Traffic Flow Theory and Models

Chapter 3 covers the theory that controls traffic flow on highways. The fundamental relationships between flow, speed, and density of vehicles on highways are also presented. Different macroscopic models are presented in this part, which describe the relationship between speed and density on the highway. Uncongested conditions as well as congested condition will also be discussed. Bottleneck conditions that accompany a sudden reduction in the capacity of the highway as a result of urgent (up normal) situations on the highway such as accidents, construction on one or more lanes of the highway, etc., will be discussed as well. And finally, the concept of gap and gap acceptance in traffic streams is introduced. The methods of determining the critical gap for merging vehicles are also discussed. The practical problems presented in the following sections will focus on the aforementioned topics. 3.1 If the traffic flow on a highway segment is estimated to be 1800 vph, compute the average time headway on the highway segment. Solution: The average space headway (d) is related to the density of vehicles (k) through the following equation: 3.1 d ¯ = 1 k https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0001.tif"/> And also, the space headway (d) is related to the time headway (h) and the space mean speed (u s ) through the following formula: 3.2 d ¯ = u s h ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0002.tif"/> And therefore, the time headway (h) can be 3.3 h ¯ = 1 k u s = 1 q https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0003.tif"/> As the flow is given in the units of vph, the units of the time headway obtained from the above formula will be h/veh. To convert that into sec/veh, the answer will be multiplied by 3600: h ¯ = 3600 q = 3600 1800 = 2 sec/veh https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0004.tif"/> A screen image of the MS Excel worksheet used to perform the computations of this problem is shown in Figure 3.1 . 20 3.2 The space mean speed (u s ) on a highway segment is 60 mph (96.6 kph) and the average time headway is 3 sec/veh. Estimate the density and the flow on this highway segment. Solution: Using Equation (3.2), the space headway (d) can be computed as shown below: d ¯ = u s h ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0005.tif"/> But before using this equation, the units must be consistent; in other words, the unit of the time headway should be in h/veh. Therefore: Time headway = 3 sec/veh = 3/3600 h/veh. d ¯ = 60 × 3 3600 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0006.tif"/> d ¯ = 0.05 mi/veh ( 0.08 km/veh ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0007.tif"/> And since: d ¯ = 1 k https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0008.tif"/> k = 1 0.05 = 20 vpm ( 12.4 veh/km ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0009.tif"/> 21 Since the space mean speed was given and the density was computed, the flow can be computed using the equation given below: q = u s k https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0010.tif"/> q = 60 × 20 = 1200 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0011.tif"/> The MS Excel worksheet used to solve this problem is shown in Figure 3.2 . 3.3 If the traffic flow and the average space headway on a highway segment are 1000 vph and 240 ft/veh, respectively, determine the space mean speed and the density on this highway segment. Solution: First the units of the time headway should be converted from ft/veh into mi/veh, the following is obtained: d ¯ = 240 5280 mi/veh https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0012.tif"/> Using Equation (3.1), the density can be obtained: d ¯ = 1 k https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0013.tif"/> ⇒ k = 5280 240 = 22 vpm ( 13.7 veh/km ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0014.tif"/> 22 Since: q = u s k https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0015.tif"/> u s = q k = 1000 22 = 45.5 mph ( 73.2 km h ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0016.tif"/> The MS Excel worksheet shown in Figure 3.3 illustrates the computed results of this problem. 3.4 The number of vehicles passing a point on a highway segment was counted to be 500 vehicles during a time interval of 15 minutes. Determine the equivalent hourly flow rate on the highway segment. Solution: The hourly flow rate is computed using the formula: q = N × 3600 T https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0017.tif"/> Where: q = equivalent hourly flow rate N = number of vehicles T = time period (seconds) Alternatively: q = N × 60 T https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0018.tif"/> 23 Where: T = time period (minutes) Therefore, q = 500 × 4 15 * 60 = 2000 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0019.tif"/> Or: q = 500 × 60 15 = 2000 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0020.tif"/> For this problem, the MS Excel worksheet used to compute the flow is shown in Figure 3.4 . 