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 ft to mile, and the time is divided by 3600 to convert from seconds to hr.

The time mean speed is then computed using the following formula as the arithmetic mean of the five speeds:

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_0033.tif"/>

⇒

u t = ∑ i = 1 5 u i 5 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_0034.tif"/>

u t = ∑ i = 1 5 u i 5 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_0035.tif"/>

u t = 257 5 = 51.4   mph ( 82.7   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_0036.tif"/>

The space mean speed is computed using the formula below:

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_0037.tif"/>

⇒

u s = 5 ∑ i = 1 5 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_0038.tif"/>

u s = 5 0.1012 = 49.4   mph ( 79.5   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_0039.tif"/>

Another solution for the space mean speed is illustrated below:27

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_0040.tif"/>

But:

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_0041.tif"/>

Therefore,

u s = n ∑ i = 1 n t i L 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_0042.tif"/>

Or:

u s = n L ∑ i = 1 n 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_0043.tif"/>

⇒

u s = 5 ( 1000 ) 69 = 72.5   ft/s = 49.4   mph ( 79.5   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_0044.tif"/>

The screen images of the MS Excel worksheets used to perform the computations in this problem to determine the time mean speed and space mean speed are shown in Figures 3.7 and 3.8.28

3.8

Traffic data were collected on a two-lane two-way highway segment between A-A and B-B 400-ft (121.9-m) apart as shown in Figure 3.9. Five vehicles passed A-A at different times. The speeds of the five vehicles were 50, 48, 45, 42, and 40 mph, respectively. At a particular time, the locations of these vehicles on the highway segment were as shown in the diagram below. If the time of arrival of vehicle #1 at A-A is t₀, compute the time at which the other four vehicles arrived at A-A and B-B.

Solution:

First, the distance from the current position of each vehicle to A-A is computed based on the distances given in the diagram as follows:

For vehicle #1: distance = 400 + 94 = 494.0 ft.29

The other distances for the other vehicles are simply given in the diagram.

Second, the travel time (seconds) from A-A to reach the current position for each vehicle is calculated using the formula given below taking into consideration the conversions of the units:

t i = Distance 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_0045.tif"/>

For vehicle #1:

t 1 = 494 50 × 5280 / 3600 = 6.74   seconds 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_0046.tif"/>

The other results are shown in Table 3.1.

Third, if the time of arrival of vehicle #1 at A-A is t₀, then the time of arrival of the other vehicles at A-A can be calculated by determining the difference in time between vehicle #1 and each of the other four vehicles from A-A to reach the current position.

Vehicle #1 took 6.74 seconds to reach its current position in the diagram. Since vehicle #2 took 4.69 seconds to reach its current position and vehicle #1 reached A-A at t₀, then vehicle #2 arrived at A-A at t₀+(6.74-4.69) = t₀+ 2.05. In a similar manner, the time of arrival at A-A for the other vehicles is computed as shown in Table 3.1.

Fourth, the time of arrival at B-B can be calculated by adding the time of arrival at A-A and the travel time required to pass from A-A to B-B for each vehicle.

Travel time from A-A to B-B (length of segment = L = 400 ft) is determined as:

t i = L 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_0047.tif"/>

Therefore, for vehicle #1:

t i = 400 50 × 5280 / 3600 = 5.45   seconds 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_0048.tif"/>

Hence, the arrival time at B-B = t₀ + 5.45

In a similar manner, the time of arrival at B-B for the other four vehicles is computed as shown in Table 3.1. The results for the five vehicles are summarized in Table 3.1.30

The screen images of the MS Excel worksheet used to perform the computations of this problem are shown in Figures 3.10 and 3.11.31

3.9

In Problem 3.8, determine the density of the highway segment at t₀+ 3.20 sec.

Solution:

At time t₀+ 3.20 sec, the number of vehicles on the highway segment are determined. Watching the times of arrival of each vehicle at A-A and B-B, it can be determined which vehicle existed on the segment at that time. For instance, vehicle #1 arrived at A-A at t₀ and reached B-B at t₀+ 5.45, that means at time t₀+ 3.20, the vehicle was still on the highway segment. In the same way, vehicle #2 arrived at A-A at t₀+ 2.05 and reached B-B at t₀+ 7.73, which means again that at time t₀+ 3.2, the vehicle was on the highway segment. Vehicle #3 also existed on the highway segment at t₀+ 3.2. The other two vehicles (#4 and #5) arrived at A-A at times t₀+ 3.40 and t₀+ 5.74, respectively; which means that at time t₀+ 3.20, both vehicles were out of the segment (specifically before A-A; i.e., they didn’t reach A-A yet). Therefore, the number of vehicles that existed on the highway segment at time t₀+ 3.2 was three vehicles. The density is computed as shown below:

3.12 k = N L 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_0049.tif"/>

Where:

N = number of vehicles on the highway segment

L = length of the highway segment

⇒

k = 3 400 / 5280 = 39.6   vpm ( 24.6   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_0050.tif"/>

The 5280 is a conversion factor to convert the unit of ft to mile.

The computation of the density is also shown in the screen images of the MS Excel worksheet of the previous problem (Problem 3.8).

3.10

Figure 3.12 shows the time–space diagram for four vehicles. Determine the following:

Space headway between vehicles #3 and #4 at time = 6 second

Speed of vehicle #3 at time = 3 seconds

Speed of vehicle #132

Solution:

From the diagram, at time = 6 seconds, vehicle #3 and vehicle #4 traveled distances of 560 ft and 280 ft, respectively. Therefore, the space headway between the two vehicles is equal to:

d = 560 − 280 = 280   ft ( 85.3   m ) 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_0051.tif"/>

At time = 3 seconds, the location of vehicle #3 is 280 ft, and the vehicle is stopping since the distance is constant. Therefore, speed = 0.

The time-distance relationship for vehicle #1 is linear as seen from this diagram and does not change all the way. The slope of this line is simply the speed. Therefore:

3.13 Speed   of   vehicle   # 1 = Distance   traveled Time   Interval 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_0052.tif"/>

Speed   of   vehicle   # 1 = ( 1010 − 600 ) ( 10 − 0 ) = 41.0   ft/sec = 28.0   mph ( 45   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_0053.tif"/>

3.11

The traffic density and speed data shown in Table 3.2 were obtained on a highway segment. Use linear regression analysis to fit the data to Greenshields model.

Solution:

The Greenshields model takes the form shown below:

3.14 u s = u f − u f k j 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_0054.tif"/>

Where:

u_s = space mean speed

u_f = mean free speed

k_j= jam density

k = density33

For a linear equation with the form: y = a 0 + a 1 x 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_0055.tif"/> , the following linear system (with matrix notation) is obtained using linear squares regression techniques and numerical methods:

[ n ∑ x ∑ x ∑ x 2 ] [ a 0 a 1 ] = [ ∑ y ∑ x y ] 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_0056.tif"/>

This system represents the two normal equations that will be used to solve for the coefficients a₀ and a₁ in the linear equation.

The solution of the above linear system is:

3.16 a 1 = ∑ i = 1 n x i y i − 1 n ( ∑ i = 1 n x i ) ( ∑ i = 1 n y i ) ∑ i = 1 n x i 2 − 1 n ( ∑ i = 1 n x i ) 2 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_0057.tif"/>

a 0 = ∑ i = 1 n y i n − a 1 ∑ i = 1 n x 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_0058.tif"/>

Or:

a 0 = y ¯ − a 1 x ¯ 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_0059.tif"/>

The required computations are done to obtain the parameters in the above linear system (see Table 3.3).34

Therefore:

3.19 [ 10 135 135 2808 ] [ a 0 a 1 ] = [ 465 5442 ] 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_0060.tif"/>

The Gauss elimination method is used to solve the above linear system. The solution provides the following values of a₀ and a₁:

a 0 = 57.95 ≅ 58.0 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_0061.tif"/>

a 1 = − 0.85 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_0062.tif"/>

The MS Excel worksheet used to apply the Gauss elimination method in order to solve the linear system is shown in Figure 3.13.

