STUDY ON THE OPTIMAL CREW SCHEDULING FOR THE GREEN LINE OF METROPOLITANO DE LISBOA. NIKHIL MENON JOAO FIALHO

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1 STUDY ON THE OPTIMAL CREW SCHEDULING FOR THE GREEN LINE OF METROPOLITANO DE LISBOA. A PROJECT REPORT PREPARED BY NIKHIL MENON JOAO FIALHO MSc (COMPLEX TRANSPORT INFRASTRUCTURE SYSTEMS) In partial fulfillment of the credit requirements for the course TRANSPORTATION SYSTEMS ANALYSIS (TSA) INSTITUTO SUPERIOR TECNICO , LISBOA DECEMBER 2011 Transportation Systems Analysis Course Project Page 1

2 CONTENTS 1. INTRODUCTION CASE STUDY 5 3. TERMINOLOGY PROBLEM DESCRIPTION MATHEMATICAL MODEL DECISION VARIABLES OBJECTIVE FUNCTION CONSTRAINTS RESULTS INFERENCE SENSITIVITY ANALYSIS SENSITIVITY ANALYSIS ON N SENSITIVITY ANALYSIS ON w CONCLUSION REFERENCES..21 APPENDIX 22 Transportation Systems Analysis Course Project Page 2

3 LIST OF FIGURES 2.1 NETWORK MAP, LISBOA METRO ROSTERING DIAGRAM DEMAND HISTOGRAM DEMAND CONSTRAINT TRAIN DRIVER SCHEDULING REQUIREMTN OF DRIVERS Vs. DRIVERS SCHEDULED SENSITIVITY ANALYSIS ON N SENSITIVITY ANALYSIS ON N TYPE OF DRIVERS SENSITIVITY ANALYSIS ON w SENSITIVITY ANALYSIS ON w 3 TYPE OF DRIVERS...20 Transportation Systems Analysis Course Project Page 3

4 LIST OF TABLES Transportation Systems Analysis Course Project Page 4

5 1. INTRODUCTION Crew scheduling is the process of assigning crews to conduct a particular activity. Crew scheduling has a wide application in transport, where it is used to assign crews to operate transportation systems. For example, Crew scheduling is a common technique adopted in railways to assign drivers to operate trains on various routes. The goal of the problem is to assign a subset of trips to each crew in such a way that no trip is left unassigned. As usual, not every possible assignment is allowed since a number of constraints must be observed. Additionally, a cost function has to be minimized. (Yunes, Moura and Souza et al, 2000) Crew scheduling problems have their great practical importance based on the fact that, in most companies, employee related expenses may rise to a very significant portion of the total expenditures. Therefore, these notoriously difficult complex combinatorial problems deserve a great deal of attention. (Yunes, Moura and Souza et al, 2000) Gomes, M., Cavique, L. and Themido, I., in their paper, The crew timetabling problem: An extension to the crew scheduling problem (2006) have dealt in depth with the aspect of preparation of performance measures developed for the crew timetabling problem, which are combined in an objective function. Through this study, it is intended to determine the optimal requirement of crew for the Green line of the Lisbon Underground, Metropolitano de Lisboa which is also abbreviated as ML in this case study. 2. CASE STUDY Metropolitano de Lisboa (ML) is the metro (subway) system of Lisbon, Portugal. It was opened on December 29, 1959 thus becoming the first subway system to be opened for the public in Portugal. As of today, it consists of four lines with 52 stations, spread over 39.6 kilometers. As of 2009, Metropolitano de Lisboa (ML) carries an estimated 177 million passengers per year. Transportation Systems Analysis Course Project Page 5

6 Figure 2.1 Network Map, Lisboa Metro The figure above shows the network map of Metropolitano de Lisboa (ML) spread over the city of Lisbon. The Green Line (Linha Verde) is being analyzed as part of the current study. The green line of the Metropolitano de Lisboa, runs from Cais do Sodre to Telheiras, a distance of 9 kms served by 13 stations. The construction started in 1972, with the link from Restauradores to Alvalade. In 1993, this was further extended from Alvalade to Campo Grande. The Blue and the Green lines were split in 1998, with the construction of the section from Baixa Chiado to Cais do Sodre and it reached the present level of completeness in 2002, with the extension from Campo Grande to Telheiras. The services in the Green line commence at 0630 from both the terminals (Cais do Sodre and Telheiras) and they run upto 0130 with varying frequencies throughout the day. The frequencies are as described below: Transportation Systems Analysis Course Project Page 6

