we wrote a simple simulation in C++. The simulation consists of a 5,000-element long (circular) array of bits, with a 1 representing a car and a 0 representing a (car-sized) empty space. The array is initialized randomly based on a value for the occupancy n: An element is initialized to 1 with probability n, or to 0 with probability 1 − n. The array is iterated over 5,000 time cycles: On each cycle, a car moves forward with probability p if the square in front of it is empty. The flow q is calculated as the number of cars N passing the end of the array divided by the number of time cycles (i.e., q = N/5000), and thus the average speed of an individual car in cells per time cycle is v = q/n = N/5000n
Traffic Flow Models 279 Table 2 Comparison of simulation and equation values for speed P=3/4 n Simulation Equation Simulation Equation 0.356 0.592 0.589 6 0.234 0.392 0.392 0.8 0.108 0.171 0.174 Table 2 shows results for various values of the occupancy n and probability , verifying the accuracy of the one-dimensional cellular automata equation(2) The value for p should be related to the mean and standard deviation of cruise. The mean and standard deviation of a binary random variable are p and p(1-p). We have p(-p) 1+ For u(cruise)=60 mph and o(cruise)=5 mph, we get p= 144/145. Now, L=utp; so assuming L=10 ft, p= 144/145, and u=60 mph=88 ft/s,we obtain a time step of 0. 113 s. We now use the model to predict how fast(on average)a car moves in a ingle lane as a function of the occupancy. We consider the" relative speed "Urel the average speed divided by the(mean) cruise speed. The average speed is given by the one-dimensional cellular automata equation, and the cruise speed is p cells per time cycle, so this gives us √1-4n(1-n)p Cruise Using p= 144/145, we calculate vrel and vavg as a function of n(Table 3) The model predicts that for low occupancy the average speed will be near the cruise speed but for occupancies greater than 0.5 the average speed will be significantly lower. The cellular automata model does not take following distance into account; thus, it tends to overestimate vavg for high speeds and is most accurate when occupancy is high and speed is low. Table 3 also shows flow rate q= nUavg/L, in cars/s, as a function of occu- pancy. The flow rate is symmetric about n=0.5. Each car movement can be thought of as switching a car with an empty space, so the movement of cars to the right is equivalent to the movement of holes to the left The model fails, however, to give a reasonable value for the maximum flow rate:4.1 cars/sx 14, 600 cars/h, about seven times a reasonable maximum rate [ Rothery 1992]. The reason is that the cell size equals the car length, a correct
Traffic Flow Models 279 Table 2. Comparison of simulation and equation values for speed. Speed p = 1/2 p = 3/4 n Simulation Equation Simulation Equation 0.2 0.433 0.438 0.694 0.697 0.4 0.356 0.349 0.592 0.589 0.6 0.234 0.232 0.392 0.392 0.8 0.108 0.110 0.171 0.174 Table 2 shows results for various values of the occupancy n and probability p, verifying the accuracy of the one-dimensional cellular automata equation (2). The value for p should be related to the mean and standard deviation of v cruise. The mean and standard deviation of a binary random variable are p and p(1 − p). We have p p(1 − p) = µ σ −→ p = 1 1 + σ µ 2 . For µ(vcruise) = 60 mph and σ(vcruise)=5 mph, we get p = 144/145. Now, L = µtp; so assuming L = 10 ft, p = 144/145, and µ = 60 mph = 88 ft/s, we obtain a time step of 0.113 s. We now use the model to predict how fast (on average) a car moves in a single lane, as a function of the occupancy. We consider the ìrelative speedî vrel, the average speed divided by the (mean) cruise speed. The average speed is given by the one-dimensional cellular automata equation, and the cruise speed is p cells per time cycle, so this gives us vrel = vavg vcruise = 1 − 1 − 4n(1 − n)p 2pn . Using p = 144/145, we calculate vrel and vavg as a function of n (Table 3). The model predicts that for low occupancy the average speed will be near the cruise speed, but for occupancies greater than 0.5 the average speed will be significantly lower. The cellular