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CE 361 Introduction to Transportation Engineering |Posted: Sun. 9 September 2007 | |

|Homework 3 (HW 3) Solutions |Due: Mon. 17 September 2007 |

HIGHWAY DESIGN FOR PERFORMANCE

• If submitted by more than one student, HW must bear the signatures and printed names of all students in the group at the top of the front page.

• Identify every problem by its number and name, etc.

|Poisson models. |Table 1 |

|(10 points) Assuming that opponents’ goals per match is a Poisson process, … | |

|λ = 54 goals/16 matches = 3.375 goals/match. Trial or “time period” = one match.| |

|Events n = goals allowed. P(n) values calculated using FTE (2.24) are shown in | |

|Table 1. Sample calc: | |

|[pic][pic] = 0.219 for n = 3. P(n>5) = 1 – P(n1200, a second iteration is needed. See values at right. |1st v(p) = V/PHF |

| | |

| |0.93 |

| |f(G) Exhibit 20-7 |

| | |

| |1.9 |

| |E(T) Exhibit 20-9 |

| | |

| |1.1 |

| |E(R) Exhibit 20-9 |

| | |

| |0.948 |

| |f(HV) Equation 3.1 |

| | |

| |1268 |

| |v(p) Equation 3.2 |

| | |

| |Iter. 2 |

| |Average Travel Speed |

| | |

| |1268 |

| |v(p) for lookups |

| | |

| |0.99 |

| |f(G) Exhibit 20-7 |

| | |

| |1.5 |

| |E(T) Exhibit 20-9 |

| | |

| |1.1 |

| |E(R) Exhibit 20-9 |

| | |

| |0.970 |

| |f(HV) Equation 3.1 |

| | |

| |1164 |

| |v(p) Equation 3.2 |

| | |

|(5 points) Field measurement of speeds. |54.5 |

|[pic]= 0.914 |S(FM) field measured speed |

|(3.4) FFS [pic]= 55.9 mph | |

| |175 |

| |V(f) observed volume |

| | |

| |0.063 |

| |P(T) |

| | |

| |0.00 |

| |P(R) |

| | |

| |2.5 |

| |E(T) Exhibit 20-9 |

| | |

| |1.1 |

| |E(R) Exhibit 20-9 |

| | |

| |0.914 |

| |f(HV) Equation 3.1 |

| | |

| |55.9 |

| |FFS free-flow speed Eqn 3.4 |

| | |

|(5 points) Average Travel Speed. The value of fnp = 1.47 comes from | |

|Exhibit 20-11 by a 2-stage linear interpolation. See the table at right.|40 |

|By (3.5), ATS = FFS – 0.00776 vp – fnp = 55.9 – (0.00776 * 1164) – 1.47 =|50 |

|45.4 mph. This ATS corresponds to LOS C. |60 |

| | |

| |1000 |

| |1.6 |

| |1.8 |

| |2.0 |

| | |

| |1164 |

| | |

| |1.47 |

| | |

| | |

| |1200 |

| |1.2 |

| |1.4 |

| |1.6 |

| | |

|(10 points) Adjusted flow rate for PTSF. After one iteration, vp = 1225|Iter. 1 |

|pc/hr. Because 1225>1200, a second iteration is needed. See values at |PTSF |

|right. | |

| |1118 |

| |1st v(p) = V/PHF |

| | |

| |0.94 |

| |f(G) Exhibit 20-8 |

| | |

| |1.5 |

| |E(T) Exhibit 20-10 |

| | |

| |1.0 |

| |E(R) Exhibit 20-10 |

| | |

| |0.971 |

| |f(HV) Equation 3.1 |

| | |

| |1225 |

| |v(p) Equation 3.2 |

| | |

| |Iter. 2 |

| |PTSF |

| | |

| |1225 |

| |v(p) for lookups |

| | |

| |1.00 |

| |f(G) Exhibit 20-8 |

| | |

| |1.0 |

| |E(T) Exhibit 20-10 |

| | |

| |1.0 |

| |E(R) Exhibit 20-10 |

| | |

| |1.00 |

| |f(HV) Equation 3.1 |

| | |

| |1118 |

| |v(p) Equation 3.2 |

| | |

|E. (5 points) BPTSP and PTSF. fd/np = 9.11 comes from Exhibit 20-12 by |50/50 |

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

|9.44 and 8.79 is 9.11. By (3.7), BPTSF = 100 [pic] = 100 (1-0.374) = |  |

|62.6. By (3.8), PTSF = BPTSF + fd/np = 62.6 + 9.1 = 71.7%. This PTSF |40 |

|corresponds to LOS D. Despite the better ATS value, the roadway’s LOS is|50 |

|D. |60 |

| | |

| |800 |

| |12.3 |

| |13.2 |

| |14.1 |

| | |

| |1118 |

| | |

| |9.44 |

| |  |

| | |

| |1400 |

| |5.5 |

| |6.1 |

| |6.7 |

| | |

| |60/40 |

| | |

| |  |

| |40 |

| |50 |

| |60 |

| | |

| |800 |

| |10.3 |

| |11.65 |

| |13.0 |

| | |

| |1118 |

| | |

| |8.79 |

| |  |

| | |

| |1400 |

| |5.4 |

| |6.25 |

| |7.1 |

| | |

4. I-96 Incident and Queueing Analysis. FTE Exercise 3.19.

A. (10 points) Draw a queueing diagram that shows the buildup and dissipation of the queue.

[pic]

B. (10 points) The queue dissipates when the departure curve reaches the arrival curve at CV = 2383 cumulative vehicles. The departure curve has the equation CV = 628 + (t-63)[pic]. The arrival curve has the equation CV = [pic]t. The two curves cross when 628 + 60(t-63) = 25.83 t. t = [pic]= 92.2 minutes.

C. (10 points) The longest vehicle queue occurs at t = 63 minutes (1:00PM), when 1628 vehicles have arrived and 628 vehicles have been served. Therefore, 1000 vehicles (the length of the vertical dashed line) are in the queue at t = 63 minutes. The longest vehicle delay also occurs at t = 63 minutes (1:00PM), and applied to the 628th vehicle to arrive. The 628th vehicle arrived at t = [pic]60 = 24.3 minutes. It left the queue at t = 63 minutes, after having been in the queue 38.7 minutes (the length of the horizontal dashed line).

5. Analyzing a Stable Queue. A traffic signal is timed with a constant cycle, so that 960 vph can enter the intersection. Normally, 400 vph approach the intersection during the morning peak period.

A. (10 points) What type of queueing system (x/y/z)? Explain your decision. M/D/1, because traffic arrives randomly, is served by a signal with constant rate, and all lanes are served at once. ρ = 400/960 = 0.417.

B. Assuming that a non-persistent queue situation exists, answer the following questions.

a. (5 points) To the nearest 0.01 vehicles, what is the average queue length?

(3.15) [pic]=[pic] = 0.149 vehicles

b. (5 points) To the nearest 0.1 second, what is the average time spent waiting in a queue?

(3.16) [pic] -= [pic] = 0.000372 hr = 1.3 seconds

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