sm ch (6)

Chapter 6 Fundamental Principles of Traffic Flow 6-1 Observers stationed at two sections XX and YY, 500 ft apart on a highway, record the arrival times of five vehicles as shown in the accompanying table. If the total time of observation at XX was 15 sec, determine (a) the time mean speed, (b) the space mean speed, and (c) the flow at section XX. Vehicle Section XX Section YY A To To + 7.6 sec B To + 3.4 sec To + 9.9 sec C To + 7.9 sec To + 14.6 sec D To + 12.0 sec To + 20.4 sec E To + 14.9 sec To + 21.7 sec

First, calculate speeds of individual vehicles: u1 = L/t = 500 ft / (7.6 sec – 0 sec) = 65.78 ft/sec u2 = L/t = 500 ft / (9.9 sec – 3.4 sec) = 76.92 ft/sec u3 = L/t = 500 ft / (14.6 sec – 7.9 sec) = 74.63 ft/sec u4 = L/t = 500 ft / (20.4 sec – 12.0 sec) = 59.52 ft/sec u5 = L/t = 500 ft / (21.7 sec – 14.9 sec) = 73.53 ft/sec a) Time mean speed (TMS), using Equation 6.2, u t =

1

n ut = (65.78 + 76.92 + 74.63 + 59.52 + 73.53) / 5 ut = 70.08 ft/sec ut = 70.08/1.47 = 47.7 mi/h

b) Space mean speed (SMS), using Equation 6.3, u s

=

n

u

i

i =1

nL n

 t i

i =1

us = (500 ft)(5) / (7.6 + 6.5 + 6.7 + 8.4 + 6.8) sec us = 69.44 ft/sec us = 47.2 mi/h c) Flow at XX, using Equation 6.1, q = n (3600) / T q = 5 (3600 sec/h) / 15 sec q = 1200 veh/h

77 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or dup licated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow 6-2 Data obtained from aerial photography showed six vehicles on a 700 ft-long section of road. Traffic data collected at the same time indicated an average time headway of 3.7 sec. Determine (a) the density on the highway, (b) the flow on the road, and (c) the space mean speed.

a) Density, k =

n

L k = 6 veh / 700 ft k = 0.0086 veh/ft k = 0.0086 × 5280 = 45.4 veh/mi

b) Flow, q

=

1

t q = 1 / (3.7 sec/veh) q = 0.27 veh/sec q = 0.27 × 3600 = 972 veh/h

c) Space mean speed (SMS), u s

=

q

k us = (972 veh/h) / (45.4 veh/mi) us = 21.4 mi/h

6-3 Two sets of students are collecting traffic data at two sections, xx and yy, of a highway 1500 ft apart. Observations at xx show that five vehicles passed that section at intervals of 3, 4, 3, and 5 sec, respectively. If the speeds of the vehicles were 50, 45, 40, 35, and 30 mi/hr respectively, draw a schematic showing the locations of the vehicles 20 sec after the first vehicle passed section xx. Also determine (a) the time mean speed, (b) the space mean speed, and (c) the density on the highway.

The distance traversed by each vehicle 20 seconds after crossing section x-x is calculated as follows: Vehicle A: x A

=

50mi / h ×

1.47 ft / sec 1mi / h

× 20 sec = 1470 ft

Similarly: xB = (45)(1.47)(20 – 3) = 1125 ft xC = (40)(1.47)(20 – 3 – 4) = 764 ft xD = (35)(1.47)(20 – 3 – 4 – 3) = 515 ft xE = (30)(1.47)(20 – 3 – 4 – 3 – 5) = 221 ft

78 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow |---------:------------:---------:---------------:-------------:---| x-x E D C B A y-y 221' 515' 764' 1125' 1470' a) Time mean speed (TMS), using Equation 6.2, u t

=

1 n

n

u

i

i =1

ut = (50 + 45 + 40 + 35 + 30) / 5 ut = 40 mi/h b) Space mean speed (SMS), using Equation 6.3, u s

=

n n

1

u i =1

i

us = 5 / (1/50 + 1/45 + 1/40 + 1/35 + 1/30) us = 5 / 0.1291 us = 38.7 mi/h c) Density, k =

q

u s k = (5 veh / 15 sec)(3600 sec / hr) / (38.7 mi/hr) k = 31.0 veh/mi

6-4 Determine the space mean speed for the data given in Problem 6-3 using Equation 6.5. Compare your answer with that obtained in Problem 6-3 for the space mean speed and discuss the results.