7 Time [hh:mm] Frequency [mm:ss] 06:30-07:30 07:30 07:30-09:30 03:50 09:30-17:00 04:40 17:00-20:00 03:50 20:00-22:30 05:45 22:30-01:30 09:15 Table 2.1: Frequency of Trains on the Green Line Nikhil Menon Joao Fialho As can be seen from the table, the frequencies of the trains on the Green line are classified into 6 periods dictated by the demand required to be met. The morning rush hours between 07:30 and 09:30 and the evening rush hours between 17:00 and 20:00 demand a frequency of 03:50 between each train plying along the Green line. The frequency is lesser during the other periods of the day, especially during the last period (22:30 01:30) when it is 09:15 between two trains. 3. TERMINOLOGY For the current study, the following terminologies are adopted: Timetable: document showing all the trips undertaken by the entire number of trains on a given day. Trip: one way movement of a train between two terminuses. Trips are divided into increments of work for the purpose of crew scheduling. Increment of work / Period: the portion of work between two adjacent relief times (or relief points). This is the smallest period into which the time table can be divided. Relief Point: point along a line where a crew may leave a train and another crew takes over. This can be done either at the line terminuses or at any intermediate point along the line. Rostering: the process of allocation of trains to trips. Break: a rest period in a crew duty. Transportation Systems Analysis Course Project Page 7

8 4. PROBLEM DESCRIPTION The railway company decides to look into the aspect of optimization for the driver s timetable. Prior to describing the problem at hand, the following assumptions are to be stated: Based on the data obtained from ML, the timetable of the trains plying along the green line is created. The timetable of the trains created is attached on the appendix for reference. For the generation of the timetable, the turnaround times of the train are not accounted for. In saying so, it is meant that when a train arrives at one of the terminal stations, it is assumed that the same train can perform the return trip as long as the departure time is after 2 minutes 30 seconds, which might not be the standard turnaround times observed. With the timetable prepared, a possible rostering of trains was planned and implemented for the course of the study. The rostering implemented for the current study is as shown below: h00 01h00 02h00 03h00 04h00 05h00 06h00 07h00 08h00 09h00 10h00 11h00 12h00 13h00 14h00 15h00 16h00 17h00 Figure 4.1 Rostering Diagram 18h00 19h00 20h00 21h00 22h00 23h00 24h00 Based on the inputs obtained from the rostering shown above, the demand for the drivers for each period of time is obtained. It is as shown below in the histogram: Transportation Systems Analysis Course Project Page 8

9 h00 01h00 02h00 03h00 04h00 05h00 06h00 07h00 08h00 09h00 10h00 11h00 12h00 13h00 14h00 15h00 16h00 17h00 18h00 19h00 20h00 21h00 22h00 23h00 24h00 Figure 4.2 Histogram of the demand As can be seen from the above two figures, the increment of work adopted for the current study is 30 minutes. By saying so, it is meant that the smallest period of time into which the timetable is divided into, is 30 minutes. Considering the timetable generated in which the first train starts at 0630 and the last train reaches the terminus at 0152 (adopted as 0200 in this study), there are 39 increments of work on a given day. The demand depicted by the histogram can be alternatively represented on a table as shown below: Increment of Work Increment of Work d t Start End t Start End t 06:30 07: :30 17: :00 07: :00 17: :30 08: :30 18: :00 08: :00 18: :30 09: :30 19: :00 09: :00 19: :30 10: :30 20: :00 10: :00 20: :30 11: :30 21: :00 11: :00 21: :30 12: :30 22: :00 12: :00 22: :30 13: :30 23: :00 13: :00 23: :30 14: :30 00: :00 14: :00 00: :30 15: :30 01: :00 15: :00 01: :30 16: :30 02: :00 16: Table 4.1: Representation of the demand for all increments of work d t Transportation Systems Analysis Course Project Page 9

10 With the increments of work and the corresponding demand defined, the number of drivers required at each instant of time is determined. As far as the drivers are concerned, three major categories of drivers are resorted to, in the course of this study. The first type is the full time driver. By saying so, it is meant the type of driver who works for 8 hours (16 increments of work) during the course of the day, all at one stretch. The full time driver does not take a break during the course of his working day. The premium wages of the full time driver is fixed at 64 /day. The second type of the driver is the split shift driver. The working style of the split shift driver is different from that of the Full time driver. In case of the split shift driver, each driver works for a period of 4 hours (8 increments of work) at a stretch, then has a break of 2 hours (4 increments of work) and resumes the work for another 4 hours (8 increments of work) during a working day. The premium wages of the split shift driver is fixed at 80 /day. The third type of drivers are the extra drivers. They differ from both the full time and the split shift drivers in the sense that, they are pressed into service only during situations which demand their presence. They are not direct employees of the railway company and thus are hired by the railway company only when required to cover up. The extra drivers work only for a period of 4 hours (8 increments of work) in a working day. Their wages are fixed at 48 /day. As can be seen, the split shift wage is more than the full time wage. The full and split shift drivers are drawn from the same pool of persons. A driver of any class can work at most one shift per day. The data shown above is aggregated in the table below: 1 st Work Period Break Period 2 nd Work Period Type of Driver Wage [ /day] Increments of Increments of Increments of Nr of hours [h] Nr of hours [h] Nr of hours [h] Work Work Work Full Time Driver Split Shift Driver Extra Driver Table 4.2: Extensive Driver data So to sum up, the problem aims at determining the number of drivers required from each class so as to meet the demand. The goal is to minimize the total wage cost. The railway company currently has total employee strength of 24. The said 24 employees can work either as a full time Transportation Systems Analysis Course Project Page 10

