Line Planning in Public Transportation Anita Schbel Institut fr - PowerPoint PPT Presentation

Cost-oriented models Cost-oriented models Anita Schöbel (NAM) 27. September 2006 13 / 78

Cost-oriented models Cost-oriented models Anita Schöbel (NAM) 27. September 2006 13 / 78

Cost-oriented models Simple cost model Contents Line planning in public transportation 1 Basic model 2 Cost-oriented models 3 Simple cost model Algorithms Extended cost model Customer-oriented models 4 Direct travelers approach Minimizing traveling time Other (more recent) models 5 Anita Schöbel (NAM) 27. September 2006 14 / 78

Cost-oriented models Simple cost model The costs of a line concept costs of a line concept ( L , f ) : � cost l l ∈L where the costs cost l of a line l ∈ L 0 depend on the frequency of l the length of l , the time needed for a complete trip along line l , and the costs per kilometer and per minute driving (Fixed costs are neglected.) Anita Schöbel (NAM) 27. September 2006 15 / 78

Cost-oriented models Simple cost model The costs of a line concept costs of a line concept ( L , f ) : � cost l l ∈L where the costs cost l of a line l ∈ L 0 depend on the frequency of l the length of l , the time needed for a complete trip along line l , and the costs per kilometer and per minute driving (Fixed costs are neglected.) → How can cost l be estimated ? Anita Schöbel (NAM) 27. September 2006 15 / 78

Cost-oriented models Simple cost model Estimating the costs of a line Notation cost time = time-dependent costs for running a vehicle (per minute) cost km = distance-dependent costs for running a vehicle (per km) Length l = the length of line l (in kilometers) Time l = time needed for driving a complete run of line l. Anita Schöbel (NAM) 27. September 2006 16 / 78

Cost-oriented models Simple cost model Estimating the costs of a line Notation cost time = time-dependent costs for running a vehicle (per minute) cost km = distance-dependent costs for running a vehicle (per km) Length l = the length of line l (in kilometers) Time l = time needed for driving a complete run of line l. The costs are dependent on the following questions: How many vehicles are necessary to run line l ? 1 How much time is spent by all vehicles serving line l within I ? 2 How many kilometers are driven by all vehicles on line l within I ? 3 Anita Schöbel (NAM) 27. September 2006 16 / 78

Cost-oriented models Simple cost model Estimating the costs of a line Notation cost time = time-dependent costs for running a vehicle (per minute) cost km = distance-dependent costs for running a vehicle (per km) Length l = the length of line l (in kilometers) Time l = time needed for driving a complete run of line l. The costs are dependent on the following questions: How many vehicles are necessary to run line l ? 1 How much time is spent by all vehicles serving line l within I ? 2 How many kilometers are driven by all vehicles on line l within I ? 3 Then: cost l = Time on line l ∗ cost time + Kilometers on line l ∗ cost km . Anita Schöbel (NAM) 27. September 2006 16 / 78

Cost-oriented models Simple cost model How many vehicles are needed? Special case: Time l = I . Then the number of vehicles needed for l is f l . Anita Schöbel (NAM) 27. September 2006 17 / 78

Cost-oriented models Simple cost model How many vehicles are needed? Special case: Time l = I . Then the number of vehicles needed for l is f l . For arbitrary Time l : number of vehicles on line l = ⌈ Time l · f l ⌉ I due to rule of proportion. Anita Schöbel (NAM) 27. September 2006 17 / 78

Cost-oriented models Simple cost model How many vehicles are needed? Special case: Time l = I . Then the number of vehicles needed for l is f l . For arbitrary Time l : number of vehicles on line l = ⌈ Time l · f l ⌉ I due to rule of proportion. Example: I = 60 min , Time l = 120 min , f l = 4 Anita Schöbel (NAM) 27. September 2006 17 / 78

Cost-oriented models Simple cost model How many vehicles are needed? Special case: Time l = I . Then the number of vehicles needed for l is f l . For arbitrary Time l : number of vehicles on line l = ⌈ Time l · f l ⌉ I due to rule of proportion. Example: I = 60 min , Time l = 120 min , f l = 4 = ⇒ 8 vehicles needed. Anita Schöbel (NAM) 27. September 2006 17 / 78

