WEBVTT 00:00:00.670 --> 00:00:03.010 The following content is provided under a Creative 00:00:03.010 --> 00:00:04.430 Commons license. 00:00:04.430 --> 00:00:06.640 Your support will help MIT OpenCourseWare 00:00:06.640 --> 00:00:10.730 continue to offer high quality educational resources for free. 00:00:10.730 --> 00:00:13.270 To make a donation or to view additional materials 00:00:13.270 --> 00:00:17.230 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.230 --> 00:00:18.103 at ocw.mit.edu. 00:00:22.670 --> 00:00:24.930 PROFESSOR: All right, so let's go ahead and proceed 00:00:24.930 --> 00:00:28.810 with vehicle scheduling. 00:00:28.810 --> 00:00:31.950 What we try to do in vehicle scheduling, 00:00:31.950 --> 00:00:36.370 and that's up to this point, we have frequently 00:00:36.370 --> 00:00:40.330 determined for different routes, identify 00:00:40.330 --> 00:00:44.860 to deliver the service that this specified on the timetable. 00:00:44.860 --> 00:00:49.360 And then vehicle scheduling, which is the main part, 00:00:49.360 --> 00:00:54.460 is basically how we are using the existing fleet 00:00:54.460 --> 00:00:58.750 to deliver the service that is specified on the timetable, how 00:00:58.750 --> 00:01:04.680 we are basically defining the duties for vehicles 00:01:04.680 --> 00:01:10.580 from the time that they pull out from depot till the time 00:01:10.580 --> 00:01:13.700 that they pull back in to depot again. 00:01:13.700 --> 00:01:19.820 What is the sequences of the revenue trips or revenue 00:01:19.820 --> 00:01:24.320 movements, and where that involve 00:01:24.320 --> 00:01:28.250 deadhead arcs, layover arcs, and et cetera. 00:01:28.250 --> 00:01:30.770 We will discuss this in this lecture 00:01:30.770 --> 00:01:32.720 and we will get to that. 00:01:37.790 --> 00:01:41.060 So the task of developing the timetable 00:01:41.060 --> 00:01:45.500 is basically the task of translating frequencies 00:01:45.500 --> 00:01:49.190 into a timetable. 00:01:49.190 --> 00:01:53.330 We may all be familiar with what a bus service or train service 00:01:53.330 --> 00:01:55.550 timetable is. 00:01:55.550 --> 00:01:59.930 It basically includes the list of departures and arrivals 00:01:59.930 --> 00:02:05.470 to different stops by different routes. 00:02:05.470 --> 00:02:11.390 There are alternative ways of determining the timetable. 00:02:11.390 --> 00:02:15.950 The simplest way is to just use equal head ways. 00:02:15.950 --> 00:02:21.770 And the frequency that you allocated to the route the time 00:02:21.770 --> 00:02:23.900 period, during the time period, just 00:02:23.900 --> 00:02:29.770 distribute them evenly over the time period. 00:02:33.400 --> 00:02:38.560 If you look at the figure on the bottom left here, 00:02:38.560 --> 00:02:45.560 we will have an example of even head ways here. 00:02:45.560 --> 00:02:53.470 However, if the demand is not uniformly distributed during 00:02:53.470 --> 00:02:58.750 the period, for which the example could be peak of peak, 00:02:58.750 --> 00:03:01.310 that is usually happening-- depending on the network, 00:03:01.310 --> 00:03:02.260 it may vary-- 00:03:02.260 --> 00:03:06.370 but usually happens between 7:00 and 8:00, 7:30 and 8:30. 00:03:06.370 --> 00:03:11.020 Then we want to work out the head ways, 00:03:11.020 --> 00:03:13.930 and work out the timetable in a way 00:03:13.930 --> 00:03:21.580 to result in balanced loads and equal loads. 00:03:21.580 --> 00:03:24.040 So in order to do this, what you need to do 00:03:24.040 --> 00:03:30.910 is to generate the cumulative passenger flow arrivals. 00:03:30.910 --> 00:03:38.300 And then if you distribute your headways on the time axis, 00:03:38.300 --> 00:03:45.080 you will get even headways or equal headway service. 00:03:45.080 --> 00:03:47.270 If you distribute your head weighs 00:03:47.270 --> 00:03:50.360 on the cumulative passenger flow evenly, 00:03:50.360 --> 00:03:55.550 then you would get balanced loads. 00:03:55.550 --> 00:03:59.570 As you can see, the headways which is here, 00:03:59.570 --> 00:04:03.500 one, two, three, four, five balancing loads 00:04:03.500 --> 00:04:06.060 for time table development. 00:04:06.060 --> 00:04:09.470 The other alternative way of developing timetables 00:04:09.470 --> 00:04:13.860 would be trying to design a weight clock-face service. 00:04:17.200 --> 00:04:22.480 Why would we want to design a clock-face time table? 00:04:22.480 --> 00:04:25.100 Does anyone have any suggestions? 00:04:25.100 --> 00:04:25.656 Yep. 00:04:25.656 --> 00:04:28.730 AUDIENCE: Sometimes it makes scheduling easier 00:04:28.730 --> 00:04:33.700 if we have some sort of cyclical pattern. 00:04:33.700 --> 00:04:36.640 It's easier to assign drivers that way. 00:04:36.640 --> 00:04:38.940 PROFESSOR: Right, yes. 00:04:38.940 --> 00:04:41.950 So that is actually one reason-- it may eventually, 00:04:41.950 --> 00:04:44.380 in the next step, which is crew scheduling, 00:04:44.380 --> 00:04:50.259 makes the service more efficient and cost efficient. 00:04:50.259 --> 00:04:51.300 Is there any other-- yep. 00:04:51.300 --> 00:04:52.966 AUDIENCE: It's easier for the passengers 00:04:52.966 --> 00:04:54.170 to remember the schedule. 00:04:54.170 --> 00:04:55.450 PROFESSOR: Exactly. 00:04:55.450 --> 00:04:59.800 If the waiting time estimations that we are doing, 00:04:59.800 --> 00:05:03.100 with respect to frequencies and with respect to headways, 00:05:03.100 --> 00:05:06.140 if passengers are able to coordinate with the service, 00:05:06.140 --> 00:05:11.290 the expected waiting time would be less than the situation 00:05:11.290 --> 00:05:13.990 where they cannot coordinate with the service. 00:05:13.990 --> 00:05:19.420 That's only the case when headway is 10, 15, 20, 30 00:05:19.420 --> 00:05:23.080 or minutes or 60 minutes. 00:05:23.080 --> 00:05:29.270 And then we would have service every-- 00:05:29.270 --> 00:05:31.090 on minute one, for example, that that 00:05:31.090 --> 00:05:36.410 would be 7-1, 7-11, 7-21, and so on and so forth. 00:05:36.410 --> 00:05:41.680 So it's easier for passengers to remember the service that 00:05:41.680 --> 00:05:44.060 is designed clock face. 00:05:44.060 --> 00:05:51.850 So again, for the two approaches to headway distribution, 00:05:51.850 --> 00:05:55.480 equal heavy distribution is basically, 00:05:55.480 --> 00:05:59.920 as it can be imagined, each headway 00:05:59.920 --> 00:06:05.140 would be calculated as peak period duration divided 00:06:05.140 --> 00:06:08.110 by number of vehicles or frequency 00:06:08.110 --> 00:06:12.280 that we determined for the route at the time period. 00:06:12.280 --> 00:06:15.400 And in terms of balance load solution, 00:06:15.400 --> 00:06:22.750 we basically need to first realize what the loads are, 00:06:22.750 --> 00:06:24.640 the balance loads are, by dividing 00:06:24.640 --> 00:06:28.150 the passenger flows to n and then work out, 00:06:28.150 --> 00:06:33.790 on the cumulative passenger arrivals, 00:06:33.790 --> 00:06:39.250 to realize what the timetable arrivals and departures are. 00:06:39.250 --> 00:06:40.670 Is there any questions about this? 00:06:40.670 --> 00:06:41.295 AUDIENCE: Yeah. 00:06:41.295 --> 00:06:44.410 Looking at the graph on the left, 00:06:44.410 --> 00:06:47.510 let's say PE4 minus PE [INAUDIBLE] 00:06:47.510 --> 00:06:51.265 is greater than the capacity of a bus, then we have a problem. 00:06:51.265 --> 00:06:55.790 Because if the number of passengers there is [INAUDIBLE] 00:06:55.790 --> 00:06:57.580 problem with the clock faced-- 00:06:57.580 --> 00:06:58.250 PROFESSOR: Mhm. 00:06:58.250 --> 00:07:04.380 Yeah, that's-- however, what we basically did in previous step, 00:07:04.380 --> 00:07:07.220 in frequency determination, that's one of the checks that 00:07:07.220 --> 00:07:10.570 we do, the maximum load. 00:07:10.570 --> 00:07:14.060 And we basically make sure that maximum load is 00:07:14.060 --> 00:07:18.627 within the allowed standards. 