Truck Driver Scheduling in Canada

Transcription

1 Truck Driver Scheduling in Canada Avin Goel 1,2, Loui-Martin Roueau 3 1 MIT-Zaragoza International Logitic Program, Zaragoza Logitic Center, Zaragoza, Spain, avin@mit.edu 2 Applied Telematic/e-Buine Group, Department of Computer Science, Univerity of Leipzig, Leipzig, Germany, avin.goel@uni-leipzig.de 3 École Polytechnique de Montréal and CIRRELT, Montréal, Québec H3C 3AT, Canada, loui-martin.roueau@cirrelt.ca December 8, 2010 Abtract Thi paper preent and tudie the Canadian Truck Driver Scheduling Problem (CAN-TDSP) which i the problem of determining whether a equence of location can be viited within given time window in uch a way that driving and working activitie of truck driver comply with Canadian Commercial Vehicle Driver Hour of Service Regulation. Canadian regulation comprie the proviion found in U.S. hour of ervice regulation a well a additional contraint on the maximum amount of driving and the minimum amount of off-duty time on each day. We preent two heuritic and an exact approach for olving the CAN-TDSP. Computational experiment demontrate the effectivene of our approache and indicate that Canadian regulation are ignificantly more permiive than U.S. hour of ervice regulation. 1 Introduction According to European Tranport Safety Council (2001), Federal Motor Carrier Safety Adminitration (2008), and Williamon et al. (2001), driver fatigue i a ignificant factor in approximately fifteen 1 Electronic copy available at:

2 to twenty per cent of commercial road tranport crahe. In Europe it i etimated that one out of two long haul driver ha fallen aleep while driving. One out of five long ditance road tranport driver in Autralia reported at leat one fatigue related incident on their lat trip and one out of three driver reported breaking road rule on at leat half of their trip. Many driver feel that fatigue i a ubtantial problem for the indutry and feel that their companie hould eae unreaonably tight chedule and hould allow more time for break and ret during their trip. According to a urvey conducted by McCartt et al. (2008), one out of ix truck driver admitted to have dozed at wheel in the month prior to the urvey and revealed that le than one out of two truck driver reported that delivery chedule are alway realitic. Truck driver who reported that they are frequently given unrealitic delivery chedule are approximately three time a likely to violate the work rule a driver who rarely or never have to deal with unrealitic delivery chedule. In their effort to increae road afety and improve working condition of truck driver, government in Autralia, Canada, Europe and the United State have independently adopted new regulation concerning driving and working hour of truck driver. Thee regulation impoe maximum limit on the amount of driving and working within certain time period and minimum requirement on the amount and duration of break and ret period which mut be taken by the truck driver. Compulory break and ret period have a ignificant impact on total travel time which are typically more than twice a long a the pure driving time required in long ditance haulage. Ignoring compulory break and ret period when generating chedule for truck driver can lead to unrealitic expectation, large delay, and or violation of driving and working hour regulation. Thi reult in poor working condition of truck driver, reduced road afety, and low cutomer atifaction. One of the firt reearch work explicitly conidering compulory break period within vehicle cheduling i preented by Savelbergh and Sol (1998) who conider a problem in which lunch break and night break mut be taken within fixed time interval. Hour of ervice rule impoed by the U.S. Department of Tranportation are firt tudied by Xu et al. (2003) who preent a column generation approach for combined vehicle routing and cheduling. Archetti and Savelbergh (2009) how that truck driver cheduling problem conidering U.S. hour of ervice regulation can be olved polynomial time. Goel and Kok (2010) preent an improved algorithm which can olve truck driver cheduling problem in the Unite State in quadratic time. Recently, everal work con- 2 Electronic copy available at:

3 idering the generation of truck driver chedule complying with European Union regulation have been preented. The firt method which i guaranteed to find a truck driver chedule complying with European Union regulation if uch a chedule exit i preented by Goel (2010). Goel (2009), Kok et al. (2010), and Precott-Gagnon et al. (2010) olve combined vehicle routing and truck driver cheduling problem in the European Union by heuritically determining truck driver chedule. To the bet of our knowledge Canadian Commercial Vehicle Driver Hour of Service Regulation have not yet been tudied. In thi paper we tudy and preent the Canadian Truck Driver Scheduling Problem, which i the problem of determining whether it i poible to chedule driving and working hour of truck driver in uch a way that a equence of location can be viited within given time window and that Canadian Commercial Vehicle Driver Hour of Service Regulation are complied with. We evaluate the performance of an exact cheduling algorithm and compare it with two heuritic approache which require ignificantly le computation time. Futhermore, we compare the impact on route feaibility of Canadian and United State regulation and analye the impact of the additional contraint found in Canadian regulation. The remainder of thi paper i organied a follow. Section 2 decribe Canadian Commercial Vehicle Driver Hour of Service Regulation which are dicued in Section 3. In Section 4 we preent the notation required in thi paper and give a mathematical formulation of the Canadian Truck Driver Scheduling Problem. In Section 5 we preent ome dominance criteria which help u in effectively tackling the Canadian Truck Driver Scheduling Problem. Section 6 preent a olution framework which i ued by the heuritic and exact olution approache preented in Section 7. In Section 8 we report on computational experiment demontrating the effectivene of our method. Finally, Section 9 give ome concluding remark. 2 Canadian Commercial Vehicle Driver Hour of Service Regulation Canadian regulation concerning driving and working hour of commercial vehicle are decribed in Tranport Canada (2005) and interpreted in Canadian Council of Motor Tranport Adminitrator (2007). In Canada two et of regulation exit, one of which applie to driving conducted outh of 3

4 latitude 60 N and one to driving north of latitude 60 N. In the remainder of thi paper we focu on the ubet of regulation applicable for driving outh of latitude 60 N becaue thi i the area of major economic concern. The regulation define on-duty time a the period that begin when a driver begin work and end when the driver top work and include any time during which the driver i driving or conducting any other work. Off-duty time i defined by any period other than on-duty time. The regulation impoe retriction on the maximum amount of on-duty time and the minimum amount of off-duty time during a day. According to the regulation a day mean a 24-hour period that begin at ome time deignated by the motor carrier. For implicity and w.l.o.g. let u aume in the remainder that thi time i midnight. The regulation demand that a driver mut not drive after accumulating 13 hour of driving time, after accumulating 14 hour of on-duty time, or after 16 hour of time have elaped ince the end of the lat period of at leat 8 conecutive hour of off-duty time. In any of thee cae the driver may only commence driving again after taking another period of at leat 8 conecutive hour of off-duty time. The regulation demand that a driver doe not drive for more than 13 hour in a day and that a driver accumulate at leat 10 hour of off-duty time in a day. At leat 2 of thee hour mut not be part of a period of 8 conecutive hour of off-duty time a required by the proviion decribed in the previou paragraph. However, if a period of more than 8 conecutive hour of off-duty time i cheduled, the amount exceeding the 8th hour may contribute to thee 2 hour. Period of le than 30 minute, in which the driver i neither driving nor working, do not count toward the minimum off-duty time requirement given by the regulation, even though they are conidered a off-duty time by the definition. The regulation impoe additional contraint on the amount of on-duty time within a period of 7 day and give ome extra flexibility in cheduling off-duty period. For the ake of conciene, however, we will not conider the correponding proviion of the regulation in the remainder of thi paper. 4

5 3 Dicuion Canadian Commercial Vehicle Driver Hour of Service Regulation hare ome imilaritie with U.S. hour of ervice regulation which are tudied in Xu et al. (2003), Archetti and Savelbergh (2009), and Goel and Kok (2010). In Canada a driver may accumulate 13 hour of driving time before taking an off-duty period of 8 conecutive hour, wherea in the United State a driver mut take 10 hour of conecutive off-duty time after accumulating 11 hour of driving time. After taking 10 hour of offduty time, a driver in the United State may commence driving again. Thu, a driver may accumulate a total of 14 hour of driving within a ingle day. In Canada the maximum amount of driving on a ingle day i limited to 13 hour. Canadian regulation enure that on any day at leat 10 hour of offduty time are taken. However, the minimum amount of continuou off-duty time i only 8 hour. In a tudy on leeping pattern of truck driver by Mitler et al. (1997), Canadian truck driver regularly drove up to 13 hour between ret period of 8 conecutive hour. The tudy revealed that on average driver leep le than 5 hour per day which i 2 hour le than the average ideal reported by the driver. More than half of the driver had at leat one ix-minute interval of drowine while driving within the five-day tudy. Thi indicate that the required amount of 8 hour continuou off-duty time may not be ufficient to guarantee that driver get enough leep. Undoubtedly, regular leep deficit may be a contributor to fatigue related road accident. Day 1 Day 2 DRIVE BREAK WORK DRIVE REST DRIVE WORK DRIVE WORK BREAK REST 8h 2h 1h 4h 8h 6h 1h 7h 1h 2h 8h Figure 1: Schedule complying with the regulation Figure 1 how a driver chedule complying with Canadian regulation. The driver begin with a driving period and reache the next cutomer location at 8.00 AM. Then, the driver take two hour of off-duty time before beginning to load the vehicle. After one hour of tationary work time, the driver continuou driving toward the next cutomer location. From 3.00 PM to PM the driver take a ret period. A the driver ha not yet reached the maximum allowed amount of driving on day 1, the 5