Or:

a 1 = ∑ i = 1 n x i y i − 1 n ( ∑ i = 1 n x i ) ( ∑ i = 1 n y i ) ∑ i = 1 n x i 2 − 1 n ( ∑ i = 1 n x i ) 2 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_0063.tif"/> 35

a 1 = 5442 − 1 10 ( 135 ) ( 465 ) 2808 − 1 10 ( 135 ) 2 = − 0.85 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_0064.tif"/>

a 0 = ∑ i = 1 n y i n − a 1 ∑ i = 1 n x 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_0065.tif"/>

a 0 = 465 10 − ( − 0.85 ) 135 10 = 58 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_0066.tif"/>

Therefore,

The Greenshields model that describes this data takes the following form:

u s = 58.0 − 0.85   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_0067.tif"/>

The screen image of the MS Excel worksheet used to conduct the computations and the linear regression analysis of this problem is shown in Figure 3.14.

3.12

For the Greenshields model determined in Problem 3.11, compute the following statistical parameters that quantify the “goodness” of fit of the model to the data and plot the density–speed relationship:

Total sum of squares of errors around the mean (S_t)

Total sum of squares of errors around the regression line (S_r)36

Standard error of the estimate (S_y/x)

Coefficient of determination (r²)

Coefficient of correlation (r)

Solution:

3.21 S t = ∑ i = 1 n ( y i − y ¯ ) 2 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_0068.tif"/>

3.22 S r = ∑ i = 1 n ( y i − y i -calculated ) 2 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_0069.tif"/>

3.23 S y / x = S r n − 2 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_0070.tif"/>

3.24 r 2 = S t − S r S 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_0071.tif"/>

3.25 r = r 2 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_0072.tif"/>

Where:

S_t = total sum of squares of errors around the mean

S_r = total sum of squares of errors around the regression line

S_y/x = standard error of the estimate

r² = coefficient of determination

r = coefficient of correlation

y_i = speed i

y ¯ 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_0073.tif"/>  = mean of speeds

y_i-predicted = speed calculated from the model

n = number of data points

n−2 = degrees of freedom37

The following computations are performed using the MS Excel worksheet shown in Table 3.4.

S t = ∑ i = 1 n ( y i − y ¯ ) 2 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_0074.tif"/>

⇒ S t = 756.5 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_0075.tif"/>

S r = ∑ i = 1 n ( y i − y i -calculated ) 2 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_0076.tif"/>

⇒ S r = 48.17 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_0077.tif"/>

S y / x = S r n − 2 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_0078.tif"/>

⇒

S y / x = 48.17 10 − 2 = 2.454 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_0079.tif"/>

r 2 = S t − S r S 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_0080.tif"/>

⇒

r 2 = 756.5 − 48.17 756.5 = 0.94 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_0081.tif"/> 38

r = r 2 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_0082.tif"/>

⇒   = 0.94 = 0.97 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_0083.tif"/>

The MS Excel worksheet used to perform the computations and to determine the linear regression parameters in this problem is shown in Figure 3.15.

Using the MS Excel worksheet, the density–speed relationship is also plotted as shown in Figure 3.16.39

3.13

In Problem 3.11, determine the mean free speed and the jam density for the traffic on the highway.

Solution:

By comparing the linear form of Greenshields model in Equation (3.12) with the linear form y = a 0 + a 1 x 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_0084.tif"/> , the following are obtained:

a 0 = u f 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_0085.tif"/>

⇒ u f = a 0 = 58   mph 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_0086.tif"/>

a 1 = − u f k j 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_0087.tif"/>

⇒

k j = − u f a 1 = − 58 − 0.85 = 68.3   vpm ≅ 69   vpm 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_0088.tif"/>

The MS Excel worksheet used to perform the computations in this problem is shown in Figure 3.14, in Problem 3.11.

3.14

If the model that describes the relationship between speed (mph) and density (vpm) on a highway segment is given as u s = 62 − 1.22   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_0089.tif"/> , determine the following:

Mean free speed

Jam density

Density at maximum flow

Speed at maximum flow

Maximum flow rate (capacity) of the highway segment

Solution:

Since the given model that describes the speed-density relationship on the highway is linear, this is the Greenshields model, which takes the form:

u s = u f − u f k j 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_0090.tif"/>

Therefore, the mean free speed is equal to the intercept (62) in the linear model.

u f = 62   mph ( ≅ 100   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_0091.tif"/>

The jam density is computed using the slope (−1.22) in the linear model, therefore:

− u f k j = − 1.22 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_0092.tif"/>

⇒

k j = 62 − 1.22 = 50.8   vpm ≅ 51   vpm ( 32   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_0093.tif"/> 40

The density at maximum flow is determined by taking the derivative of the flow with respect to density and equating the result to zero.

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_0094.tif"/>

But:

u s = u f − u f k j 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_0095.tif"/>

Hence:

q = ( u f − u f k j k ) 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_0096.tif"/>

Or:

q = u f k − u f k j k 2 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_0097.tif"/>

By deriving the above equation and equating it to zero, the following formula is obtained:

d q d k = u f − 2 u f k j k = 0 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_0098.tif"/>

⇒

k m = k j 2 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_0099.tif"/>

This is called the density at maximum flow.

Density   at   maximum   flow = 51 2 = 25.5 ≅ 26   vpm ( 16   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_0100.tif"/>

The speed at maximum flow is determined following the same procedure: by taking the derivative of the flow with respect to speed and equating the result to zero.

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_0101.tif"/>

But:

u s = u f − u f k j 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_0102.tif"/>

Reformulating the above formula to obtain k, the following formula for the density (k) is obtained:

k = k j − k j u f u s 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_0103.tif"/>

Hence:

3.31 q = ( k j − k j u f u s ) u s 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_0104.tif"/> 41

Or:

3.32 q = k j u s − k j u f u s 2 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_0105.tif"/>

By deriving the above equation and equating it to zero, the following formula is obtained:

d q d u s = k j − 2 k j u f u s = 0 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_0106.tif"/>

⇒

u m = u f 2 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_0107.tif"/>

This is called the speed at maximum flow.

Speed   at   maximum   flow = 62 2 = 31   mph ( 49.9   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_0108.tif"/>

The maximum flow rate (capacity) of the highway segment is simply equal to the density at maximum flow multiplied by the speed at maximum flow. Therefore,

q max = 26 × 31 ≅ 806   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_0109.tif"/>

A screen image of the MS Excel worksheet used to perform the computations of this problem is shown in Figure 3.17.42

3.15

If the relationship between the density and the space mean speed for a traffic stream is given as shown in Figure 3.18, determine the following:

Jam density

Mean free speed

Density at maximum flow

Speed at maximum flow

Maximum flow

Solution:

From the speed–density relationship in the figure above, the jam density is obtained at a space mean speed of zero. Therefore:

k j = 55   vpm 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_0110.tif"/>

On the other hand, the space mean speed is obtained at a density of zero, Hence:

u f = 60   mph . 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_0111.tif"/>

Since the speed–density relationship in this problem is linear, this relationship is simply the Greenshields model; and in the Greenshields model as seen in an earlier problem, the density at maximum flow is equal to the mean free speed divided by two. Therefore,

Density   at   maximum   flow = 55 2 = 27.5   vpm ( 17.1   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_0112.tif"/>

In the same way, the speed at maximum flow is equal to the mean fee speed divided by two in the Greenshields model. Therefore,

Speed   at   maximum   flow = 60 2 = 30   mph ( 48.3   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_0113.tif"/>

The maximum flow is equal to the density at maximum flow multiplied by the speed at maximum flow.

q max = 27.5 × 30 = 825   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_0114.tif"/>

The MS Excel worksheet used to obtain the results of this problem is shown in the screen image in Figure 3.19.43

3.16

The traffic density–speed data shown in Table 3.5 is obtained on a highway segment. If the data can be described by the Greenberg model, use linear regression analysis to determine the following:

