320 CEE Spring Quarter 2009
Instructor: Yinhai Wang
CEE 320 Assignment #5 Solution
Problems from Textbook
7.2 An approach to a pretimed signal has 30 seconds of effective red, and D/D/1 queuing holds. The total delay at the approach is 83.3 veh-s/cycle and the saturation flow rate is 1000 veh/h. If the capacity of the approach equals the number of arrivals per cycle, determine the approach flow rate and cycle length. [15 points] Solution Given r = 30, Dt = 83.33, = 1000/3600 = 0.278 determine cycle length Since the capacity of the approach equals the number of arrivals in C, we have tc = g from the area of a triangle r * * (r + t c ) r * * (r + g ) r * * C Dt = = = 2 2 2 and *g * C = * g or = C and g = C r use equation(7.5) substituting * (C - r ) *r = =- C C r * ( * C - * r ) Dt = 2 solving for C yields C = 50 sec determine approach flow *r =- = 0.111 *3600 = 400 veh/hr C Approach flow rate is 400 veh/hr.
7.12 Vehicles arrive at an approach to a pretimed signalized intersection. The arrival rate over the cycle is given by the function (t) = 0.22 + 0.012*t [(t) is in veh/s and t is in seconds]. There are no vehicles in the queue when the cycle (effective red) begins. The cycle length is 60 seconds and the saturation flow rate is 3600 veh/h. Determine the effective green and red times that will allow the queue to clear exactly at the end of the cycle (the end of the effective green), and determine the total vehicle delay over the cycle (assuming D/D/1 queuing). [10 points]
CEE 320 Spring Quarter 2009 Solution Given C = 60 sec (t ) = 0.22 + 0.012 * t Arrivals(t) = (t )dt 0.22 * t + 6.00*10-3 * t2 Arrivals(C) = 34.8 veh To clear at the end of the cycle Arrivals(C ) = * g 3600 = = 1 veh/s (given) 3600 Arrival(C ) = 34.8 sec g=
Instructor: Yinhai Wang
r = C g = 25.2 sec use equation(7.5) determine total delay 60 1 Delay = Arrival(t )dt - * g * ( * g ) = 222.5 veh-sec 0 2
7.17 An intersection has a four-phase signal with the movements allowed in each phase, and corresponding analysis and saturation flow rates shown in Table 7.9 (Data for Problem 7.17). Calculate the sum of the flow ratios for the critical lane groups. [10 points] Solution Phase 1 EBL = 245/1750 = 0.14 WBL = 230/1725 = 0.133 Phase 2 EBTR = 975/3350 = 0.291 WBTR = 1030/3400 = 0.303 Phase 3 SBL = 255/1725 = 0.148 SBTR = 235/1750 = 0.134 Phase 4 NBL = 225/1700 = 0.132 NBTR = 215/1750 = 0.123
Yc=v/s = EBL + WBTR + SBL + NBL = 0.723
CEE 320 Spring Quarter 2009
Instructor: Yinhai Wang
7.28 For Problem 7.17, calculate the minimum cycle length and the effective green time for each phase (balancing v/c for the critical movements). Assume lost time is 4 seconds per phase and a critical intersection v/c of 0.95 is desired. [15 points] Solution From 7.17 solution, Yc=v/s = 0.723 L = 4 * 4 = 16 seconds Xc = 0.95 is given Using Equation (7.20) L X c 16 0.95 C min = = = 66.96 Sec 4 v 0.95 - 0.723 X c - ( ) ci i =1 s Rounding up to the nearest 5 seconds, Cmin = 70 sec This corresponds to Y C min 0.723 70 = 0.937 = Xc = c 70 - 16 C min - L Equation Using (7.22), the effective green times for the three phases can be calculated as v C 70 C g1 = ( ) c1 ( ) = EBL = 0.14 = 10.46 sec -> 10 sec s X1 0.937 Xc v C 70 C g 2 = ( ) c 2 ( ) = WBTR = 0.303 = 22.64 sec -> 23 sec s X2 0.937 Xc v C C 70 g 3 = ( ) c 3 ( ) = SBL = 0.148 = 11.06 sec -> 11 sec s X3 Xc 0.937 v C C 70 g 4 = ( ) c 4 ( ) = NBL = 0.132 = 9.86 sec -> 10 sec s X4 Xc 0.937
Check: g1 + g2 + g3 + g4 + L = 70 sec
Other Problem 1. An isolated intersection is controlled by a two-phase pre-timed signal with the movements allowed in each phases, and corresponding analysis and saturation flow rates shown in Table 1-1. Assume the startup loss time is 2 seconds per phase and the clearance loss time is 3 seconds per phase. The traffic flow accounts for the peak 15min period and that there is no initial queue at the start of analysis period. Progression adjustment factor PF=1.0; Upstream filtering/metering adjustment factor I = 1.0. Please answer the following questions: (1) What are the optimal cycle length (round up to nearest 5 seconds) using Webster's optimum cycle length formula and effective green times (based on lane group v/c equalization)? [10 points]
CEE 320 Spring Quarter 2009
Instructor: Yinhai Wang
(2) What is the northbound approach delay and level of service? [10 points] Table 1-1 Phase and Flow Data for the Intersection Phase 1 Allowed movements NB T/R/L, SB T/R/L Analysis flow rate 800, 820 Saturation flow rate 2800, 2900 Solution: (1) Calculate flow ratios: Phase 1: Northbound: 800 / 2800 = 0.286 Southbound: 820 / 2900 = 0.283 Northbound is critical. Phase 2: Eastbound: 1120 / 3000 = 0.373 Westbound: 960 / 3200 = 0.300 Eastbound is critical. Yc = 0.286 + 0.373 = 0.659 L = (2+3)*2 = 10 sec Xc = 0.90 is given. Using Equation (7.21), we get 1.5L + 5 1.5 10 + 5 = = 58....