Equation 6.5 presents the following relationship: ut = 0.966us + 3.541 First, the individual vehicle speeds must be converted to units of km/h and then averaged to determine time mean speed, which is 64.4 km/h. Using Eq. 6.5, 64 .4 = 0.966 u s + 3.541 u s = 63 .0 km/h The value for space mean speed found in Problem 6-3 is 38.7 mi/h, which is equal to 62.7 km/h. Equation 6.5 predicted a value of 63.0 km/h, 0.3 km/h higher than the space mean speed calculated directly from the data. This difference is insignificant for most purposes.


Chapter 6: Fundamental Principles of Traffic Flow 6-5 The following dataset consists of 30 observations of vehicle speed and length taken from a 6-ft by 6-ft inductive loop detector during a 60-second time period. Determine the occupancy, density, and flow rate.

Vehicle

Speed (mi/h)

Length (ft)

1

61

18

2

66

17

3

62

19

4

70

21

5

65

16

6

69

26

7

72

21

8

66

19

9

65

20

10

64

20

11

67

25

12

68

70

13

65

35

14

66

20

15

71

65

16

64

24

17

59

23

18

58

22

19

64

65

20

64

30

21

68

24

22

58

21

23

66

56

24

57

21

25

64

20

26

61

50

27

69

19

28

63

23

29

63

17

30

66

18


Chapter 6: Fundamental Principles of Traffic Flow a) Occupancy is is given by Equation 6.13:

 = 1   +   1

 = 601   + 606  1  = 601 ×8.86+ 606 ×0.32  = 0.1797 = 18% b) Density can be calculated by Equation 6.16:

 = 1   +  0.1797 = 601 × 8.86 + 6 ×  1 ×8.86 0.179 0.17977 − 60 = = 0.0053 ℎ/ 6k= 28.2 veh/mi

c) Flow is calculated by Equation 6.1:

 =  × 3600 3600 00  = 30 ×6036  = 1800 ℎ/ℎ

6-6 Data from a 6-ft by 6-ft inductive loop detector collected during a 30-second time period indicate that the mean speed of traffic is 50 mi/h among the 16 vehicles counted. Assume an average vehicle length of 19 ft. Determine the density and occupancy.

a) Density can be calculated by Equation 6.7:

 =    = 

16 × 3600  = 30 50 = 38.4 ℎ/ 81 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or dup licated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow b) Assuming the lengths of the vehicles are approximately the same, Occupancy can be calculated by Equation 6.19:

 = ( +  )  = ( = (19 19 + 6) 6) × 38.4 5280   = 0.182 = 18.2% 6-7 The data shown below were obtained by time-lapse photography on a highway. Use regression analysis to fit these data to the Greenshields model and determine (a) the mean free speed, (b) the jam density, (c) the capacity, and (d) the speed at maximum flow. Speed (mi/h)

Density (veh/mi)

14.2

85

24.1

70

30.3

55

40.1

41

50.6

20

55.0

15

A linear regression analysis can be applied to the given data to estimate parameters in Greenshields’ model of traffic flow. Greenshields’ model (Equation 6.20) is: u f u s = u f − k k j A linear regression model takes the form y = a + bx; therefore, in this case, the given data us and k correspond to y and x respectively. The linear regression analysis can be performed using equations 6.21, 6.22, and 6.23, or using a computer software spreadsheet package. Linear regression analysis yields values of a = 62.8124 and b = –0.56845. Therefore, a) Mean free flow speed, uf = a = 62.8 mi/h b) Jam density, k j In the regression model, b = uf / k j b = 0.56845 k j = 62.8 / 0.56845 = 110.49 k j = 110 veh/mi


Chapter 6: Fundamental Principles of Traffic Flow c) Capacity, qmax Capacity occurs at maximum flow. State flow in terms of density. u f q = k × u s = k (u f − k ) k j 2

q = 62.8k – (62.8/110.49)k 2 q = 62.8k – 0.5684k

Take the derivative and set equal to zero to maximize flow; solve for density. 0 = 62.8 – 1.1368k k = 55.25 when q = qmax Solve for q 2 qmax = 62.8(55.25) – 0.5684(55.25) qmax = 1735 veh/h d) Speed at maximum flow Solve for mean speed using k when when q = qmax u f u s = u f − k k j us = 62.8 – 0.5684(55.25) us = 31.4 mi/h

6-8 Under what traffic conditions will you be able to use the Greenshields model but not the Greenberg model? Give the reason for your answer.