6 driver may continue to drive after the ret. After the end of the lat on-duty time, the driver mut take 10 hour of off-duty time to reach the minimum amount of off-duty time on day 2. Let u now aume that the chedule in Figure 1 i the planned chedule of a driver. If, for any reaon, the driver deviate from the plan and ret until PM on day 1, thirty minute of driving time are puhed from day 1 into day 2 and thirty minute of off-duty time are puhed from day 2 into day 3. Conequently, the driver doe not take the minimum amount of off-duty time required on day 2 and violate againt the regulation. We can ee that in thi example, taking additional off-duty time i actually penalied by the regulation and the driver may be tempted to take a hort ret period regardle of hi or her actual fitne for duty. Thi i unique to Canadian regulation and we believe that thi i not intended by the legilator. In the remainder of thi paper we will thu only conider chedule in which the duration of ret period can be extended by any amount without violating the regulation. 4 The Truck Driver Scheduling Problem In thi ection we preent a formal model of the Canadian Truck Driver Scheduling Problem which i the problem of viiting a equence of λ location within given time window in uch a way that driving and working activitie of truck driver comply with Canadian Commercial Vehicle Driver Hour of Service Regulation. Let u firt preent the notation ued throughout the paper and decribe under which condition a truck driver chedule complie with the regulation. The parameter impoed by the regulation are ummaried in Table 1. In order to repreent truck driver chedule, let u denote with DRIVE any on-duty time during which the driver i driving, with WORK any on-duty time during which the driver i not driving. Furthermore, let u denote with REST any period of 8 hour or more of off-duty time which reet the accumulated amount of driving, the accumulated amount of on-duty time, and the time elaped ince the lat period of 8 conecutive hour of off-duty time. Let u denote with BREAK any period contributing to the off-duty requirement which i not interpreted a ret period. Note that it i poible to take a BREAK period immediately before or after a REST period. If a BREAK period i taken immediately before or after a REST period, the duration of the BREAK period may be horter than 6

7 Notation Value Decription t ret 8 h The minimum duration of a ret period t drive 13 h The maximum accumulated driving time between two conecutive ret period and the maximum accumulated driving time on a day t on-duty 14 h The maximum accumulated on-duty time until a driver may drive between two conecutive ret period t elaped 16 h The maximum time ince the end of the off-duty period commencing with the lat ret period t day 24 h The duration of a day t off-duty 10 h The minimum amount of off-duty time on a day t break 2 h The minimum amount of off-duty time on a day which i not part of a ret period t length 1 2 h The minimum length of an off-duty period period to be counted Table 1: Parameter impoed by the regulation 1 2 hour, a both period can be interpreted a one continuou block of off-duty time. For the ame reaon, the time elaped ince the end of the lat period of 8 conecutive hour of off-duty time doe not increae if a BREAK period i taken immediately after a REST period. Let u denote with IDLE any other off-duty time of le than 1 2 hour duration which doe not contribute to the off-duty requirement of the regulation. Note, that idle period are only required between on-duty period, becaue the duration of a preceding or uceeding off-duty period could be increaed otherwie. A truck driver chedule can be pecified by a equence of activitie to be performed by the driver. Let A := a =(a type,a length ) a type {DRIVE, WORK, REST, BREAK, IDLE},a length > 0 denote the et of driver activitie to be cheduled. Let «.» be an operator which concatenate different activitie. Thu, a 1.a a k denote a chedule in which for each i {1, 2,...,k 1} activity a i+1 i performed immediately after activity a i. During concatenation the operator merge conecutive driver activitie of the ame type. That i, for a given chedule := a 1.a a k and an activity a with a type k = a type we have.a = a a k 1.(a type k,a length k + a length ). For a given chedule := a 1.a a k and 1 i k let 1,i := a 1.a a i denote the partial chedule compoed of activitie a 1 to a i. We aume that at the beginning of the planning horizon, the driver return from 7