The regression constants in the Greenberg model

The coefficient of determination (r²) for the model

Plot the relationship between density and speed44

Solution:

To use the linear regression analysis, linearization of the Greenberg model is to be done since the model is not linear. The Greenberg model takes the following form:

u s = C ln k j 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_0115.tif"/>

Where:

u_s = space mean speed

k = density

C = regression constant

k_j = jam density

The model can be linearized and reformulated as in the following form:

3.36 u s = C ln k j − C ln 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_0116.tif"/>

This form is compatible with the linear line equation y = a 0 + a 1 x 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_0117.tif"/> , such that:

y = u s 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_0118.tif"/>

a 0 = C ln k j 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_0119.tif"/>

a 1 = − C 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_0120.tif"/>

x = ln 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_0121.tif"/>

Therefore, the normal equations (the linear system with matrix notation) that was used earlier for linear models can be used for the Greenberg non-linear model by replacing x and y with lnk and u_s, respectively as shown below:

[ n ∑ x ∑ x ∑ x 2 ] [ a 0 a 1 ] = [ ∑ y ∑ x y ] 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_0122.tif"/>

[ n ∑ ln k ∑ ln k ∑ ( ln k ) 2 ] [ a 0 a 1 ] = [ ∑ u s ∑ ( ln k ) u s ] 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_0123.tif"/>

The solution of the above linear system is:

a 1 = ∑ i = 1 n x i y i − 1 n ( ∑ i = 1 n x i ) ( ∑ i = 1 n y i ) ∑ i = 1 n x i 2 − 1 n ( ∑ i = 1 n x i ) 2 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_0124.tif"/>

a 0 = ∑ i = 1 n y i n − a 1 ∑ i = 1 n x 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_0125.tif"/> 45

Therefore,

3.38 a 1 = ∑ i = 1 n ( ln k i ) u s i − 1 n ∑ i = 1 n ( ln k i ) ∑ i = 1 n u s i ∑ i = 1 n ( ln k i ) 2 − 1 n ( ∑ i = 1 n ( ln k i ) ) 2 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_0126.tif"/>

3.39 a 0 = ∑ i = 1 n u s i n − a 1 ∑ i = 1 n ( ln k 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_0127.tif"/>

The computations of the results needed to determine the regression coefficients a₀ and a₁ are performed using the MS Excel worksheet and are shown in Table 3.6.

a 1 = 1651.69 − 1 12 ( 48.08 ) ( 429.00 ) 195.22 − 1 12 ( 48.08 ) 2 = − 26.3 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_0128.tif"/>

a 0 = 429.00 12 − ( − ) 48.08 12 = 141.1 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_0129.tif"/>

− C = a 1 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_0130.tif"/> 46

⇒

C = − a 1 = 26.3 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_0131.tif"/>

C ln k j = a 0 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_0132.tif"/>

⇒

k j = e ( a 0 C ) = e ( 141.1 26.3 ) = 214   vpm ( 133   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_0133.tif"/>

Therefore, the Greenberg model that describes this data takes the form:

u s = 26.3 ln 214 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_0134.tif"/>

To compute the coefficient of determination (r²) for the model, the speed calculated from the model should be determined. The computations are shown in Table 3.7.47

S t = 2001.00 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_0135.tif"/>

S r = 16.83 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_0136.tif"/>

r 2 = S t − S r S 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_0137.tif"/>

⇒

r 2 = 2001.00 − 16.83 2001.00 = 0.99 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_0138.tif"/>

The density-speed relationship described by the Greenberg model is plotted in Figure 3.20.

An image of the MS Excel worksheet used to conduct the computations and the linear regression analysis of this problem is shown in Figures 3.21 and 3.22. Note that linearization is done for the non-linear model (Greenberg model) so that linear regression techniques can be used.

48 49

3.17

In Problem 3.16, determine the density at maximum flow, the speed at maximum flow, and the maximum flow (capacity) of the highway.

Solution:

u s = C ln k j 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_0139.tif"/>

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_0140.tif"/>

But:

u s = C ln k j 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_0141.tif"/>

Hence:

q = ( C ln k j k ) 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_0142.tif"/>

By deriving the above equation and equating it to zero, the following formula is obtained:

d q d k = ( C ln k j k ) + k ( − c k ) = 0 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_0143.tif"/>

⇒

ln k j k = 1 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_0144.tif"/>

k j k = e 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_0145.tif"/>

⇒

k m = k j e 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_0146.tif"/>

This is called the density at maximum flow.

Density   at   maximum   flow = 214 2 = 107   vpm ( ≅ 67   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_0147.tif"/>

The speed at maximum flow is determined following the same procedure: by taking the derivative of the flow with respect to speed and equating the result to zero.

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_0148.tif"/>

But:

u s = C ln k j 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_0149.tif"/> 50

Reformulating the above formula to obtain k, the following formula for the density (k) is obtained:

3.45 k = k j e ( u s C ) 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_0150.tif"/>

Hence:

3.46 q = ( k j e ( u s C ) ) u s 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_0151.tif"/>

By deriving the above equation and equating it to zero, the following formula is obtained:

3.47 d q d u s = k j e ( u s C ) − 1 C k j u s e ( u s C ) = 0 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_0152.tif"/>

⇒

k j e ( u s C ) ( 1 − u s C ) = 0 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_0153.tif"/>

⇒

u m = C 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_0154.tif"/>

This is called the speed at maximum flow.

Speed   at   maximum   flow = 26.3   mph ( 42.3   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_0155.tif"/>

And therefore, the maximum flow is equal to the density at maximum flow multiplied by the speed at maximum flow.

q = 107 × 26.3 = 2814   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_0156.tif"/>

3.18

If the model that describes the relationship between speed (mph) and density (vpm) on a highway is given as u s = 65 e ( − k 52 ) 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_0157.tif"/> , determine the following:

Density at maximum flow

Speed at maximum flow

Capacity of the highway

Solution:

To determine the density at maximum flow, the flow as a function of the density is determined and derived with respect to the density.

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_0158.tif"/> 51

But according to the model given in this problem between speed and density, it is expressed in the following generalized form:

u s = A e ( − k B ) 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_0159.tif"/>

Hence:

3.51 q = A k e ( − k B ) 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_0160.tif"/>

By deriving the above equation and equating it to zero, the following formula is obtained:

3.52 d q d k = − 1 B A k e ( − k B ) + A e ( − k B ) = 0 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_0161.tif"/>

⇒

A e ( − k B ) ( 1 − k B ) = 0 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_0162.tif"/>

⇒

k m = B 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_0163.tif"/>

This is called the density at maximum flow.

Density   at   maximum   flow = 52   vpm ( ≅ 33   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_0164.tif"/>

To determine the speed at maximum flow, the flow as a function of the speed is determined and derived with respect to the density.

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_0165.tif"/>

But:

u s = A e ( − k B ) 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_0166.tif"/>

Reformulating the above equation to obtain k as a function of u_s:

3.55 k = − B ln ( u s A ) 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_0167.tif"/>

And therefore,

3.56 q = − B u s ln ( u s A ) 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_0168.tif"/> 52

By deriving the above equation and equating it to zero, the following formula is obtained:

d q d u s = − B − B ln ( u s A ) = 0 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_0169.tif"/>

⇒

− B ( 1 + ln ( u s A ) ) = 0 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_0170.tif"/>

⇒

1 + ln ( u s A ) = 0 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_0171.tif"/>

⇒

u m = A e 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_0172.tif"/>

This is called the speed at maximum flow.

Speed   at   maximum   flow = 65 e = 23.9   mph ( 38.5   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_0173.tif"/>

The maximum flow is determined by multiplying the density at maximum flow by the speed at maximum flow.

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_0174.tif"/>

q = 23.9 × 52 = 1243   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_0175.tif"/>

3.19

The relationship between the density and the space mean speed for a traffic stream is described by the Greenberg model. If the density at maximum flow is 32 vpm (20 veh/km), determine the jam density.