In light flow conditions, in which the mean speed of traffic is near the mean free flow speed, the Greenberg model is not appropriate. This is due to mean free flow speed approaching infinity as density approaches zero in the Greenberg model.


Chapter 6: Fundamental Principles of Traffic Flow 6-9 In a freeway traffic stream, the capacity flow was observed to be 2200 veh/h/ln, and the jam density at this location had been observed to be 125 veh/ln/mi. If the traffic stream is modeled using Greenberg’s model, determine the optimum speed and optimum density. If the traffic stream is modeled using Greenshields’ model, determine the free flow speed, optimum density, and optimum speed.

a) For the Greenshields model: i. The mean free flow speed can be calculated by Equation 6.25:

 = 4  =  × 4 2200 × 4  = 22 125  = 70.4 /ℎ

ii. For maximum flow, the optimum speed is given by Equation 6.23:

 = 2  = 35.2 /ℎ iii. For maximum flow, the optimum optimum density is given by Equation 6.24:  = 2  = 62.5 ℎ// b) For the Greenberg model: i. For maximum flow, the optimum density can be calculated by Equation 6.27:

  = 1 ln ln  = 1 + +   ln125 ln 125 = 1 +   4.828 4.828 = 1 + +   k = 45.97 97 ℎ// ℎ//

ii. For maximum maximum flow, the speed will be:

 =   =    = 2200 45.97 = 47.8 /ℎ 84

© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow 6-10 The table below shows data on speeds and corresponding densities on a section of a rural collector road. If it can be assumed that the traffic flow characteristics can be described by the Greenberg model, develop an appropriate relationship between the flow and density. Also determine the capacity of this section of the road. Speed (mi/hr)

Density (veh/mi)

58.8

21

48.8

31

41.4

38

39.1

41

36.7

44

35.1

48

30.8

53

29.3

55

26.5

63

24.3

68

A linear regression analysis can be applied to a logarithmic transformation the given data to estimate parameters in Greenberg’s model of traffic flow. Greenberg’s model (Equation 6.26) is:

u s

=c

ln k j − c ln k

A linear regression model takes the form y = a + bx; therefore, in this case, the given data u s and ln k correspond correspond to y to y and and x x respectively, respectively, c ln k j is represented by a, and c and c is is represented by – b. The b. The linear regression analysis can be performed using Equations 6.28, 6.29, and 6.30, or using a computer software spreadsheet package. Linear regression analysis yields values of a = 151.2 and b = –30.17. Therefore, the jam the jam density, k j can be found as follows: In the regression model, a = c ln k j and b = –c c ln k j = 151.2 c = 30.17 ln k j = 5.01 k j = 149.9 veh/mi


Chapter 6: Fundamental Principles of Traffic Flow The fitted model is therefore: u s = c ln k j − c ln k = 30 .17 ln 151.2 − 30.17 ln k = 151.4 − 30.17 ln k A relationship between flow and density can be developed by multiplying both sides by k : q = 151.4k − 30.17k ln k Using the properties of Greenberg’s model, capacity can be found: uo = c and k o = k j /e uo = 30.17 mi/h = 149.9/2.718 = 55.15 veh/mi k o = 149.9/2.718 qmax = k ouo = (30.17)(55.15) = 1664 veh/h

6-11 Researchers have used analogies between the flow of fluids and the movement of vehicular traffic to develop mathematical algorithms describing the relationship among traffic flow elements. Discuss in one or two paragraphs the main deficiencies in this approach.

The main deficiency in using fluid flow theory to describe traffic flow is that, unlike fluids which are continuous, traffic streams are made up of discrete elements which have the ability to act independently of one another. In using fluid flow theory, the traffic stream is analyzed at a macroscopic level, without consideration given to the microscopic interaction between vehicles and the effects of one vehicle on others in the stream. Average values are used to describe the traffic stream over a given section (e.g. mean speed); characteristics and attributes of individual vehicles are not considered.


Chapter 6: Fundamental Principles of Traffic Flow 6-12 Assuming that the expression:

u s

=

u f e

−k

/ k j

can be used to describe the speed-density relationship of a highway, determine the capacity of the highway from the data below using regression analysis. ū

k (veh/mi)

s

(mi/h)

43

38.4

50

33.8

8

53.2

31 42.3 Under what flow conditions is the above model valid?

