Problem #4 (20 points) a.A bus route having seven scheduled stops has the following number of passengers alighting and boarding the bus Sto Alighting 0 Boarding 20 0 per passenger 1.8 seconds and boarding Passengers board and alight through the front door. Alighting time p time per passenger 3 seconds. Compute the dwell time at each stop and the total dwell time. Assume the clearance ime 5 seconds. Briefly describe the four categories of transit system characteristics. c. Briefly describe the various transit modes of transportation End of Problem #4 Problemi #5 (20 points) Transportation Engineering is a multidisciplinary area of study. It is a profession that carries distinct societal responsibilities. What would a comprehensive training include? Provide examples of each training tem a) b) c) d) Explain the difference between direct elasticity and cross elasticity. Give one example of each What does CBD mean? Give an example of a CBD A transportation network having the following characteristics Travel time (minutes) between nodes Nodes From 10 10 10 0 16 12 Each node represents an activity center, and each link represents travel time in minutes. Draw a link/node diagrams for the travel times and distances, if the average travel speed on all links is 30 miles per hour End of Problem #5 Solution 4. (a). There are 2 ways in general to model the dwell time at a particular stop which depends on whether the boarding and alighting processes are sequential or simultaneous. Assuming in the question given that sequential boarding and alighting takes place since only one front door is available for flow of passengers: Dw = a * A + b*B + td ; where A and B are the number of boarding and alighting passengers respectively a and b are time intervals for boarding and alighting per passenger. td is the average clearance time for opening and closing the doors. Here a= 3 sec, b=1.8sec Therefore, at Stop 1: Dw1 = 3*20 + 1.8*0 + 5 = 65 sec Stop 2: Dw2 = 3*5 + 1.8*3 + 5 = 25.4 sec Stop 3: Dw3 = 3*7 + 1.8*5 + 5 = 35 sec Stop 4: Dw4 = 3*3 + 1.8*6 + 5 = 24.8 sec Stop 5: Dw5 = 3*4 + 1.8*8 + 5 = 31.4 sec Stop 6: Dw6 = 3*2 + 1.8*7 + 5 = 23.6 sec Stop 7: Dw7 = 3*0 + 1.8*12 + 5 = 26.6 sec Total dwell time= 231.8 seconds (b). Transit system characteristics can be classified into four main categories: i. System performance elements: Contains entire set of physical performance related elements which decides the raw efficiency of the central system employed as the primary functional core unit of the transit model. This includes Service frequency, operating speed, reliability and safety measures and line and productive capacities. ii. Level of Service (LOS) is an important service parameter that affects the users drastically due to differential change in its delivery. It is a simple basic most powerful characteristic that can attract the users or provide bad experiences to them if miniscule error occurs in provider’s implementation. The factors that affect LOS are performance.