8 a ret period which i long enough uch that previou driving and working activitie do not have any influence on the driving and working hour within the planning horizon. We will thu only conider chedule := a 1.a a k which begin with a ret period, i.e. a type 1 = REST. For a given chedule := a 1.a a k with a type 1 = REST let l end denote the completion time of the chedule, let l drive and let l on-duty denote the accumulated driving time ince the end of the lat ret period, denote the accumulated on-duty time ince the end of the lat ret period, and let l idle denote the accumulated idle time ince the end of the lat ret period. Thee value can be recurively computed during chedule generation by etting l end 1,1 and := a length 1, l drive 1,1 := 0, l on-duty 1,1 := 0, l idle 1,1 := 0, l drive.a := l on-duty.a := l idle.a := l end.a := l end + a length, 0 if a type = REST l drive + a length ele if a type = DRIVE l drive ele 0 if a type = REST l on-duty + a length ele if a type {DRIVE, WORK} l on-duty ele 0 if a type = REST l idle + a length ele if a type = IDLE l idle ele The retriction that a driver mut not drive after accumulating 13 hour of driving time and that a driver mut not drive after accumulating 14 hour of on-duty time are atified if l drive 1,i t drive for any 1 <i k with a type i = DRIVE and l on-duty 1,i t on-duty for any 1 <i k with a type i = DRIVE 8

9 Let l elaped denote the amount of time which ha elaped ince the end of the off-duty period commencing with the lat ret period. Thi value can be recurively computed during chedule generation by etting l elaped 1,1 := 0 and l elaped.a := 0 if a type = REST 0 if a type = BREAK and l elaped =0 l elaped + a length ele Note, that we can append any amount of break to a ret period without increaing l elaped. Furthermore, note that for notational reaon, we do not reet l elaped, if a break period of 8 hour or more i cheduled. In order to reet l elaped, a ret period can be cheduled intead of a break period of 8 hour or more. A chedule atifie the retriction that no driving i conducted after 16 hour of time have elaped ince the end of the off-duty period commencing with the lat ret period if l elaped 1,i t elaped for any 1 <i k with a type i = DRIVE We mut aure that ret activitie have the minimum duration required by the regulation, i.e. Let l length a length i t ret for any 1 <i k with a type i = REST denote the minimum duration a break period which i appended to chedule mut have to be counted a off-duty time. Thi value can be recurively computed during chedule generation by etting l length 1,1 := 0 and l length.a 0 if a type {REST, BREAK} := t length ele Each break period that contribute to the 2 hour of off-duty time during a day mut have a duration of at leat 30 minute or mut precede or ucceed a ret period, i.e. each chedule mut atify l drive d a length i l length 1,i 1 for any 1 <i<kwith a type i = BREAK,a type i+1 = REST Let u aume that the planning horizon begin at day 1 and end at the day denoted by d max. Let denote the accumulated driving time on day d. Thi value can be recurively computed for all d {1, 2,...,d max } by etting l drive d 1,1 := 0 and l drive d.a := l drive d l drive d if a type = DRIVE or l.a end (d 1) t day or l end d t day +min{d t day,l end.a } max{(d 1) t day,l end } ele 9

10 A chedule atifie the retriction that no driving i conducted after accumulating 13 hour of driving during a day if l drive d t drive for any d {1, 2,...,d max } Let l off-duty d denote the accumulated off-duty time on day d. Thi value can be recurively computed for all d {1, 2,...,d max } by etting l off-duty d 1,1 l.a off-duty d := l off-duty d l off-duty d := 0 and if a type {BREAK, REST} or l.a end (d 1) t day or l end d t day +min{d t day,l end.a } max{(d 1) t day,l end } ele A chedule atifie the retriction that at leat 10 hour of off-duty time are cheduled during a day if l off-duty d t off-duty for any d {1, 2,...,d max } Let l break d denote the accumulated off-duty time on day d which i not part of a ret period. Thi value can be recurively computed for all d {1, 2,...,d max } by etting l break d 1,1 l.a break d := l break d l break d := 0 and if a type = BREAK or l.a end (d 1) t day or l end d t day +min{d t day,l end.a } max{(d 1) t day,l end } ele A chedule atifie the retriction that at leat 2 hour of off-duty time are cheduled during a day which are not part of a ret period if l break d t break for any d {1, 2,...,d max } A we only conider chedule in which the duration of ret period can be extended without violating the regulation, we demand that l on-duty 1,i + l idle 1,i t day t off-duty for any 1 <i k Under thi condition ret period can be extended by any value while leaving enough time for the required off-duty time on the ubequent day. Let u now give formal model of the Canadian Truck Driver Scheduling Problem. Let u conider a equence of location denoted by n 1,n 2,...,n λ which hall be viited by a truck driver. At each location n µ ome tationary work of duration w µ hall be conducted. Thi work mut be a continuou period which hall begin within a time window denoted by [t min µ,t max µ ]. We aume that n 1 correpond 10