11 driver or a split shift driver. The rest of the drivers (if required), can be hired as extra drivers. There is no restriction on the number of extra drivers hired. The extra drivers are always available as long as they are required to cover up. 5. MATHEMATICAL MODEL In this section, the formulation of the model is explained, which works as a basis towards obtaining the solution to the problem at hand. 5.1 DECISION VARIABLES As is known, the primary objective of this mathematical model is to identify the decision variables. The main concern in this study is to determine the number of drivers of each type to be pressed into service at each instant of time. Therefore, the decision variables chosen are described in the general form, below: where, and { } All the decision variables are integer numbers because they represent the requirement of train drivers at each instant of time. 5.2 OBJECTIVE FUNCTION The objective of this problem is to minimize the total wage spent by the railway company in hiring the train drivers, irrespective of their type to fulfill the requirements of the timetable. This is achieved by summing the product of all train drivers hired with their respective wages over all the increments of work. This is as shown below: Transportation Systems Analysis Course Project Page 11

12 where, w i is the wage/day for each train driver of type i. [ ] [ ] Nikhil Menon Joao Fialho 5.3 CONSTRAINTS - Demand: The demand of train drivers must be fulfilled at all times. This is achieved by formulating the following demand constraint, as shown below: For every increment of work t, the number of train drivers must be equal to or higher than the demand. In order to ensure this, it is necessary to check how many full time drivers have started to work in the past 8 hours (16 increments of work), or until the beginning of the service [06:30]. Also in case of split shift drivers, it is necessary to check how many drivers have started to work in the past 4 hours (8 increments of work), or until the beginning of the service [06:30]. It is also necessary to check how many drivers started to work in the period between the last 6 and 10 hours, (since they work on a split shift) or until the beginning of the service [06:30]. And in case of extra drivers, it is necessary to check how many extra drivers have started to work in past 4 hours. This description is illustrated by the figure below. As can be seen from the figure below, the requirement for a driver is pressed in at the 18 th increment of work, corresponding to the period between 15:00 and 15:30. In case of the full time driver, it is necessary to check how many drivers have started to work in the past 16 increments (3 rd increment of work) or until the beginning of service [06:30], which happens earlier in this case. For the split shift driver, it is necessary to check into the past 8 increments (11 th ) of work and follow it up with a check on the past 20 increments from the past 12 increments (6 th ) or until the beginning of service [06:30], which in this case is the key. For the extra time driver, it is necessary to check into the past 8 increments (till the 11 th increment of work). Transportation Systems Analysis Course Project Page 12

13 Figure 5.1 Demand Constraint - Number of drivers who are employees of the company: The railway company has discretion on the number of drivers who are employees of the company. Extra drivers do not belong the employee class of the company and they are hired only when there is a need for cover up. And thus, they have a different pay scale from the employees of the railway company. For the purpose of the model formulation, the number of drivers who are employees of the company are fixed at 24. The mathematical formulation of this fact has been described below. - subject to the following condition: Transportation Systems Analysis Course Project Page 13

14 6. RESULTS The mathematical model illustrated above was input on XPRESS 7.2 IVE and used to find the optimal solution. The results obtained give an idea on the number of drivers that are required of each type to fulfill demand at each instant of time. The XPRESS model used is attached alongside, on the Appendix. The results obtained on XPRESS are summarized as below: Period Start End t x 1t x 2t x 3t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: Period Start End t 16:30 17:00 21 x 1t x 2t x 3t 17:00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02:00 39 Σ Table 6.1: Result of the Mathematical Model The objective function obtained for the above result is 1696 /day. Transportation Systems Analysis Course Project Page 14