Cost-oriented models Simple cost model How much time is spent on line l ? total time spent on l within I number of vehicles on line l · I = Time l · f l = ∗ I I = f l Time l (neglecting the rounding in the number of vehicles) Anita Schöbel (NAM) 27. September 2006 18 / 78

Cost-oriented models Simple cost model How many kilometers are spent on line l ? kilometers on line within I l I number of vehicles on line l · Length l · = Time l Time l · f l I = · Length l · I Time l = f l Length l Anita Schöbel (NAM) 27. September 2006 19 / 78

Cost-oriented models Simple cost model Finally: The costs of line l Time on line l · cost time + Kilometers on line l · cost km cost l = = Time l f l cost time + Length l f l cost km = f l ( Time l cost time + Length l cost km ) � �� � =: cost l = f l cost l . Anita Schöbel (NAM) 27. September 2006 20 / 78

Cost-oriented models Simple cost model Finally: The costs of line l Time on line l · cost time + Kilometers on line l · cost km cost l = = Time l f l cost time + Length l f l cost km = f l ( Time l cost time + Length l cost km ) � �� � =: cost l = f l cost l . Note: cost l , defined by the equation above, does not depend on the frequency f l . Anita Schöbel (NAM) 27. September 2006 20 / 78

Cost-oriented models Simple cost model Cost-oriented model (LP1) (Cost-oriented line concept) Given a PTN, a set L 0 of potential lines, lower and upper frequencies f min ≤ f max for all e ∈ E , and parameters cost l for all l ∈ L 0 , find a e e feasible line concept ( L , f ) with minimal overall costs � cost ( L , f ) = f l cost l . l ∈L Anita Schöbel (NAM) 27. September 2006 21 / 78

Cost-oriented models Simple cost model Complexity of (LP1) Theorem (LP1) is NP-hard. Proof: Special case of (LP0) Anita Schöbel (NAM) 27. September 2006 22 / 78

Cost-oriented models Simple cost model Complexity of (LP1) Theorem (LP1) is NP-hard. Proof: Special case of (LP0) Note that (LP0) is trivially solvable without upper frequencies. What about this case in the cost model? Anita Schöbel (NAM) 27. September 2006 22 / 78

Cost-oriented models Simple cost model Complexity of (LP1) Theorem (LP1) is NP-hard. Proof: Special case of (LP0) Note that (LP0) is trivially solvable without upper frequencies. What about this case in the cost model? Theorem (LP1) is NP-hard, even without considering upper frequencies (i.e. without constraints (2)) and with cost l = 1 for all l ∈ L 0 and f min = 1 for e all e ∈ E. Anita Schöbel (NAM) 27. September 2006 22 / 78

Cost-oriented models Simple cost model Integer programming formulation variables: f l � min f l cost l l ∈L 0 � f min s.t. (1) f l ≥ ∀ e ∈ E e l ∈L 0 , e ∈ l � f max (2) ≤ ∀ e ∈ E f l e l ∈L 0 , e ∈ l ∈ f l IN 0 . Note: solution f l for all l ∈ L 0 , then line concept ( L , f ) is given through L = { l ∈ L 0 : f l > 0 } . Anita Schöbel (NAM) 27. September 2006 23 / 78

Cost-oriented models Simple cost model Cost model without upper frequencies Observations: (LP1) without upper frequencies is a multi covering problem . (MC) (Multi covering problem): Given an M × N -matrix A = ( a ij ) with elements a ij ∈ { 0 , 1 } , a vector b ∈ IN 0 M and some integer K , does there exist x ∈ IN 0 N with Ax ≥ b and � N n = 1 x n ≤ K ? Anita Schöbel (NAM) 27. September 2006 24 / 78

Cost-oriented models Algorithms Contents Line planning in public transportation 1 Basic model 2 Cost-oriented models 3 Simple cost model Algorithms Extended cost model Customer-oriented models 4 Direct travelers approach Minimizing traveling time Other (more recent) models 5 Anita Schöbel (NAM) 27. September 2006 25 / 78

