CIV2282: Transport and Traffic Engineering

Group

Participating Group Members

MONASH UNIVERSITY

DEPARTMENT OF CIVIL ENGINEERING

CIV2282: Transport and Traffic Engineering

Practical Class 11: Traffic assignment 2 and some example exam questions

Part 1: Traffic Assignment. This exercise aims to improve your understanding of the formulation of network optimization and the use of fmincon to solve the problem. This will help you in group assignment 2.

Consider the following network with the free-flow travel time given for each link (page 4, workshop week 10). The arrow shows the traffic direction.

Link 1: node 1 to node 2

Link 2: node 2 to node 4

Link 3: node 1 to node 3

Link 4: node 3 to node 4

Link 5: node 2 to node 3

Link 6: node 3 to node 2

1. Formulate the UE flow pattern problem and solve it using fmincon.

There are 4 possible paths as shown in figure above.

Look at the path patterns:

• link 1 is shared by path 2 and path 3 so
• link 2 is shared by path 2 and path 4 so
• link 3 is shared by path 1 and path 4 so
• link 4 is shared by path 1 and path 3 so
• link 5 is fully used by path 3 so
• link 6 is fully used by path 4 so

Path flow constraints:

To use fmincon:

Step 1: rewrite z(x) by z(y)

Step 2: write the constraints in the matrix form:

A*y <= 0

Aeq*y = 2000

y = [y1, y2, y3, y4]

Initial guess y0 = [500; 500; 500; 500]

Optimal solution: y = [2000 0 0 0];

Then link flow

x = [0 0 2000 2000 0 0];

total system cost: 12000 veh-minutes

2. Formulate the SO flow pattern problem and solve it using fmicon.

Similar to above

3. Compare the flow patterns obtained from AON, UE and SO in terms of the total system travel cost.

Total system cost of all three flow patterns are identical, = 12000 veh-min.

Now, assuming that the network is getting congested so that the link travel time is a function of the flow as below:

4. Solving the congested network for UE flow patterns.

Following the above steps for fmincon to get

Optimal solution: y = [1000.5 999.5 0 0];

Tst = 4014000 veh-min

5. Solving the congested network for SO flow patterns.

Following the above steps for fmincon to get

Optimal solution: y = [1000.25 999.75 0 0];

Tst = 4013999.75veh-min

6. Compare the total system cost for UE and SO flow patterns obtained from 4 and 5.

The SO flow patterns have a smaller total system travel cost, though the difference is very small.

Part 2: Mock exam questions.

Question 1 (2 marks)

Figure 2a displays a freeway stretch including a bottleneck. The fundamental diagrams of the bottleneck section and non-bottleneck section are presented in Figure 2b, where the unit for flow is veh/h, and for density is veh/km. Assume that the demand coming upstream of the stretch jumps from 6000 veh/h to 7000 veh/h. As a consequence, congestion is created at the start of the bottleneck and spills back into the non-bottleneck section. While the congestion propagates upstream, what is the traffic density within the bottleneck?

1. 65 veh/km.
2. 72 veh/km
3. 88 veh/km.
4. 100 veh/km.
5. Between 65 veh/km and 100 veh/km.

(a)

72

(b)

Figure 2: (a) a freeway stretch including a bottleneck; (b) fundamental diagrams.

Data from a section of highway has been used to calibrate the following relationship s = 0.77 / (96-v), where s is the average spacing (km/veh) and v is the speed (km/h).

Question 2 (2 marks)

The critical density is:

1. 42 veh/km
2. 52 veh/km
3. 62 veh/km
4. 72 veh/km
5. 82 veh/km

Kj = 96/0.77 = 124.7 veh/km

V0=96km/h

Kc = kjam/2= ~ 62 veh/km

Question 3 (2 marks)

The jam density is:

1. 55 veh/km
2. 75 veh/km
3. 101 veh/km
4. 125 veh/km
5. 145 veh/km

Question 4 (2 marks)

The minimum time headway is:

1. 0.1 sec/veh
2. 0.5 sec/veh
3. 0.8 sec/veh
4. 1.0 sec/veh
5. 1.2 sec/veh

Qc = kj*v0/4 = 3000 veh/h

Hmin = 1/Qc = 1.2 sec /veh

Question 5 (6 marks)

The Negative Exponential Distribution is commonly used to represent the distribution of headways between vehicles.

1. Under what conditions does the Negative Exponential Distribution NOT give the best fit to headway data?
2. Describe a probability distribution that would be better suited to fitting of headway data under those conditions.