First, to use linear regression, the equation given in the problem must be converted to the form y = a + bx; this can be done by taking the natural logarithm of each side of the equation as follows: –k/kj

us = uf e

becomes ln(us) = ln(uf ) – k/kj

uf and and k j are constants; therefore, in the regression, a = ln(uf ); b = –1/k j; x = k; and y = ln(us) The only transformation needed for the input values is: ln(us) us 38.4 3.648057 33.8 3.520461 53.2 3.974058 42.3 3.744787 The linear regression analysis can be performed using equations 6.28, 6.29, and 6.30, or using a computer software spreadsheet package. Linear regression analysis yields values of a = 4.0626 and b = –0.0103. mean free flow speed, uf a 4.0626 uf = e = e uf = 58.1 mi/h b) jam density, k j k j = –1/b = –1/–0.0103 k j = 97.0 veh/mi c) capacity, qmax Capacity occurs at maximum flow. State flow in terms of density.

a)

q

=

k × u s

=

k (u f e

− k / k j

)

–k/97.0

q = (58.1)(k)(e

)


Chapter 6: Fundamental Principles of Traffic Flow Take the derivative and set equal to zero to maximize flow; solve for density. Take the natural logarithm of each side of the equation as follows: ln(q) = (ln 58.1) – k/97.0 + ln(k) Then take the derivative and set equal to zero. 1/q = –0.0103 + 1/k 0 = 1/k – 0.0103 k = 97.0 when q = qmax Solve for q q

=

k × u s

=

k (u f e

− k / k j

)

–1

q = 97.0(58.1e ) qmax = 2073 veh/h Because k, at qmax, is approaching k j, this model is valid for high density conditions only. 6-13 Results of traffic flow studies on a highway indicate that the flow-density relationship can be described by the expression: q = u f k −

u f k j

k2

If speed and density observations give the data shown below, develop an appropriate expression for speed versus density for this highway, and determine the density at which the maximum volume will occur as well as the value of the maximum volume. Also plot speed versus density and volume versus speed for both the expression developed and the data shown. Comment on the differences between the two sets of curves. Speed (mi/h) Density (veh/mi) 55

18

52

25

43

41

38

58

26

71

19

88

17

99

Note that the given traffic flow relationship, u f 2 q = u f k − k k j 88 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow is equivalent to u f q = k (u f − k) k j

and u s

= u f −

u f k j

k which is Greenshields’ model.

A linear regression model takes the form y = a + bx; therefore, in this case, the given data us and k correspond to y and x respectively, while a = uf and b = uf /k j. The linear regression analysis can be performed using Equations 6.28, 6.29, and 6.30, or using a computer software spreadsheet package. Linear regression analysis yields values of a = 63.949 and b = –0.4941. Therefore, a)

mean free flow speed, uf uf = a = 63.9 mi/h jam density, k j uf / k j = b = –0.4941 k j = 63.9 / 0.4941 = 129.33 jam density, k j = 129.33 veh/mi

b)

density at maximum volume State flow in terms of density. u f q = k × u s = k (u f − k ) k j 2

q = 63.9k – (63.9/129.33)k 2 q = 63.9k – 0.4941k

Take the derivative and set equal to zero to maximize flow; solve for density. 0 = 63.9 – 0.9882k k = 64.66 veh/mi when q = qmax Solve for q

2

q = 63.9(64.66) – 0.4941(64.66) q = 2066 veh/hr

(a) maximum volume, qmax = 2066 veh/h (b) density at maximum volume, k = 65 veh/mi (c) Plot speed vs. density for data points and equation


Chapter 6: Fundamental Principles of Traffic Flow

Speed vs. Densit Density y - Data 60 50

) h 40 / i m ( 30 d e e p S 20 10 0 0

20

40

60

80

100

Density (veh/mi)

Speed vs. Density Density - Equation Equation 60 50

) h 40 / i m ( 30 d e e p S 20 10 0 0

20

40

60

80

100

Density (veh/mi)

(d) Comment on the differences between the two curves.

The equation used to describe the speed-density relationship is linear, thereby yield a straight-line relationship as shown above. The data follow an approximately linear relationship, the extent of which could be quantified using 2 the R value that would be obtained through regression analysis or by calculating the sum of squared errors (SSE) between the equation and the data points.


Chapter 6: Fundamental Principles of Traffic Flow 6-14 Traffic on the eastbound approach of a signalized intersection is traveling at 35 mi/hr, with a density of 46 veh/mi/ln. The duration of the red signal indication for this approach is 30 sec. If the saturation flow is 1900 veh/h/ln with a density of 52 veh/mi/ln, and the jam density is 125 veh/mi/ln, determine the following: (i)

The length of the queue at the end of the red phase

(ii) The maximum queue length (iii) The time it takes for the queue to dissipate after the end of the red indication.