15 6.1 INFERENCE Nikhil Menon Joao Fialho 1. The first point of importance, based on the results obtained is shown in the figure above. The total number of drivers required for the complete working day is estimated to be 26, by the model. 2. The total number of drivers of type 1 and 2 number 24, which is the number of employees of the railway company. The remaining 2 employees are belonging to the type of extra drivers. 3. The next point of inference is given by the figure below. It gives the total requirement of drivers at all increments of work, the number of drivers available for the same and also the number of excess drivers present. Period Total Excess of dt Start End t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: Table 6.2: Comprehensive Analysis Driver Scheduling Numerically stating, there is a requirement of 367 increments of work which require drivers in a day. The optimal solution of the problem obtained indicates an availability of 400, which are 33 more than that required. This means that there are a proportion of drivers who are non - utilized during the course of the working day. It is given by the following relation: Period Start End t Excess of 18:00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: dt Total Transportation Systems Analysis Course Project Page 15

16 Nikhil Menon Joao Fialho 4. The results are also used to cross analyze with the rostering planned, done in chapter 3, to generate a possible schedule for the train drivers. The representation is as shown below: R R R R h00 01h00 02h00 03h00 04h00 05h00 06h00 07h00 08h00 09h00 10h00 11h00 12h00 13h00 14h00 15h00 16h00 17h00 18h00 19h00 20h00 21h00 22h00 23h00 24h00 Legend x y z Full Time Driver number x Split Shift Driver number y Extra Driver number z Figure 6.1 Train Driver Scheduling h00 01h00 02h00 03h00 04h00 05h00 06h00 07h00 08h00 09h00 10h00 11h00 12h00 13h00 14h00 15h00 16h00 17h00 18h00 19h00 20h00 21h00 22h00 23h00 24h00 Figure 6.2 Requirement of drivers Vs Train drivers scheduled Transportation Systems Analysis Course Project Page 16

17 From the figures above, the rate of non - utilization of drivers is also evident, as is the split of the drivers according to the type. 6.2 SENSITIVITY ANALYSIS Sensitivity Analysis is the study of how the variation (uncertainty) in the output of a mathematical model can be attributed to different variations in the inputs of the model. Put another way, it is a technique for systematically changing the variables in a model to determine the effects of such changes. In the current study undertaken, sensitivity analysis is done on two parameters that are under the influence of the railway company: Sensitivity Analysis on the number of drivers who are employees of the company (N). Sensitivity Analysis on the wages of the type extra drivers (w 3 ) Sensitivity Analysis on N Through this process, it is desired to determine the influence of the employee dimensions of the company on the output, which in this case is the total wage spent on the drivers. It also is an indicator on the rate of non-utilized drivers. The Sensitivity Analysis on N would thus indicate the optimal dimension of the work force of the company and also indicate the corresponding levels of efficiency with each value of N. It is to be noted that the constraint on the number of employees belonging to the company is imposed to be equal to N rather than it being lesser than or equal to N, as is the case during the model formulation. The observations of the sensitivity analysis obtained, are as shown below: N W Rate of non - utilized 8,3% 8,3% 8,3% 8,3% 11,8% x 1t x 2t x 3t N Table 6.3: Sensitivity Analysis on N From the table above, it can be seen that the output, which in this case is the total wage is the highest (1728 ) when N=22 and N=26. It is the lowest (1696 ) when N=24 and N=25. In case of the rate of non-utilized workers, it is the highest (11.8%) Transportation Systems Analysis Course Project Page 17

18 Train /day Nikhil Menon Joao Fialho at N=26 and the lowest (8.3%) for all other values of N. The split of drivers based on the various types are also shown. It is of importance to note here that N=x 1t + x 2t. Sensitivity Analysis on N ,0% ,3% ,3% 8,3% 8,3% ,8% ,0% 9,0% 6,0% 3,0% W Rate of non - utilized ,0% N Figure 6.3: Sensitivity Analysis on N Sensitivity Analysis on N - Type of train drivers x1t x2t x3t N N Figure 6.4: Sensitivity Analysis on N - type of train drivers As a conclusion, it can be seen that the company will minimize the total wage paid to the drivers, by employing N=24 or N=25 employees. Thus, the company currently works under the ideal size, in accordance with the problem description stated earlier Sensitivity Analysis on w 3 Through this process, it is desired to understand the influence of the variation in the wages of the extra time drivers to the output of the railway company, which in this case is the total wage paid to the drivers. The variation on w 3 is more likely to occur since it involves outsourcing of drivers and thus has a better scope for obtaining a Transportation Systems Analysis Course Project Page 18

19 /day Nikhil Menon Joao Fialho good bargain. It also gives an idea of the work force dimensions of the company and the corresponding split of drives based on their types. The observations are as shown under: w 3 [ /day] 80% 85% 90% 95% 100% W ,6 1686,4 1691, Rate of non - utilized 8,3% 8,3% 8,3% 8,3% 8,3% x 1t x 2t x 3t N Table 6.4: Sensitivity Analysis on w 3 From the table above, it can be seen that the output, which in this case is the total wage is the highest (1696 ) when the wage is as present now. It is the lowest (1664 ) when the revised wage is 80% the existing wage. The split of drivers based on the various types are also shown. It is of importance to note here that N=x 1t + x 2t. Sensitivity Analysis on w ,3% ,3% 8,3% 8,3% 8,3% 15,0% 12,0% 9,0% W ,6 1686,4 1691, ,0% 3,0% Rate of non - utilized ,0% 80% 85% 90% 95% 100% % change in w 3 Figure 6.5: Sensitivity Analysis on w 3 Transportation Systems Analysis Course Project Page 19