Traffic becomes congested (slide 33 workshop week 4), used other distributions such as shifted negative exponential distribution.

Question 6 (19 marks)

Consider a urban road. It has a triangular-shape fundamental diagram with a free flow speed of 80 km/h, a critical density of 50 veh/km and a jam density of 300 veh/km. The arrival flow is constant at 2000 veh/h. Consider a tractor driving slowly is entering the road. There are no overtaking possibilities. This creates congestion, of which the tail happens to stay at the same position. After 5km, the tractor drives out of the road.

• Density) and vertical (Flow) axes labelled [1 mark, 1/2 mark each]
• Four regions identified [2 marks, 1/2 mark each]
• Shockwave correctly illustrated [2 marks, 1/2 mark each]

1. Calculate the flow and density in the congestion

• Describe the condition and how to obtain arrival flow. Since the tail is at the same place, the flow in the queue should be equal to the arrival flow [2 mark, 1 mark each]
• Using the fundamental diagram solving the density (e.g. solve by geometry or write the functional form of the FD) [1 mark]
• Flow being 2000 veh/h, correct density being 175 veh/km [1 mark]

2. Calculate the speed of the tractor

• Describe the tractor’s speed is the speed of the moving bottleneck region/state [2 mark]
• Correct equation and final result, Vs=q/k=2000/175=11.4km/h [1 mark, 1/2 mark each]

3. Construct the time-space diagram of this situation, from before the moment the tractor vehicle enters the road to after the moment the traffic situation recovers. (4 marks)

• Calculate states A,C,D: 1/2 mark

Flow A, 2000veh/h, Density A, 2000/80=25veh/km

Flow C, 0veh/h, Density C, 0 veh/km

Flow D, 4000veh/h, Density D, 50 veh/km

• Horizontal (Time) and vertical (Distance) axes labelled: 1/2 mark

The shock waves between 4 regimes and trajectories:

• Stationary moving shock wave between A and B: 1/2 mark
• Forwards moving shock wave between B and C: 1/2 mark
• Backwards moving shock wave between B and D: 1/2 mark
• Forwards moving wave between AC, CD, AD: 1/2 mark
• Vehicle trajectories in A,B,C, D: 1 mark

4. At the maximum queue, how many vehicles are in congestion (2 marks)

• Describe: From the time-space diagram, the maximum queue happens at the point where the tractor leaves (1 mark)
• Obtain number of vehicles: density*distance which is 175*5=875veh (1 mark: ½ mark for equation, ½ mark for answer)

Question 7 (14 marks)

A four-approach roundabout, having one entry lane on each approach and one circulating lane in front of each approach, has the following turning movements, given in Figure 3.

Figure 3 – Roundabout Intersection Turning Movement Volumes (vehicles per hour)

The road authority had received several complaints about excessive queuing and delays at one of the approaches to the roundabout.

1. Which approach is the one most likely to have the excessive queueing and delays, and why does it have these large queues and delays?

South Approach, having a large circulating volume giving few gaps, and a large entry volume

• Identify South Approach as having excessive queuing and delays
• Explain why South Approach would have queues and delays

Total for Question 7(a)

2. What is the circulating volume in front of this approach?

835 + 200 + 200 = 1235 veh/h

• Describe calculation of circulating volume as sum of 3 turning movement volumes
• Obtain correct answer of 1235 vehicles per hour

Total for Question 7(b)

3. Assuming the approach has a critical gap of 3 seconds and a follow-up headway of 2 seconds, what is the practical capacity (90% of the theoretical capacity) in vehicles per hour?

Opposing flow, q = circulating volume in front of approach = 1235 / 3600 = 0.343 veh/s

Theoretical Capacity, C = 0.343 exp(-0.343×3) / (1–exp(-0.343×2) = 0.25 veh/s = 888 veh/h

Practical Capacity, C_p = 0.9 C = 800 veh/h

• Correct method of calculating practical capacity using supplied formula
• Obtain correct answer of 799.94 vehicles per hour
  (accept answers between 780 and 810 veh/h)

Total for Question 7(c)

4. What is the total entry volume at this intersection approach?

Total entry volume = 360 + 500 + 300 = 1160 veh/h

• Correct method of calculating entry volume as sum of 3 turning movements
• Obtain correct answer of 1160 vehicles per hour

Total for Question 7(d)

5. What is the utilisation factor of the intersection approach?

Utilisation factor, ρ = λ / µ

= arrival rate / service rate

= entry volume / practical capacity

= 1160 / 800 = 1.45 (oversaturated)

• Correct method of calculating utilisation factor as ratio of arrival rate to service rate
• Obtain correct answer of 1.45

Total for Question 7(e)

6. Given that the traffic stream approaching the back of the queue at this approach has a speed of 50 km/h, what is the density of the traffic stream?