11 to the driver current location and that the driver complete her or hi work week after finihing work at location n λ. The work to be conducted at location n 1 and n λ can include loading and unloading activitie a well a time for getting ready or cleaning the vehicle. The (poitive) driving time required for moving from node n µ to node n µ+1 hall be denoted by δ µ,µ+1. Let u aume that all value repreenting driving time, working time, and time window are a multiple of 15 minute. For a given equence of location n 1,n 2,...,n λ and a chedule = a 1.a a k with a type 1 = REST, let u denote with i(µ) the index correponding to the µth tationary work period, i.e. a i(µ) correpond to the work performed at location n µ. The Canadian Truck Driver Scheduling Problem (CAN-TDSP) i the problem of determining whether a chedule := a 1.a a k with a type 1 = REST exit which atifie a length i 1 j k a type =WORK j 1=λ (1) a length i(µ) = w µ for each µ {1, 2,...,λ} (2) t min µ l end 1,i(µ) 1 t max µ for each µ {1, 2,...,λ} (3) a length j = δ µ,µ+1 for each µ {1, 2,...,λ 1} (4) i(µ) j i(µ+1) a type =DRIVE j l on-duty 1,i l elaped 1,i a length i l drive 1,i t drive for any 1 <i k (5) t on-duty for any 1 <i k with a type i = DRIVE (6) t elaped for any 1 <i k with a type i = DRIVE (7) t ret for any 1 <i k with a type i = REST (8) l length 1,i 1 for any 1 <i<kwith a type i = BREAK,a type i+1 = REST (9) l drive d t drive for any d {1, 2,...,d max } (10) l off-duty d t off-duty for any d {1, 2,...,d max } (11) l break d t break for any d {1, 2,...,d max } (12) l on-duty 1,i + l idle 1,i t day t off-duty for any 1 <i k (13) 11

12 Condition (1) demand that the number of work activitie in the chedule i λ. Condition (2) demand that the duration of the µth work activity matche the pecified work duration at location n µ. Condition (3) demand that each work activity begin within the correponding time window. Condition (4) demand that the accumulated driving time between two work activitie matche the driving time required to move from one location to the other. Condition (5) to (12) are the contraint impoed by the regulation. violating the regulation. Condition (13) guarantee that ret period can be extended without In the remainder of thi paper, we will ay that a chedule := a 1.a a k with a type 1 = REST i feaible if it atifie condition (1) to (13). The olution approache preented in thi paper will generate partial chedule which are iteratively modified in order to find a feaible chedule for the CAN-TDSP. All partial chedule generated in the olution approache will atify condition (1) to (4) given that λ i replaced with λ, where λ i the number of work activitie in the partial chedule. Furthermore, the accumulated driving time ince the lat work activity in the partial chedule will not exceed δ λ,λ +1. Let u ay that the day d {1, 2,...,d max } i the current day of a partial chedule if (d 1) t day l end <d t day. A partial chedule complie with the regulation if condition (5) to (9) are atified, condition (10) to (12) are atified for all day prior to the current day, and for the current day condition (10) i atified and condition (11) and (12) can be atified by appending a break period of ufficient duration. In the remainder of thi paper we will ay that a partial chedule i feaible if above condition are atified. 5 Dominance In general there are too many different alternative feaible partial chedule to be conidered in order to fully enumerate the earch pace uing the «.» operator. In thi ection we preent dominance criteria that help u reducing the number of partial chedule that need to be conidered when olving the CAN-TDSP. For thi, let u firt define ome additional operator which allow for inerting break activitie into the chedule. Thee operator allow u to only append ret and break period of minimal duration to a chedule becaue we can increae the off-duty time later on. 12

13 Let «R» be an operator which inert a break activity of given duration after the lat activity of type REST and let «B» be an operator which inert a break activity of given duration after the lat activity of type BREAK. If no break i cheduled after the lat ret period, «B» inert the break after the lat activity of type REST. That i, if i i the index of the lat ret activity and j i the index of the lat break or ret activity in chedule := a 1.a a k and > 0, then we have R = a 1.a a i.(break, ).a i a k. and B =a 1.a a j.(break, ).a j a k. Inerting a break activity after the lat ret or break activity can reult in a violation of time window contraint. Furthermore, inerting a break activity after the lat break could puh the end of ome driving activity to a value exceeding the maximum time that may elape after the end of the lat off-duty period commencing with the lat ret activity. For any chedule := a 1.a a k, let u denote with µ() the index of the next work location to be viited. Let l puh R and l puh B denote the maximum amount by which activitie ucceeding the break activity which i inerted by «R» or «B» may be puhed into the future without violating time window contraint or condition (7). Thee value can be recurively computed during chedule generation by etting l puh R 1,1 :=, if a type = REST l.a puh R := min{l puh R,t max µ() lend } if a type = WORK l puh R ele and l puh B 1,1 :=, l puh B.a := if a type {REST, BREAK} min{l puh B,t max µ() lend } if a type = WORK min{l puh B,t elaped l.a elaped } if a type = DRIVE l puh B ele A partial chedule obtained by applying R or B i feaible if and only if l puh R or l puh B. If a feaible chedule exit for an intance of the contrained CAN-TDSP, we can efficiently determine thi chedule by applying a equence of operator move to the initial chedule := (REST,t ret ). Here, each operator move either append an activity to the chedule, inert a break period after the lat ret period, or inert a break period after the lat break period. 13