00:07:18.627 --> 00:07:20.210 AUDIENCE: But basically, so what we're 00:07:20.210 --> 00:07:25.340 saying is if we do clock faced, then 00:07:25.340 --> 00:07:30.310 we look where we have the maximum flow of passengers 00:07:30.310 --> 00:07:34.700 and take that headway and replicate that headway all 00:07:34.700 --> 00:07:38.540 across the rest of the time period, 00:07:38.540 --> 00:07:41.810 is what you're saying if I understand correctly. 00:07:41.810 --> 00:07:46.880 I'm saying I have to design a headway to serve the maximum, 00:07:46.880 --> 00:07:50.190 and then the rest of the time period, which has lower flows, 00:07:50.190 --> 00:07:52.550 will have, you know, excess capacity. 00:07:52.550 --> 00:07:53.450 PROFESSOR: Right. 00:07:53.450 --> 00:07:55.350 Yeah. 00:07:55.350 --> 00:07:56.610 So yeah, that makes sense. 00:07:56.610 --> 00:08:01.450 It really depends if we design based on the time period 00:08:01.450 --> 00:08:07.750 and we distribute the frequencies evenly 00:08:07.750 --> 00:08:15.270 and make sure that we are checking the requirements that 00:08:15.270 --> 00:08:18.570 are related to maximum load factors, 00:08:18.570 --> 00:08:22.020 then we won't have any problems. 00:08:22.020 --> 00:08:25.740 However, the other issue that could actually 00:08:25.740 --> 00:08:30.030 be the counterpart of the issue that you suggested 00:08:30.030 --> 00:08:35.860 is that when we're designing the timetable based on balance load 00:08:35.860 --> 00:08:40.289 solution, we may actually end up having 00:08:40.289 --> 00:08:46.240 to use more vehicles required for that solution. 00:08:46.240 --> 00:08:49.200 And the reason for that-- 00:08:49.200 --> 00:08:54.250 does anyone have any suggestions for the reason for [INAUDIBLE] 00:08:54.250 --> 00:08:55.030 if the number-- 00:08:57.820 --> 00:08:59.500 why the number of vehicles that we 00:08:59.500 --> 00:09:02.170 may want to require for balance [INAUDIBLE] 00:09:02.170 --> 00:09:03.540 could be [INAUDIBLE]? 00:09:07.950 --> 00:09:11.280 AUDIENCE: Because if you have a peak right at a certain time, 00:09:11.280 --> 00:09:13.455 the vehicle would be able to make one trip, 00:09:13.455 --> 00:09:16.120 and then you'll be sending out a bus just to make a single trip 00:09:16.120 --> 00:09:18.661 and go back, which adds a lot of deadhead time on either end. 00:09:18.661 --> 00:09:19.661 PROFESSOR: Right. 00:09:19.661 --> 00:09:20.160 Yeah. 00:09:20.160 --> 00:09:23.200 So that could be one reason. 00:09:23.200 --> 00:09:25.290 The other reason would be if we have 00:09:25.290 --> 00:09:30.720 a peak of peak in our passenger arrivals, 00:09:30.720 --> 00:09:34.310 then we would have really short time intervals or short head 00:09:34.310 --> 00:09:40.410 ways during peak of the peak, and rather longer headways 00:09:40.410 --> 00:09:42.480 during the off peak. 00:09:42.480 --> 00:09:45.750 And then what happens is that if these longer headways exceed 00:09:45.750 --> 00:09:48.450 the maximum headway that is allowed, then 00:09:48.450 --> 00:09:52.710 we would have to basically throw in more vehicles to make sure 00:09:52.710 --> 00:09:56.590 that we do not exceed the maximum headway allowed, 00:09:56.590 --> 00:09:58.440 which is one of the considerations related 00:09:58.440 --> 00:10:00.990 to maximum waiting time for passengers and level 00:10:00.990 --> 00:10:02.260 of service. 00:10:02.260 --> 00:10:04.560 So that's one of the issues. 00:10:04.560 --> 00:10:08.580 So as a result, in general, it is better 00:10:08.580 --> 00:10:13.080 to assume that design of timetable 00:10:13.080 --> 00:10:15.870 is going to be based on equal headways, 00:10:15.870 --> 00:10:17.970 and when we do the frequency computations, 00:10:17.970 --> 00:10:20.132 be consistent with that. 00:10:20.132 --> 00:10:33.360 AUDIENCE: So when we [INAUDIBLE] and they are, like, strictly 00:10:33.360 --> 00:10:39.644 bound to follow the movement they are set on [INAUDIBLE]?? 00:10:39.644 --> 00:10:43.606 PROFESSOR: I'm not sure I understand your comment. 00:10:43.606 --> 00:10:44.730 Can you repeat that please? 00:10:44.730 --> 00:10:46.422 AUDIENCE: If there's [INAUDIBLE]?? 00:10:46.422 --> 00:10:47.130 PROFESSOR: Right. 00:10:47.130 --> 00:10:51.855 AUDIENCE: [INAUDIBLE] then the number of vehicles 00:10:51.855 --> 00:10:53.370 is going to go up. 00:10:53.370 --> 00:10:56.840 But since the number of vehicles required during off-peak 00:10:56.840 --> 00:11:03.936 [INAUDIBLE] then it necessarily [INAUDIBLE] 00:11:03.936 --> 00:11:06.060 you need more vehicles during service [INAUDIBLE].. 00:11:08.596 --> 00:11:09.804 AUDIENCE: That's interesting. 00:11:09.804 --> 00:11:11.160 I have a question-- 00:11:11.160 --> 00:11:13.840 why do you think that a balance load 00:11:13.840 --> 00:11:16.560 will require more vehicles? 00:11:16.560 --> 00:11:18.540 I tend to believe that it's somewhat 00:11:18.540 --> 00:11:24.970 more efficient, capacity wise, than the alternative. 00:11:24.970 --> 00:11:27.136 Why do you think that it has more vehicles? 00:11:27.136 --> 00:11:29.260 AUDIENCE: I thought it was because the frequency is 00:11:29.260 --> 00:11:30.880 higher. 00:11:30.880 --> 00:11:33.135 AUDIENCE: Only at certain points. 00:11:33.135 --> 00:11:35.180 It's higher only-- 00:11:35.180 --> 00:11:37.770 AUDIENCE: You have enough vehicles to cover [INAUDIBLE].. 00:11:37.770 --> 00:11:39.186 AUDIENCE: But what you were saying 00:11:39.186 --> 00:11:41.880 is that you can't cut service so short that you 00:11:41.880 --> 00:11:45.130 don't need those load factors. 00:11:45.130 --> 00:11:48.320 So in the end, you end up using more vehicles. 00:11:48.320 --> 00:11:51.000 PROFESSOR: So the reason that we would basically 00:11:51.000 --> 00:11:53.660 require more vehicles, in balanced load, 00:11:53.660 --> 00:11:56.850 is that for some of these headways, 00:11:56.850 --> 00:11:59.040 we need to stretch the headway in a way 00:11:59.040 --> 00:12:03.630 that it basically meets the balance 00:12:03.630 --> 00:12:06.900 load on the other headways. 00:12:06.900 --> 00:12:10.100 And then, once we do that, we may actually go over 00:12:10.100 --> 00:12:16.350 the maximum allowed headway, that 00:12:16.350 --> 00:12:19.800 is basically trying to take care of maximum waiting 00:12:19.800 --> 00:12:23.490 time and level of service in the end. 00:12:23.490 --> 00:12:26.490 In that case, then we need to throw in more vehicles 00:12:26.490 --> 00:12:29.960 during the same period of the route. 00:12:29.960 --> 00:12:32.240 And if we need to do this for all the routes 00:12:32.240 --> 00:12:34.880 at the same time, then we would eventually 00:12:34.880 --> 00:12:41.720 observe more deficit, and more vehicles to be used. 00:12:41.720 --> 00:12:44.420 Is this the point that you tried to-- 00:12:44.420 --> 00:12:46.680 AUDIENCE: Yes, but [INAUDIBLE]. 00:13:02.500 --> 00:13:04.189 PROFESSOR: On reverse? 00:13:04.189 --> 00:13:05.686 AUDIENCE: [INAUDIBLE]. 00:13:30.531 --> 00:13:32.530 PROFESSOR: Well, these are two different factors 00:13:32.530 --> 00:13:35.270 and both are important in terms of level of service. 00:13:35.270 --> 00:13:40.870 Headway consideration is related to maximum waiting time. 00:13:40.870 --> 00:13:44.200 You don't want to have like reading the long headways that 00:13:44.200 --> 00:13:46.800 would force people to wait longer. 00:13:46.800 --> 00:13:49.420 And at the same time, you don't want 00:13:49.420 --> 00:13:52.870 buses that are overcrowded, and there's 00:13:52.870 --> 00:13:57.910 discomfort resulting from the crowding effect. 00:13:57.910 --> 00:14:03.070 So these are two different sources of this utility, 00:14:03.070 --> 00:14:07.870 basically, in the service, that we try to balance somehow. 