Solution: (i) The length of the queue at the end of the red phase. Determine speed of backward forming shock wave ω13 when signals turn to red. Use Equation 6.40.

uw

=

u13

=

q2 − q1 k2 − k 1 q1 − q3 k1 − k 3

q1 = (35 mi/hr)(46 veh/mi/ln) = 1610 veh/h/ln q3 = 0 veh/h/ln k 1 = 46 veh/mi/ln u13 =

1610 − 0 46 − 125

mi/h = 20.4 mi/hr

= 20.4 mi/h × 1.47 ft/sec/mi/h = 30.0 ft/sec Length of queue at end of red phase = 30 × 30.0 = 900 ft Determine speed of backward recovery wave velocity. Use Equation 6.45:

u34

=

q3 − q4 k3 − k 4

=

0 − 1900 125 − 52

= 26.0 mi/h = 38.2 ft/sec

(ii) The maximum queue length. Use Equation 6.47: Maximum queue length =

r ω13ω 34 ω34 − ω 13

=

(30)(30.0)(38.2) 38.2 − 30.0

= 4193 ft

(iii) The time it takes for the queue to dissipate after the end of the red indication. Use Equation 6.48: ω13 r ω (30)(30.0) = = 110 seconds ω13 − ω 34 38.2 − 30.0


Chapter 6: Fundamental Principles of Traffic Flow 6-15 A developer wants to provide access to a new building from a driveway placed 1000 ft upstream of a busy intersection. He is concerned that queues developing during the red phase of the signal at the intersection will block access. If the speed on the approach averages 35 mi/hr, the density is 50 veh/mi, and the red phase is 20 sec, determine if the driveway will be affected. Assume that the traffic flow has a jam density density of 110 veh/mi veh/mi and can be described described by the the Greenshields Greenshields model. model.

The red phase of the traffic signal creates a stopping shock wave. The speed of a stopping shock wave is given by Equation 6.47, as follows:

uw

= −u f η 1

where η1 = k/k j uw = –(35)(50/110) = -15.9 mi/h uw = –23.4 ft/s In 20 seconds, the wave will have traveled backward (toward the driveway in question) (23.4)(20) = 468 ft/s, therefore not reaching the driveway.

6-16 Studies have shown that the traffic flow on a two-lane road adjacent to a school can be described by the Greenshields model. A length of 0.5 mi adjacent to a school is described as a school zone (see Figure 6.19) and operates for a period of 30 min just before the start of school and just after the close of school. The posted speed limit for the school zone during its operation is 15 mi/h. Data collected at the site when the school zone is not in operation show that the jam density and mean free speed for each lane are 125 veh/mi and 57 mi/h. If the demand flow on the highway at the times of operation of the school zone is 90% of the capacity of the highway, determine: (i) The speeds of the shock waves created by the operation of the school zone (ii) The number of vehicles affected by the school zone during this 30-minute operation

(i) k j = 125, u f = 57 The traffic flow on this road section can be described by the Greenshields model So, the capacity of this road section is given by Equation 6.25:

1255 ××57 57 = 1781.25 ℎ/ℎ  = 4 = 12 4 92 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow The demand flow is 90% of the capacity, so q A = 0.90 qmax = 1603.125

 =   −    57   1603.125 = 57 − 125   0.456 − 57 + 1603.125 = 0 k A = 42.73

Given the speed at the school zone u B = 15 mph

 =  −   57  15 = 57 − 125 k B = 92.1

 =   = 92.1×15 = 1381.5 Therefore, the backward shock wave speed is

1603.125 = −4.5 /ℎ  =  − −  = 1381.5− 92.10− 42.73

(ii) In the school zone, the mean speed of traffic flow is 15 mi/h (forward), and the shockwave speed is 4.5 mi/h (backward). Therefore, the relative speed of the traffic flow to the end of shockwave is 15 + 4.5 = 19.5 mph. During the 30-minute school zone operation, the affected traffic flow travels 19.5 × 30/60 = 9.75 miles. The density of the school zone traffic is k B = 92.1 veh/mile Therefore, the number of vehicles affected by the school zone operation is 9.75 × 92.1 = 898 vehicles 6-17 Briefly describe the different shock waves that can be formed and the traffic conditions that will result in each of these shock waves.

