<p>CE 361 Introduction to Transportation Engineering Posted: Thurs. 7 September 2006 Homework 3 (HW 3) Solutions Due: Mon. 18 September 2006</p><p>HIGHWAY DESIGN FOR PERFORMANCE</p><p> You will be permitted to submit this HW with as many as three other CE361 students.  For every problem, identify the problem by its number and name, be clear, be concise, cite your sources, attach documentation (if appropriate), and let your methodology be known.</p><p>1. Poisson models. Soccer goals per match. A.  = 17/17 = 1.00. Trial or “time period” = one match. n P(n) P(N  n) Events n = goals scored. P(n) values calculated using FTE 0 0.368 0.368 tn et 1 0.368 0.736 (2.24) are shown in table at right. P(n)  n! 2 0.184 0.920 3 0.061 0.981 1.0*12 e1.0*1  = 0.184 for n=2. P(n  5) = 1 – P(n  4 0.015 0.996 2! 4) = 1 – 0.996 = 0.004. B. FTE Figure 2.25 uses Frequency, not P(n), on the vertical axis. In the plot below, Poisson frequency = 17 * P(n). P(5+) = 0.06 or 0.07, depending on roundoff. The Poisson assumption seems to be reasonable.</p><p>Figure 3.1 Frequency histogram for Soccer Goals per Match</p><p>Observed Freq. Poisson Freq.</p><p>8</p><p>6.25 6.25 y c n e</p><p> u 4 q e r 3.13 F 3</p><p>1 1.04 1 0.26 0 0.06</p><p>0 1 2 3 4 5+ Goals per Match</p><p>2. Time between events. FTE Exercise 2.46, with part (c) reworded as: At what vehicle flow rate would bikes have a 90 percent chance of being able to cross the road during the next gap in traffic? A. (5 pts) Mean time between arrivals (headway) = 1/q = 1/(465 veh/hr) * 3600 sec/hr = 7.74 sec B. (10 pts) Arrival rate= 465 veh/hr * 1 hr/3600 sec = 0.1292 veh/sec; P (x > 7.3 s) = e -0.1292 * 7.3 = 0.389 C. (10 pts) Pr(T > t) = e -t = 0.90; therefore, t =0.1054. With *7.3 sec = 0.0144 veh/sec or 52 veh/hr average or 69.3 sec between vehicles. 3. LOS on State Road 361. A. (10 points) Adjusted flow rate for ATS. Iter. 1 Average Travel Speed Begin with Initial vp = V/PHF = 740/0.86 = 860. 860 1st v(p) = V/PHF 1 0.93 f(G) Exhibit 20-7 fHV  (3.1) 1 PT (ET 1)  PR (ER 1) 1.9 E(T) Exhibit 20-9 1 1.1 E(R) Exhibit 20-9 fHV  = 0.932 0.932 f(HV) Equation 3.1 1 0.08(1.9 1)  0.01(1.11) 993 v(p) Equation 3.2 After one iteration, vp = 993 pc/hr. Because 993<1200, a second iteration is not needed. B. (5 points) Field measurement of speeds. 65.0 S(FM) field measured speed 1 72 V(f) observed volume fHV  = 0.842 1 (9 / 72)(2.5 1)  0.01(1.11) 0.125 P(T) 0.00 P(R) Vf FFS  SFM  0.00776 (3.4) 2.5 E(T) Exhibit 20-9 f HV 1.1 E(R) Exhibit 20-9 72  65.0  0.00776 = 65.66 mph 0.842 f(HV) Equation 3.1 0.842 65.7 FFS free-flow speed Eqn 3.4 40 50 60 C. (5 points) Average Travel Speed. The value of fnp = 1.43 comes from Exhibit 20-11 by a 2-stage linear 800 1.9 2.15 2.4 interpolation. See the table at right. By (3.5), ATS = FFS 993 1.81 – 0.00776 vp – fnp = 65.66 – (0.00776 * 993) – 1.81 = 56.14 1000 1.6 1.80 2.0 mph. This ATS corresponds to LOS A. D. (10 points) Adjusted flow rate for PTSF. After one 0.94 f(G) Exhibit 20-8 iteration, vp = 952 pc/hr. Because 952<1200, a second 1.5 E(T) Exhibit 20-10 iteration is not needed. 