00:14:07.870 --> 00:14:12.220 The other issue that you brought up first, in terms of the fact 00:14:12.220 --> 00:14:15.340 that the service may be directional, and demand 00:14:15.340 --> 00:14:19.150 may be directional during peak, and there 00:14:19.150 --> 00:14:23.590 could be possibility of using vehicles in other directions, 00:14:23.590 --> 00:14:27.070 or using deadheading, and situations 00:14:27.070 --> 00:14:29.800 like that, we are actually going to cover that. 00:14:29.800 --> 00:14:35.707 And that's true for both of the two scenarios. 00:14:35.707 --> 00:14:37.207 AUDIENCE: It seems like there should 00:14:37.207 --> 00:14:38.620 be some kind of hybrid way to say, 00:14:38.620 --> 00:14:40.536 this is the amount of vehicles I want to use-- 00:14:40.536 --> 00:14:44.370 and there's a way to not have to go over schedule 00:14:44.370 --> 00:14:49.090 based on equal headways, and still [INAUDIBLE] demand. 00:14:49.090 --> 00:14:50.870 PROFESSOR: Exactly, yeah. 00:14:50.870 --> 00:14:54.760 And so there is all sorts of tricks in these steps, where 00:14:54.760 --> 00:14:58.210 you tweak some of the solutions from one level 00:14:58.210 --> 00:15:02.650 and go back to the previous levels, and adjust those. 00:15:02.650 --> 00:15:05.110 There is all sorts of engineering judgments, 00:15:05.110 --> 00:15:09.400 and the planning judgments that are involved, and can actually 00:15:09.400 --> 00:15:10.070 take place. 00:15:10.070 --> 00:15:13.960 We will see a couple of those in this lecture, too. 00:15:13.960 --> 00:15:15.160 OK, so let's go. 00:15:18.040 --> 00:15:24.080 So now, once we know what the timetable is, 00:15:24.080 --> 00:15:28.050 the sequence of arrivals and departures for different trips, 00:15:28.050 --> 00:15:30.300 for all routes, then it's a matter 00:15:30.300 --> 00:15:33.720 of determining what's the fleet size that 00:15:33.720 --> 00:15:41.250 is required for delivering the service that is timetabled. 00:15:41.250 --> 00:15:46.650 There is this famous theorem in the literature 00:15:46.650 --> 00:15:51.360 that deals with an estimation of number of fleet 00:15:51.360 --> 00:15:52.300 that is required. 00:15:52.300 --> 00:15:54.840 It is basically an upper bound to the number 00:15:54.840 --> 00:16:00.460 of vehicles that are required to deliver the service. 00:16:00.460 --> 00:16:02.100 The reason that it's an upper bound 00:16:02.100 --> 00:16:05.610 is that it basically assumes that we 00:16:05.610 --> 00:16:10.540 are going to stick to the original timetable, 00:16:10.540 --> 00:16:13.020 and we are not allowed to do any trip shifting, 00:16:13.020 --> 00:16:16.800 or we are not doing any trip shifting in this step 00:16:16.800 --> 00:16:20.520 to accommodate some of [INAUDIBLE] that 00:16:20.520 --> 00:16:27.180 are one or two minutes, or a few minutes basically conflicting. 00:16:27.180 --> 00:16:29.760 And then the other assumption that 00:16:29.760 --> 00:16:35.070 is made here to compute the upper bound of fleet size 00:16:35.070 --> 00:16:39.090 is that we are not allowing deadheading trips 00:16:39.090 --> 00:16:42.360 to balance the deficit among the terminals. 00:16:42.360 --> 00:16:45.150 So let's start with the definitions. 00:16:45.150 --> 00:16:55.890 The function l of k, t, and s-- k is the index of terminal, 00:16:55.890 --> 00:17:01.640 t is the index of time, and s is the index of the schedule that 00:17:01.640 --> 00:17:05.410 is already developed. 00:17:05.410 --> 00:17:10.560 So l is basically the cumulative number 00:17:10.560 --> 00:17:17.099 of departures from terminal by time t following the schedule. 00:17:17.099 --> 00:17:20.010 And a would be the cumulative number 00:17:20.010 --> 00:17:25.319 of arrivals at terminal k by time t following that schedule. 00:17:25.319 --> 00:17:27.960 Then we define a function, which is 00:17:27.960 --> 00:17:31.230 the difference between the two, it's 00:17:31.230 --> 00:17:33.840 basically the deficit function which 00:17:33.840 --> 00:17:40.620 is the cumulative departures minus cumulative arrivals. 00:17:40.620 --> 00:17:43.290 We use this deficit function to estimate 00:17:43.290 --> 00:17:46.500 what's the required to deliver the service. 00:17:46.500 --> 00:17:56.040 So the fleet size theorem says that, if we basically 00:17:56.040 --> 00:18:02.700 compute the maximum number of deficit 00:18:02.700 --> 00:18:12.010 over all time intervals for each terminal, and we add those up, 00:18:12.010 --> 00:18:14.850 we will get the total number that is required-- 00:18:14.850 --> 00:18:16.320 number of vehicles that is required 00:18:16.320 --> 00:18:18.070 to deliver the service. 00:18:18.070 --> 00:18:19.860 So it's basically this equation. 00:18:19.860 --> 00:18:22.710 The summation over all terminals, maximum 00:18:22.710 --> 00:18:27.520 over t of the deficit function. 00:18:27.520 --> 00:18:30.530 So in this figure, we basically on top, 00:18:30.530 --> 00:18:34.370 we see different trips that are scheduled, 00:18:34.370 --> 00:18:39.690 and each trip is basically identified with a number 00:18:39.690 --> 00:18:45.210 and a pair of terminals. 00:18:45.210 --> 00:18:49.080 That's, for example, trip one, it starts from terminal m, 00:18:49.080 --> 00:18:52.540 and ends at terminal u. 00:18:52.540 --> 00:18:57.510 And then, as you can see, these trips all start at terminal m, 00:18:57.510 --> 00:19:00.060 some of them go to u, some of them go to v, 00:19:00.060 --> 00:19:02.970 and there is also trips between u and v, 00:19:02.970 --> 00:19:06.730 and basically some of them come back to m again. 00:19:06.730 --> 00:19:12.940 So if we basically work out the deficit function 00:19:12.940 --> 00:19:16.970 for terminal m, what we see is that, for example, 00:19:16.970 --> 00:19:21.010 at time around 5:00 AM, we will have one trip scheduled here, 00:19:21.010 --> 00:19:22.690 trip number six. 00:19:22.690 --> 00:19:25.630 And at 6:00, there is another one. 00:19:25.630 --> 00:19:31.930 Our deficit is basically growing, until at noon, 00:19:31.930 --> 00:19:36.550 where the first trip comes back to terminal m. 00:19:36.550 --> 00:19:43.420 And then, at that point, it continues at a deficit level 00:19:43.420 --> 00:19:49.480 three, where there is one departure and one arrival 00:19:49.480 --> 00:19:57.660 scheduled at 17, at 5:00 PM. 00:19:57.660 --> 00:20:03.180 And then it continues until the last trip arrivals, at a little 00:20:03.180 --> 00:20:04.710 be before 10:00 PM. 00:20:04.710 --> 00:20:10.550 So in this case, if we look at the maximum deficit 00:20:10.550 --> 00:20:16.130 over time for this terminal, it would be four. 00:20:16.130 --> 00:20:20.680 And that's for terminal m. 00:20:20.680 --> 00:20:22.750 When we look at all terminals in the network, 00:20:22.750 --> 00:20:24.820 and some of the maximum deficits, 00:20:24.820 --> 00:20:26.830 the summation will give us the number of fleet 00:20:26.830 --> 00:20:28.640 that is required. 00:20:28.640 --> 00:20:31.760 However, there are a couple tricks 00:20:31.760 --> 00:20:36.200 that we can basically improve the efficiency. 00:20:36.200 --> 00:20:37.790 One of the tricks would be shifting 00:20:37.790 --> 00:20:40.040 the departures or arrivals a little, 00:20:40.040 --> 00:20:43.680 bit so we accommodate more trips. 00:20:43.680 --> 00:20:45.270 If they're required, for example, 00:20:45.270 --> 00:20:53.270 in this case, like the deficit part that actually gets 00:20:53.270 --> 00:21:02.200 to level four, is as the result of trip nine being scheduled 00:21:02.200 --> 00:21:05.710 around like 11:15, or so. 00:21:05.710 --> 00:21:08.460 If we can manage to shift trip nine a little bit 00:21:08.460 --> 00:21:15.490 forward, up to noon, then we may actually 00:21:15.490 --> 00:21:19.420 get rid of the fourth unit of deficit 00:21:19.420 --> 00:21:27.640 here, and make the total deficit, or maximum deficit, 00:21:27.640 --> 00:21:31.340 here equal to three. 