20 Nr. Train Nikhil Menon Joao Fialho Sensitivity Analysis on w 3 - Type of train drivers x1t x2t x3t N 80% 85% 90% 95% 100% % change in w 3 Figure 6.6: Sensitivity Analysis on w 3 - type of train drivers As expected, if the wage of the extra driver is lower, it is possible to obtain a better outcome in terms of the total wage paid to the drivers. If the wages of the extra driver are reduced to 80% of that is prevailing at the present, the size of the railway company can be reduced from the existing 24 to the revised strength of 20 employees. Transportation Systems Analysis Course Project Page 20

21 7. CONCLUSION The objectives proposed during the commencement of the study have been achieved. The total wage paid by the railway company towards the drivers is minimized. Some of the simplifications employed during the current study may not be relevant to what is the standard observed on a global scale. For instance, the relief points are assumed to be either of the line terminuses or any intermediate point along the line where the drivers change their duty. Another aspect is the non-accounting for the break/rest periods for the full time drivers, which are not bound to be in line with the common practices adopted. So, as a suggestion for improving the model, some of the above given aspects could be taken into account which will work in the betterment of the accuracy of this study. Even with the simplifications adopted, the mathematical model used in this study could be applied to any real world situation of mass transit planning, even though this study has been restricted to Linha Verde of Metropolitano de Lisboa. 8. REFERENCES 1. Yunes, T., Moura, A. and Souza, C., Solving Very Large Crew Scheduling Problems to Optimality ACM /97/05., pp Gomes, M., Cavique, L. and Themido, I., (2006). The crew timetabling problem: An extension of the crew scheduling problem Springer Science + Business Media, LLC 2006, pp 2-2. Transportation Systems Analysis Course Project Page 21

22 APPENDIX Nikhil Menon Joao Fialho 1. Timetable and Rostering Generation Transportation Systems Analysis Course Project Page 22

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25 Transportation Systems Analysis Course Project Page 25

26 2. XPRESS IVE Model Nikhil Menon Joao Fialho model "Crew_Schedule" uses "mmxprs","mmive" declarations Period = TypeDriver = 1..3!Number of Time Periods!Type of Train Driver D: array(period) of real!demand w: array(typedriver) of real!wage for type of driver x: array(typedriver, Period) of mpvar!decisions Variables end-declarations D:: [ 6, 6, 12, 12, 12, 12, 11, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 12, 11, 8, 8, 8, 8, 7, 6, 6, 6, 6, 6, 4] w:: [ 64, 80, 45.6 ] N:=24!Number of drivers who are employed of the company!constraints!demand D(1)<=x(1,1)+x(2,1)+x(3,1)!t=1 D(2)<=x(1,2)+x(1,1)+x(2,2)+x(2,1)+x(3,2)+x(3,1)!t=2 D(3)<=x(1,3)+x(1,2)+x(1,1)+x(2,3)+x(2,2)+x(2,1)+x(3,3)+x(3,2)+x(3,1)!t=3 D(4)<=x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,4)+x(2,3)+x(2,2)+x(2,1)+x(3,4)+x(3,3)+x(3,2)+x(3,1)!t=4 D(5)<=x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,5)+x(2,4)+x(2,3)+x(2,2)+x(2,1)+x(3,5)+x(3,4)+x(3,3)+x(3,2 )+x(3,1)!t=5 D(6)<=x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,6)+x(2,5)+x(2,4)+x(2,3)+x(2,2)+x(2,1)+x(3,6)+x(3,5 )+x(3,4)+x(3,3)+x(3,2)+x(3,1)!t=6 D(7)<=x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,7)+x(2,6)+x(2,5)+x(2,4)+x(2,3)+x(2,2)+x(2,1 )+x(3,7)+x(3,6)+x(3,5)+x(3,4)+x(3,3)+x(3,2)+x(3,1)!t=7 D(8)<=x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,8)+x(2,7)+x(2,6)+x(2,5)+x(2,4)+x(2,3 )+x(2,2)+x(2,1)+x(3,8)+x(3,7)+x(3,6)+x(3,5)+x(3,4)+x(3,3)+x(3,2)+x(3,1)!t=8 D(9)<=x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(2,5 )+x(2,4)+x(2,3)+x(2,2)+x(3,9)+x(3,8)+x(3,7)+x(3,6)+x(3,5)+x(3,4)+x(3,3)+x(3,2)!t=9 D(10)<=x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,10)+x(2,9)+x(2,8)+x (2,7)+x(2,6)+x(2,5)+x(2,4)+x(2,3)+x(3,10)+x(3,9)+x(3,8)+x(3,7)+x(3,6)+x(3,5)+x(3,4)+x(3,3)!t=10 Transportation Systems Analysis Course Project Page 26