Total entry flow = 1160 veh/h (calculated in part d)
Density, k = q / v = 1160 / 50 = 23.2 veh/km

• Correct method of calculating density as entry flow divided by speed
• Obtain correct answer of 23.2 vehicles per kilometre

Total for Question 7(f)

7. The queue of slow-moving vehicles trying to get the through limited gaps in the circulating traffic moves forward at a speed of 10 km/h. What is the density of this congested traffic stream?

Flow through congested traffic = practical capacity = 800 veh/h (calculated in part c)
Density, k = q / v = 800 / 10 = 80 veh/km

• Correct method of calculating density as practical capacity divided by speed
• Obtain correct answer of 80 vehicles per kilometre

Total for Question 7(g)

8. The position of the rear of the queue of congested vehicles moves at a speed given by a backward moving shock wave between these two regions. At what speed does this rear of the queue move backwards (in km/h and in m/s)?

Shock wave speed given by w_AB = (q_A – q_B) / (k_A – k_B) = (1160 – 800) / (23.2 – 80)
= -6.34 km/h or -1.76 m/s

• Correct method of calculating shock wave speed from formula provided
• Obtain correct answer of -6.34 km/h or -1.76 m/s

Total for Question 7(h)

9. How much time would it take (in seconds) for the back of the queue to move backwards a distance of 100 metres?

Shock wave travels backwards a distance of 100 metres in time = 100 m ÷ 1.76 m/s
= 56.8 seconds

• Correct method of calculating time as distance divided by speed
• Obtain correct answer of 56.8 seconds
  (accept answers between 55 and 58 seconds)

Total for Question 7(i)

In response to the complaints about excessive queueing delays, the road authority has decided to install roundabout metering signals to stop traffic on the metered approach (i.e. the approach responsible for the heavy circulating flow).

These signals are activated whenever the back-of-queue reaches a distance of 100 metres.

10. What is the circulating traffic volume in front of the approach when the metering signals show red?

Only the right turn volume from the North approach (= 200 veh/h)

• Mention of right turn movement from North approach
• Correct answer of 200 vehicles per hour

Total for Question 7(j)

11. What is the practical capacity (in vehicles per hour) of the approach now that the circulating traffic has been reduced?

Opposing flow, q = circulating volume in front of approach = 200 / 3600 = 0.056 veh/s
Theoretical Capacity, C = 0.056 exp(-0.056×3) / (1–exp(-0.056×2) = 0.447 veh/s = 1610 veh/h
Practical Capacity, C_p = 0.9 C = 1449 veh/h

• Correct method of calculating practical capacity using supplied formula
• Obtain correct answer of 1449 vehicles per hour
  (accept answers between 1400 and 1500 veh/h)

Total for Question 7(k)

12. What is the utilisation factor of the approach now that the circulating traffic has been reduced?

Utilisation factor, ρ = λ / µ
= arrival rate / service rate
= entry volume / practical capacity
= 1160 / 1449 = 0.80 (undersaturated)

• Correct method of calculating utilisation factor as ratio of arrival rate to service rate
• Obtain correct answer of 0.80

Total for Question 7(l)

13. Assuming the intersection arrival and service rates are randomly distributed with this reduced circulating flow, what is the average number of vehicles either in the queue, or deciding whether it is safe to proceed through the intersection?

Average number of customers in M/M/1 queueing system = ρ / (1 – ρ) = 0.8 / (1 – 0.8) = 4

• Correct method of calculating average number using formula provided
• Obtain correct answer of 4 vehicles

Total for Question 7(m)

14. Using Little’s Law, what now is the average time (in seconds) spent by vehicles in the queue while the metering signals show red?

Little’s Law: N = λ × T

N = 4 (from part m) and λ = total entry volume (from part d) = 1160 veh/h

T = 1160 / 4 = 0.0035 hours
= 0.0035 / 3600 = 12.5 seconds

• Correct method of calculating average time from average number and arrival rate
• Obtain correct answer of 12.5 seconds

Total for Question 7(n)