00:21:31.340 --> 00:21:34.190 There's also the issue of reliability 00:21:34.190 --> 00:21:37.670 that you may want to consider for these trip arrivals 00:21:37.670 --> 00:21:38.930 and departures. 00:21:38.930 --> 00:21:42.800 For example, if you look at this point 00:21:42.800 --> 00:21:48.620 here, where trip two arrives, and it's basically 00:21:48.620 --> 00:21:50.780 based on this method, it is going 00:21:50.780 --> 00:21:54.230 to be assigned to trip eight. 00:21:54.230 --> 00:22:02.584 However, what if trip two is late by a few minutes? 00:22:02.584 --> 00:22:04.000 What should be doing in that case? 00:22:04.000 --> 00:22:07.520 Are we going to delay trip eight to accommodate that? 00:22:07.520 --> 00:22:13.170 One way of looking at this is that instead of working out 00:22:13.170 --> 00:22:19.160 the deficit functions with trip times, and then schedule trips, 00:22:19.160 --> 00:22:23.270 we can work it out with half cycles that already includes 00:22:23.270 --> 00:22:27.980 layovers, and 95 percentiles at the terminus, 00:22:27.980 --> 00:22:35.170 to make sure that we have a leeway at the terminals. 00:22:35.170 --> 00:22:42.560 That will definitely require and result in higher deficits. 00:22:42.560 --> 00:22:46.520 But again, if we are allowed to shift the trips a little bit 00:22:46.520 --> 00:22:49.580 back and forth, then that's something 00:22:49.580 --> 00:22:52.400 that would be recommended-- to work with half cycles, 00:22:52.400 --> 00:22:55.640 instead of working with the scheduled trips. 00:22:55.640 --> 00:22:57.830 Is there any questions about fleet 00:22:57.830 --> 00:23:00.680 size required computation here? 00:23:03.440 --> 00:23:12.122 AUDIENCE: [INAUDIBLE] to reduce fleet size deficits? 00:23:12.122 --> 00:23:15.950 Do they actually employ the techniques of shifting? 00:23:15.950 --> 00:23:17.900 PROFESSOR: Yeah, of course. 00:23:17.900 --> 00:23:21.890 That's a part in timetable-- 00:23:21.890 --> 00:23:29.480 it's going back to one of the steps that was already decided 00:23:29.480 --> 00:23:31.870 and optimized in the past. 00:23:31.870 --> 00:23:35.990 But there are all sorts of these revisiting that is going on 00:23:35.990 --> 00:23:38.900 in transit agencies. 00:23:38.900 --> 00:23:44.630 Because we do not optimize all the components 00:23:44.630 --> 00:23:48.520 in an integrated manner, all at the same time, 00:23:48.520 --> 00:23:52.310 and we basically do a sequential optimization, 00:23:52.310 --> 00:23:54.620 there is always this requirement to go back 00:23:54.620 --> 00:23:57.140 to some of those decisions that you thought 00:23:57.140 --> 00:24:03.050 were optimal at the upper level, and adjust those to account 00:24:03.050 --> 00:24:06.926 for gains in the lower level. 00:24:06.926 --> 00:24:11.930 AUDIENCE: I assume-- suppose that this graph represents 00:24:11.930 --> 00:24:18.490 an entire agency, and we'll reduce that agency's fleet size 00:24:18.490 --> 00:24:21.860 from four to three, saving money on another bus. 00:24:21.860 --> 00:24:24.410 That's a lot of money in savings, 00:24:24.410 --> 00:24:26.150 by not having that fourth bus. 00:24:29.050 --> 00:24:32.770 Do they look-- 00:24:32.770 --> 00:24:34.990 PROFESSOR: When they design timetables, 00:24:34.990 --> 00:24:39.130 they try to have some sort of coordination 00:24:39.130 --> 00:24:42.670 with arrivals and departures to the terminals. 00:24:42.670 --> 00:24:44.680 There is an example at the end of this lecture 00:24:44.680 --> 00:24:47.920 that we're going to talk about, how you can coordinate back 00:24:47.920 --> 00:24:51.640 and forth trips, so it will result in lower deficit 00:24:51.640 --> 00:24:53.510 in the first place. 00:24:53.510 --> 00:25:00.400 But yeah, sometimes it may actually 00:25:00.400 --> 00:25:04.610 be at the level of the fleet size determination, 00:25:04.610 --> 00:25:07.410 that you do this these shiftings again 00:25:07.410 --> 00:25:11.630 to improve the efficiency. 00:25:11.630 --> 00:25:15.130 Any other questions? 00:25:15.130 --> 00:25:17.690 It's a bit faster on these parts. 00:25:17.690 --> 00:25:23.850 So when it comes to matter of vehicle scheduling problems, 00:25:23.850 --> 00:25:26.580 it is a very interesting problem to optimize. 00:25:26.580 --> 00:25:29.970 The objective of this problem is to define vehicle blocks, 00:25:29.970 --> 00:25:35.970 or sequence of duties for vehicles that includes revenue 00:25:35.970 --> 00:25:38.970 trips and non-revenue trips. 00:25:38.970 --> 00:25:43.650 In other words, deadheads to rebalance, or to balance 00:25:43.650 --> 00:25:45.180 the deficit at terminals. 00:25:47.870 --> 00:25:51.620 Or layovers that would actually take place 00:25:51.620 --> 00:25:55.280 for reliability considerations. 00:25:55.280 --> 00:26:01.610 So the objective is basically to define these vehicle blocks, 00:26:01.610 --> 00:26:05.510 given that we are going to cover all the trips that are already 00:26:05.510 --> 00:26:10.090 scheduled, in a way to minimize the fleet 00:26:10.090 --> 00:26:13.750 size as a measure of the capital cost, 00:26:13.750 --> 00:26:17.530 and minimize the non-revenue time and mileage. 00:26:17.530 --> 00:26:20.240 As a measure of a non-revenue time, 00:26:20.240 --> 00:26:25.480 could actually account for crew cost, 00:26:25.480 --> 00:26:30.010 and mileage could account for both crew cost and fuel 00:26:30.010 --> 00:26:35.460 and basic operational costs. 00:26:35.460 --> 00:26:39.385 One important observation is that these are proxies for us. 00:26:39.385 --> 00:26:42.220 Again, in the [INAUDIBLE] step of the work when 00:26:42.220 --> 00:26:44.840 we are doing the scheduling for crew, 00:26:44.840 --> 00:26:47.890 we will realize some of the costs 00:26:47.890 --> 00:26:55.630 actually depends on how these vehicle blocks are defined. 00:26:55.630 --> 00:27:00.130 And probably the important components of cost 00:27:00.130 --> 00:27:01.790 that is related to-- 00:27:01.790 --> 00:27:04.000 they're related to crew duties are going 00:27:04.000 --> 00:27:05.770 to be determined in data stage. 00:27:05.770 --> 00:27:09.370 So these are only proxies for cost, fleet size, 00:27:09.370 --> 00:27:13.000 and non-revenue time and mileages. 00:27:13.000 --> 00:27:16.780 Input into the vehicles scheduling problem 00:27:16.780 --> 00:27:21.100 would be [INAUDIBLE] vehicle revenue trips is the timetable 00:27:21.100 --> 00:27:24.220 that we are to deliver. 00:27:24.220 --> 00:27:27.790 Each is characterized by trip stop time, and the end time 00:27:27.790 --> 00:27:29.470 and point. 00:27:29.470 --> 00:27:33.050 Then the other input would be possible layover arcs 00:27:33.050 --> 00:27:38.440 to account for reliability of these trips, and delay trips. 00:27:38.440 --> 00:27:44.260 And deadhead arcs connecting depots to trip starting 00:27:44.260 --> 00:27:49.170 points-- we call those pull-out arcs from [? depots. ?] 00:27:49.170 --> 00:27:52.270 And then pull-ins back into depots from trip end 00:27:52.270 --> 00:27:53.620 points to depots. 00:27:53.620 --> 00:28:01.180 And then the deadhead, or balancing deficit arcs-- 00:28:01.180 --> 00:28:03.760 these are basically trips that bring vehicles 00:28:03.760 --> 00:28:08.140 from terminals with lower deficit than what 00:28:08.140 --> 00:28:12.820 is scheduled to terminals with higher deficit. 00:28:12.820 --> 00:28:16.090 So what we try to do is that, from these arcs, 00:28:16.090 --> 00:28:20.110 we certainly need to choose the optimal combination 00:28:20.110 --> 00:28:22.630 of these arcs to deliver the service, 00:28:22.630 --> 00:28:27.280 and bring our vehicles out from depot and the same number 00:28:27.280 --> 00:28:32.810 back to people depot again, after the day is over. 00:28:32.810 --> 00:28:35.960 So another important observation, or quick note, 00:28:35.960 --> 00:28:41.470 here, is that there may be many possible number of deadhead 00:28:41.470 --> 00:28:43.250 and layover arcs. 00:28:43.250 --> 00:28:46.050 It's actually from a combinatorial nature. 