Cost-oriented models Algorithms Algorithms General idea: for (LP1-Pool) without upper frequencies Basic idea: Let f min be the required frequency for edge e . e While there is an uncovered edge e (i.e. with f min > 0) 1 e Choose a line l , such that g ( l ) minimal 2 Choose frequency of line as f l := f ( l ) 3 Update f min accordingly 4 e Anita Schöbel (NAM) 27. September 2006 26 / 78

Cost-oriented models Algorithms Applying Dobson, 1982 While there is an uncovered edge e (i.e. with f min > 0) 1 e Choose a line l , such that g ( l ) minimal 2 Choose frequency of line as f l := f ( l ) 3 Update f min accordingly 4 e with c l g ( l ) = |{ e ∈ l : f min > 0 }| e f ( l ) = f l + 1 Anita Schöbel (NAM) 27. September 2006 27 / 78

Cost-oriented models Algorithms Applying Dobson, 1982 While there is an uncovered edge e (i.e. with f min > 0) 1 e Choose a line l , such that g ( l ) minimal 2 Choose frequency of line as f l := f ( l ) 3 Update f min accordingly 4 e with c l g ( l ) = |{ e ∈ l : f min > 0 }| e f ( l ) = f l + 1 Theorem HEU ≤ OPT H ( max l | l | ) , where H ( d ) = � d 1 i = 1 i Anita Schöbel (NAM) 27. September 2006 27 / 78

Cost-oriented models Algorithms Applying van Slyke/Xiaoming, 1984 While there is an uncovered edge e (i.e. with f min > 0) 1 e Choose edge e with maximal f min 2 e Choose a line l , covering e such that g ( l ) minimal 3 Choose frequency of line f l = f ( l ) 4 Update f min accordingly 5 e with g ( l ) = c l f ( l ) = f l + f min e Anita Schöbel (NAM) 27. September 2006 28 / 78

Cost-oriented models Algorithms Applying van Slyke/Xiaoming, 1984 While there is an uncovered edge e (i.e. with f min > 0) 1 e Choose edge e with maximal f min 2 e Choose a line l , covering e such that g ( l ) minimal 3 Choose frequency of line f l = f ( l ) 4 Update f min accordingly 5 e with g ( l ) = c l f ( l ) = f l + f min e Theorem HEU ≤ OPT max e ∈ E |L ( e ) | Anita Schöbel (NAM) 27. September 2006 28 / 78

Cost-oriented models Algorithms Line planning heuristic While there is an uncovered edge e (i.e. with f min > 0) 1 e Choose edge e with maximal f min 2 e Choose a line l , covering e such that g ( l ) minimal 3 Choose frequency of line f l = f ( l ) 4 Update f min accordingly 5 e with c l g ( l ) = |{ e ∈ l : f min > 0 f min > 0 }| min e ∈ l : f min e e e > 0 f min f ( l ) = f l + min e ∈ l : f min e e Anita Schöbel (NAM) 27. September 2006 29 / 78

Cost-oriented models Algorithms Two combinations While there is an uncovered edge e (i.e. with f min > 0) 1 e Choose edge e with maximal f min 2 e Choose a line l , covering e such that g ( l ) minimal 3 Choose frequency of line f l = f ( l ) 4 Update f min accordingly 5 e with c l g ( l ) = |{ e ∈ l : f min > 0 f min > 0 }| min e ∈ l : f min e e e f ( l ) = f l + 1 or c l g ( l ) = |{ e ∈ l : f min > 0 }| e f ( l ) = f l + min e ∈ l f min e Anita Schöbel (NAM) 27. September 2006 30 / 78

Cost-oriented models Algorithms Numerical results Data: Simulated data with 1000 stations, 4000 edges and 30, 60, 90, 120, . . . , 1350, 1500, 3000 possible lines Algorithms tested on 500 instances. Anita Schöbel (NAM) 27. September 2006 31 / 78

Cost-oriented models Algorithms Numerical results size 30 60 90 120 150 180 210 240 Heuristic Dobson 620 1470 2060 2720 3340 4090 4730 5640 6100 Sly/Xia 880 2640 4290 6210 8500 10680 13440 16330 17330 LiPla 650 1520 2150 2730 3490 4220 4980 5800 6430 Comb 1 610 1390 1950 2620 3190 3940 4610 5500 Comb 2 580 1380 1910 2640 3220 3910 4690 5450 6240 Anita Schöbel (NAM) 27. September 2006 31 / 78