3.5 At a particular time on a highway, the speeds of three vehicles were 48.2, 44.6, and 38.2 mph (77.6, 71.8, and 61.5 kph). Compute the time mean speed and the space mean speed of the vehicles. Solution: The time mean speed (u t ) represents the arithmetic average of the speed of vehicles. Hence, it is computed using the following formula: 3.7 u t = ∑ i = 1 n u i n https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0021.tif"/> Where: u i = speed of vehicle i n = number of vehicles Therefore, u t = ∑ i = 1 3 u i 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0022.tif"/> 24 u t = 48.2 + 44.6 + 38.2 3 = 43.7 mph ( 70.3 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0023.tif"/> The space mean speed (u s ) is the harmonic mean of the speeds of vehicles. In other words, it is estimate d using the following formula: 3.8 u s = n ∑ i = 1 n 1 u i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0024.tif"/> Therefore, u s = 3 ∑ i = 1 3 1 u i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0025.tif"/> u s = 3 ∑ i = 1 3 1 48.2 + 1 44.6 + 1 38.2 = 43.3 mph ( 69.6 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0026.tif"/> The computations of this problem are also performed using the MS Excel worksheet shown in Figure 3.5 . 25 3.6 If the space mean speed for three vehicles on a highway segment is 40.3 mph (64.9 kph), and the individual speeds for two vehicles are 45.0 and 40.4 mph (72.4 and 65.0 kph), then what is the speed for the third vehicle? Solution: u s = n ∑ i = 1 n 1 u i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0027.tif"/> Therefore, u s = 3 ∑ i = 1 3 1 u i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0028.tif"/> u s = 3 ∑ i = 1 3 1 45.0 + 1 40.4 + 1 u 3 = 40.3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0029.tif"/> ⇒ 1 u 3 = 0.0275 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0030.tif"/> ⇒ u 3 = 36.4 mph ( 58.6 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0031.tif"/> The MS Excel worksheet shown in Figure 3.6 is used to compute the required results in this problem. 26 3.7 Five vehicles pass a 1000-ft (304.8-m) highway segment in time periods of 10, 14, 18, 15, 12 seconds, respectively. Determine the time mean speed and the space mean speed of the vehicles. Solution: The speeds of the five vehicles are computed using the following formula: u i = L t i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH003_eqn_0032.tif"/> Where: u i = speed of vehicle i L = length of segment t = time for vehicle i to pass the segment The length is divided by 5280 to convert from to and the time is divided by 3600 to convert from to The time mean speed is then computed using the following formula as the arithmetic mean of the five u t = ∑ i = 1 n u i n ⇒ u t = ∑ i = 1 u i u t = ∑ i = 1 u i u t = = mph ( kph ) The space mean speed is computed using the formula below: u s = n ∑ i = 1 n 1 u i ⇒ u s = ∑ i = 1 1 u i u s = = mph ( kph ) for the space mean speed is below: u s = n ∑ i = 1 n 1 u i u i = L t i Therefore, u s = n ∑ i = 1 n t i L Or: u s = n L ∑ i = 1 n t i ⇒ u s = ( 1000 ) = = mph ( kph ) The screen of the MS Excel used to perform the computations in this problem to determine the time mean speed and space mean speed are shown in 3.7 and 3.8 . 3.8 were on a highway segment between and as shown in Figure . Five vehicles The speeds of the five vehicles were and respectively. At a particular the of vehicles on the highway segment were as shown in the If the time of of vehicle is t compute the time which the other vehicles and Solution: the from the of vehicle to is computed on the given in the as For vehicle = + = The other for the other vehicles are given in the the time (seconds) from to the for vehicle is using the formula given into the of the t i = u i For vehicle t 1 = × 5280 3600 = The other results are shown in 3.1 . the time of of vehicle is t then the time of of the other vehicles can be by determining the in time between vehicle and of the other vehicles from to the to in the Since vehicle to and vehicle t then vehicle t = t + In a the time of for the other vehicles is computed as shown in 3.1 . the time of can be by the time of and the time required to pass from to for time from to of segment = L = is t i = L u i Therefore, for vehicle t i = × 5280 3600 = Hence, the