Shock waves include frontal include frontal stationary, backward forming, backward recovery, rear stationary and forward recovery shock waves . Frontal stationary shock waves occur when capacity is reduced to zero and upstream demand continues. Backward forming shock waves occur when capacity is reduced below the demand flow rate but not to zero. Backward recovery shock waves form when capacity is restored or increased to a value greater than the upstream demand. A rear stationary shockwave occurs when a restricted downstream capacity is increased to a value above the queued demand, thereby dissipating the queue from through a forward recovery shock wave. 93 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow 6-18 Traffic flow on a three-lane (one direction) freeway can be described by the Greenshields model. One lane of the three lanes on a section of this freeway will have to be closed to undertake an emergency bridge repair that is expected to take 2 hours. It is estimated that the capacity at the work zone will be reduced by 30% of that of the section just upstream stream of the work zone. The mean free flow speed of the highway is 55 mph and the jam density is 135 veh/mi/ln. If it is estimated that the demand flow on the highway during the emergency repairs is 90% of the capacity, using the deterministic approach, determine: (i) The maximum queue length that will be formed (ii) The total delay (iii) The number of vehicles that will be affected by the incident (iv) The average individual delay

(i) u f = 55 mph, and k j = 135 veh/mi/ln Traffic flow can be described by the Greenshields model, therefore, the capacity of this freeway is

1355 = 5569 ℎ/ℎ  = 3 × 4 = 3 × 55 ×413

The demand flow is 90% of the capacity, so it is

 = 0.90× = 0.90×5569 = 5012 ℎ/ℎ Due to the work zone, the capacity is reduced by 30%, so the reduced capacity is

 = 0.70 0.70 × 55 5569 69 = 3899 3899 ℎ/ℎ ℎ/ℎ

The duration of work zone is 2 hours. The maximum queue length is determined by Equation 6.65.

 = ( − ) = (5012 − 3899) × 2 = 2226 ℎ ℎ (ii) The total delay is determined by Equation 6.68.

( −  )( −  ) 2(501  − 38 3899 99))(5569 (5569 − 3899) 3899) = 6674 ℎ  = 2( − ) = 501222(5569 2(5569 −501 − 5012) 2) (iii) The number of vehicles affected by the work zone = v × t = = 5012 × 2 = 10024 vehicles (iv) The average individual delay = d T T / n = 6674 / 10024 = 0.6658 hours

≈49 min.


Chapter 6: Fundamental Principles of Traffic Flow 6-19 Repeat Problem 6-18 for the expected repair periods of 1 hr, 1.5 hr, 2.5 hr, 2.75 hr, and 3 hr. Plot a graph of average individual delay vs the repair period and use this graph to discuss the effect of the expected repair time on the average delay.

Average Delay vs. Duration Time 1.2

) s r 1.0 h ( y 0.8 a l e D0.6 e g 0.4 a r e v A0.2 0.0 0

0.5

1

1.5

2

2.5

3

3.5

Duration Time (hrs)

The increase in expected repair time causes an increase on average delay; the relationship is linear. 6-20 Repeat Problem 6-18 for the expected demand flows of 70%, 75%, 80%, and 85% of the capacity of the highway. Plot a graph of average individual delay vs the expected demand flow and use this graph to discuss the effect of the expected demand flow on the average delay.

Average Delay vs. Expected Demand Flow 0.8

) 0.7 s r h ( 0.6 y a 0.5 l e D0.4 e g 0.3 a r e 0.2 v A0.1 0.0 70%

75%

80%

85%

90%

95%

Expected Demand Flow (as Percentage of Capacity)

The increase in expected demand flow causes an increase on average delay, the relationship is non-linear. 95 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow 6-21 Traffic flow on a section of a two-lane highway can be described by the Greenshields model, with a mean free speed of 55 mph and a jam density of 145 veh/mi/ln. At the time when the flow was 90% of the capacity of the highway, a large dump truck loaded with heavy industrial machinery from an adjacent construction site joins the traffic stream and travels at a speed of 15 mi/hr for a length of 3.5 mi along the upgrade before turning off onto a dump site. Due to the relatively high flow in the opposite direction, it is impossible for any car to pass the truck. Determine how many vehicles will be in the platoon behind the truck by the time the truck leaves the highway.

