Intersection_Analysis

Intersection_Analysis - Introduction to Transportation...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Introduction to Transportation Engineering Applications of Queueing Theory to Intersection Analysis Level of Service Dusan Teodorovic and Antonio A. Trani Civil and Environmental Engineering Virginia Polytechnic Institute and State University Blacksburg, Virginia Spring 2005 Virginia Tech Material Covered • Application of deterministic queueing models to study intersection level of service • Study various types of intersection controls schemes used in transportation engineering • Most of the material applies to ground transportation modes (highways) Virginia Tech 2 Basic Ideas • Traffic control represents a surveillance of the motion of vehicles and pedestrians in order to secure maximum efficiency and safety of conflicting traffic movements. • Traffic lights or traffic signals are the basic devices used in traffic control of vehicles on roads. • They are located at road intersections and/or pedestrian crossings. • The first traffic light was installed even before there was automobile traffic (London on December 10, 1868). The current traffic lights were invented in USA (Salt Lake City, (1912), Cleveland (1914), New York and Detroit (1920)). Virginia Tech 3 Basic Definitions • Drivers move toward the intersection from different approaches • Every intersection is composed of a number of approaches and the crossing area (see Figure) • Each approach can have one, or more lanes. The traffic stream is composed of all drivers who cross the intersection from the same approach • During green time, vehicles from the observed approach can leave the stop line and cross the intersection • The corresponding average flow rate of vehicles that cross the stop line is known as a saturation flow. Virginia Tech 4 Intersection Geometry Approach Crossing area Figure 1. Typical Road Intersection. Virginia Tech 5 Flow Conditions at an Intersection • In most cases, queues of vehicles are established exclusively during the red phases, and are terminated during the green phases. • Such traffic conditions are known as a undersaturated traffic conditions. • An intersection is considered an unsaturated intersection when all of the approaches are undersaturated. • Traffic conditions in which queue of vehicles can arrive at the upstream intersection are known as a oversaturated traffic conditions. Virginia Tech 6 Traffic Control Techniques • Traffic engineers apply various traffic control strategies (see the Figure) in order to minimize the total delay at the intersection and/or maximize the intersection capacity. Phase 2 Phase 1 Cycle Phase 1 Phase 2 Cycle Time Figure 2. Intersection with Two Phases. Virginia Tech 7 Traffic Signal Control Strategies • Many isolated intersections operate under the fixed-time control strategies. • These strategies assume existence of the signal cycle that represents one execution of the basic sequence of signal combinations at an intersection. • A phase represents part of the signal cycle, during which one set of traffic streams has right of way. Figure 2 shows two-phase traffic operations for the intersection. • The cycle contains only two phases. Phase 1 is related to the movement of the north-southbound vehicles through the intersection. Phase 2 represents the movement of the east-westbound vehicles. Virginia Tech 8 Traffic Control Strategies • The cycle length c represents the duration of the cycle measured in seconds. The sum of the phase lengths represents the cycle length. • For example, in