27 D(11)<=x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,11)+x(2,10) +x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(2,5)+x(2,4)+x(3,11)+x(3,10)+x(3,9)+x(3,8)+x(3,7)+x(3,6)+x(3,5)+x(3,4)!t=11 D(12)<=x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,12) +x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(2,5)+x(3,12)+x(3,11)+x(3,10)+x(3,9)+x(3,8)+x(3,7)+x(3,6 )+x(3,5)!t=12 D(13)<=x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2)+x(1,1) +x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(2,1)+x(3,13)+x(3,12)+x(3,11)+x(3,10)+x (3,9)+x(3,8)+x(3,7)+x(3,6)!t=13 D(14)<=x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(1,2 )+x(1,1)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(2,7)+x(2,2)+x(2,1)+x(3,14)+x(3,13)+x( 3,12)+x(3,11)+x(3,10)+x(3,9)+x(3,8)+x(3,7)!t=14 D(15)<=x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(1, 3)+x(1,2)+x(1,1)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(2,3)+x(2,2)+x(2,1)+x( 3,15)+x(3,14)+x(3,13)+x(3,12)+x(3,11)+x(3,10)+x(3,9)+x(3,8)!t=15 D(16)<=x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x( 1,4)+x(1,3)+x(1,2)+x(1,1)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,4)+x(2,3) +x(2,2)+x(2,1)+x(3,16)+x(3,15)+x(3,14)+x(3,13)+x(3,12)+x(3,11)+x(3,10)+x(3,9)!t=16 D(17)<=x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x (1,5)+x(1,4)+x(1,3)+x(1,2)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,5)+x(2, 4)+x(2,3)+x(2,2)+x(2,1)+x(3,17)+x(3,16)+x(3,15)+x(3,14)+x(3,13)+x(3,12)+x(3,11)+x(3,10)!t=17 D(18)<=x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+ x(1,6)+x(1,5)+x(1,4)+x(1,3)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,6)+x(2,5)+x(2,4)+x(2,3)+x(2,2)+x(2,1)+x(3,18)+x(3,17)+x(3,16)+x(3,15)+x(3,14)+x(3,13)+x(3,12)+x(3,11)!t=18 D(19)<=x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(1,8) +x(1,7)+x(1,6)+x(1,5)+x(1,4)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,7)+x (2,6)+x(2,5)+x(2,4)+x(2,3)+x(2,2)+x(2,1)+x(3,19)+x(3,18)+x(3,17)+x(3,16)+x(3,15)+x(3,14)+x(3,13)+x(3,1 2)!t=19 D(20)<=x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(1,9 )+x(1,8)+x(1,7)+x(1,6)+x(1,5)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,8)+ x(2,7)+x(2,6)+x(2,5)+x(2,4)+x(2,3)+x(2,2)+x(2,1)+x(3,20)+x(3,19)+x(3,18)+x(3,17)+x(3,16)+x(3,15)+x(3,1 4)+x(3,13)!t=20 D(21)<=x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(1,1 0)+x(1,9)+x(1,8)+x(1,7)+x(1,6)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,9) +x(2,8)+x(2,7)+x(2,6)+x(2,5)+x(2,4)+x(2,3)+x(2,2)+x(3,21)+x(3,20)+x(3,19)+x(3,18)+x(3,17)+x(3,16)+x(3, 15)+x(3,14)!t=21 Transportation Systems Analysis Course Project Page 27