Solution:

Based on the Greenberg model, the density at maximum flow is given by:

k m = k j e 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_0176.tif"/>

Therefore, the jam density can be computed as:

k j = e k m 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_0177.tif"/>

k j = e × 32 = 87   vpm ( 54   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_0178.tif"/> 53

3.20

If the model ln u s = ln u f − k k j 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_0179.tif"/> can be used to describe the relationship between speed (mph) and density (vpm) on a highway segment; and using regression analysis, the constants of the model Y = A + BX (where Y = lnu _s and X = k) are A = 3.9 and B = −0.018, then estimate the mean free speed (u _f ) and the jam density (k _j ).

Solution:

Since the model is given in the following mathematical expression:

ln u s = ln u f − k k j 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_0180.tif"/>

⇒

3.62 A = ln u f 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_0181.tif"/>

⇒

u f = e A = e 3.9 = 49.4   mph ( 79.5   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_0182.tif"/>

And:

3.63 B = − 1 k j 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_0183.tif"/>

⇒

k j = − 1 B = − 1 − 0.018 ≅ 56   vpm ( 35   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_0184.tif"/>

3.21

If the model that describes the relationship between speed (mph) and density (vpm) on a highway is given as u s = 45 ln 80 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_0185.tif"/> , estimate the speed and the density at maximum flow.

Solution:

This relationship is simply the Greenberg model. In Problem 3.17, the derivations showed that the speed at maximum flow is equal to the constant, and the density at maximum flow is equal to the jam density divided by e.

Since:

3.64 u s = 45 ln 80 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_0186.tif"/>

C = 45 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_0187.tif"/>

And

k j = 80   vpm 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_0188.tif"/>

Therefore:

Speed   at   maximum   flow = C = 45   mph ( 72.4   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_0189.tif"/> 54

Density   at   maximum   flow = k j e = 80 e = 29.4   vpm ( 18.3   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_0190.tif"/>

A screen image of the MS Excel worksheet used to perform the computations of this problem is shown in Figure 3.23.

3.22

The density and speed of traffic on a two-lane two-way highway segment are 35 vpm (21.7 veh/km) and 50 mph (80.5 kph), respectively. A loaded truck traveling at a speed of 25 mph joins the traffic stream on the highway from an aggregate quarry in the vicinity of the area. The truck remains on the highway for a time period of 15 minutes before it exits the highway. The highway segment is located in a no-passing zone; and hence, vehicles are not permitted to make passing. If this situation creates a platoon of vehicles behind the truck traveling at the same speed of the truck and having a density of 120 vpm (74.6 veh/km), determine the following (see Figure 3.24):55

The speed of the shockwave

The length of the platoon that will be created during the existence of the truck on the highway

The number of vehicles that will be affected by the incident

Solution:

The speed of the shockwave is calculated using the formula shown below:

3.65 u w = q 2 − q 1 k 2 − k 1 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_0191.tif"/>

u w = ( 25 ) ( 120 ) − ( 50 ) ( 35 ) 120 − 35 = 14.7   mph ( 23.7   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_0192.tif"/>

The length of the platoon of vehicles that will be created is equal to the relative speed multiplied by the time of the incident; it is computed using the following formula:

3.66 L = ( u r 2 ) 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_0193.tif"/>

Where:

L = length of platoon

u_r2 = relative speed after the shockwave condition = (u_s2-u_w)

t = time period of incident

Therefore:

3.67 L = ( u s 2 − u w ) 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_0194.tif"/>

L = ( 25 − 14.7 ) 15 60 = 2.57   mi ( 4.14   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_0195.tif"/>

The number of vehicles that will be affected by this incident is simply the number of vehicles in the platoon created behind the truck. Hence, it is calculated as:

Number   of   vehicles   in   platoon = L k 2 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_0196.tif"/>

Or:

Number   of   vehicles   in   platoon = ( u s 2 − u w ) t k 2 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_0197.tif"/>

Therefore,

Number   of   vehicles   in   platoon = 2.57 × 120 ≅ 309   vehicles 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_0198.tif"/>

The screen images of the MS Excel worksheet used to conduct the computations of this problem are shown in Figures 3.25 and 3.26. 56 57

3.23

The density on a two-lane two-way highway is 40 vpm (24.9 veh/km). An accident occurs on the highway blocking one lane completely. The incident remains on the highway for 20 minutes before it is cleared. This incident creates a queue of vehicles stopping behind the accident due to the heavy traffic volume in the other direction. If the traffic flow on this highway can be described by the Greenshields model with a mean free speed of 60 mph (96.6 kph) and a jam density of 100 vpm (62.1 veh/km), determine how many vehicles are affected by the occurrence of the accident (see Figure 3.27).

Solution:

The speed of the shockwave is calculated using the formula shown below:

Using the Greenshields model, the speed of traffic can be computed as follows:

u s = u f − u f k j 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_0199.tif"/>

⇒

u s = 60 − 60 100 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_0200.tif"/>

At k₁ = 40 vpm, the speed is equal to:

u s 1 = 60 − 60 100 ( 40 ) = 36   mph ( 57.9   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_0201.tif"/>

Therefore,

u w = q 2 − q 1 k 2 − k 1 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_0202.tif"/>

u w = 0 − ( 36 ) ( 40 ) 100 − 40 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_0203.tif"/>

u w = − 24   mph ( 38.6   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_0204.tif"/>

The number of vehicles that will be affected by this incident is equal to the length of the queue multiplied by the density of the traffic in the queue:

Number   of   vehicles   in   platoon = ( u s 2 − u w ) t k 2 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_0205.tif"/>

Number   of   vehicles   in   platoon = ( 0 − ( − 24 ) ) 20 60 × 100 = 800   vehicles 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_0206.tif"/>

The screen images of the MS Excel worksheet used to perform the computations of this problem are illustrated in Figures 3.28 and 3.29.58 59

3.24

The traffic flow on a two-lane two-way highway segment is described by the Greenberg model with a speed at maximum flow of 50 mph (80.5 kph) and a jam density of 100 vpm (62.1 veh/km). The traffic speed on this highway segment is 45 mph (72.4 kph) before a loaded truck traveling at a speed of 20 mph (32.2 kph) joins the traffic stream on the highway for a time period of 20 minutes until it exits the highway. Vehicles are prohibited to make passing due to the fact that the highway segment is located in a no-passing zone. If this situation creates a platoon of vehicles behind the truck traveling at the same speed of the truck, determine the following (see Figure 3.30):

The speed of the shockwave

The length of the platoon that will be created during the existence of the truck on the highway

The number of vehicles that will be affected by the incident

Solution:

The speed of the shockwave is calculated using the formula shown below:

u w = q 2 − q 1 k 2 − k 1 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_0207.tif"/>

Using the Greenberg model, the density of traffic can be computed as follows:

u s = C ln k j 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_0208.tif"/>

In Greenberg model, the speed at maximum flow is equal to the constant C in the model as shown in Problem 3.17; in other words:

C = 50 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_0209.tif"/>

And therefore:

u s = 50 ln 100 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_0210.tif"/>

At u_s1 = 45 mph, the density is equal to:

k 1 = 100 e ( 45 50 ) = 40.7   vpm ( 25.3   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_0211.tif"/>

And at u_s2 = 20 mph, the density is equal to:

k 2 = 100 e ( 20 50 ) = 67   vpm ( 41.7   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_0212.tif"/> 60

Hence:

u w = ( 20 ) ( 67 ) − ( 45 ) ( 40.7 ) 67 − 40.7 = − 18.5   mph ( − 29.8   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_0213.tif"/>

The length of the platoon of vehicles that will be created is equal to the relative speed multiplied by the time of the incident; it is computed using the following formula:

L = ( u s 2 − u w ) 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_0214.tif"/>

L = ( 20 − ( − 18.5 ) ) 20 60 = 12.8   mi   ( 20.7   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_0215.tif"/>