time = t + In a the time of for the other vehicles is computed as shown in 3.1 . The results for the five vehicles are in 3.1 . The screen of the MS Excel worksheet used to perform the computations of this problem are shown in and . In determine the density of the highway segment t + Solution: At time t + the number of vehicles on the highway segment are the of of vehicle and it can be which vehicle on the segment that For vehicle t and t + that time t + the vehicle was on the highway segment. In the vehicle t + and t + which that time t + the vehicle was on the highway segment. also on the highway segment t + The other two vehicles and t + and t + which that time t + vehicles were of the segment before Therefore, the number of vehicles that on the highway segment time t + 3.2 was three vehicles. The density is computed as shown below: k = N L Where: N = number of vehicles on the highway segment L = length of the highway segment ⇒ k = 3 5280 = vpm ( veh/km ) The 5280 is a to convert the unit of to The of the density is also shown in the screen of the MS Excel worksheet of the problem Figure the for vehicles. Determine the headway between vehicles and time = of vehicle time = 3 of vehicle Solution: the time = seconds, vehicle and vehicle of and respectively. Therefore, the space headway between the two vehicles is d = = ( ) At time = 3 seconds, the of vehicle is and the vehicle is the is Therefore, speed = The relationship for vehicle is as from this and the The of this is the Therefore: of vehicle 1 = Time of vehicle 1 = ( ) ( ) = = mph ( kph ) The traffic density and speed shown in 3.2 were obtained on a highway segment. to the to Solution: The the shown below: u s = u u k k Where: u s = space mean speed u = mean speed k = density k = density For a equation the = a + a 1 the following is obtained using and n ∑ ∑ ∑ 2 a a 1 = ∑ ∑ represents the two that will be used to solve for the a and a 1 in the The of the above a 1 = ∑ i = 1 n i i 1 n ( ∑ i = 1 n i ) ( ∑ i = 1 n i ) ∑ i = 1 n i 2 1 n ( ∑ i = 1 n i ) 2 a = ∑ i = 1 n i n a 1 ∑ i = 1 n i n Or: a = ¯ a 1 ¯ The required computations are to the in the above 3.3 Therefore: a a 1 = The is used to solve the above The the following of a and a 1 a = a 1 = The MS Excel worksheet used to the in to solve the is shown in Figure . Or: a 1 = ∑ i = 1 n i i 1 n ( ∑ i = 1 n i ) ( ∑ i = 1 n i ) ∑ i = 1 n i 2 1 n ( ∑ i = 1 n i ) 2 a 1 = 1 ( ) ( ) 1 ( ) 2 = a = ∑ i = 1 n i n a 1 ∑ i = 1 n i n a = ( ) = Therefore, The that this the following u s = k The screen image of the MS Excel worksheet used to the computations and the of this problem is shown in Figure . For the in compute the following that the of of the to the and the of of the mean t ) of of the ) of the estimate ) of 2 ) of Solution: t = ∑ i = 1 n ( i ¯ ) 2 = ∑ i = 1 n ( i i ) 2 = n 2 2 = t t = 2 Where: t = of of the mean = of of the = of the estimate 2 = of = of i = speed i ¯ = mean of speeds = speed from the n = number of = of The following computations are performed using the MS Excel worksheet shown in 3.4 . t = ∑ i = 1 n ( i ¯ ) 2 ⇒ t = = ∑ i = 1 n ( i i ) 2 ⇒ = = n 2 ⇒ = 2 = 2 = t t ⇒ 2 = = = 2 ⇒ = = The MS Excel worksheet used to perform the computations and to determine the in this problem is shown in Figure . Using the MS Excel the relationship is also as shown in Figure . In determine the mean speed and the density for the traffic on the highway. Solution: the of in Equation the = a + a 1 the following are obtained: a = u ⇒ u = a = mph a 1 = u k ⇒ k = u a 1 = = vpm vpm The MS Excel worksheet used to perform the computations in this problem is shown in Figure in If the that the relationship between speed and density on a highway segment is given as u s = k determine the speed density flow flow flow rate of the highway segment Solution: Since the given that the relationship on the highway is this is the which the u s = u u k k . Therefore, the mean speed is to the in the u = mph ( kph ) The density is computed using the in the u k = ⇒ k = = vpm vpm ( veh/km ) The density flow is by the of the flow to density and the result to q = u s k u s = u u k k q = ( u u k k ) k Or: q = u k u k k 2 the above equation and it to the following formula is obtained: d q d k = u 2 u k k = ⇒ k = k 2 is the density flow = 2 = 26 vpm ( veh/km ) The speed flow is following the by the of the flow to speed and the result to q = u s k u s = u u k k the above formula to k the following formula for the density ( k ) is obtained: k = k k u u s q = ( k k u u s ) u s Or: q = k u s k u u s 2 the above equation and it to the following formula is obtained: d q d u s = k 2 k u u s = ⇒ u = u 2 is the speed flow = 2 = mph ( kph ) The flow rate of the highway segment is to the density flow multiplied by the speed Therefore, q = 26 × vph A screen image of the MS Excel worksheet used to perform the computations of this problem is shown in Figure . If the relationship between the density and the space mean speed for a traffic is given as shown in Figure determine the density speed flow flow flow Solution: the relationship in the the density is obtained a space mean speed of Therefore: k = vpm the other the space mean speed is obtained a density of u = 60 mph . Since the relationship in this problem is this relationship is the and in the as in problem, the density flow is to the mean speed divided by Therefore, flow = 2 = vpm ( veh/km ) In the the speed flow is to the mean speed divided by two in the Therefore, flow = 60 2 = mph ( kph ) The flow is to the density flow multiplied by the speed q = × = vph The MS Excel worksheet used to the results of this problem is shown in the screen image in Figure . The traffic shown in 3.5 is obtained on a highway segment. If the can be by the to determine the The in the The of 2 ) for the the relationship between density and speed Solution: To the of the is to be the is The the following u s = k k Where: u s = space mean speed k = density = k = density The can be and as in the following u s = k k is the equation = a + a 1 such = u s a = k a 1 = = k Therefore, the that was used for models can be used for the by and and u s as shown below: n ∑ ∑ ∑ 2 a a 1 = ∑ ∑ n ∑ k ∑ k ∑ ( k ) 2 a a 1 = ∑ u s ∑ ( k ) u s The of the above a 1 = ∑ i = 1 n i i 1 n ( ∑ i = 1 n i ) ( ∑ i = 1 n i ) ∑ i = 1 n i 2 1 n ( ∑ i = 1 n i ) 2 a = ∑ i = 1 n i n a 1 ∑ i = 1 n i n Therefore, a 1 = ∑ i = 1 n ( k i ) u s i 1 n ∑ i = 1 n ( k i ) ∑ i = 1 n u s i ∑ i = 1 n ( k i ) 2 1 n ( ∑ i = 1 n ( k i ) ) 2 a = ∑ i = 1 n u s i n a 1 ∑ i = 1 n ( k i ) n The computations of the results to determine the a and a 1 are performed using the MS Excel worksheet and are shown in 3.6 . a 1 = 1 12 ( ) ( ) 1 12 ( ) 2 = a = 12 ( ) 12 = = a 1 ⇒ = a 1 = k = a ⇒ k = ( a ) = ( ) = vpm ( veh/km ) Therefore, the that this the u s = k To compute the of 2 ) for the the speed from the should be The computations are shown in 3.7 . t = = 2 = t t ⇒ 2 = = The relationship by the is in Figure . image of the MS Excel worksheet used to the computations and the of this problem is shown in and . that is for the that can be In determine the density flow, the speed flow, and the flow of the highway. Solution: u s = k k q = u s k u s = k k q = ( k k ) k the above equation and it to the following formula is obtained: d q d k = ( k k ) + k ( k ) = ⇒ k k = 1 k k = ⇒ k = k is the density flow = 2 = vpm ( veh/km ) The speed flow is following the by the of the flow to speed and the result to q = u s k u s = k k the above formula to k the following formula for the density ( k ) is obtained: k = k ( u s ) q = ( k ( u s ) ) u s the above equation and it to the following formula is obtained: d q d u s = k ( u s ) 1 k u s ( u s ) = ⇒ k ( u s ) ( 1 u s ) = ⇒ u = is the speed flow = mph ( kph ) And therefore, the flow is to the density flow multiplied by the speed q = × = vph If the that the relationship between speed and density on a highway is given as u s = ( k ) determine the flow flow of the highway Solution: To determine the density flow, the flow as a of the density is and to the q = u s k But to the given in this problem between speed and it is in the following u s = A ( k ) q = A k ( k ) the above equation and it to the following formula is obtained: d q d k = 1 A k ( k ) + A ( k ) = ⇒ A ( k ) ( 1 k ) = ⇒ k = is the density flow = vpm ( veh/km ) To determine the speed flow, the flow as a of the speed is and to the q = u s k u s = A ( k ) the above equation to k as a of u s k = ( u s A ) And therefore, q = u s ( u s A ) the above equation and it to the following formula is obtained: d q d u s = ( u s A ) = ⇒ ( 1 + ( u s A ) ) = ⇒ 1 + ( u s A ) = ⇒ u = A is the speed flow = = mph ( kph ) The flow is by the density flow by the speed q = u s k q = × = vph The relationship between the density and the space mean speed for a traffic is by the If the density flow is vpm determine the Solution: on the the density flow is given k = k Therefore, the density can be computed k = k k = × = vpm ( veh/km ) If the u s = u k k can be used to describe the relationship between speed and density on a highway and using the of the = A + = s and = k ) are A = and = then estimate the mean speed ( u ) and the density ( k Solution: Since the is given in the following u s = u k k ⇒ A = u ⇒ u = A = = mph ( kph ) = 1 k ⇒ k = 1 = 1 vpm ( veh/km ) If the that the relationship between speed