Cost-oriented models Algorithms Numerical results 540 600 750 900 1200 1350 1500 3000 12810 14470 18520 22560 30760 35000 39960 85100 40980 47050 61240 75490 111030 125130 143430 318920 12990 14650 18430 22470 30800 35020 39890 85310 11130 12580 14080 18030 22020 30360 34680 84950 12730 14300 18330 22180 30380 34710 39180 84890 Anita Schöbel (NAM) 27. September 2006 31 / 78

Cost-oriented models Algorithms Results of heuristics Winner: The two combinations (where the first combination is slightly better) then Dobson and line planing heuristic Far away: van Slyke and Xiaoming Anita Schöbel (NAM) 27. September 2006 32 / 78

Cost-oriented models Extended cost model Contents Line planning in public transportation 1 Basic model 2 Cost-oriented models 3 Simple cost model Algorithms Extended cost model Customer-oriented models 4 Direct travelers approach Minimizing traveling time Other (more recent) models 5 Anita Schöbel (NAM) 27. September 2006 33 / 78

Cost-oriented models Extended cost model Extended cost model Evaluate costs of line concept Determine the lines and the frequencies Determine also the type of vehicle t = 1 , . . . , T which is used to run a line In rail transportation: determine also the number of cars of the trains for each line Anita Schöbel (NAM) 27. September 2006 34 / 78

Cost-oriented models Extended cost model Additional parameters ccost t fixed cost per car of type t = fix ccost t cost per kilometer with a car of type t = km vcost t cost per kilometer with a vehicle of type t = km capacity of a car of type t A t = c min min number of cars allowed for a train of type t = t c max max number of cars allowed for a train of type t = t Anita Schöbel (NAM) 27. September 2006 35 / 78

Cost-oriented models Extended cost model Recall . . . Number of vehicles needed to run line l is ⌈ f l Time l ⌉ . I Moreover, we define f max = max e ∈ E f max e as the upper bound over all upper frequencies. Anita Schöbel (NAM) 27. September 2006 36 / 78

Cost-oriented models Extended cost model Objective function variables: � 1 if line l is served by vehicles of type t with c cars X tc = . l 0 otherwise c max T t Time l � � � min ( ccost t c + Length l ccost t km c fix I t = 1 l ∈L 0 c = c min t + Length l vcost t km ) f l X tc l Anita Schöbel (NAM) 27. September 2006 37 / 78

Cost-oriented models Extended cost model But: A quadratic term f l X tc in the objective! l Solution: Substitute f l X tc by a new variable X tfc with l l if line l ∈ L 0 is served by vehicles of type t  1  X tfc with c cars and frequency f = l  0 otherwise Anita Schöbel (NAM) 27. September 2006 38 / 78

Cost-oriented models Extended cost model Extended cost model (LP2) c max f max T t fix · ⌈ Time l � � � � min ( ccost t Length l · ccost t · f ⌉ · c km · f · c + I l ∈L 0 t = 1 f = 1 c = c min t Length l · vcost t km · f ) · X tfc + l c max f max T t � � � � s.t. A t · f · c · X tfc (3) ≥ w e ∀ e ∈ E l t = 1 f = 1 l ∈L 0 : e ∈ l c = c min t c max f max T t � � � � f min f max f · X tfc (4) ≤ ≤ ∀ e ∈ E e l e l ∈L 0 : e ∈ l t = 1 f = 1 c = c min t c max f max t � � 1 ∀ l ∈ L 0 , ∀ t = 1 , . . . , T (5) X tfc ≤ l f = 1 c = c min t X tfc ∈ { 0 , 1 } (6) ∀ l , f , t , c l Anita Schöbel (NAM) 27. September 2006 39 / 78

Cost-oriented models Extended cost model Compare (LP1) with (LP2) The main changes of (LP2) compared to (LP1) are the following: The type of vehicles and the number of cars of the vehicles is determined within (LP2). The approximation of the costs is more detailed by having the new parameters. Constraints (3) are needed to make sure that all passengers can be transported. (LP2) is NP hard as special case of (LP1). Anita Schöbel (NAM) 27. September 2006 40 / 78