Knowing the traffic conditions can be described using Greenshields’ model allows use of Equation 6.53 which relates speed of the shockwave to the densities associated with flow before and during the blockage.

uw

u f [1 − (η 1 + η 2 )] in which η1 = k 1/k j and η2 = k 2/k j

=

To find k 1, use the form of Greenshields’ model expressed exp ressed in equation 6.22: u f 2 q = u f k − k k j Given that q1 = 90% of capacity, u f k j q1 = 0.90q max = 0.90 4 q1

=

u f k 1 −

u f k j

k 1

2

= 1794 =

=

0.90

55k 1 −

(55)(145) 4 55

145

k 1

= 1794

veh/h/ln

2

k 1 = 49.6 veh/mi/ln To find k 2, apply the general gene ral form of Greenshields’ model, Equation 6.20: u f u s = u f − k k j u2

= u f −

u f k j

k 2

= 15 = 55 −

55 145

k 2

k 2 = 105.5 veh/mi/ln

uw

=



 49.6 105.5   +   = –3.8 mi/h  145 145  

u f [1 − (η 1 + η 2 )]= 551 − 



The growth rate of the platoon is 15 mi/h forward and 3.8 mi/h backward, which is a total of 18.8 mi/h. The truck is on the highway for 3.5 mi / 15 mph = 0.233 hr The platoon length at the end of the duration is (18.8 mi/h)(0.233 hr) = 4.38 mi The number of vehicles in the platoon is (4.38 mi)(105.5 veh/mi) = 462 vehicles. veh icles. 96 © 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 6: Fundamental Principles of Traffic Flow 6-22 Briefly discuss the phenomenon of gap acceptance with respect to merging and weaving maneuvers in traffic streams.

Merging is the process of entering a roadway from a minor road or ramp and weaving is the process of moving across lanes of a major facility. A vehicle would merge onto a freeway and weave left to reach a left side exit. In the process of merging and weaving, drivers must assess a ssess available gaps in the traffic stream and determine if they are large enough to move into safely. A gap can be defined in terms of time or space, and the minimum time or space acceptable to drivers varies. Based on observations of accepted and rejected gap sizes on a roadway, a critical gap can be estimated and from this, the expected number of acceptable gaps can be determined. This value will allow one to calculate the storage space required on a ramp, or the point at which a signal should be installed at an unsignalized intersection.

6-23 The table below gives data on accepted and rejected gaps of vehicles on the minor road of an unsignalized intersection. If the arrival of major road vehicles can be described by the Poisson distribution, and the peak hour volume is 1100 veh/hr, determine the expected number of accepted gaps that will be available for minor road vehicles during the peak hour. Gap (t ) (s)

Number of Rejected Gaps > t

Number of Accepted Gaps < t

1.5

92

3

2.5

52

18

3.5

30

35

4.5

10

62

5.5

2

100

First, determine critical gap, tc, using the algebraic method. Determine the change in number of accepted and rejected gaps for the gap lengths given, shown in the following table. Range of Gap Lengths (s) 1.5 – 2.5 2.5 – 3.5 3.5 – 4.5 4.5 – 5.5

Change in Number of Accepted Gaps 92 – 52 = 40 52 – 30 = 22 30 – 10 = 20 10 – 2 = 8

Change in Number of Rejected Gaps 18 – 3 =15 35 – 18 = 17 62 – 35 = 27 100 – 62 = 38

Difference 40 – 15 = 25 22 – 17 = 5 27 – 20 = 7 38 – 8 = 30


Chapter 6: Fundamental Principles of Traffic Flow The critical gap occurs in the range exhibiting the smallest difference between change in number of gap accepted and change in number of gaps rejected; in this case, this is between 2.5 and 3.5 seconds. Using Equation 6.56, determine the value of the critical gap. tc = t1 + [Δ [Δt(r – m)] / [(n – p) + (r – m)] tc= 2.5 + 1(52 – 18) / [(35 – 30) + (52 – 18)] tc= 3.37 s Then, using Equation 6.61, determine the expected number of available gaps during the peak hour. V= 1100 veh. T = 3600 sec. λ = = 1100/3600 = 0.306 – λ λ t –(0.306(3.37)) –(0.306(3.37)) Freq (h≥t) = (V – 1)(e ) = (1,100 – 1)(e ) = 392 gaps

6-24 Using appropriate diagrams, describe the resultant effect of a sudden reduction of the capacity (bottleneck) on a highway both upstream and downstream of the bottleneck.

The diagrams below illustrate the impact of a bo ttleneck on traffic flow. In the first diagram, the reduction in capacity is shown (C1 reduced to C2) and the 1 2 corresponding density at capacity changes from k o to k o . The second diagram illustrates the effect of the bottleneck in terms of the shock wave that is formed. When the flow is reduced due to the bottleneck, a queue is formed and continues to grow as long as the demand flow is greater than the service flow. The rate at which the queue grows is dependent on the speed of the shock wave, uw.