14 Let u now conider two feaible partial chedule and. If for any feaible chedule that can be generated by applying a equence of operator move to chedule, we can generate a feaible chedule by applying a equence of operator move to chedule, then dominate. Lemma 1 Let and be feaible partial chedule which have the ame amount of accumulated on-duty time. Furthermore, let u denote with d the current day of chedule. If l end l drive l off-duty d l drive, l on-duty l off-duty d, l puh R l on-duty, l idle l puh R, l puh B l idle, l elaped l puh B l elaped, l drive d l drive d, l break d = l end, l break d,, and if for all [0,l elaped ] the accumulated amount of break time in the lat minute of chedule i at leat a high a the correponding value for chedule and the accumulated amount of driving time in the lat minute of chedule i not higher than the correponding value for chedule, then dominate. Lemma 2 Let and be feaible partial chedule which have the ame amount of accumulated onduty time. Furthermore, let u denote with d the current day of chedule. If l end + max{0,t break l break d } + t off-duty l end, and if l end d t day or l drive d l drive d, then dominate chedule. The proof of thee lemmata can be found in the Appendix. Note, that we can ue Lemma 1 to derive other dominance criteria. For example, under certain condition we can how dominance of a chedule over a chedule if l end and l puh R l end l end min{l end <l end. If l end +min{l end l end,l elaped l end,l elaped l elaped } l end l elaped }, we can et ŝ :=.(BREAK, min{l end l end,l elaped l elaped }) and := ŝ R (l end l end ŝ ). If the condition of Lemma 1 hold for and, then dominate. The dominance criteria can not only be ued to dicard ome of the partial chedule which are generated throughout the olution proce, but they can alo be ued to derive ome guideline allowing u to develop efficient olution approache which only generate the mot promiing partial chedule. Firt, if it i poible to drive or work no off-duty period hould be cheduled. If neceary, the off-duty period can be appended after the driving or working period or a break can later be inerted after the lat ret or break period in the chedule. Second, a ret period hould only be cheduled if l elaped > 0 and every ret period hould have a duration of exactly t ret, becaue otherwie we can chedule break time intead. Third, the amount of break time cheduled hould be a mall a poible, becaue we can append or inert additional break time later. The olution framework 14

15 preented in the next ection ue thee guideline to minimie the number of chedule which are generated. 6 Solution Framework In thi ection we preent a olution framework for olving the Canadian Truck Driver Scheduling Problem which take a et of feaible chedule for a partial tour n 1,n 2,...,n µ and extend each chedule to contruct feaible chedule for tour n 1,n 2,...,n µ,n µ+1. Let S µ denote the et of feaible chedule found for the partial tour n 1,n 2,...,n µ. In the beginning of our olution approach we et and {(REST,t ret ).(WORK,w 1 )} S 1 := if t min 1 t ret {(REST,t ret ).(BREAK,t min 1 t ret ).(WORK,w 1 )} if t min 1 >t ret S µ := for all 1 <µ λ. We et µ := 1 and determine S µ+1. Thi proce i repeated with µ := µ +1until the CAN-TDSP for tour n 1,n 2,...,n λ i olved. The cheduling framework i compoed of two part: the firt part chedule all activitie on the trip from node n µ to node n µ+1 ; the econd part chedule the tationary activitie after the (phyical) arrival at location n µ+1. The method for cheduling activitie on a trip from node n µ to node n µ+1 i illutrated in Figure 2. In the method we denote with δ the remaining driving time required to reach the next location n µ+1. The method tart by initialiing the et of partial chedule S which i et to S µ and the initially empty et S µ+1 of chedule in which no further driving i required to reach location n µ+1. Then, it chooe a partial chedule Sand remove it from S. The method determine the amount of break till required on the current day d by etting break := max{0,t break l break d }, the amount of off-duty time till required on the current day d by etting off-duty := max{0,t off-duty l off-duty d }, and the maximum amount of driving that can be appended to by etting drive := min{δ,t drive l drive,t on-duty l on-duty,t elaped l elaped,t day t off-duty l on-duty l idle }. 15