00:28:46.050 --> 00:28:51.680 However, the designer, or the person 00:28:51.680 --> 00:28:56.240 who tries to solve this problem, needs to review these arcs, 00:28:56.240 --> 00:29:03.290 and come up with feasible and plausible arcs to start with. 00:29:03.290 --> 00:29:05.750 And then, as I mentioned, the layover arcs 00:29:05.750 --> 00:29:10.910 are also important in terms of making sure we will have 00:29:10.910 --> 00:29:14.240 a certain level of reliability. 00:29:14.240 --> 00:29:18.290 There are variations to the problem that actually 00:29:18.290 --> 00:29:22.010 can make the problem more complicated, 00:29:22.010 --> 00:29:27.050 but we are going to stick to the simplest one, 00:29:27.050 --> 00:29:32.390 and show how we can formulate the problem, model it, and what 00:29:32.390 --> 00:29:34.580 are the algorithms that can solve it for us. 00:29:34.580 --> 00:29:38.420 These variations, basically, are related to-- 00:29:38.420 --> 00:29:40.370 one possible variation is related 00:29:40.370 --> 00:29:45.890 to vehicles servicing single lines, or single routes, 00:29:45.890 --> 00:29:50.880 or inter-lining between different routes. 00:29:50.880 --> 00:29:54.560 The other variation would be related to possibility 00:29:54.560 --> 00:29:59.690 of having multiple depots in the network, big, 00:29:59.690 --> 00:30:04.190 larger transit agencies usually have multiple depots. 00:30:04.190 --> 00:30:08.540 For example, for Boston, I know a few of those. 00:30:08.540 --> 00:30:11.790 I do not know what's the exact number of depots 00:30:11.790 --> 00:30:14.410 that exist in the network. 00:30:14.410 --> 00:30:16.520 Alfred Street is one-- 00:30:16.520 --> 00:30:22.605 AUDIENCE: Albany, Southhampton, two of them at Lynn, 00:30:22.605 --> 00:30:27.220 Fellsway, North Cambridge-- that's it. 00:30:27.220 --> 00:30:31.560 PROFESSOR: That's, yeah, OK, so it's almost close to 10-- 00:30:31.560 --> 00:30:32.220 eight, nine. 00:30:32.220 --> 00:30:36.920 OK, so when it comes to different depots, and having 00:30:36.920 --> 00:30:40.070 multiple depots, there will be more complexities 00:30:40.070 --> 00:30:41.390 introduced to the problem. 00:30:41.390 --> 00:30:44.480 Because you may want to make sure that the exact number 00:30:44.480 --> 00:30:49.820 of vehicles who depart at a depot, come back to depot. 00:30:49.820 --> 00:30:51.710 There are also some considerations 00:30:51.710 --> 00:30:56.850 related to the fuel that the fleet is using. 00:30:56.850 --> 00:31:00.650 You may have diesel buses, you may have CNG buses. 00:31:00.650 --> 00:31:05.270 If you have restriction in terms of fueling facilities 00:31:05.270 --> 00:31:09.750 in these depots, you may want to return back the vehicles 00:31:09.750 --> 00:31:12.470 to the particular depots that are 00:31:12.470 --> 00:31:16.640 consistent with the fueling system. 00:31:16.640 --> 00:31:21.230 Then, there are different considerations 00:31:21.230 --> 00:31:26.660 related to vehicle types, related to parking spaces 00:31:26.660 --> 00:31:29.562 at the depots, and that actually makes 00:31:29.562 --> 00:31:30.770 the problem more complicated. 00:31:30.770 --> 00:31:34.130 But we can actually make some simplifying assumptions 00:31:34.130 --> 00:31:36.080 for now, and see how the problem looks 00:31:36.080 --> 00:31:38.300 like in the most basic form. 00:31:41.590 --> 00:31:44.970 So this is the time space network representation 00:31:44.970 --> 00:31:47.400 of the vehicle scheduling problem-- 00:31:47.400 --> 00:31:49.380 we are trying, in the next three slides, 00:31:49.380 --> 00:31:53.740 we are trying to build on the time and space network 00:31:53.740 --> 00:31:57.450 structure, to realize what is the decision that we're 00:31:57.450 --> 00:32:00.510 trying to make, and how we can basically solve this. 00:32:00.510 --> 00:32:04.350 So this is time space representation 00:32:04.350 --> 00:32:11.100 with x-axis being time of day, and y-axis being space. 00:32:11.100 --> 00:32:13.200 This is only one dimensional space here, 00:32:13.200 --> 00:32:18.480 but it could be two dimensional space, too. 00:32:18.480 --> 00:32:22.920 So in this case, we have two routes 00:32:22.920 --> 00:32:25.980 that are represented here-- route one, 00:32:25.980 --> 00:32:28.920 or n routes-- route one to route n. 00:32:28.920 --> 00:32:32.370 As you can see there are trips from either direction 00:32:32.370 --> 00:32:36.570 of these routes, that are already timetabled, 00:32:36.570 --> 00:32:38.860 and we have already committed to deliver this service. 00:32:41.490 --> 00:32:45.930 So the black arcs here are revenue arcs 00:32:45.930 --> 00:32:48.060 that are the arcs that we committed, 00:32:48.060 --> 00:32:52.230 and these have to be delivered between the two 00:32:52.230 --> 00:32:54.510 terminals of every route. 00:32:54.510 --> 00:33:00.150 Then we have-- in this problem, we have only one depot. 00:33:00.150 --> 00:33:04.080 There are two nodes here that are represented as depot, 00:33:04.080 --> 00:33:06.540 but it's basically the representation 00:33:06.540 --> 00:33:09.840 of depot at the beginning of the day when vehicles pull out, 00:33:09.840 --> 00:33:13.380 and the representation at the end of the day when vehicles 00:33:13.380 --> 00:33:15.050 pull in, basically. 00:33:15.050 --> 00:33:18.120 There are also layover arcs, where 00:33:18.120 --> 00:33:24.600 you connect all possible alternatives that you may think 00:33:24.600 --> 00:33:29.270 would be plausible for connecting these trips to one 00:33:29.270 --> 00:33:31.040 another. 00:33:31.040 --> 00:33:34.220 So in this figure, we only have a few of those-- 00:33:34.220 --> 00:33:35.960 probably the closest one-- but you 00:33:35.960 --> 00:33:40.370 can work out any possible combination of bus arrival 00:33:40.370 --> 00:33:43.370 to a terminal, being connected to a different trip 00:33:43.370 --> 00:33:44.732 from that terminal. 00:33:44.732 --> 00:33:46.190 AUDIENCE: Would you mind explaining 00:33:46.190 --> 00:33:48.170 what the [INAUDIBLE]? 00:33:48.170 --> 00:33:49.480 Maybe I'm missing something. 00:33:49.480 --> 00:33:51.035 Like, for example, when we start out 00:33:51.035 --> 00:33:54.186 at the beginning of the day, [INAUDIBLE].. 00:33:56.750 --> 00:34:01.040 We have two lines emanating from the very beginning of the day, 00:34:01.040 --> 00:34:04.590 so is that two buses leaving the depot at the same time? 00:34:04.590 --> 00:34:05.840 Maybe slightly different time? 00:34:05.840 --> 00:34:10.760 PROFESSOR: Right, so let me go ahead and continue and finish 00:34:10.760 --> 00:34:12.659 all the arcs in the network-- 00:34:12.659 --> 00:34:14.870 that may clarify the situation. 00:34:14.870 --> 00:34:17.510 Because there are other arcs in the network, too. 00:34:17.510 --> 00:34:24.679 So there are arcs that are pull-out arcs, from the depot 00:34:24.679 --> 00:34:30.560 to start these trips, and arcs that return the vehicles back 00:34:30.560 --> 00:34:31.850 to the depot. 00:34:31.850 --> 00:34:37.580 There's also arcs that are there to account for possibilities 00:34:37.580 --> 00:34:41.449 of deadheading trips, from one route to another route. 00:34:41.449 --> 00:34:46.010 AUDIENCE: So now we have those two nodes going into route one, 00:34:46.010 --> 00:34:49.940 so two buses going into route one, right? 00:34:49.940 --> 00:34:51.530 PROFESSOR: That's only a link. 00:34:51.530 --> 00:34:54.440 That link may include multiple vehicles. 00:34:58.320 --> 00:35:00.840 AUDIENCE: Multiple vehicles at different times? 