The number of vehicles that will be affected by this incident is simply the number of vehicles in the platoon created behind the truck. Hence, it is calculated as:

Number   of   vehicles   in   platoon = L k 2 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_0216.tif"/>

Therefore,

Number   of   vehicles   in   platoon = 12.8 × 67 ≅ 861   vehicles 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_0217.tif"/>

The results in US customary units and SI units are shown in the following screen images of the MS Excel worksheet that is utilized to obtain the results (see Figures 3.31 and 3.32).61

3.25

Traffic flow on one of the single-lane approaches of a signalized intersection traveling at 40 mph (eastbound shown in Figure 3.33) can be described by the Greenshields model. The duration of the red phase for this approach is 25 seconds. If the jam density is 125 vpm and the mean free speed is 50 mph, determine the following:

Speed of shockwave

Length of the queue at the end of the red phase

Number of vehicles in the queue62

Solution:

Since the traffic flow follows the Greenshields model, the density of traffic is computed using the model as shown below:

u s = u f − u f k j 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_0218.tif"/>

⇒

u s = 50 − 50 125 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_0219.tif"/>

At u_s1 = 40 mph, the density is estimated as:

40 = 50 − 50 125 ( k 1 ) 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_0220.tif"/>

⇒

k 1 = 25   vpm ( 15.5   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_0221.tif"/>

The speed of the shockwave is then computed using the common formula shown below:

u w = q 2 − q 1 k 2 − k 1 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_0222.tif"/>

u w = 0 − ( 40 ) ( 25 ) 125 − 25 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_0223.tif"/>

u w = − 10   mph ( − 16.1   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_0224.tif"/>

The length of the queue is calculated using the formula below:

L = ( u s 2 − u w ) 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_0225.tif"/>

⇒

L = ( 0 − ( − 10 ) ) × 25 3600 × 5280 = 366.7   ft ( 111.8   m ) 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_0226.tif"/>

The “5280” is a conversion factor to convert from mi to ft.

The number of vehicles in the queue is equal to the length of the queue multiplied by the density of the traffic in the queue:

Number   of   vehicles   in   queue = ( u s 2 − u w ) t k 2 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_0227.tif"/>

Number   of   vehicles   in   queue = ( 0 − ( − 10 ) ) 25 3600 × 125 ≅ 9   vehicles . 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_0228.tif"/>

Figures 3.34 and 3.35 show screen images of the MS Excel worksheet used to perform the computations of this problem.63 64

3.26

In Problem 3.25, if the speed of vehicles when the traffic signal turns to green is 15 mph, estimate the time needed to dissipate the queue (created from the red phase) after the end of the red phase assuming that the traffic flow still follows the Greenshields model.

Solution:

After the queue is created due to the red phase duration, it takes time for the queue to dissipate when the traffic signal turns to green. In this case, the shockwave conditions are also applied but in reverse order; in other words, the speed before the shockwave is zero (since the vehicles are stopping at the red signal) and after the shockwave, the speed is 15 mph. Hence:

u w = q 2 − q 1 k 2 − k 1 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_0229.tif"/>

u w = q 2 − 0 k 2 − k 1 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_0230.tif"/>

The Greenshields model is used to determine the density after the shockwave:

u s = u f − u f k j 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_0231.tif"/>

⇒

u s = 50 − 50 125 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_0232.tif"/>

At u_s2 = 15 mph, the density is computed as:

15 = 50 − 50 125 ( k 2 ) 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_0233.tif"/>

⇒

k 2 = 87.5   vpm ( 54.4   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_0234.tif"/>

u w = ( 15 ) ( 87.5 ) − 0 87.5 − 125 = − 35   mph ( − 56.3   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_0235.tif"/>

From Problem 3.25:

The   length   of   the   queue = 366.7   ft 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_0236.tif"/>

Therefore, the time required to dissipate this queue is equal to the length of the queue divided by the relative speed after the shockwave; in other words:

L = ( u s 2 − u w ) 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_0237.tif"/>

⇒

t = L ( u s 2 − u w ) = ( 366.7 5280 ) ( 15 − ( − 35 ) ) × 3600 = 5   seconds 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_0238.tif"/>

The division by the conversion factor “5280” and the multiplication by the conversion factor “3600” are performed to convert from ft to mi and from hr to seconds, respectively.

The results (in US customary units and SI units) are computed using the MS Excel worksheet shown in Figures 3.36 and 3.37.65 66

3.27

A traffic signal near a school is used to allow the school kids to cross the road to the other side (as in Figure 3.38). During the green phase for school kids, vehicles must stop creating a queue on the road. If the green phase duration is 40 seconds, and the traffic flow can be described by the Greenberg model u s = 45 ln 90 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_0239.tif"/> , determine the number of vehicles that exist in the queue during this phase. Assume space mean speed of traffic within the school zone is 20 mph.

Solution:

Since the traffic flow follows the Greenberg model, the density of traffic is computed using the model as shown below:

u s = 45 ln 90 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_0240.tif"/>

At u_s1 = 20 mph, the density is calculated as:

20 = 45 ln 90 k 1 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_0241.tif"/>

⇒

k 1 = 57.7   vpm ( 35.9   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_0242.tif"/>

The speed of the shockwave is then computed using the formula shown below:

u w = q 2 − q 1 k 2 − k 1 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_0243.tif"/>

u w = 0 − ( 20 ) ( 57.7 ) 90 − 57.7 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_0244.tif"/>

u w = − 35.7   mph ( − 57.5   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_0245.tif"/>

The length of the queue is calculated using the formula below:67

L = ( u s 2 − u w ) 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_0246.tif"/>

⇒

L = ( 0 − ( − 35.7 ) ) × 40 3600 = 0.397   mi ( 0.639   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_0247.tif"/>

The number of vehicles in the queue is equal to the length of the queue multiplied by the density of the traffic in the queue (jam density):

Number   of   vehicles   in   queue = L k 2 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_0248.tif"/>

Number   of   vehicles   in   queue = 0.397 × 90 ≅ 36   vehicles 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_0249.tif"/>

The screen images of the MS Excel worksheet that is used to perform the detailed computations of the results using the US customary units as well as the SI units are shown in Figures 3.39 and 3.40. 68

3.28

An urban transportation planner is assigned to conduct a study for a parking garage in the vicinity of a signalized T-intersection as shown in Figure 3.41. Required from the study is to check if there will be any trouble for vehicles to access the garage using the entrance road that is 1100 ft (335.3 m) from the intersection. During the red phase of the signal, queues are developed at the intersection; this situation may block the access to the entrance road leading to the garage as shown in the diagram. If the traffic flow on the approach is 1200 vph and can be described by the Greenberg model u s = 35 ln 100 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_0250.tif"/> , and the red phase at the intersection is 25 seconds, determine if the entrance road will be affected by this situation.69

Solution:

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_0251.tif"/>

But:

u s = 35 ln 100 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_0252.tif"/>

Therefore:

q 1 = ( 35 ln 100 k 1 ) k 1 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_0253.tif"/>

⇒

1200 = ( 35 ln 100 k 1 ) k 1 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_0254.tif"/> 70

To solve this equation, the MS Excel Solver tool is used (see Figure 3.42) or the simple fixed-point iteration method (a numerical method) is used. Both methods are explained below.

The left side of the equation is 1200, and the right side is written in terms of k₁. The error is defined as [log (left side)−log (right side)]². The MS Excel solver does an iterative approach by changing the value of k₁ until a minimum (or zero) error is obtained; i.e., the left side and the right side are approximately equal.

There are two solutions for this equation. Based on the initial value used in Excel Solver, one of the two solutions is obtained.

k 1 = 51.2   vpm   ( or   24.1   vpm ) 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_0255.tif"/>

Using the numerical method (simple fixed-point iteration), k₁ should be separated from the other terms in the above equation as below:

According to the simple fixed-point iteration method, the (i + 1)th k₁ value is equal to the value of the function on the right side at ith k₁ value. In other words:

k 1 ( i + 1 ) = 1200 35 ln ( 100 k 1 ( 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_0256.tif"/>

Using the MS Excel worksheet, this can be done for as many iterations as possible by copying and pasting the formula down in the cells to obtain a reasonable acceptable error (approximate percent relative error, ε_a).

