1 - Purdue University



|CE 361 Introduction to Transportation Engineering |Posted: Fri. 12 September 2008 |

|Homework 3 (HW 3) |Due: Wed. 24 September 2008 |

HIGHWAY DESIGN FOR PERFORMANCE

• Permitted to submit this HW with as many as three other CE361 students.

• Identify each problem by its number and name.

|Poisson models. Vehicles at a local fast food restaurant’s drive-up window |Table 1. Squawk Box arrivals |

|“squawk box”. |Bin |

|(10 points) Value of λ? Calculate P(0), …, P(5), and P(n>5). |Frequency |

|Assume clock started at 0:00. Vehicles in data file are numbered 5-90. The 86th |P(n) |

|vehicle (#90) arrived at 62:19 = 62.32 minutes, so λ = 86/62.32 = 1.38 veh/min. |E(n) |

|P(n) values with t = 1 minute are calculated using FTE (2.24) are shown in Table | |

|1. Sample calc: |0 |

| |14 |

| |0.251 |

| |15.84 |

| | |

| |1 |

| |25 |

| |0.347 |

| |21.87 |

| | |

| |2 |

| |14 |

| |0.240 |

| |15.09 |

| | |

| |3 |

| |8 |

| |0.110 |

| |6.95 |

| | |

| |4 |

| |1 |

| |0.038 |

| |2.40 |

| | |

| |5 |

| |1 |

| |0.011 |

| |0.66 |

| | |

| |More |

| |0 |

| |0.003 |

| |0.19 |

| | |

| | |

| |63 |

| |1.000 |

| |63.00 |

| | |

[pic][pic] = 0.110 for n = 3. P(n>5) = 1 – P(n 1700? Eqn 3.3 |

| | |

| |Yes |

| |v(p) in same range? |

| | |

|B. (5 points) Field measurement of speeds. |72.0 |

|[pic]= 0.923 |S(FM) field measured speed |

|(3.4) FFS [pic]= 73.8 mph | |

| |215 |

| |V(f) observed volume |

| | |

| |0.06 |

| |P(T) |

| | |

| |0.00 |

| |P(R) |

| | |

| |2.5 |

| |E(T) Exhibit 20-9 |

| | |

| |1.1 |

| |E(R) Exhibit 20-9 |

| | |

| |0.923 |

| |f(HV) Equation 3.1 |

| | |

| |73.8 |

| |FFS free-flow speed Eqn 3.4 |

| | |

|(5 points) Average Travel Speed. The value of fnp = 1.23 comes from |Exh 20-11 |

|Exhibit 20-11 by a 2-stage linear interpolation. See the table at right. |20 |

|By (3.5), ATS = FFS – 0.00776 vp – fnp = 73.8 – (0.00776 * 979) – 1.23 = |24 |

|65.0 mph. This ATS corresponds to LOS A. |40 |

| | |

| |800 |

| |1.4 |

| | |

| |1.9 |

| | |

| |979 |

| |1.132 |

| |1.232 |

| |1.632 |

| | |

| |1000 |

| |1.1 |

| | |

| |1.6 |

| | |

|(10 points) Adjusted flow rate for PTSF. After one iteration, vp = 950 |Percent Time Spent Following |

|pc/hr. Because 950 3200? |

| | |

| |522 |

| |d(maj)*v(p) > 1700? Eqn 3.3 |

| | |

| |Yes |

| |v(p) in same range? |

| | |

|E. (5 points) BPTSP and PTSF. fd/np = 7.44 comes from Exhibit 20-12 by a |Exh 20-12 |

|3-stage linear interpolation. See the tables at right. The average of |20 |

|7.97 (for 50%) and 6.92 (for 60%) is 7.44. By (3.7), BPTSF = 100 [pic] =|24 |

|100 (1-0.434) = 56.6. By (3.8), PTSF = BPTSF + fd/np = 56.6 + 7.44 = |40 |

|64.1%. This PTSF corresponds to LOS D. Despite the better ATS value, the | |

|roadway’s LOS is D. |800 |

| |9.0 |

| | |

| |12.3 |

| | |

| |979 |

| |7.389 |

| |7.965 |

| |10.271 |

| | |

| |1400 |

| |3.6 |

| | |

| |5.5 |

| | |

| | |

| | |

| | |

| | |

| | |

| |Exh 20-12 |

| |20 |

| |24 |

| |40 |

| | |

| |800 |

| |7.6 |

| | |

| |10.3 |

| | |

| |979 |

| |6.437 |

| |6.917 |

| |8.838 |

| | |

| |1400 |

| |3.7 |

| | |

| |5.4 |

| | |

3. Analysis of Freeway Lane Blockage using Queueing Diagram. Queue builds for 15 minutes, until drivers start using offramp. Thirty minutes after the incident, full capacity is restored.

a. (15 points) Queueing diagram on next page. Note that AC2 and DC2 are parallel because of the exits to the offramp upstream from the blockage.

b. (5 points) Maximum length of queue = max length of vertical line from AC to DC = 1000–800 = 200 at t=15 minutes. Also, 1800-1600 = 200 at t=30. So queue length is 200 vehs from t=15 to t=30.

c. (5 points) Maximum time in queue = max length of horizontal line from AC to DC. The 1000th vehicle arrives at t=15. The equation of DC2 is D = 800 + [pic]t; set that equal to 1000 and solve for t. t = [pic]= 3.75 minutes. The last vehicle to wait this long leaves the queue at t = 30 minutes and entered the queue at 30 – 3.75 = 26.25 minutes. The max time in queue applies to all vehicles that arrived between t=15 and t=26.25.

d. (5 points) Queue dissipation. Starting at t=30, AC3 = [pic] and DC3 = [pic]. Set AC3 = DC3 and solve for t. [pic]= 8.57 minutes after t=30, or t = 38.6 minutes.

e. (5 points) Total delay to all vehicles time in the queue = area between ACs and DCs, [pic] Area 1 for [pic] = [pic]= 1500 veh-min. Area 2 for [pic] = 200 * 15 = 3000 veh-min. Area 3 for [pic] = [pic]= 860 veh-min. Area 1 + Area 2 + Area 3 = 5360 veh-min = 89.33 veh-hr. (Do not include the delay incurred by drivers who left the freeway at the offramp and had to use surface streets.)

[pic]

4. Analyzing a Stable Queue. Offramp traffic has a constant cycle, so that 960 vph can enter the surface streets from the ramp. Normally, 400 vph use the ramp during the morning peak period.

a. (10 points) AC1 = 4000 vph, AC2 = 3200 vph. Vehicles diverted to the offramp = (4000-3200)*[pic]= 200 vehs. Arrival rate on the offramp before was 400 vph and 400+800=1200 vph after the freeway blockage. [pic]= 960 vph, so queue is stable only before the freeway blockage.

b. (10 points) Type of queueing system (x/y/z) is M/D/1. Random traffic arrivals/pretimed or fixed traffic signal cycle/one signal for all “customer” vehicles.

c. For the cases in which a stable queue exists:

i. (5 points) Average queue length.

(3.15) with [pic]= 0.15 vehs.

ii. (5 points) Average time spent waiting in a queue.

(3.16) [pic]= 1.36 sec.

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