00:35:00.840 --> 00:35:06.785 PROFESSOR: Multiple vehicles at different times-- 00:35:10.870 --> 00:35:12.550 well, it could be-- 00:35:12.550 --> 00:35:18.050 yeah, this is not the exact representation 00:35:18.050 --> 00:35:20.630 of time-dependent service. 00:35:20.630 --> 00:35:23.210 You can actually have connections, 00:35:23.210 --> 00:35:25.420 if you consider each one of these links 00:35:25.420 --> 00:35:29.790 to carry only one vehicle or not, 00:35:29.790 --> 00:35:32.750 then you need to add vehicles from depot 00:35:32.750 --> 00:35:37.280 to other trips along the time dimension, 00:35:37.280 --> 00:35:40.530 to vehicles that are basically joining that route 00:35:40.530 --> 00:35:41.880 [INAUDIBLE] time. 00:35:41.880 --> 00:35:44.820 AUDIENCE: What is the top [INAUDIBLE]?? 00:35:44.820 --> 00:35:48.370 PROFESSOR: The top line, you can-- 00:35:48.370 --> 00:35:50.290 the top line is just-- 00:35:50.290 --> 00:35:50.850 [INAUDIBLE] 00:35:50.850 --> 00:35:54.730 AUDIENCE: [INAUDIBLE]. 00:35:54.730 --> 00:35:56.040 PROFESSOR: It could be-- 00:35:56.040 --> 00:35:57.280 yeah, this is-- 00:35:57.280 --> 00:35:58.937 AUDIENCE: [INAUDIBLE]. 00:35:58.937 --> 00:36:01.020 PROFESSOR: This is, again, a simplified situation, 00:36:01.020 --> 00:36:04.530 but you can pull out at the middle of the route, too. 00:36:04.530 --> 00:36:10.620 Where you can, for example, just serve a vehicle from here. 00:36:10.620 --> 00:36:13.890 But you may want to make sure that the scheduled trip is 00:36:13.890 --> 00:36:15.610 served. 00:36:15.610 --> 00:36:19.290 So if you have services that's basically 00:36:19.290 --> 00:36:24.510 a start from the middle of the line, then you can model that, 00:36:24.510 --> 00:36:25.200 too. 00:36:25.200 --> 00:36:30.070 This is just to show how you can model a situation like this. 00:36:30.070 --> 00:36:33.626 It may not be all comprehensive combinations, or situation. 00:36:33.626 --> 00:36:36.250 AUDIENCE: I just wanted to make sure I understand what it says. 00:36:36.250 --> 00:36:37.791 PROFESSOR: Doesn't id make sense now? 00:36:37.791 --> 00:36:39.527 AUDIENCE: Yeah, I think so. 00:36:39.527 --> 00:36:41.735 I think one of the confusing things about this model, 00:36:41.735 --> 00:36:44.440 actually, is that you have that horizontal line. 00:36:44.440 --> 00:36:46.691 And it's saying [INAUDIBLE] revenue. 00:36:46.691 --> 00:36:49.066 Really what you should have is that layover arc shouldn't 00:36:49.066 --> 00:36:51.680 be an arc, it should just be a line connecting the end of-- 00:36:51.680 --> 00:36:52.640 [INAUDIBLE] 00:36:52.640 --> 00:36:55.370 PROFESSOR: Exactly-- these horizontal lines, 00:36:55.370 --> 00:36:57.170 I should actually remove that. 00:36:57.170 --> 00:36:58.010 I will-- 00:36:58.010 --> 00:37:00.180 AUDIENCE: Yeah, just make the layover arcs lines. 00:37:00.180 --> 00:37:03.300 PROFESSOR: Yeah, exactly, layover arcs-- 00:37:03.300 --> 00:37:04.250 that's exactly true. 00:37:06.790 --> 00:37:11.440 Actually, make the connection between the two directions 00:37:11.440 --> 00:37:15.060 of the route, and then the other arcs that are involved 00:37:15.060 --> 00:37:18.670 are the deadhead arc's points in the network. 00:37:18.670 --> 00:37:22.300 AUDIENCE: But it could be noted that the layover 00:37:22.300 --> 00:37:24.800 arcs are not really-- 00:37:24.800 --> 00:37:28.120 [INAUDIBLE] there's no spatial dimension to them, 00:37:28.120 --> 00:37:29.570 because the bus isn't moving. 00:37:29.570 --> 00:37:31.210 PROFESSOR: Right, yeah, exactly. 00:37:31.210 --> 00:37:35.020 Maybe that's-- the only spatial dimension to it would be just 00:37:35.020 --> 00:37:38.290 go ahead and turn around over the corner. 00:37:38.290 --> 00:37:39.580 And then come back, yeah. 00:37:39.580 --> 00:37:40.180 AUDIENCE: It would still be the case 00:37:40.180 --> 00:37:42.250 that the bus that pulls into the terminal is the next one out. 00:37:42.250 --> 00:37:44.541 So that's when you might actually want to have the arc, 00:37:44.541 --> 00:37:47.350 to show that even though it's not really moving-- 00:37:47.350 --> 00:37:49.000 [INAUDIBLE] there is a reason why you 00:37:49.000 --> 00:37:50.429 might want to show it that way. 00:37:50.429 --> 00:37:51.220 PROFESSOR: Exactly. 00:37:51.220 --> 00:37:53.710 And then the other point is that, this 00:37:53.710 --> 00:37:59.910 is, like even front from this node to nodes further down, 00:37:59.910 --> 00:38:03.150 the reason that we are [INAUDIBLE] think 00:38:03.150 --> 00:38:06.070 the optimization problem in the network structure, 00:38:06.070 --> 00:38:08.320 like many other optimization problems, 00:38:08.320 --> 00:38:13.990 is that with networks that consists of nodes and arcs, 00:38:13.990 --> 00:38:17.980 we can represent the constraints very efficiently. 00:38:17.980 --> 00:38:20.140 And there are efficient algorithms 00:38:20.140 --> 00:38:27.050 that use this representation to solve the optimization problem. 00:38:27.050 --> 00:38:31.900 So whenever we can present an optimization problem 00:38:31.900 --> 00:38:33.640 in a network structure, we do that, 00:38:33.640 --> 00:38:37.930 because we have really elegant algorithms and network 00:38:37.930 --> 00:38:42.950 flows that can be used to solve optimization problems. 00:38:42.950 --> 00:38:45.440 So let's see what's the optimization problem we 00:38:45.440 --> 00:38:48.330 are trying to solve here. 00:38:48.330 --> 00:38:50.820 One other link that we need to add here 00:38:50.820 --> 00:38:53.100 is basically from depot to depot, 00:38:53.100 --> 00:38:57.300 to make sure that the total number of vehicles that 00:38:57.300 --> 00:39:00.792 are going out of the depot, in the morning, 00:39:00.792 --> 00:39:03.000 are going to be equal to the total number of vehicles 00:39:03.000 --> 00:39:05.474 that are coming back to depot. 00:39:05.474 --> 00:39:06.890 AUDIENCE: So we're assuming-- just 00:39:06.890 --> 00:39:08.848 to go back to that-- there are some lines there 00:39:08.848 --> 00:39:11.198 that don't have a pull-out, so we assume those 00:39:11.198 --> 00:39:13.170 are coming from other depots? 00:39:13.170 --> 00:39:15.740 PROFESSOR: Yeah, in this example, 00:39:15.740 --> 00:39:19.620 we are only simplifying it into one depot situation, 00:39:19.620 --> 00:39:20.997 but some of the-- yeah, exactly-- 00:39:20.997 --> 00:39:22.700 some of the other routes could actually be-- 00:39:22.700 --> 00:39:24.250 AUDIENCE: I guess the question is, if a route ends, 00:39:24.250 --> 00:39:26.839 the first one that's going from the bottom to the top 00:39:26.839 --> 00:39:28.380 doesn't have a pull-out, and so there 00:39:28.380 --> 00:39:29.620 should be an arrow going back. 00:39:29.620 --> 00:39:31.953 PROFESSOR: There should be an arrow, yeah, to that, too. 00:39:31.953 --> 00:39:34.040 But it's missing in this. 00:39:34.040 --> 00:39:36.290 These arrows are only representative, 00:39:36.290 --> 00:39:40.670 and they are not comprehensive. 00:39:40.670 --> 00:39:45.230 And we basically are defining integer capacities 00:39:45.230 --> 00:39:49.250 that would reflect a number of vehicles 00:39:49.250 --> 00:39:53.060 that are doing that duty, or are basically 00:39:53.060 --> 00:39:55.560 assigned to those duties. 00:39:55.560 --> 00:39:58.970 These arcs are revenue, layover, and deadhead arcs. 00:39:58.970 --> 00:40:01.970 And these arcs, again, have costs. 00:40:01.970 --> 00:40:07.040 Like revenue arcs have given amount of cost, 00:40:07.040 --> 00:40:12.620 depending on the time that they take to make the trip, and so 00:40:12.620 --> 00:40:13.640 on and so forth. 00:40:13.640 --> 00:40:18.980 Layover trips, depending on how much time they take. 00:40:18.980 --> 00:40:21.920 And deadhead arcs, depending on how much time they take, 00:40:21.920 --> 00:40:26.080 and how much mileage they require to travel, 00:40:26.080 --> 00:40:27.300 the vehicles. 