1.0 E(R) Exhibit 20-10 0.962 f(HV) Equation 3.1 952 v(p) Equation 3.2 50/50 E. (5 points) BPTSP and PTSF. The value of fd/np = 1.79 40 50 60 comes from Exhibit 20-12 by a 3-stage linear 800 12.3 13.2 14.1 interpolation. See the tables at right. The average of 952 11.40 1400 5.5 6.1 6.7 11.40 and 10.28 is 10.84. By (3.7), BPTSF = 100 60/40 40 50 60  0.000879 vp  0.000879*952 1  e  = 100 1 e  = 800 10.3 11.65 13.0     952 10.28 100 (1-0.433) = 56.7. By (3.8), 1400 5.4 6.25 7.1</p><p>PTSF = BPTSF + fd/np = 56.7 + 10.84 = 67.54%. This PTSF corresponds to LOS D. Despite the good ATS value, the roadway’s LOS is D. CE361 HW 3 Fall 2006 - 3 -</p><p>F. (5 points) Two-Way Two-Lane Highway Segment Worksheet. Must be attached. CE361 HW 3 Fall 2006 - 4 -</p><p>4. Interstate Backup Analysis using Queueing Diagrams. At noon, a fatal crash at mile marker 226. Flow 1625 veh/hr over two NB lanes. The previous exit was at mile marker 220. Average stopped vehicle occupies 30 feet of highway. A. (5 points) Arrival rate = 1625 vph. Time until queue backs up to mile marker 220 = 6mi *5280 ft / mi *2lanes = 1.30 hr = 77.98 minutes. 1625veh / hr *30 ft / veh B. (5 points) Service rate from U-turns = 2 veh/min * 60 min/hr = 120 veh/hr C. (15 points) Queueing diagram from noon until queue dissipation.</p><p>3500</p><p>AC3 = 1625: Arrivals resume 3000 AC2 = 0: No new arrivals</p><p>AC1 = 1625 vph: 2500 Arrival rate until </p><p> n diversions begin o o n</p><p>0 0 : 2</p><p>1 2000</p><p> r e t f a</p><p> s e l c i</p><p> h 1500 e v</p><p> e DC2 = 4000 vph: v i t Maximum queue length Road reopens a l u</p><p> m 1000 Maximum time in queue u C</p><p>500 DC1 = 120 vph: U-turns reduce queue</p><p>0 0 50 100 150 200 250 300 350 400 Minutes after 12:00 noon</p><p>D. Estimate the following values, showing the computations: (i) (5 points) Longest vehicle delay is from (x1=24.4,y2=660) to (x3=330,y2=660), where y2 = 660 (330/60)*120= 660, x1= 1625 = 24.4 and x3 – x1 = 330 - 24.4 = 305.6 minutes.  60 CE361 HW 3 Fall 2006 - 5 -</p><p>(ii) (5 points) Longest vehicle queue is from (x2=77.98,y1=155.96) to (x2=77.98,y3=2112), where y1=77.98 min * 2 veh/min = 155.96 veh. y2-y1=2112-155.96=1956.04 veh. (iii)(5 points) When will queue dissipate? Equation for DC2 is y = (330*2) + ((4000/60)*(x-330)); Equation for AC3 is y = 2112 + ((1625/60)*(x-330)). These equations cross at x=366.62 minutes or at about 6:07 PM.</p><p>5. Queueing Analysis using Equations. Transit magnetic stripe cards. A. (5 pts) Because the “transaction time” for a Mag stripe card Cash/token turnstile is very nearly constant, and lambda 16.0 lambda 16.0 because passenger arrivals are random at mu 20.0 mu 30.0 the time scale being used (minutes), an M/D/1 queue system is most appropriate. rho 0.80 rho 0.53 B. (15 pts) Use  = 16,  = 20 or 30,  = 16/20 rho^2 0.640 rho^2 0.284 = 0.8 and 16/30 = 0.53. Calcs for mag stripe 1-rho 0.20 1-rho 0.47  2 (0.8)2 Q bar (pax) 1.600 Q bar (pax) 0.305 card: (3.15) Q   2(1 ) 2(1 0.8) W bar (min) 0.100 W bar (min) 0.019 = 1.60 passengers  0.8 (3.16) W   = 0.100 minutes = 6 seconds. Repeat for Cash/token. Q  0.305 2(1  ) 2*20*(1 0.8) passengers, W  0.019 minutes = 1.14 seconds</p>