28 D(22)<=x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(1,1 1)+x(1,10)+x(1,9)+x(1,8)+x(1,7)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,1 0)+x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(2,5)+x(2,4)+x(2,3)+x(3,22)+x(3,21)+x(3,20)+x(3,19)+x(3,18)+x(3,17)+x( 3,16)+x(3,15)!t=22 D(23)<=x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(1,1 2)+x(1,11)+x(1,10)+x(1,9)+x(1,8)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2, 11)+x(2,10)+x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(2,5)+x(2,4)+x(3,23)+x(3,22)+x(3,21)+x(3,20)+x(3,19)+x(3,18) +x(3,17)+x(3,16)!t=23 D(24)<=x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(1,1 3)+x(1,12)+x(1,11)+x(1,10)+x(1,9)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(2,5)+x(3,24)+x(3,23)+x(3,22)+x(3,21)+x(3,20)+x(3,19 )+x(3,18)+x(3,17)!t=24 D(25)<=x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(1,1 4)+x(1,13)+x(1,12)+x(1,11)+x(1,10)+x(2,25)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x( 2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(2,7)+x(2,6)+x(3,25)+x(3,24)+x(3,23)+x(3,22)+x(3,21)+x(3, 20)+x(3,19)+x(3,18)!t=25 D(26)<=x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(1,1 5)+x(1,14)+x(1,13)+x(1,12)+x(1,11)+x(2,26)+x(2,25)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x( 2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(2,7)+x(3,26)+x(3,25)+x(3,24)+x(3,23)+x(3,22)+x( 3,21)+x(3,20)+x(3,19)!t=26 D(27)<=x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(1,1 6)+x(1,15)+x(1,14)+x(1,13)+x(1,12)+x(2,27)+x(2,26)+x(2,25)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x( 2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(2,8)+x(3,27)+x(3,26)+x(3,25)+x(3,24)+x(3,23)+x (3,22)+x(3,21)+x(3,20)!t=27 D(28)<=x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(1,1 7)+x(1,16)+x(1,15)+x(1,14)+x(1,13)+x(2,28)+x(2,27)+x(2,26)+x(2,25)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x( 2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(2,9)+x(3,28)+x(3,27)+x(3,26)+x(3,25)+x(3,24)+ x(3,23)+x(3,22)+x(3,21)!t=28 D(29)<=x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(1,1 8)+x(1,17)+x(1,16)+x(1,15)+x(1,14)+x(2,29)+x(2,28)+x(2,27)+x(2,26)+x(2,25)+x(2,24)+x(2,23)+x(2,22)+x( 2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(2,10)+x(3,29)+x(3,28)+x(3,27)+x(3,26)+x(3,25) +x(3,24)+x(3,23)+x(3,22)!t=29 D(30)<=x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(1,1 9)+x(1,18)+x(1,17)+x(1,16)+x(1,15)+x(2,30)+x(2,29)+x(2,28)+x(2,27)+x(2,26)+x(2,25)+x(2,24)+x(2,23)+x( 2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(2,11)+x(3,30)+x(3,29)+x(3,28)+x(3,27)+x(3,26) +x(3,25)+x(3,24)+x(3,23)!t=30 Transportation Systems Analysis Course Project Page 28

29 D(31)<=x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(1,2 0)+x(1,19)+x(1,18)+x(1,17)+x(1,16)+x(2,31)+x(2,30)+x(2,29)+x(2,28)+x(2,27)+x(2,26)+x(2,25)+x(2,24)+x( 2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(2,12)+x(3,31)+x(3,30)+x(3,29)+x(3,28)+x(3,27) +x(3,26)+x(3,25)+x(3,24)!t=31 D(32)<=x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(1,2 1)+x(1,20)+x(1,19)+x(1,18)+x(1,17)+x(2,32)+x(2,31)+x(2,30)+x(2,29)+x(2,28)+x(2,27)+x(2,26)+x(2,25)+x( 2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(2,13)+x(3,32)+x(3,31)+x(3,30)+x(3,29)+x(3,28) +x(3,27)+x(3,26)+x(3,25)!t=32 D(33)<=x(1,33)+x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(1,2 2)+x(1,21)+x(1,20)+x(1,19)+x(1,18)+x(2,33)+x(2,32)+x(2,31)+x(2,30)+x(2,29)+x(2,28)+x(2,27)+x(2,26)+x( 2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(2,14)+x(3,33)+x(3,32)+x(3,31)+x(3,30)+x(3,29) +x(3,28)+x(3,27)+x(3,26)!t=33 D(34)<=x(1,34)+x(1,33)+x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(1,2 3)+x(1,22)+x(1,21)+x(1,20)+x(1,19)+x(2,34)+x(2,33)+x(2,32)+x(2,31)+x(2,30)+x(2,29)+x(2,28)+x(2,27)+x( 2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(2,15)+x(3,34)+x(3,33)+x(3,32)+x(3,31)+x(3,30) +x(3,29)+x(3,28)+x(3,27)!t=34 D(35)<=x(1,35)+x(1,34)+x(1,33)+x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,25)+x(1,2 4)+x(1,23)+x(1,22)+x(1,21)+x(1,20)+x(2,35)+x(2,34)+x(2,33)+x(2,32)+x(2,31)+x(2,30)+x(2,29)+x(2,28)+x( 2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(2,16)+x(3,35)+x(3,34)+x(3,33)+x(3,32)+x(3,31) +x(3,30)+x(3,29)+x(3,28)!t=35 D(36)<=x(1,36)+x(1,35)+x(1,34)+x(1,33)+x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,26)+x(1,2 5)+x(1,24)+x(1,23)+x(1,22)+x(1,21)+x(2,36)+x(2,35)+x(2,34)+x(2,33)+x(2,32)+x(2,31)+x(2,30)+x(2,29)+x( 2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(2,17)+x(3,36)+x(3,35)+x(3,34)+x(3,33)+x(3,32) +x(3,31)+x(3,30)+x(3,29)!t=36 D(37)<=x(1,37)+x(1,36)+x(1,35)+x(1,34)+x(1,33)+x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,27)+x(1,2 6)+x(1,25)+x(1,24)+x(1,23)+x(1,22)+x(2,37)+x(2,36)+x(2,35)+x(2,34)+x(2,33)+x(2,32)+x(2,31)+x(2,30)+x( 2,25)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(2,18)+x(3,37)+x(3,36)+x(3,35)+x(3,34)+x(3,33) +x(3,32)+x(3,31)+x(3,30)!t=37 D(38)<=x(1,38)+x(1,37)+x(1,36)+x(1,35)+x(1,34)+x(1,33)+x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,28)+x(1,2 7)+x(1,26)+x(1,25)+x(1,24)+x(1,23)+x(2,38)+x(2,37)+x(2,36)+x(2,35)+x(2,34)+x(2,33)+x(2,32)+x(2,31)+x( 2,26)+x(2,25)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(2,19)+x(3,38)+x(3,37)+x(3,36)+x(3,35)+x(3,34) +x(3,33)+x(3,32)+x(3,31)!t=38 D(39)<=x(1,39)+x(1,38)+x(1,37)+x(1,36)+x(1,35)+x(1,34)+x(1,33)+x(1,32)+x(1,31)+x(1,30)+x(1,29)+x(1,2 8)+x(1,27)+x(1,26)+x(1,25)+x(1,24)+x(2,39)+x(2,38)+x(2,37)+x(2,36)+x(2,35)+x(2,34)+x(2,33)+x(2,32)+x( 2,27)+x(2,26)+x(2,25)+x(2,24)+x(2,23)+x(2,22)+x(2,21)+x(2,20)+x(3,39)+x(3,38)+x(3,37)+x(3,36)+x(3,35) +x(3,34)+x(3,33)+x(3,32)!t=39!number of drivers who are employed of the company sum(t in Period) x(1, t)+ sum(t in Period) x(2, t)<= N Transportation Systems Analysis Course Project Page 29