automata model does not take following distance into account; thus, it tends to overestimate vavg for high speeds and is most accurate when occupancy is high and speed is low. Table 3 also shows flow rate q = nvavg/L, in cars/s, as a function of occupancy. The flow rate is symmetric about n = 0.5. Each car movement can be thought of as switching a car with an empty space, so the movement of cars to the right is equivalent to the movement of holes to the left. The model fails, however, to give a reasonable value for the maximum flow rate: 4.1 cars/s ≈ 14,600 cars/h, about seven times a reasonable maximum rate [Rothery 1992]. The reason is that the cell size equals the car length, a correct
280 The UMAP Journal 22.3 (2001) Relative speed, average speed, and flow rate as a function of occupancy flow rate 0.6658 88888582 (ft/s)(ft/s)(cars/s) 0.9.111 10 0.9 approximation only as car speeds approach zero and occupancy approaches 1 For 20.5, we should have a larger cell size; so we assume that cell size equals car length plus following distance and that following distance is proportional to speed. Assuming a l s following distance, we obtain cell size as C=L+vavg×(1 But we do not know the value of vavg until we use the cell size to obtain it! For m large, we can assume that vavg a Cruise and find an upper bound on cell size C=L+Cruise X(1 sec)=98 ft We divide the original flow rate by the increased cell size to obtain a more reasonable flow rate 4.063 cars/s 98 ft/10 f=0415 cars/s x 1,500 cars/h This is likely to be an underestimate; for greater accuracy, we must find a method to compute the correct cell size before finding the speed. We address this problem later Two Lanes We expand the one-dimensional model. The simulation consists of a two- dimensional (1000 x 2)array of bits. The array is initialized randomly and chen iterated over 1,000 time cycles: On each cycle, a car moves forward with probability p if the cell in front of it is empty. If not, provided the cells beside it and diagonally forward from it are unoccupied, with probability p the car changes lanes and moves one cell forward Like the one-lane simulation, the two-lane one is correct only for high den- sities and low speeds, since it uses cell size equal to car length. Hence, we do not use the two-lane simulation to compute the maximum flow rate. However
280 The UMAP Journal 22.3 (2001) Table 3. Relative speed, average speed, and flow rate as a function of occupancy. n vrel vavg flow rate (ft/s) (ft/s) (cars/s) 0.1 .999 88 0.9 0.2 .998 88 1.8 0.3 .995 88 2.6 0.4 .987 87 3.5 0.5 .923 81 4.1 0.6 .658 58 3.5 0.7 .426 38 2.6 0.8 .249 22 1.8 0.9 .111 10 0.9 approximation only as car speeds approach zero and occupancy approaches 1. For n ≥ 0.5, we should have a larger cell size; so we assume that cell size equals car length plus following distance and that following distance is proportional to speed. Assuming a 1 s following distance, we obtain cell size as C = L + vavg × (1 sec). But we do not know the value of vavg until we use the cell size to obtain it! For n large, we can assume that vavg ≈ vcruise and find an upper bound on cell size: C = L + vcruise × (1 sec) = 98 ft. We divide the original flow rate by the increased cell size to obtain a more reasonable flow rate: q = 4.063 cars/s 98 ft/10 ft = 0.415 cars/s ≈ 1,500 cars/h. This is likely to be an underestimate; for greater accuracy, we must find a method to compute the correct cell size before finding the speed. We address this problem later. Two Lanes We expand the one-dimensional model. The simulation consists of a twodimensional (1000 × 2) array of bits. The array is initialized randomly and then iterated over 1,000 time cycles: On each cycle, a car moves forward with probability p if the cell in front of it is empty. If not, provided the cells beside it and diagonally forward from it are unoccupied, with probability p the car changes lanes and moves one cell forward. Like the one-lane simulation, the two-lane one is correct only for high densities and low speeds, since it uses cell size equal to car length. Hence, we do not use the two-lane simulation to compute the maximum flow rate. However