The simple fixed-point iteration method converges to a value of k₁ equal to 24.1 vpm (see Table 3.8).71

Since the Greenberg model satisfies the traffic conditions of this intersection, and the Greenberg model can be applied only under congested conditions; therefore, the highest value of k₁ is selected (51.2 vpm).

When the traffic signal is red, the traffic flow, q₂ = 0 and the density, k₂ = k_j = 100 vpm. Hence:

u w = q 2 − q 1 k 2 − k 1 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_0257.tif"/>

u w = 0 − 1200 100 − 51.2 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_0258.tif"/>

u w = − 24.6   mph ( − 39.5   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_0259.tif"/>

The length of the queue is calculated using the formula below:

L = ( u s 2 − u w ) 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_0260.tif"/>

⇒

L = ( 0 − ( − 24.6 ) ) × 25 3600 × 5280 = 901   ft ( 274.6   m ) 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_0261.tif"/>

Since the length of the queue stopping at the signalized intersection during the red phase (901 ft) < 1000 ft (the distance from the entrance road to the intersection), the entrance road will not be affected.72

An image of the MS Excel worksheet used for the computations of the results (represented by the two systems of units: US Customary Units and SI Units) of this problem is shown in Figures 3.43 and 3.44.73

3.29

The accepted and rejected gaps of vehicles on the minor approach of an unsignalized intersection are given as shown in Table 3.9. If the arrival of vehicles on the major approach (see Figure 3.45) can be described by the Poisson distribution, and the peak-hour volume is 1200 vph, determine the following:

The critical gap using the algebraic method.

The critical gap using the graphical method

The number of accepted gaps that will be available for the minor approach vehicles during the peak hour

The number of rejected gaps during the peak hour

Solution:

To determine the critical gap using the algebraic method, the change in accepted gaps between two successive gaps and the change in rejected gaps between two successive gaps are computed. Then, the difference between the two changes is determined. The critical gap is located within the interval that has the smallest difference between the two changes as shown in Table 3.10. 74

The lowest difference is 10, which refers to the gap interval of 2.0–3.0 seconds. Using the original data for the number of accepted gaps and the number of rejected gaps at 2.0 and 3.0 seconds, respectively; the critical gap (t_c) and the number of gaps (n = accepted gaps = rejected gaps) can be estimated using interpolation or similarity in triangles as follows (see Table 3.11):

32 − 22 3 − 2 = 32 − n 3 − t c 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_0262.tif"/>

36 − 58 3 − 2 = 36 − n 3 − t c 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_0263.tif"/>

Solving these two equations simultaneously provides the following solution:

32 − 22 32 − n = 36 − 58 36 − 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_0264.tif"/>

⇒

n = 33.25 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_0265.tif"/>

t c = 3.125   seconds 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_0266.tif"/>

To determine the critical gap using the graphical method, the cumulative number of accepted gaps smaller the gap (t) and the cumulative number of rejected gaps greater than the gap (t) are calculated as shown in Table 3.12.75

After that, the two sets of data are plotted against the gap (t) to obtain two curves in one figure. The intersection point of the two curves refers to the critical gap on the x-axis as shown in Figure 3.46.

From the graph, t_c = 2.95 seconds.

The number of accepted gaps that will be available for the vehicles in the minor approach during the peak hour is estimated using the following formula:

Freq . ( h ≥ t ) = ( V − 1 ) e − λ 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_0267.tif"/>

Where :

V = peak hour volume (total number of vehicles arriving in a time period T)

λ = average number of vehicles arriving per time period in seconds = V/T; in other words, it is given by the following formula:

λ = V 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_0268.tif"/> 76

h = any gap

e^−λt = probability of h ≥ t

Assuming that the arrival of the major approach traffic follows Poisson distribution, then:

p ( x ) = μ x e − μ x ! 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_0269.tif"/>

Where:

p(x) = probability of x vehicles arriving in time t

μ = average number of vehicles arriving in time t

μ = λ 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_0270.tif"/>

A vehicle in the minor approach will be able to merge into the major approach when there is a gap available on the major approach that is equal to or greater than the critical gap. In other words, this will happen when there is no vehicles (zero vehicles) arrive during this gap. The probability of this to occur is the probability of x = 0 in Equation (3.80).

⇒

3.82 p ( 0 ) = p ( h ≥ t ) = μ 0 e − μ 0 ! = e − μ = e − λ 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_0271.tif"/>

Since the addition of the two probabilities (p(h ≥ t), p(h < t)) must equal 1.0:

3.83 p ( h &lt; t ) = 1 − p ( h ≥ t ) = 1 − e − λ 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_0272.tif"/>

The total number of gaps between V vehicles on the major approach is equal to V−1.

Therefore, the predicted number of gaps equal to or greater than the critical gap (number of accepted gaps) during the peak hour is given by:

Freq . ( h ≥ t ) = ( V − 1 ) e − λ 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_0273.tif"/>

V = 1200 vph

λ = 1200/3600 = 0.333 veh/sec

t = critical gap = 3.125 sec

Therefore:

Freq .   ( h ≥ t ) = ( V − 1 ) e − λ 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_0274.tif"/>

Freq .   ( h ≥ t ) = ( 1200 − 1 ) e − ( 0.333 ) ( 3.125 ) = 423 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_0275.tif"/>

In a similar manner, the predicted number of gaps smaller than the critical gap (the number of rejected gaps) during the peak hour is calculated by multiplying the probability of its occurrence by the total number of gaps (V−1).

Freq .   ( h &lt; t ) = ( V − 1 ) ( 1 − e − λ 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_0276.tif"/> 77

Therefore:

Freq .   ( h &lt; t ) = ( 1200 − 1 ) ( 1 − e − ( 0.333 ) ( 3.125 ) ) = 776 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_0277.tif"/>

Or simply:

The number of rejected gaps = the total number of gaps–the number of accepted gaps

Freq .   ( h &lt; t ) = 1199 − 423 = 776 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_0278.tif"/>

Figures 3.47 and 3.48 show the screen images of the MS Excel worksheet used to conduct the computations and analyze the gap acceptance data for this problem.78

3.30

The numbers of accepted and rejected gaps of vehicles on an on-ramp that leads to a four-lane highway (shown in Figure 3.49) are given in Table 3.13.79

If the traffic flow rate during the peak hour on the highway is 1,800 vph, and the arrival of vehicles on the highway can be described by the Poisson distribution, determine the following:

The critical gap using the algebraic method

The critical gap using the graphical method

The number of accepted gaps that will be available for the on-ramp vehicles during the peak hour

The number of rejected gaps during the peak hour

Solution:

Using the algebraic method (see Table 3.14):

Using interpolation (similarity in triangles) between the two gaps 3 and 4 seconds, the critical gap along with the number of accepted/rejected gaps can be determined (see Table 3.15):

65 − 32 4 − 3 = 65 − n 4 − t c 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_0279.tif"/>

28 − 51 4 − 3 = 28 − n 4 − t c 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_0280.tif"/>

Solving the above two equations simultaneously provides the solution shown below:

65 − 32 65 − n = 28 − 51 28 − 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_0281.tif"/> 80

⇒

n = 43.20 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_0282.tif"/>

t c = 3.339   seconds 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_0283.tif"/>

Using the graphical method (see Table 3.16 and Figure 3.50)

The number of accepted gaps is computed as follows:

Freq .   ( h ≥ t ) = ( V − 1 ) e − λ 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_0284.tif"/>

λ = V 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_0285.tif"/> 81

λ = 1800 3600 = 0.5   veh/sec 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_0286.tif"/>

V = 1200 vph

t = critical gap = 3.34 sec

Therefore:

Freq .   ( h ≥ t ) = ( V − 1 ) e − λ 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_0287.tif"/>

⇒

Freq .   ( h ≥ t ) = ( 1800 − 1 ) e − ( 0.5 ) ( 3.34 ) = 339 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_0288.tif"/>

The number of rejected gaps is calculated as shown below:

Freq .   ( h &lt; t ) = ( V − 1 ) ( 1 − e − λ 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_0289.tif"/>

⇒

Freq .   ( h &lt; t ) = ( 1800 − 1 ) ( 1 − e − ( 0.5 ) ( 3.34 ) ) = 1460 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_0290.tif"/>

Note that the summation of the number of accepted gaps and the number of rejected gaps is equal to the total number of gaps within 1800 vehicles in the peak hour = 1800−1 = 1799.