00:40:27.300 --> 00:40:29.770 However, what's interesting is that [INAUDIBLE] 00:40:29.770 --> 00:40:33.710 arcs costs of zero for those, because it's not relevant. 00:40:33.710 --> 00:40:39.170 We eventually would have to deliver those revenue arc 00:40:39.170 --> 00:40:40.630 services. 00:40:40.630 --> 00:40:43.400 So it's a constant in the optimization problem. 00:40:43.400 --> 00:40:46.890 We might as well just remove them. 00:40:46.890 --> 00:40:48.410 But what's important is that when we 00:40:48.410 --> 00:40:51.200 want to see what the layovers-- 00:40:51.200 --> 00:40:55.600 what's the optimal combination of layovers and deadheads 00:40:55.600 --> 00:40:56.660 in this problem? 00:40:56.660 --> 00:41:04.190 So again, the bounds that we have for revenue arcs 00:41:04.190 --> 00:41:08.160 is basically lower bound is one, upper bound is one. 00:41:08.160 --> 00:41:11.680 Why is it like this? 00:41:11.680 --> 00:41:14.260 Does anyone have any suggestion, why 00:41:14.260 --> 00:41:18.910 the lower bound and upper bound are both equal to one 00:41:18.910 --> 00:41:19.945 for revenue arcs? 00:41:19.945 --> 00:41:21.820 AUDIENCE: Because you said we have to operate 00:41:21.820 --> 00:41:23.050 the trips, no matter what. 00:41:23.050 --> 00:41:25.930 PROFESSOR: And we only have to operate to deliver them once, 00:41:25.930 --> 00:41:28.120 right? 00:41:28.120 --> 00:41:34.030 Remember, these are trips in the time space representation. 00:41:34.030 --> 00:41:36.610 So each trip is basically one arc-- 00:41:36.610 --> 00:41:38.290 we need to deliver it once. 00:41:38.290 --> 00:41:43.320 So it's basically the matter of choosing [INAUDIBLE] 00:41:43.320 --> 00:41:47.620 or configuration of layover arcs between zero and one, 00:41:47.620 --> 00:41:49.770 and deadhead arcs between zero and one. 00:41:49.770 --> 00:41:53.200 A maximum block length is a guarantee-- 00:41:53.200 --> 00:41:55.300 this is usually not a problem. 00:41:55.300 --> 00:41:58.120 If the fuel situation, especially 00:41:58.120 --> 00:42:02.630 for electric vehicles, is restricting, 00:42:02.630 --> 00:42:07.360 we may want to put additional restrictions on maximum length 00:42:07.360 --> 00:42:08.290 of the vehicle. 00:42:08.290 --> 00:42:13.940 But with the existing diesel and CNG vehicles, 00:42:13.940 --> 00:42:17.380 these are usually fine to deliver the whole service day 00:42:17.380 --> 00:42:20.314 services without having to refuel. 00:42:20.314 --> 00:42:21.730 It's [INAUDIBLE] that is something 00:42:21.730 --> 00:42:28.300 that is already taken care of in the vehicle scheduling problem. 00:42:28.300 --> 00:42:30.910 However, shifting heuristic needs 00:42:30.910 --> 00:42:35.480 to take place before we get to vehicle scheduling, 00:42:35.480 --> 00:42:37.930 at the level of timetable development, 00:42:37.930 --> 00:42:44.240 as we discussed after the deficit functions. 00:42:44.240 --> 00:42:47.860 So the matter would be the-- 00:42:47.860 --> 00:42:51.190 if we have multiple depot, the problem 00:42:51.190 --> 00:42:53.590 becomes a little bit more complicated. 00:42:53.590 --> 00:42:57.010 For single depot situation, we solved the problem 00:42:57.010 --> 00:43:02.290 with network flow algorithms, minimum cost flow algorithm, 00:43:02.290 --> 00:43:07.330 that is very efficient and very, very fast. 00:43:07.330 --> 00:43:13.370 When it comes to [INAUDIBLE] flow with integer, [INAUDIBLE] 00:43:13.370 --> 00:43:16.450 which is [INAUDIBLE] in the problem that we have, 00:43:16.450 --> 00:43:20.650 that problem becomes really difficult. It becomes NP-hard, 00:43:20.650 --> 00:43:24.790 and it basically is very difficult to scale. 00:43:24.790 --> 00:43:28.270 That basically means that we do not 00:43:28.270 --> 00:43:34.570 use the network flow algorithms for multi depot situation. 00:43:34.570 --> 00:43:39.750 We use some of the existing [INAUDIBLE] stakes. 00:43:39.750 --> 00:43:41.990 Are there are any questions about this so far? 00:43:41.990 --> 00:43:44.990 So this is another network representation 00:43:44.990 --> 00:43:49.430 of the multiple vehicle scheduling problem. 00:43:49.430 --> 00:43:54.590 This is not a time space representation anymore, 00:43:54.590 --> 00:44:00.650 because it would become really complicated with time space 00:44:00.650 --> 00:44:01.700 representation. 00:44:01.700 --> 00:44:05.120 As you can see, there are different depots 00:44:05.120 --> 00:44:11.390 in the network, and there are different services 00:44:11.390 --> 00:44:13.810 that need to actually-- or revenue arcs-- 00:44:13.810 --> 00:44:16.970 that need to be served. 00:44:16.970 --> 00:44:18.650 There may be some constraints on number 00:44:18.650 --> 00:44:23.090 of vehicles that pull out from each one of these depots, 00:44:23.090 --> 00:44:25.790 and the pull in each one of these depots. 00:44:25.790 --> 00:44:29.750 And again, if we didn't have the constraint 00:44:29.750 --> 00:44:36.770 of having integer solution, this could be solved in linear time, 00:44:36.770 --> 00:44:37.490 too. 00:44:37.490 --> 00:44:40.610 However, with integer constraints, it's not possible. 00:44:44.430 --> 00:44:48.630 So is there any questions about this? 00:44:48.630 --> 00:44:52.410 It basically-- in order to be able to implement this, 00:44:52.410 --> 00:44:56.820 we need to make a definition on trips, 00:44:56.820 --> 00:44:59.750 and identify compatible trips. 00:44:59.750 --> 00:45:05.850 Two compatible trips at a terminal, I and J, two trips I 00:45:05.850 --> 00:45:08.330 and J are compatible at the terminal, 00:45:08.330 --> 00:45:13.230 if J is arriving to terminal, I is departing from terminal, 00:45:13.230 --> 00:45:15.960 and the time difference between the two 00:45:15.960 --> 00:45:20.160 is actually greater than the layover that we want. 00:45:20.160 --> 00:45:26.910 And then, also, less than twice the time of traveling to depot. 00:45:26.910 --> 00:45:31.230 The solution may be only as good as just sending the vehicle 00:45:31.230 --> 00:45:36.460 back to depot, and then pull it out from depot 00:45:36.460 --> 00:45:38.250 to another part of the network. 00:45:38.250 --> 00:45:42.930 So we only restrict the search for compatible trips 00:45:42.930 --> 00:45:44.370 to these two bounds. 00:45:44.370 --> 00:45:49.770 Like MK here is the minimum recovery time 00:45:49.770 --> 00:45:52.080 that we define as a lower bound of time 00:45:52.080 --> 00:45:56.790 difference between the arrival of trip I minus departure 00:45:56.790 --> 00:46:01.020 of trip J. And then, we use 2 times 00:46:01.020 --> 00:46:06.510 DK, which is the deadhead time from terminal to depot. 00:46:06.510 --> 00:46:09.300 Then the algorithm is really simple-- 00:46:09.300 --> 00:46:11.470 it has only four steps. 00:46:11.470 --> 00:46:17.000 We start with the earliest arrival at terminal. 00:46:17.000 --> 00:46:20.450 If there is no earliest arrival, we go to step D. 00:46:20.450 --> 00:46:25.040 And basically, in step D, we need to pull out from depot 00:46:25.040 --> 00:46:28.320 to serve that service. 00:46:28.320 --> 00:46:31.010 But if there is one earliest arrival, 00:46:31.010 --> 00:46:32.810 then we link that earliest arrival 00:46:32.810 --> 00:46:37.760 to compatible trip departure, based on the compatibility 00:46:37.760 --> 00:46:41.240 constraint, or definition that we had. 00:46:41.240 --> 00:46:44.900 Again, if there is no trips that is compatible with that, 00:46:44.900 --> 00:46:54.410 we need to go back to step A, and basically use 00:46:54.410 --> 00:47:01.020 the next arrival at terminal. 00:47:01.020 --> 00:47:05.820 If there is one trip that is compatible with the trip that 00:47:05.820 --> 00:47:10.440 arrives, then we will check the vehicle block 00:47:10.440 --> 00:47:14.700 length against the existing constraint. 00:47:14.700 --> 00:47:19.810 Again, that could actually be related to the fuel, 00:47:19.810 --> 00:47:21.970 fueling constraint, and so on and so forth. 00:47:21.970 --> 00:47:23.790 If it's constraining, then we need 00:47:23.790 --> 00:47:28.020 to do to cancel that and return the vehicle to depot. 00:47:28.020 --> 00:47:32.790 Otherwise, we would basically make the connection 00:47:32.790 --> 00:47:37.250 between the compatible trips, and consider 00:47:37.250 --> 00:47:44.310 the second trip a new arrival to the other [? extreme ?] 00:47:44.310 --> 00:47:46.570 [INAUDIBLE] basically. 00:47:46.570 --> 00:47:51.160 After that, we serve, again, all remaining unlinked trips 00:47:51.160 --> 00:47:53.428 from depot. 00:47:57.100 --> 00:48:02.650 We need to work out in your mind these steps once. 00:48:02.650 --> 00:48:05.980 It's very easy to practice, to realize how this works. 00:48:05.980 --> 00:48:09.230 But it's basically using the strict first in, 00:48:09.230 --> 00:48:14.890 first out rule, based on compatibility of trips 00:48:14.890 --> 00:48:17.012 and over the compatible trips. 00:48:17.012 --> 00:48:17.956 Is there any question? 00:48:17.956 --> 00:48:21.008 AUDIENCE: Kind of going back a little bit, [INAUDIBLE] 00:48:21.008 --> 00:48:23.468 the single depot [INAUDIBLE]? 00:48:28.890 --> 00:48:32.340 PROFESSOR: For single depot situation, 00:48:32.340 --> 00:48:35.820 the existing algorithms that solve minimum cost 00:48:35.820 --> 00:48:39.840 flow, like successive shortest path algorithm, or some 00:48:39.840 --> 00:48:42.870 of the negative cycle canceling algorithms, 00:48:42.870 --> 00:48:46.620 can give you integer solution. 00:48:46.620 --> 00:48:48.920 However, when it comes to multi-depot, 00:48:48.920 --> 00:48:52.550 and when it comes to multi-commodity flow, 00:48:52.550 --> 00:48:54.120 that's not always true. 00:48:54.120 --> 00:48:58.070 So you may actually get a solution with a linear-- 00:48:58.070 --> 00:49:03.710 or with a network flow algorithm that is in linear time, 00:49:03.710 --> 00:49:05.360 but it's not integer. 00:49:05.360 --> 00:49:12.380 So you may have you decimal, or partial vehicles going back 00:49:12.380 --> 00:49:14.060 to different depots. 00:49:14.060 --> 00:49:16.390 That's the difference between the two cases. 00:49:21.000 --> 00:49:23.210 So other questions? 00:49:23.210 --> 00:49:27.970 All right, so after this, we will have just one single route 00:49:27.970 --> 00:49:33.260 of scheduling practice, with two nodes, A and B, 00:49:33.260 --> 00:49:36.500 or with two terminals, A and B. This 00:49:36.500 --> 00:49:42.540 is one route that we want to schedule for AM peak period, 00:49:42.540 --> 00:49:43.820 and base period. 00:49:43.820 --> 00:49:47.030 AM peak is defined from 6:00 to 9:00. 00:49:47.030 --> 00:49:49.090 And base period is after 9:00. 00:49:49.090 --> 00:49:54.340 Headways are during peak period, 20 minutes. 00:49:54.340 --> 00:49:58.690 And scheduled trip times are 40 minutes for both directions. 00:49:58.690 --> 00:50:00.070 Layover is 10. 00:50:02.590 --> 00:50:05.550 During off-peak, headways are 30. 00:50:05.550 --> 00:50:07.540 Scheduled trip time is 35. 00:50:07.540 --> 00:50:09.910 And minimum layover is, again, 10. 00:50:09.910 --> 00:50:14.890 Dominant direction of travel is AM peak. 00:50:14.890 --> 00:50:17.860 And it's in the direction of A to B. 00:50:17.860 --> 00:50:21.460 So then, what we do is that we will [INAUDIBLE] direction A 00:50:21.460 --> 00:50:26.470 to B, and given the headways and running times, 00:50:26.470 --> 00:50:30.970 we work out the [INAUDIBLE] for all the trips 00:50:30.970 --> 00:50:34.210 that are scheduled during peak from 6:00 to 9:00, 00:50:34.210 --> 00:50:37.330 and then after that up to 11:00, and the rest 00:50:37.330 --> 00:50:39.100 could be actually similar. 00:50:39.100 --> 00:50:42.100 So as you can see, the headways from 6:00 to 9:00 00:50:42.100 --> 00:50:45.170 are 20 minutes, after that it becomes 30 minutes. 00:50:45.170 --> 00:50:50.760 The running time is 40 minutes in the peak, and 35 00:50:50.760 --> 00:50:51.440 [INAUDIBLE]. 00:50:51.440 --> 00:50:56.260 First thing that we do is that we tie the two directions 00:50:56.260 --> 00:50:57.400 together. 00:50:57.400 --> 00:51:01.780 This is the kind of coordination that we previously 00:51:01.780 --> 00:51:06.580 discussed with the comment that [? Aton ?] made 00:51:06.580 --> 00:51:10.700 in terms of shifting these trips together. 00:51:10.700 --> 00:51:15.520 So this is one example of shifting some of the trips 00:51:15.520 --> 00:51:18.260 to make better efficiency in the next steps. 00:51:18.260 --> 00:51:23.410 In this case, we tie the trips in the first direction 00:51:23.410 --> 00:51:25.460 with trips in the reverse direction. 00:51:25.460 --> 00:51:29.590 So first trip that departs at 6:00 arrives at 6:40. 00:51:29.590 --> 00:51:32.800 Considering 10 minutes layover, we 00:51:32.800 --> 00:51:36.580 will schedule the first trip for the reverse direction 00:51:36.580 --> 00:51:38.320 from 6:50. 00:51:38.320 --> 00:51:41.890 And it's anticipated to arrive at 7:30. 00:51:41.890 --> 00:51:44.830 And this is so on and so forth for all the timetable, 00:51:44.830 --> 00:51:50.040 given the travel times between the two ends, and the headways. 00:51:50.040 --> 00:51:52.060 AUDIENCE: Which means we need five buses? 00:51:52.060 --> 00:51:55.760 PROFESSOR: Yes, but you will work this out. 00:51:55.760 --> 00:52:00.310 Right, we need five buses, but some of them are partial only. 00:52:00.310 --> 00:52:04.326 They can pull in in the middle, and pull out. 00:52:04.326 --> 00:52:05.700 But we will see that in a second. 00:52:08.360 --> 00:52:12.880 So we will start with first pull-out-- 00:52:12.880 --> 00:52:16.900 that is the bus that is going to serve the first round trip. 00:52:16.900 --> 00:52:20.110 After one cycle, it's going to be 00:52:20.110 --> 00:52:24.370 able to serve the 7:40 trip for us, 00:52:24.370 --> 00:52:26.950 after 10 minutes of layover. 00:52:26.950 --> 00:52:30.070 It will end this trip at 9:10. 00:52:30.070 --> 00:52:32.110 Given 10 minutes layover, it will 00:52:32.110 --> 00:52:38.890 be able to serve 9:30 for us, and also 11:00 again. 00:52:38.890 --> 00:52:42.880 The next bus that we will have to pull out 00:52:42.880 --> 00:52:47.140 can serve the trip from 6:20, and from 8:00. 00:52:47.140 --> 00:52:51.580 After 8:00, once it does the cycle, it's going to be 9:30. 00:52:51.580 --> 00:52:54.070 And 9:30 is already-- 00:52:54.070 --> 00:52:55.810 trip is already taken. 00:52:55.810 --> 00:53:00.700 Then we will have to choose between sending 00:53:00.700 --> 00:53:07.450 this back to the depot, or using it for trip at 10:00. 00:53:07.450 --> 00:53:12.400 This trip at 10:00 basically means like 30 more minutes, 00:53:12.400 --> 00:53:14.770 or 20 minutes of waiting, we decide 00:53:14.770 --> 00:53:18.130 to send it back to depot. 00:53:18.130 --> 00:53:23.810 We will have five blocks for five different buses. 00:53:23.810 --> 00:53:26.620 Block one basically gives the itinerary 00:53:26.620 --> 00:53:32.320 of duties for bus one, or for whichever 00:53:32.320 --> 00:53:34.780 bus that this is assigned to. 00:53:34.780 --> 00:53:41.030 Then so on and so forth for all the buses. 00:53:41.030 --> 00:53:44.790 You mainly will have to work with this method 00:53:44.790 --> 00:53:47.790 for the homework. 00:53:47.790 --> 00:53:51.470 But it's valuable to know what are the [INAUDIBLE] 00:53:51.470 --> 00:53:55.920 approaches in practice, and what's 00:53:55.920 --> 00:54:01.200 going on in research that deals with problems like this. 00:54:01.200 --> 00:54:03.440 Any questions? 00:54:03.440 --> 00:54:05.550 All right, OK. 00:54:05.550 --> 00:54:07.650 Thank you very much.