the case shown in Figure 2., the cycle length could be 90 seconds, length of the Phase 1 could be 50 seconds, while the length of the Phase 2 could be equal to 40 seconds. • The cycle length is a design parameter of the intersection as well as the green times allocated to each phase. Traffic engineers can modify the settings of intersection controllers based on demand needs at the intersection. Virginia Tech 9 Control Strategies Phase 1 Phase 2 Phase 3 Cycle Figure 3. Intersection with Three Phases. Virginia Tech 10 Control Strategies • Higher number of phases is usually caused by traffic engineer’s wish to protect some movements (usually leftturning vehicles) • “Protection” assumes avoiding potential conflicts with the opposing traffic movement, and/or pedestrians • There is always a certain amount of lost time (few seconds) during phase change. For example, when the green light changes to red there is am amber light period to warn drivers of an impending change • Obviously, the higher the number of phases, the better the protection, and the higher the value of the lost time associated with a phase change. Virginia Tech 11 Control Strategies • Traffic signals are control devices. The typical sequence of lights at the intersection approach could be: “Red, Red All, Green, Amber, Red, Red All,...”. Flow [veh/h] Saturation flow Time 0 Red Green Red All Amber Effective green Figure 4. Definition of Green, Amber and Red Times. Virginia Tech 12 Control Strategies • “Green time”, “effective green”,”red time”, and “effective red” are linguistic expressions frequently used by traffic engineers • In theory, all drivers should cross the intersection during the green light. In reality, no one driver starts his/her car exactly in a moment of the green light appearance • Similarly, at the end of a green light, some drivers speed up, and cross the intersection during the amber light • “Green Time” represents the time interval within the cycle when observed approach has green indication. On the other hand, “Effective Green” represents the time interval during which observed vehicles are crossing the intersection. Virginia Tech 13 Vehicle delays at signalized intersections: Uniform Vehicle Arrivals • For simplicity, let us assume for the moment that observed signalized intersection could be treated as a D/ D/1 (deterministic) queueing system with one server (hence the notation (D/D/1)) • We assume uniform arrivals, and uniform departure rate (see Figure 5). Virginia Tech 14 Queueing Theory Short-hand Nomenclature • Queues come in different flavors as demonstrated so far • Kendall developed a simple scheme to designate queues back in the early 50s. His nomenclature has been widely adopted • Typically 6 parameters: • • • • a/b/c/d/e/f a = inter-arrival time distribution (arrivals) b = service time distribution c = number of servers Virginia Tech (A.A. Trani) 14a Queueing Theory Short-hand Nomenclature • Typically 6 parameters: • • • • a/b/c/d/e/f d = service order (i.e., FIFO, LIFO, etc) e = Max. number of customers f =Size of the arrival population Virginia Tech (A.A. Trani) 14b Queueing Theory Short-hand Nomenclature • • • • • Possible outcomes for (a) and (b) M = Times are neg. exponential (i.e., Poisson arrivals) D = Deterministic distribution Ek = Erlang distribution G = general distribution Virginia Tech (A.A. Trani) 14c Example Queueing Systems we Have Studied • M/M/1/FIFO/∞/∞ • Stochastic queue with neg. exponential time between arrivals • Neg. exponential service times • 1 server • First in-first out • Infinite no. of customers in system • Infinite arrival population Virginia Tech (A.A. Trani) 14d Example Queueing Systems we Have Studied • M/M/2/FIFO/15/15 • Stochastic queue with neg. exponential time between arrivals • Neg. exponential service times • 2 servers (2 pavers) • First in-first out • Up to 15 no. of trucks in system • 15 trucks population Virginia Tech (A.A. Trani) 14e Definition of Queueing Terms for Intersection Analysis Cumulative number of vehicles Cumulative arrivals Cumulative departures D(t) C A(t) g0 h A r g B c Time Red Green Figure 5. Arrivals and Departures at an Intersection. Virginia Tech 15 Deterministic Queueing Analysis Let us denote by λ vehicles arrival rate, and by µ vehicles departure rate during the green time period. In the deterministic case, the cumulative number of arrivals A ( t ) and the cumulative number of departures D ( t ) are: A(t) = λ ⋅ t (1) D(t) = µ ⋅ t (2) where: c - the duration of the signal cycle r - effective red g - effective green Virginia Tech 16 Deterministic Queueing Analysis The duration of the signal cycle equals: c = r+g (3) The formed queue is the longest at the beginning of effective green. The queue decreases at the beginning of effective green. We denote by g the time necessary for queue to dissipate (Figure 5). The queue must dissipate before the end of effective green. In the opposite case, the queue would escalate indefinitely. In other words, queue dissipation will happen in every cycle if the following relation is satisfied: 0 Virginia Tech 17 Deterministic Queueing Analysis (4) g0 ≤ g The relation (4) will be satisfied if the total number of vehicle arrivals during cycle length c is less than or equal to the total number of vehicle departures during effective green g , i.e.: c g (5) ∫ λ dt ≤ ∫ µdt 0 0 λ⋅t c≤µ⋅t 0 (6) g 0 (7) λ⋅c≤µ⋅g Virginia Tech 18 Deterministic Queueing Analysis Finally, we get: λ -- ≤ g - -µc (8) Let us note the triangle ABC (Figure 5). Vehicles arrive during time period ( r + g ) . Vehicles depart during time period g . The total number of vehicle arrivals equals the total number of vehicle departures, i.e.: 0 0 λ ⋅ ( r + g0 ) = µ ⋅ g0 (9) ( µ – λ ) ⋅ g0 = λ ⋅ r (10) Virginia Tech 19 Deterministic Queueing Analysis The time period g required for queue to dissipate equals: 0 λ⋅r g 0 = ----------µ–λ (11) We divide both numerator and denominator by µ. We get: λ -- ⋅ r µ g 0 = ----------λ 1 – -µ (12) Define the utilization factor ( ρ ) (or traffic intensity) of the λ intersection as ρ = --- , we can write: µ ρ⋅r g 0 = ----------1–ρ (13) Virginia Tech 20 Deterministic Queueing Analysis The area A of the triangle ABC represents the total delay d of all vehicles arrived during the cycle. This area equals: ∆ ABC 1 A ∆ABC = -- ⋅ r ⋅ h 2 (14) where h is the height of the triangle (ABC). The ratio h ----------------( r + g0 ) represents the slope λ , i.e.: h λ = ----------------( r + g0 ) (15) Virginia Tech 21 Deterministic Queueing Analysis The height h of the triangle ABC is: h = λ ⋅ ( r + g0 ) The area of the triangle ABC equals: 1 λ⋅r 1 A ∆ABC = -- ⋅ r ⋅ h = -- ⋅ r ⋅ λ ⋅ ( r + g 0 ) = --------- ⋅ ( r + g 0 ) 2 2 2 (16) The total delay d of all vehicles arriving during the cycle equals: λ⋅r λ⋅r λ ⋅ r2 ρ ρ ⋅ r d = --------- ⋅ ( r + g 0 ) = --------- ⋅ r + ----------- = ----------- ⋅ 1 + ----------- 2 2 2 1 – ρ 1 – ρ Virginia Tech (17) 22 Deterministic Queueing Analysis λ ⋅ r2 d = ----------------------2 ⋅ (1 – ρ) (18) The average delay per vehicle d represents the ratio between the total delay d and the total number of vehicles per cycle. The total number of vehicles per cycle equals λ ⋅ c . Therefore the average delay per vehicle d is: dd = --------λ⋅c (19) or λ ⋅ r2 ----------------------2 ⋅ (1 – ρ) d = -----------------------λ⋅c (20) Virginia Tech 23 Simplifying the previous expression, average delay per vehicle is the average: r2 d = ------------------------------2 ⋅ c ⋅ (1 – ρ) (21) Virginia Tech 24 Example Problem 1 The cycle length at the signalized intersection is 90 seconds. The considered approach has the saturation flow of 2200 [veh/hr], the green time duration of 27 seconds, and flow rate of 600 [veh/hr]. Analyze traffic conditions in the vicinity of the intersection. Calculate average delay per vehicle. Assume that the D/D/1 queueing system adequately describes considered intersection approach. Virginia Tech 25 Problem 1 - Solution The corresponding values of the cycle length and the green time are: c = 90 [ s ] ;g = 27 [ s ] The flow rate ( λ ) and the service rate ( µ ) are: veh 600 veh veh λ = 600 -------- = ----------- -------- = 0.167 -------hr 3600 s s veh 2200 veh veh µ = 2200 -------- = ----------- -------- = 0.611 -------hr 3600 s s Traffic intensity ρ equals: Virginia Tech 26 Problem 1 - Solution veh 0.167 -------s λ ρ = -- = --------------------------- = 0.273 µ veh 0.611 -------s The duration of the red light for the considered approach is: r = c – g = 90 – 27 = 63 [ s ] The number of arriving vehicles per cycle is: veh λ ⋅ c = 0.167 -------- ⋅ 90 [ s ] = 15.03 [ veh ] s Virginia Tech 27 Problem 1 - Solution The number of departing vehicles during green light is: veh µ ⋅ g = 0.611 -------- ⋅ 27 [ s ] = 16.497 [ veh ] s We conclude that the following relation is satisfied: λ⋅c≤µ⋅g This means that the traffic conditions in the vicinity of the intersection are undersaturated traffic conditions. The average delay per vehicle is estimated using: r2 d = ------------------------------2 ⋅ c ⋅ (1 – ρ) Virginia Tech 28 Problem 1 - Solution 63 2 d = -------------------------------------------- = 30.33 [ s ] 2 ⋅ 90 ⋅ ( 1 – 0.273 ) Virginia Tech 29 Example Problem 2 The cycle length at the signalized intersection is 60 seconds. The considered approach has the saturation flow of 2200 [veh/hr], the green time duration of 15 seconds, and flow rate of 400 [veh/hr]. Analyze traffic conditions in the vicinity of the intersection. Assume that the D/D/1 queueing system adequately describes the intersection approach considered. Calculate: (a) the average delay per vehicle; (b) the longest queue length; (c) percentage of stopped vehicles. Virginia Tech 30 Problem 2 - Solution: (a) The corresponding values of the cycle length and the green time are: c = 60 [ s ] ;g = 20 [ s ] The red time is: r = c – g = 60 – 20 = 40 [ s ] The flow rate and the service rate are: veh 400 veh veh λ = 400 -------- = ----------- -------- = 0.111 -------hr 3600 s s Virginia Tech 31 Problem 2 - Solution veh 2200 veh veh µ = 2200 -------- = ----------- -------- = 0.611 -------hr 3600 s s The utilization factor for the queue ρ is: veh 0.111 -------s λ ρ = -- = --------------------------- = 0.182 µ veh 0.611 -------s The average delay per vehicle equals: r2 d = ------------------------------2 ⋅ c ⋅ (1 – ρ) Virginia Tech 32 Problem 2 - Solution 40 2 d = -------------------------------------------- = 16.3 [ s ] 2 ⋅ 60 ⋅ ( 1 – 0.182 ) (b) The longest queue length L happens at the end of a red light (Figure 5). The quantity L is calculated as follows: max max veh L max = λ ⋅ r = 0.111 -------- ⋅ 40 [ s ] = 4.44 [ vehicles ] s (c) Vehicles arrive all the time during the cycle. The total number vehicles arrived A during the cycle equals: veh A = λ ⋅ c = 0.111 -------- ⋅ 60 [ s ] = 6.66 [ vehicles ] s Virginia Tech 33 Problem 2 - Solution All vehicles that arrive during time interval ( r + g ) are stopped. The total number of stopped vehicles S equal: 0 S = λ ⋅ ( r + g0 ) The time period g required for queue to dissipate is estimated using equation: 0 λ⋅r g 0 = ----------µ–λ We get: λ⋅r 0.111 ⋅ 40 S = λ ⋅ ( r + g 0 ) = λ ⋅ r + ----------- = 0.111 ⋅ 40 + -------------------------------- µ – λ 0.611 – 0.111 Virginia Tech 34 S = 5.43 [ vehicles ] The percentage of stopped vehicles equal: S 5.43 P = -- ⋅ 100 = --------- ⋅ 100 = 81.53 [%] A 6.66 Virginia Tech 35 Example Problem 3 A simple “T” intersection is signalized. There are two approaches indicated in the figure. The cycle length at the signalized intersection (Figure) is 50 seconds. Phase 2 Phase 1 Approach 1 Approach 2 Cycle Virginia Tech 36 Example Problem 3 Approach 1 has the saturation flow of 2200 [veh/hr], the effective green time duration of 35 seconds, and the flow rate of 600 [veh/hr]. Approach 2 has the saturation flow of 2000 [veh/hr], the effective green time duration of 15 seconds, and the flow rate of 550 [veh/hr]. Assume that the D/D/1 queueing system adequately describes considered intersection approach. Calculate: (a) the average delay per vehicle for every approach; (b) Allocate effective red and green time among approaches in such a way to minimize the total delay of the “T” intersection. Virginia Tech 37 Problem 3 -Solution (a) Approach 1: The corresponding values of the cycle length and the green time are: c = 50 [ s ] ;g 1 = 35 [ s ] The red time equals: r 1 = c – g 1 = 50 – 35 = 15 [ s ] The flow rate and the service rate are respectively equal: veh 600 veh veh λ 1 = 600 -------- = ----------- -------- = 0.167 -------hr 3600 s s Virginia Tech 38 Problem 3 -Solution veh 2200 veh veh µ 1 = 2200 -------- = ----------- -------- = 0.611 -------hr 3600 s s The utilization factor for the queue in approach 1 ρ is: 1 veh 0.167 -------s λ1 ρ 1 = ---- = --------------------------- = 0.273 µ1 veh 0.611 -------s The average delay per vehicle equals: r12 15 2 d 1 = -------------------------------- = -------------------------------------------- = 3.09 [ s ] 2 ⋅ c ⋅ ( 1 – ρ1 ) 2 ⋅ 50 ⋅ ( 1 – 0.273 ) Virginia Tech 39 Problem 3 -Solution Approach 2: The corresponding values of the cycle length and the green time are: c = 50 [ s ] ;g 2 = 15 [ s ] The red time is: r 2 = c – g 2 = 50 – 15 = 35 [ s ] The flow rate and the service rate are: veh 550 veh veh λ 2 = 550 -------- = ----------- -------- = 0.153 -------hr 3600 s s Virginia Tech 40 Problem 3 -Solution veh 2000 veh veh µ 2 = 2000 -------- = ----------- -------- = 0.555 -------hr 3600 s s The utilization factor for the queue ρ equals: 2 veh 0.153 -------s λ2 ρ 2 = ---- = --------------------------- = 0.276 µ2 veh 0.555 -------s The average delay per vehicle equals: r22 35 2 d 2 = -------------------------------- = -------------------------------------------- = 16.92 [ s ] 2 ⋅ c ⋅ ( 1 – ρ2 ) 2 ⋅ 50 ⋅ ( 1 – 0.276 ) Virginia Tech 41 Problem 3 -Solution (b) The total delay per cycle of all vehicles on both approaches is the sum of the delays of every approach: TD = λ 1 ⋅ d 1 + λ 2 ⋅ d 2 substituting the definitions of d and d (see equation 21) 1 2 r12 r22 TD = λ 1 ⋅ -------------------------------- + λ 2 ⋅ -------------------------------2 ⋅ c ⋅ ( 1 – ρ1 ) 2 ⋅ c ⋅ ( 1 – ρ2 ) Since: r1 + r2 = c after substitution, we get: Virginia Tech 42 r12 ( c – r1 )2 TD = λ 1 ⋅ -------------------------------- + λ 2 ⋅ -------------------------------2 ⋅ c ⋅ ( 1 – ρ1 ) 2 ⋅ c ⋅ ( 1 – ρ2 ) The total delay is minimal when: d[ TD ] --------------- = 0 dr 1 After substitution, we get: r12 ( c – r1 )2 d λ 1 ⋅ -------------------------------- + λ 2 ⋅ -------------------------------2 ⋅ c ⋅ ( 1 – ρ1 ) 2 ⋅ c ⋅ ( 1 – ρ2 ) ---------------------------------------------------------------------------------------------------- = 0 dr 1 r1 ( c – r1 ) λ 1 ⋅ ------------------------- – λ 2 ⋅ ------------------------- = 0 c ⋅ ( 1 – ρ1 ) c ⋅ ( 1 – ρ2 ) Virginia Tech 43 Problem 3 -Solution ( 50 – r 1 ) r1 0.167 ⋅ ------------------------------------- – 0.153 ⋅ ------------------------------------- = 0 50 ⋅ ( 1 – 0.273 ) 50 ⋅ ( 1 – 0.276 ) After solving the equation, we get: r 1 = 24 [ s ] g 1 = c – r 1 = 50 – 24 = 26 [ s ] r 2 = 26 [ s ] g 2 = 24 [ s ] These are the optimal values of green and red times to minimize the intersection delay. Virginia Tech 44 Problem 3 -Solution We can recalculate the average delays per vehicle: Approach 1 r12 24 2 d 1 = -------------------------------- = -------------------------------------------- = 7.92 [ s ] 2 ⋅ c ⋅ ( 1 – ρ1 ) 2 ⋅ 50 ⋅ ( 1 – 0.273 ) Approach 2: r22 26 2 d 2 = -------------------------------- = -------------------------------------------- = 9.34 [ s ] 2 ⋅ c ⋅ ( 1 – ρ2 ) 2 ⋅ 50 ⋅ ( 1 – 0.276 ) These delays compare favorably with those obtained before (3.09 and 16.92 seconds, respectively for approaches 1 and 2). Virginia Tech 45 Vehicle Delays at Signalized Intersections: Random Vehicle Arrivals • Traffic flows are characterized by random fluctuations • The delay that a specific vehicle experiences depends on the probability density function of the interarrival times, as well as on signal timings and the time of a day when the vehicle shows up • Obviously, individual vehicles experience at a signalized approach various delay values. Virginia Tech 46 Intersection with Random Arrivals Cumulative number Cumulative arrivals of vehicles Overflow Delay Uniform delay Time τ Red Green Figure 6. Intersection with Random Arrivals. Virginia Tech 47 Intersection with Random Arrivals • Let us calculate the delay D for the vehicle arriving at time τ (Figure 6). The overall delay D is composed of the uniform delay d and the overflow delay d , i.e.: R D = d + dR (22) • The uniform delay d represents delay that would be exp rienced by a vehicle when all vehicle arrive uniformly and when traffic conditions are unsaturated (see Equations in previous sections). • Due to the random nature of vehicle arrivals, the arrival rate during some time periods can go over the capacity, causing overflow queues. Virginia Tech 48 Considering Random Arrivals • The overflow delay d represents the delay that is caused by short-term overflow queues. This delay can be easily calculated using queueing theory techniques. R Queueing System Crossing area γ Virginia Tech λ 49 Intersection with Random Arrivals We assume that vehicle interarrival times are exponentially distributed. The service rate is deterministic (we denote by γ departure rate from the artificial queue into the signal), and there is only one server. This means that the artificial Queueing System is M/D/1 queueing system (single server with Poisson arrivals and deterministic service times). The average delay per customer in the M/D/1 queueing system equals: α2 d R = -------------------------------2 ⋅ λ ⋅ (1 – α) (23) Virginia Tech 50 Intersection with Random Arrivals where: α - utilization ratio in the M/D/1 queueing system The utilization ratio α in the M/D/1 queueing system equals: α=λ -γ (24) The departure rate from the artificial queue into the signal γ can be