Cost-oriented models Extended cost model Solving (LP2) see Claessens, M.T. and van Dijk, N.M. and Zwaneveld, P .J., EJOR, 1996 get rid of many of the X tfc variables l use commercial IP-solver Anita Schöbel (NAM) 27. September 2006 41 / 78

Cost-oriented models Extended cost model Literature Claessens (1994), Zwaneveld, Claessens, van Dijk (1996), Claessens, van Dijk, Zwaneveld (1996), Bussieck and Zimmermann (1997), Claessens, van Dijk and Zwaneveld (1998), Goessens, Hoesel, and Kroon (2001, 2002), Bussieck, Lindner, and Lübbecke (2003) Anita Schöbel (NAM) 27. September 2006 42 / 78

Customer-oriented models Customer-oriented models Anita Schöbel (NAM) 27. September 2006 43 / 78

Customer-oriented models Customer-oriented models Anita Schöbel (NAM) 27. September 2006 43 / 78

Customer-oriented models Direct travelers approach Contents Line planning in public transportation 1 Basic model 2 Cost-oriented models 3 Simple cost model Algorithms Extended cost model Customer-oriented models 4 Direct travelers approach Minimizing traveling time Other (more recent) models 5 Anita Schöbel (NAM) 27. September 2006 44 / 78

Customer-oriented models Direct travelers approach Goal Notation Given a line concept ( L , f l ) . A customer does not change the line on his/her journey is called a direct traveler Goal of direct travelers approach: design the lines in such a way that as many customers as possible have a direct connection, i.e. maximize the number of direct travelers. Anita Schöbel (NAM) 27. September 2006 45 / 78

Customer-oriented models Direct travelers approach Preferable paths Assumption: Customers use preferable paths (e.g. shortest paths) which are known beforehand for each OD-pair i , j ∈ V × V . Notation P ij denotes a shortest (or preferable) path between station i and station j. P ij ⊆ l if there exists a shortest (or preferable) path between i and j which is completely contained in line l. Anita Schöbel (NAM) 27. September 2006 46 / 78

Customer-oriented models Direct travelers approach Example Graph with 5 nodes, all edge weights are 1. line l=(1,2,3,4,5) 5 4 3 l 2 1 Anita Schöbel (NAM) 27. September 2006 47 / 78

Customer-oriented models Direct travelers approach Example Graph with 5 nodes, all edge weights are 1. line l=(1,2,3,4,5) 5 4 3 l 2 1 P contained in line l 23 Anita Schöbel (NAM) 27. September 2006 47 / 78

Customer-oriented models Direct travelers approach Example Graph with 5 nodes, all edge weights are 1. line l=(1,2,3,4,5) 5 4 3 l 2 1 P contained in line l 14 Anita Schöbel (NAM) 27. September 2006 47 / 78

Customer-oriented models Direct travelers approach Example Graph with 5 nodes, all edge weights are 1. line l=(1,2,3,4,5) 5 4 3 l 2 1 P NOT contained in line l 15 Anita Schöbel (NAM) 27. September 2006 47 / 78

Customer-oriented models Direct travelers approach Formulation of objective Variables: d ijl for all i , j ∈ V and all lines l ∈ L 0 , denotes the number of direct travelers between i and j that use line l ∈ L 0 . Anita Schöbel (NAM) 27. September 2006 48 / 78

Customer-oriented models Direct travelers approach Formulation of objective Variables: d ijl for all i , j ∈ V and all lines l ∈ L 0 , denotes the number of direct travelers between i and j that use line l ∈ L 0 . Objective: � max d ijl i , j ∈ V , l ∈L 0 : P ij ⊆ l Anita Schöbel (NAM) 27. September 2006 48 / 78

Customer-oriented models Direct travelers approach Recall . . . A denotes the (fixed) capacity of a vehicle. W ij denotes the number of travelers between station i and station j . Anita Schöbel (NAM) 27. September 2006 49 / 78

Customer-oriented models Direct travelers approach Model (LP3) � � max d ijl l ∈L 0 i , j ∈ V : P ij ⊆ l � s.t. W ij for all i , j ∈ V (7) d ijl ≤ l ∈L 0 : P ij ⊆ l � A · f l for all e ∈ E , l ∈ L 0 (8) ≤ d ijl i , j ∈ V : e ∈ P ij ⊆ l � f min f max for all e ∈ E (9) ≤ f l ≤ e e l ∈L 0 : e ∈ l IN 0 for all i , j ∈ V , l ∈ L 0 ∈ d ijl , f l Anita Schöbel (NAM) 27. September 2006 50 / 78