30 !Objective function: Minimize total wages cost Wage:= sum(i in TypeDriver, t in Period) w(i)*x(i,t) minimize(wage) end-model Transportation Systems Analysis Course Project Page 30

31 3. Sensitivity Analysis on N Results Nikhil Menon Joao Fialho Period N=22 x 1t x 2t x 3t d t Total Excess of Start End t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1728 Transportation Systems Analysis Course Project Page 31

32 Period N=23 x 1t x 2t x 3t d t Total Nikhil Menon Joao Fialho Excess of Start End t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1712 Transportation Systems Analysis Course Project Page 32

33 Period N=25 Nikhil Menon Joao Fialho Total Excess of Start End t x 1t x 2t x 3t dt 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1696 Transportation Systems Analysis Course Project Page 33

34 Period N=26 Nikhil Menon Joao Fialho Total Excess of Start End t x 1t x 2t x 3t dt 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1728 Transportation Systems Analysis Course Project Page 34

35 4. Sensitivity Analysis on w 3 Results Nikhil Menon Joao Fialho Period 80% w 3 x 1t x 2t x 3t d t Total Excess of Start End t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1664 Transportation Systems Analysis Course Project Page 35

36 Period 85% w 3 x 1t x 2t x 3t d t Total Nikhil Menon Joao Fialho Excess of Start End t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1681,6 Transportation Systems Analysis Course Project Page 36

37 Period 90% w 3 x 1t x 2t x 3t d t Total Nikhil Menon Joao Fialho Excess of Start End t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1686,4 Transportation Systems Analysis Course Project Page 37

38 Period 95% w 3 x 1t x 2t x 3t d t Total Nikhil Menon Joao Fialho Excess of Start End t 06:30 07: :00 07: :30 08: :00 08: :30 09: :00 09: :30 10: :00 10: :30 11: :00 11: :30 12: :00 12: :30 13: :00 13: :30 14: :00 14: :30 15: :00 15: :30 16: :00 16: :30 17: :00 17: :30 18: :00 18: :30 19: :00 19: :30 20: :00 20: :30 21: :00 21: :30 22: :00 22: :30 23: :00 23: :30 00: :00 00: :30 01: :00 01: :30 02: :00 02: :30 03: :00 03: :30 04: :00 04: :30 05: SUM W 1691,2 Transportation Systems Analysis Course Project Page 38