Figures 3.51 and 3.52 show screen images of the MS Excel worksheet used to conduct the computations for this problem.82

Figure 3.1 A screen image of the MS Excel worksheet used for the computations of Problem 3.1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_1_B.tif"/> Figure 3.2 A screen image of the MS Excel worksheet used for the computations of Problem 3.2. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_2_B.tif"/> Figure 3.3 A screen image of the MS Excel worksheet used for the computations of Problem 3.3. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_3_B.tif"/> Figure 3.4 A screen image of the MS Excel worksheet used to compute the flow rate for Problem 3.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_4_B.tif"/> Figure 3.5 A screen image of the MS Excel worksheet used to compute the time mean speed and the space mean speed for Problem 3.5. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_5_B.tif"/> Figure 3.6 A screen image of the MS Excel worksheet used for the computations of Problem 3.6. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_6_B.tif"/> Figure 3.7 A screen image of the MS Excel worksheet used to compute the time mean speed and the space mean speed for Problem 3.7 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_7_B.tif"/> Figure 3.8 A screen image of the MS Excel worksheet used to compute the time mean speed and the space mean speed for Problem 3.7 (SI Units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_8_B.tif"/> Figure 3.9 Traffic data on a two-lane two-way highway for five vehicles for Problems 3.8 and 3.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_9_B.tif"/> Table 3.1 The Results for the Arrival Times at A-A and B-B for Problem 3.8 Vehicle # Speed (mph) Distance from A-A (ft) Travel Time from A-A to reach current position (sec) Difference in Time between vehicle #1 and the others (sec) Time of Arrival at A-A (sec) Travel Time from A-A to B-B (sec) Time of Arrival at B-B (sec) 1 50.0 494.0 6.74 0.00 t₀ 5.45 t₀+ 5.45 2 48.0 330.0 4.69 2.05 t₀+ 2.05 5.68 t₀+ 7.73 3 45.0 246.4 3.73 3.00 t₀+ 3.00 6.06 t₀+ 9.06 4 42.0 205.5 3.34 3.40 t₀+ 3.40 6.49 t₀+ 9.89 5 40.0 58.7 1.00 5.74 t₀+ 5.74 6.82 t₀+ 12.56 Figure 3.10 Screen images of the MS Excel worksheet for the computations of Problem 3.8 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_10_B.tif"/> Figure 3.11 Screen images of the MS Excel worksheet for the computations of Problem 3.8 (SI units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_11_B.tif"/> Figure 3.12 Time–space diagram for four vehicles for Problem 3.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_12_B.tif"/> Table 3.2 Traffic Density and Speed Data on a Highway Segment for Problem 3.11 Density, k (vpm) Speed, u_s (mph) 37.5 30.0 22.5 35.0 18.0 40.0 15.0 44.0 12.0 47.0 9.0 50.0 7.5 52.0 6.0 54.0 4.5 55.0 3.0 58.0 Table 3.3 Computations Needed for the Determination of the Linear Regression Coefficients a₀ and a₁ for Problem 3.11 Density, k (vpm) Speed, u_s (mph) k² (u_s)(k) 37.5 30.0 1406.3 1125.0 22.5 35.0 506.3 787.5 18.0 40.0 324.0 720.0 15.0 44.0 225.0 660.0 12.0 47.0 144.0 564.0 9.0 50.0 81.0 450.0 7.5 52.0 56.3 390.0 6.0 54.0 36.0 324.0 4.5 55.0 20.3 247.5 3.0 58.0 9.0 174.0 SUM 135 465 2808.0 5442.0 Figure 3.13 A screen image of the MS Excel worksheet used to apply the Gauss elimination method in the linear regression analysis for Problem 3.11. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_13_B.tif"/> Figure 3.14 A screen image of the MS Excel worksheet used for the computations and the linear regression analysis for Problem 3.11. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_14_B.tif"/> Table 3.4 Linear Regression Analysis and Computations of Linear Regression Parameters for Problem 3.11 Density, k (vpm) Speed, u_s (mph) k² (u_s)(k) (u_s-u_{s avg.})² u_{s calculated} (mph) (u_s-u_{s calculated})² 37.5 30.0 1406.3 1125.0 272.3 26.2 14.80 22.5 35.0 506.3 787.5 132.3 38.9 14.98 18.0 40.0 324.0 720.0 42.3 42.7 7.21 15.0 44.0 225.0 660.0 6.3 45.2 1.51 12.0 47.0 144.0 564.0 0.3 47.8 0.60 9.0 50.0 81.0 450.0 12.3 50.3 0.10 7.5 52.0 56.3 390.0 30.3 51.6 0.17 6.0 54.0 36.0 324.0 56.3 52.9 1.30 4.5 55.0 20.3 247.5 72.3 54.1 0.76 3.0 58.0 9.0 174.0 132.3 55.4 6.75 SUM 135 465 2808.0 5442.0 756.5 48.17 Figure 3.15 A screen image of the MS Excel worksheet used for determining the linear regression parameters for Problem 3.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_15_B.tif"/> Figure 3.16 Density–speed relationship described by the Greenshields model for Problem 3.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_16_B.tif"/> Figure 3.17 A screen image of the MS Excel worksheet used for the computations of Problem 3.14. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_17_B.tif"/> Figure 3.18 Density and space mean speed relationship for a traffic stream for Problem 3.15. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_18_B.tif"/> Figure 3.19 A screen image of the MS Excel worksheet used to obtain the results of Problem 3.15. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_19_B.tif"/> Table 3.5 Traffic Density–Speed Data for a Highway Segment for Problem 3.16 Density, k (vpm) Speed, u_s (mph) 100 20 92 21 84 25 75 27 70 29 64 33 60 35 54 38 42 41 35 46 28 53 22 61 Table 3.6 Computations needed for the Determination of the Regression Coefficients a₀ and a₁ for Problem 3.16 Density, k (vpm) Speed, u_s (mph) lnk (lnk)² (u_s)(lnk) 100 20 4.61 21.21 92.1 92 21 4.52 20.45 95.0 84 25 4.43 19.63 110.8 75 27 4.32 18.64 116.6 70 29 4.25 18.05 123.2 64 33 4.16 17.30 137.2 60 35 4.09 16.76 143.3 54 38 3.99 15.91 151.6 42 41 3.74 13.97 153.2 35 46 3.56 12.64 163.5 28 53 3.33 11.10 176.6 22 61 3.09 9.55 188.6 SUM 726.00 429.00 48.08 195.22 1651.69 Table 3.7 Regression Analysis and Computations of Linear Regression Parameters for Problem 3.16 Density, k (vpm) Speed, u_s (mph) lnk (lnk)² (u_s)(lnk) (u_s-u_{s avg.})² u_{s calculated} (mph) (u_s-u_{s calculated})² 100 20 4.61 21.21 92.1 132.3 20.0 0.0007 92 21 4.52 20.45 95.0 110.3 22.2 1.4818 84 25 4.43 19.63 110.8 42.3 24.6 0.1536 75 27 4.32 18.64 116.6 20.3 27.6 0.3438 70 29 4.25 18.05 123.2 6.3 29.4 0.1596 64 33 4.16 17.30 137.2 2.3 31.8 1.5511 60 35 4.09 16.76 143.3 12.3 33.5 2.4004 54 38 3.99 15.91 151.6 42.3 36.2 3.1698 42 41 3.74 13.97 153.2 90.3 42.8 3.3279 35 46 3.56 12.64 163.5 210.3 47.6 2.6106 28 53 3.33 11.10 176.6 462.3 53.5 0.2304 22 61 3.09 9.55 188.6 870.3 59.8 1.3974 SUM 726.00 429.00 48.08 195.22 1651.69 2001.00 429.00 16.83 Figure 3.20 Density–speed relationship described by the Greenberg model for Problem 3.16. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_20_B.tif"/> Figure 3.21 A screen image of the MS Excel worksheet for the computations and the linear regression analysis for Problem 3.16. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_21_B.tif"/> Figure 3.22 A screen image of the MS Excel worksheet used for determining the linear regression parameters for Problem 3.16. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_22_B.tif"/> Figure 3.23 A screen image of the MS Excel worksheet used to compute the speed and density at maximum flow for Problem 3.21. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_23_B.tif"/> Figure 3.24 A shockwave on a two-lane two-way highway due to a truck for Problem 3.22. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_24_B.tif"/> Figure 3.25 A screen image of the MS Excel worksheet used for the computations of Problem 3.22 (US Customary Units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_25_B.tif"/> Figure 3.26 A screen image of the MS Excel worksheet used for the computations of Problem 3.22 (SI units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_26_B.tif"/> Figure 3.27 A shockwave on a two-lane two-way highway due to an accident for Problem 3.23. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_27_B.tif"/> Figure 3.28 A screen image of the MS Excel worksheet used for the computations of Problem 3.23 (US Customary Units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_28_B.tif"/> Figure 3.29 A screen image of the MS Excel worksheet used for the computations of Problem 3.23 (SI Units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_29_B.tif"/> Figure 3.30 A shockwave on a two-lane two-way highway due to a truck for Problem 3.24. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_30_B.tif"/> Figure 3.31 A screen image of the MS Excel worksheet used for the computations of Problem 3.24 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_31_B.tif"/> Figure 3.32 A screen image of the MS Excel worksheet used for the computations of Problem 3.24 (SI units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_32_B.tif"/> Figure 3.33 A shockwave on a single-lane approach at a signalized intersection for Problem 3.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_33_B.tif"/> Figure 3.34 A screen image of the MS Excel worksheet used for the computations of Problem 3.25 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_34_B.tif"/> Figure 3.35 A screen image of the MS Excel worksheet used for the computations of Problem 3.25 (SI units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_35_B.tif"/> Figure 3.36 A screen image of the MS Excel worksheet used for the computations of Problem 3.26 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_36_B.tif"/> Figure 3.37 A screen image of the MS Excel worksheet used for the computations of Problem 3.26 (SI Units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_37_B.tif"/> Figure 3.38 A shockwave and vehicles queue due to a signal near a school for Problem 3.27. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_38_B.tif"/> Figure 3.39 A screen image of the MS Excel worksheet used for the computations of Problem 3.27 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_39_B.tif"/> Figure 3.40 A screen image of the MS Excel worksheet used for the computations of Problem 3.27 (SI units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_40_B.tif"/> Figure 3.41 A parking garage in the vicinity of a signalized intersection with shockwave conditions for Problem 3.28. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_41_B.tif"/> Figure 3.42 Image of MS Excel Solver result for Problem 3.28. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_42_B.tif"/> Table 3.8 The Solution for k₁ Using the Simple Fixed-Point Iteration Method after 16 Iterations for Problem 3.28 i k ₁ ε_a (%) 0 20 1 21.3 6.116 2 22.2 3.921 3 22.8 2.587 4 23.2 1.740 5 23.4 1.186 6 23.6 0.816 7 23.8 0.565 8 23.9 0.393 9 23.9 0.274 10 24.0 0.191 11 24.0 0.134 12 24.0 0.094 13 24.0 0.066 14 24.1 0.046 15 24.1 0.032 16 24.1 0.023 Figure 3.43 A screen image of the MS Excel worksheet used for the computations of Problem 3.28 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_43_B.tif"/> Figure 3.44 A screen image of the MS Excel worksheet used for the computations of Problem 3.28 (SI units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_44_B.tif"/> Table 3.9 The Number of Accepted and Rejected Gaps on the Minor Approach of an Unsignalized T-Intersection for Problem 3.29 Gap, t (sec) Number of Accepted Gaps < t Number of Rejected Gaps > t 1.0 4 100 2.0 22 58 3.0 32 36 4.0 66 15 5.0 96 3 Figure 3.45 Gap acceptance at an unsignalized T-intersection for Problem 3.29. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_45_B.tif"/> Table 3.10 The Change in Accepted Gaps and Rejected Gaps between Two Successive Gaps for Problem 3.29 Successive Gaps (sec) Change in Number of Accepted Gaps < t Change in Number of Rejected Gaps > t Difference Between the Two Changes 1.0–2.0 18 42 24 2.0–3.0 10 22 12 3.0–4.0 34 21 13 4.0–5.0 30 12 18 Table 3.11 The Gap Interval for Critical Gap for Problem 3.29 Length of Gap, t (sec) Number of Accepted Gaps (less than t) Number of Rejected Gaps (more than t) 2.0 22 58 t_c n n 3.0 32 36 Table 3.12 The Cumulative Number of Accepted Gaps and Rejected Gaps for Problem 3.29 Gap, t (sec) Cumulative Number of Accepted Gaps < t Cumulative Number of Rejected Gaps > t 1.0 4 212 2.0 26 112 3.0 58 54 4.0 124 18 5.0 220 3 Figure 3.46 The graphical method for critical gap for Problem 3.29. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_46_B.tif"/> Figure 3.47 A screen image of the MS Excel worksheet used to compute the critical gap by the algebraic method for Problem 3.29. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_47_B.tif"/> Figure 3.48 A screen image of the MS Excel worksheet used to determine the critical gap by the graphical method and to conduct the other computations for Problem 3.29. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_48_B.tif"/> Figure 3.49 Gap acceptance on an on-ramp on a four-lane highway for Problem 3.30. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_49_B.tif"/> Table 3.13 The Number of Accepted and Rejected Gaps on the Minor Approach of an Unsignalized T-Intersection for Problem 3.30 Length of Gap, t (sec) Number of Accepted Gaps (less than t) Number of Rejected Gaps (more than t) 1.0 2 125 2.0 20 88 3.0 32 51 4.0 65 28 5.0 105 11 6.0 145 3 Table 3.14 The Change in Accepted Gaps and Rejected Gaps between Two Successive Gaps for Problem 3.30 Successive Gaps (sec) Change in Number of Accepted Gaps < t Change in Number of Rejected Gaps > t Difference Between the Two Changes 1.0–2.0 18 37 19 2.0–3.0 12 37 25 3.0–4.0 33 23 10 4.0–5.0 40 17 23 5.0–6.0 40 8 32 Table 3.15 The Gap Interval for Critical Gap for Problem 3.30 Length of Gap, t (sec) Number of Accepted Gaps (less than t) Number of Rejected Gaps (more than t) 3.0 32 51 t_c n n 4.0 65 28 Table 3.16 The Cumulative Number of Accepted Gaps and Rejected Gaps for Problem 3.30 Gap, t (sec) Cumulative Number of Accepted Gaps < t Cumulative Number of Rejected Gaps > t 1 2 306 2 22 181 3 54 93 4 119 42 5 224 14 6 369 3 Figure 3.50 The graphical method for critical gap for Problem 3.30. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_50_B.tif"/> Figure 3.51 A screen image of the MS Excel worksheet used to compute the critical gap by the algebraic method for Problem 3.30. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_51_B.tif"/> Figure 3.52 A screen image of the MS Excel worksheet used to determine the critical gap by the graphical method and to conduct the other computations for Problem 3.30. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig3_52_B.tif"/>