16 S := S µ, S µ+1 := [ele] [S = ] chooe S and et S := S \{} et d to current day of break off-duty drive := max{0,t break l break d } := max{0,t off-duty l off-duty d } := min {δ,t drive l drive,t on-duty l on-duty,t elaped l elaped,t day t off-duty l on-duty If break > 0 then drive := min{ drive,d t day l end break } If off-duty > 0 then drive := min{ drive,d t day l end off-duty } If t drive l drive d <d t day l end then drive := min{ drive,t drive l drive d } l idle } [ele] [ drive > 0] :=.(DRIVE, drive ) [ele] [ele] l elaped l end [δ =0] > 0, ( break =0or + t ret + break d t day ) S µ+1 := S µ+1 {} S := S {.(REST,t ret )} l drive l on-duty l elaped l on-duty +l idle = t drive or t on-duty or t elaped or =t day t off-duty [ele] [ele] [ele] [ele] lelaped l end > 0, ( break > 0and + t ret + break >d t day ) [ break > 0] [ off-duty > 0] S := S {.(BREAK, break ).(REST,t ret )} addoffduty(s,, break ) addoffduty(s,, off-duty ) wait(s,,d t day l end ) Figure 2: Method for cheduling activitie on a trip from n µ to n µ+1 16

17 If break > 0 and/or off-duty > 0, ome additional off-duty time i required on the current day and the method et drive := min{ drive,d t day l end l end break } and/or drive := min{ drive,d t day off-duty } to guarantee that enough time for taking the required amount of off-duty remain. If t drive l drive d <d t day l end, then the method et drive := min{ drive,t drive l drive d } to guarantee that the accumulated driving time on the current day will not exceed t drive. If drive > 0, the method append a driving period of duration drive to the chedule. Any other chedule that could be generated by applying ome other operator move would be dominated by thi chedule generated. If δ =0after cheduling the driving activity, the next location i reached and the chedule i included to the et S µ+1. If drive 0 or δ > 0 after cheduling the driving activity, ome off-duty time mut be added to chedule before another driving activity may be cheduled. If l elaped > 0, ome on-duty time ha been cheduled ince the lat ret period. Thu, it may be beneficial to reinitialie l drive Thi, however, can only be made if break =0or l end, l on-duty, l ilde, and l elaped by appending a ret period to the chedule. + t ret + break d t day, becaue otherwie the accumulated amount of break time during day d would not be achieved. If break l end > 0 and + t ret + break >d t day, we mut chedule the required amount of break before we can append a ret period to the chedule. The cheduling method generate a new chedule by appending a ret period which, if neceary, i preceded by a break period of appropriate length and add the chedule to the et S. If l drive = t drive, l on-duty t on-duty, l elaped t elaped, or l on-duty + l idle = t day t off-duty a ret period i required before further driving can be conducted. A there i no better way to generate a chedule by adding a ret period a decribed above, the method continue with the next loop. Otherwie, we have break = d t day l end, off-duty = d t day l end, or l drive d = t drive. We can modify the chedule in uch a way that thee contraint are no longer binding without cheduling a ret period. If break > 0 or off-duty > 0 the method determine new chedule by increaing the amount of off-duty time on day d. Poible implementation of the method addoffduty(,, ) which i ued to increae the amount of off-duty time by break or off-duty are decribed in the next ection. After increaing the accumulated amount of off-duty time the method continuou with the next loop. If break =0and off-duty =0we have l drive d = t drive. Thu, no further driving on the current day d i allowed. The method generate new chedule by increaing the completion time of the 17