Chapter 6: Fundamental Principles of Traffic Flow

6-25 The capacity of a highway is suddenly reduced to 50% of its normal capacity due to closure of certain lanes in a work zone. If the Greenshields model describes the relationship between speed and density on the highway, the jam density of the highway is 112 veh/mi, and the mean free speed is 68 mi/hr, determine by what percentage the space mean speed at the vicinity of the work zone will be reduced if the flow upstream is 70% of the capacity of the highway.

In Greenshields’ traffic flow model, qmax = k juf /4 /4 qmax = (112)(68)/4 qmax = 1904 veh/h Therefore, capacity at the site is normally 1904 veh/h. Upstream flow is 70% of capacity, qupstream = (0.70)(1904) = 1333 veh/h Upstream density can be found the following fo llowing form of Greenshields’ model:

q = u f k −

u f k j

k

2

2

q = (68)k – (68/112)k 2 0.6071k – 68k + 1333 = 0 Using the quadratic formula, k = 25.3 veh/mi Upstream speed can then be found using q = kus uupstream = q/k = 1333/25.3 uupstream = 52.7 mi/h


Chapter 6: Fundamental Principles of Traffic Flow At the work zone, flow is at 50% of capacity, qworkzone = (0.50)(1904) = 952 veh/h Upstream density can be found as: 2 q = (68)k – (68/112)k 2 0.6071k – 68k + 952 = 0 Using the quadratic formula, k = 95.6 veh/mi Upstream speed can then be found using q = kus uupstream = q/k = 952/95.6 uupstream = 10 mi/h The percentage reduction in speed due to the work zone is: (52.7 – 10)/52.7 = 81.02%

6-26 The arrival times of vehicles at the ticket gate of a sports stadium may be assumed to be Poisson with a mean of 30 veh/hr. It takes an average of 1.5 min for the necessary tickets to be bought for occupants of each car. (a) What is the expected length of queue at the ticket gate, not including the vehicle being served? (b) What is the probability that there are no more than 5 cars at the gate, including the vehicle being served? (c) What will be the average waiting time of a vehicle?

q = 30 veh/h (arrival rate) Q = 40 veh/h (service rate) a) Expected queue length Using Equation 6.71, 2 2 E(m) = q /[Q(Q – q)] = (30) /[40(40 – 30)] = 2.25 vehicles b) Probability of no more than 5 cars Using Equation 6.76, N+1 P(n>N) = (q/Q) (Probability of more than N) P(n>5) = (30/40)6 = 0.178 for P(n<=5) = 1 – 0.178 = 0.822 c) Average waiting time per vehicle Using Equations 6.73 and 6.72, E(v) = 1/(Q – q) = 0.1 hr = 6 minutes wait time including queue que ue time and service time E(w) = q/(Q(Q – q)) = 0.075 hr = 4.5 minutes wait time in the queue


Chapter 6: Fundamental Principles of Traffic Flow 6-27 An expressway off-ramp consisting of a single lane leads directly to a tollbooth. The rate of arrival of vehicles at the expressway can be considered to be Poisson with a mean of 45 veh/hr, and the rate of service to vehicles can be assumed to be exponentially distributed with a mean of 1 min. (a) What is the average number of vehicles waiting to be served at the booth (that is, the number of vehicles in queue, not including the vehicle being served)? (b) What is the length of the ramp required to provide storage for all exiting vehicles 90% of the time? Assume the average length of a vehicle is 18 ft and that there is an average space of 10 ft between consecutive vehicles waiting to be served. (c) What is the average waiting waiting time a driver waits before being served at the tollbooth (that is, the average waiting time in the queue)?

a) Expected queue length Using Equation 6.71, 2 2 E(m) = q /[Q(Q – q)] = (45) /[60(60 – 45)] = 2.25 vehicles b) Ramp length 1.00 – 0.90 = (q/Q)^(N+1) 0.10 = (45/60)^(N+1) 0.10 = (0.75)^N+1 ln(0.10) = ln(0.75)(N+1) 8 = N+1 N = 7 vehicles Ramp length = 7 veh (18 ft/veh) + 6 spaces (10 ft/space) = 186 feet c) Average waiting time per vehicle Using Equations 6.73 and 6.72, E(v) = 1/(Q – q) = 1/(60 – 45) = 0.067 hr = 4 minutes wait time including queue time and service time E(w) = q/(Q(Q – q)) = 45/(60(60 – 45)) = 0.05 hr = 3 minutes wait time in the queue


sm ch (6)

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