Customer-oriented models Direct travelers approach Complexity of (LP3) Theorem (LP3) is NP-hard. Proof: For A and W ij sufficiently large, the feasible set of (LP3) coincides with (LP0), hence (LP3) is NP-hard. Anita Schöbel (NAM) 27. September 2006 51 / 78

Customer-oriented models Direct travelers approach Simplifying (LP3) Problem: O( | V | 2 · |L 0 | ) variables. Idea: aggregate the d ijl variables to D ij = � l ∈L 0 : P ij ⊆ l d ijl . Anita Schöbel (NAM) 27. September 2006 52 / 78

Customer-oriented models Direct travelers approach Simplifying (LP3) Problem: O( | V | 2 · |L 0 | ) variables. Idea: aggregate the d ijl variables to D ij = � l ∈L 0 : P ij ⊆ l d ijl . (LP3’) � max D ij i , j ∈ V s.t. W ij for all i , j ∈ V D ij ≤ � f l for all i , j ∈ V D ij ≤ A l ∈L 0 : P ij ⊆ l � f min f max for all e ∈ E ≤ ≤ f l e e l ∈L 0 : e ∈ l IN 0 for all i , j ∈ V , e ∈ E ∈ D ij , f e Anita Schöbel (NAM) 27. September 2006 52 / 78

Customer-oriented models Direct travelers approach Relation between (LP3) and (LP3’) Lemma The optimal objective value of (LP3’) is an upper bound for the optimal objective value of (LP3). Anita Schöbel (NAM) 27. September 2006 53 / 78

Customer-oriented models Direct travelers approach Relation between (LP3) and (LP3’) Lemma The optimal objective value of (LP3’) is an upper bound for the optimal objective value of (LP3). feas. sol. of (LP3’) feas. sol. of (LP3) I.e. (LP3’) is a relaxation of (LP3). Anita Schöbel (NAM) 27. September 2006 53 / 78

Customer-oriented models Direct travelers approach Solving (LP3’) see M.R. Bussieck and P . Kreuzer and U.T. Zimmermann, EJOR, 1996 further relax integrality constraints of the D ij variables of (LP3’) use a cutting-plane approach Anita Schöbel (NAM) 27. September 2006 54 / 78

Customer-oriented models Minimizing traveling time Contents Line planning in public transportation 1 Basic model 2 Cost-oriented models 3 Simple cost model Algorithms Extended cost model Customer-oriented models 4 Direct travelers approach Minimizing traveling time Other (more recent) models 5 Anita Schöbel (NAM) 27. September 2006 55 / 78

Customer-oriented models Minimizing traveling time Goal Observation: Customers choose a path P with minimal “traveling time” I ( P ) , including riding time on trains (proportional to length of trip) and time for transfers (dependent on number of transfers). Anita Schöbel (NAM) 27. September 2006 56 / 78

Customer-oriented models Minimizing traveling time Goal Observation: Customers choose a path P with minimal “traveling time” I ( P ) , including riding time on trains (proportional to length of trip) and time for transfers (dependent on number of transfers). Goal: Design the lines in such a way that the sum of all traveling times over all customers is minimal. Anita Schöbel (NAM) 27. September 2006 56 / 78

Customer-oriented models Minimizing traveling time Illustration Examle: Anita Schöbel (NAM) 27. September 2006 57 / 78

Customer-oriented models Minimizing traveling time Illustration Minimizing riding times . . . Anita Schöbel (NAM) 27. September 2006 57 / 78

Customer-oriented models Minimizing traveling time Illustration Minimize number of transfers . . . Anita Schöbel (NAM) 27. September 2006 57 / 78

Customer-oriented models Minimizing traveling time Objective function Idea: Take both efects into account! Inconvenience= k 1 · Riding Time + k 2 · number of transfers K 2 is an estimate for the average waiting time when changing (since timetable is not known in the phase of line planning). Anita Schöbel (NAM) 27. September 2006 58 / 78