expressed in terms of departure rates from the traffic signal µ . The departure rate equals µ during green time. During red time, departure rate equals zero (see Figure 6). Virginia Tech 51 Intersection with Random Arrivals Service rate g [veh/h] µ r Time 0 Green Red Cycle Figure 7. Service Rate Definition at a Traffic Intersection. Virginia Tech 52 Intersection with Random Arrivals Departure rate γ during whole cycle : γ = 0⋅r+µ⋅g -------------------------c (25) g γ = µ ⋅ -c (26) The utilization ratio α in the M/D/1 queueing system : λα = λ = ----------γ g µ ⋅ -c (27) i.e.: Virginia Tech 53 λ⋅c α = ---------µ⋅g (28) The quantity α is known as a volume to capacity ratio. The average vehicle delay is: D = d + dR (29) r2 α2 D = ------------------------------- + -------------------------------2 ⋅ c ⋅ (1 – ρ) 2 ⋅ λ ⋅ (1 – α) (30) It has been shown by simulation that Equation (30) overestimate the average vehicle delay. The following two formulas for average vehicle delay calculation were proposed as a corrections of the equation (30): Virginia Tech 54 Webster’s formula: 1 -3 5 ⋅ g 2 + --------cr α ---- ⋅ α c D = ------------------------------- + -------------------------------- – 0.65 ⋅ 2 λ 2 ⋅ c ⋅ (1 – ρ) 2 ⋅ λ ⋅ (1 – α) 2 2 (31) Allsop’s formula: 9 α2 r2 D = ----- ⋅ ------------------------------- + -------------------------------10 2 ⋅ c ⋅ ( 1 – ρ ) 2 ⋅ λ ⋅ ( 1 – α ) Virginia Tech (32) 55 Example Problem 4 Using data given in the Example Problem 1, calculate: (a) average delay per vehicle using Allsop’s formula. (b) Calculate duration of the green time necessary to achieve average delay per vehicle of 40 seconds. Solution: (a) The cycle length, green time, arrival rate, departure rate, traffic intensity, volume to capacity ratio, and red time duration are: c = 90 [ s ] g = 27 [ s ] Virginia Tech 56 veh 600 veh veh λ = 600 -------- = ----------- -------- = 0.167 -------hr 3600 s s veh 2200 veh veh µ = 2200 -------- = ----------- -------- = 0.611 -------hr 3600 s s veh 0.167 -------s λ ρ = -- = --------------------------- = 0.273 µ veh 0.611 -------s veh 0.167 -------s --------------------------veh λ 0.611 --------s µ α = -- = ---------------------------- = 0.273 = 0.91 -----------g 27 [ s ] 0.3 ------------c 90 [ s ] Virginia Tech 57 Solution - Problem 4 r = c – g = 90 – 27 = 63 [ s ] The average delay per vehicle based on Allsop’s formula equals: 9 α2 r2 D = ----- ⋅ ------------------------------- + -------------------------------10 2 ⋅ c ⋅ ( 1 – ρ ) 2 ⋅ λ ⋅ ( 1 – α ) 90.91 2 63 2 D = ----- ⋅ -------------------------------------------- + ------------------------------------------------10 2 ⋅ 90 ⋅ ( 1 – 0.273 ) 2 ⋅ 0.167 ⋅ ( 1 – 0.91 ) D = 52.083 [ s ] (b) The average delay per vehicle is: Virginia Tech 58 9 α2 r2 D = ----- ⋅ ------------------------------- + -------------------------------10 2 ⋅ c ⋅ ( 1 – ρ ) 2 ⋅ λ ⋅ ( 1 – α ) 10 r2 α2 ------------------------------- = ----- ⋅ D – -------------------------------9 2 ⋅ c ⋅ (1 – ρ) 2 ⋅ λ ⋅ (1 – α) r= 10 α2 [ 2 ⋅ c ⋅ ( 1 – ρ ) ] ⋅ ----- ⋅ D – -------------------------------9 2 ⋅ λ ⋅ (1 – α) r= 10 0.91 2 [ 2 ⋅ 90 ⋅ ( 1 – 0.273 ) ] ⋅ ----- ⋅ 40 – ------------------------------------------------9 2 ⋅ 0.167 ⋅ ( 1 – 0.91 ) r = 47 [ s ] g = c – r = 90 – 47 g = 43 [ s ] Green time to achieve a delay of 40 seconds per vehicle. Virginia Tech 59 ...
View Full Document

  • Fall '08
  • KATZ
  • Atlantic Coast Conference, Queueing theory, Virginia Polytechnic Institute and State University, Blacksburg, Virginia, T. Marshall Hahn

{[ snackBarMessage ]}

Ask a homework question - tutors are online