MODULE 2 CE 416 PRINCIPLES OF TRANSPORTATION ENGINEERING March 22, 2022 Trip Generation Trip Distribution Construction Engineering & Management Modal Split BSCE-3205 Trip Assignment 19-02157 Answered & Prepared by: Kayessel Grace B. Binay 19-02157 Modular Assessment

Trip Generation Local study shows that the trip generation from zones A and B is a function of Population (P), Auto Ownership (C), and Household Income (I). This is represented by equation. Zone Population Table 1.1 Household Income A 2500 Auto Ownership 300 B 3300 400 150 100 Formula Ti = 0.20P + 0.1C + 0.2I Solution/s Ti (ZONE A) = 0.20 (2500) + 0.1 (400) + 0.2 (300) Ti (ZONE B) = 0.20 (3300) + 0.1 (100) + 0.2 (150) Zone Total Trips A 600 B 700 For Zones C and D, there is no available regression model but there is an existing data on Trip Rates and Trip Generation Forecast. These are summarized in the following tables. Household Size Table 1.2 Trip Rates 2+ Auto Ownership 2.29 01 3.3 3.58 1 1.96 2.45 2 3.25 2.81 3 + 3.2 3.04 Note: In estimating, the no. of trips for zone C & D, multiply the values from Tb. 1.3 & Tb.1.4 to Tb. 1.2, respectively. Household Size Table 1.3.1 Forecasted number of household in Zone C 2+ Auto Ownership 33 01 26 27 1 56 47 2 24 13 3 + 21 12

Table 1.3.2 Contents of Tb. 1.2 multiplied to Contents of Tb. 1.3.1 Auto Ownership 01 2+ 75.57 Household Size 1 109.76 115.15 85.8 Household Size 96.66 2 78 36.53 2+ 3 + 67.2 36.48 76 86 Table 1.3.3 Rounded Off Contents of Tb. 1.3.2 97 Auto Ownership 01 2+ 37 1 110 115 14 2 78 37 15 3 + 67 36 2+ Total Trips (ZONE C) = 702 TRIPS 84.73 46.2 Table 1.4 Forecasted number of household in Zone D 53.7 Auto Ownership 2+ 85 01 46 54 Household Size 1 24 28 Household Size Household Size 2 8 10 3+ 7 8 Table 1.4.2 Contents of Tb. 1.4.3 multiplied to Contents of Tb. 1.2 Auto Ownership 01 1 47.04 68.6 2 26 28.1 3 + 22.4 24.32 Table 1.4.3 Forecasted number of household in Zone D Auto Ownership 01 1 47 69 2 26 28 3 + 22 24 Total Trips (ZONE D) = 401 TRIPS Total Trips Zone Trips A 600 B 700 C 702 D 401

Trip Distribution From trip generation, the following were observed in the base condition: 1. Trips from Zone A will be divided between Zones B, C and D. Two thirds of the trips will go to Zone B and the rest will be divided equally between Zones C and D. Zone A = 600 trips Zone B = 2 (600)= 400 trips 3 (������������������������ ������ −������������������������ ������) 200 Zone C = 2 = 2 = 100 trips Zone D = 100 trips 2. The destination of trips from Zone B is limited to either A or C, with the number of trips going to A one hundred trips more than the trips going to C. Zone B = 700 trips Zone A = 400 trips Zone C = 300 trips 3. Trips from Zone C will be divided between Zones A, B and D. One hundred trips will go to Zone A and the rest will be divided equally between B and D. Zone C = 702 trips Zone A = 100 trips Zone B = (������������������������ ������ −������������������������ ������) = 602 = 301 trips 2 2 Zone D = 301 trips 4. The destination of trips from Zone D is limited to either A or C, with the number of trips going to Zone C three times the number of the trips going to A. Zone D = 401 trips 401 4 = 100.25 Zone C = 100.25 × 3 = 300.75 Zone A = 100.25

Modal Split Three modes are available: car, bus and walk. The utility factor for each mode is given by: Where: Uauto = 1.0 − 0.2(Tauto) − 0.05(Cauto) Ubus = −0.2(Tbus) − 0.05(Cbus) Uwalk = −0.05 − 0.2Twalk T = Travel time in minutes C = Travel cost in pesos Tauto = 7 minutes Tbus = 10 minutes Twalk = 20 minutes Cauto = 20 Cbus = 8 Solution/s: Solve for Uauto , Uauto = 1.0 − 0.2(7) − 0.05(20) = - 1.40 Solve for Pij−auto , eUauto Pij−auto = eUauto + eUbus + eUwalk e−1.40 Pij−auto = e−1.40 + e−2.4 + e−4.05 Pij−auto = 0.69515376475 or 0.6915 Pij−auto = 0.69515376475 × 100 Pij−auto = 69.52% Solve for Ubus , Ubus = − 0.2(10) − 0.05(8) = - 2.4

Solve for Pij−bus , eUbus Pij−bus = eUauto + eUbus + eUwalk e−2,4 Pij−bus = e−1.40 + e−2.4 + e−4.05 Pij−bus = 0.25573277851 or 0.2557 Pij−bus = 0.25573277851 × 100 Solve foPrijU−wbaulks, = ������������. ������������% Uwalk = − 0.05 − 0.2(20) = - 4.05 Solve for Pij−walk , eUwalk Pij−walk = eUauto + eUbus + eUwalk e−4.05 Pij−walk = e−1.40 + e−2.4 + e−4.05 Pij−walk = 0.04911345674 or 0.0491 Pij−walk = 0.04911345674 × 100 Pij−walk = ������. ������������%

Trip Assignment Assign the trips coming from Zone A towards all destinations using All or Nothing Trip Assignment Method. Assign only the car and bus (assume that 1car = 1 bus). Enclosed in parentheses are the travel resistances. Solution/s: Solve for the shortest travel resistance starting from Zone A to the following: Zone B = 2+3+2 = 7 Zone C = 2+1+5 = 8 Zone D = 2+1+2+3+2+2+3 = 15 Solve for the probabilities of car & bus only from Zone A: Pij−auto + Pij−bus = 0.69515376475 + 0.25573277851 Pij−auto + Pij−bus = 0.95088654333 Therefore, TA(Pij−auto + Pij−bus) = 600 × 0.95088654333 TA Pij−auto + Pij−bus = 570.531926 TA Pij−auto + Pij−bus = 571 Trips; considering bus & cars only

Zone A to Zone B: 571 × 2 = 380.6666667 = 381 Trips ; from Zone C to Zone D, 2 × 2 = 1 3 3 6 Zone A to Zone C: 571 × 2 = 95.16666667 = 95 Trips 3 Zone A to Zone D: 571 × 2 = 95.16666667 = 95 Trips 3 A 381 2 95 95 571 2 3 4 381 381 B 1 95 3 95 190 2 3 15 14 95 95 7 5 95 2 23 D 95 95 C 8 11 Figure 4.2

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