18 chedule to at leat d t day uing the method wait(,, ). Poible implementation of thi method are alo decribed in the next ection. After that, the method continuou with the next loop. The econd part of the cheduling framework, i.e. the method for cheduling tationary activitie at node n µ+1, i illutrated in Figure 3. The method tart by initialiing the et of partial chedule S which i et to Sµ+1. Then, it remove all chedule from S which have a completion time exceeding the time window of n µ+1. If S i empty after removing thee chedule, the method prematurely terminate becaue no feaible chedule can be found. Otherwie, it chooe a partial chedule S and remove it from S. The method determine the amount of break till required on the current day d by etting break := max{0,t break l break d }, and the amount of off-duty time till required on the current day d by etting off-duty := max{0,t off-duty l off-duty d }. If l end t min µ+1, lon-duty + l idle + w µ+1 t day t off-duty and if break =0or l end + w µ+1 + break d t day and if off-duty =0 or l end + w µ+1 + off-duty d t day, the method add the chedule.(work,w µ+1 ) to the et S µ+1 and continue with the next loop a all other chedule that could be generated by applying different operator move to would be dominated by thi chedule. Otherwie, ome off-duty time i required before the work activity can be appended to the chedule. The off-duty time required may be a ret period, a break period, an idle period, or a combination of thee period. If l elaped =0, no on-duty time ha been cheduled ince the lat ret period and there i no reaon to add another ret period to the chedule. Otherwie, we may append a ret period to the chedule if break =0or l end +t ret + break d t day. If l elaped > 0, break > 0, and l end +t ret + break >d t day, we mut chedule the required amount of break before we may append a ret period to the chedule. The cheduling method generate a new chedule by appending a ret period which, if neceary, i preceded by a break period of appropriate length and add the chedule to the et S. If l on-duty + l idle + w µ+1 >t day t off-duty, or if l elaped > 0 and l end + break + t off-duty t min µ+1, no better way to increae the completion time of the chedule exit and the method continue with the next loop. If break > 0 and either t min µ+1 d tday or l end + w µ+1 + break >d t day we mut add break of break time to the chedule before the work activity can be cheduled. If off-duty > 0 and either t min µ+1 d tday or l end + w µ+1 + off-duty > d t day we mut add off-duty of off-duty time to 18

19 S := S µ+1 S := { S l end t max µ+1 } [ele] [S = ] chooe S and et S := S \{} et d to current day of break off-duty := max{0,t break l break d } := max{0,t off-duty l off-duty d } [ele] [ele] l on-duty l end t min µ+1 and + l idle + w µ+1 t day t off-duty and ( break =0or l end + w µ+1 + break d t day ) and ( off-duty =0or l end + w µ+1 + off-duty d t day ) l elaped l end > 0, ( break =0or + t ret + break d t day ) S µ+1 := S µ+1 {.(WORK,w µ+1 )} S := S {.(REST,t ret )} l on-duty + l idle + w µ+1 > [ele] t day t off-duty or (l elaped > 0and + break +t off-duty t min [ele] l end µ+1 ) [ele] [ele] lelaped l end > 0, ( break > 0and + t ret + break >d t day ) break > 0and(t min µ+1 d t day or l end + w µ+1 + break >d t day ) off-duty > 0and(t min µ+1 d t day or l end + w µ+1 + off-duty >d t day ) S := S {.(BREAK, break ).(REST,t ret )} addoffduty(s,, break ) addoffduty(s,, off-duty ) wait(s,,t min µ+1 lend ) Figure 3: Method for cheduling tationary activitie after arrival at node n µ+1 19

20 the chedule before the work activity can be cheduled. In either cae the method generate new chedule by increaing the amount of off-duty time uing the method addoffduty(,, ). After increaing the accumulated amount of off-duty time the method continue with the next loop. If neither of thee cae hold, the method generate new chedule by increaing the completion time of the chedule to t min µ+1 uing the method wait(,, ). After that the method continue with the next loop. In each iteration we remove dominated chedule from S µ in order to reduce the computational effort required by the cheduling method. 7 Solution Approache In thi ection we preent olution approache uing the framework preented in the previou ection. The firt olution approach i a heuritic in which the method addoffduty(s,, ) and wait(s,, ) imply add a chedule.(break, max{,l length }) to the et S. If l elaped =0, every other chedule that could be generated would be dominated by.(break, ). l elaped However, if > 0 the time elaped ince the end of the off-duty period commencing with the lat ret period increae when appending a break period. Thu, it may be beneficial to inert a break after the lat ret period, intead of appending a break period. Figure 4 illutrate another heuritic method for addoffduty(s,, ) which can generate two alternative chedule: the firt chedule i the chedule which i obtained by appending a break period; the econd chedule i a chedule obtained by inerting off-duty time after the lat ret period. The method firt determine the current day d of chedule. If l end + >d t day then it i impoible to increae the amount of off-duty time on day d by the required amount and the method terminate. Otherwie, the method generate a new chedule by appending a break activity of duration max{,l length } to chedule and include the new chedule in the et S. If l elaped method terminate. l puh R =0, there i no better way to increae the amount of off-duty time and the Otherwie, it et := and := and trie to inert off-duty time after the lat ret. If =0or l end + >d t day the method terminate becaue either it cannot inert off-duty time after the lat ret or there i not enough time to add enough off-duty time on the current day. 20

21 et d to current day of [ele] [l end + >d t day ] S := S {.(BREAK, max{,l length })} [ele] [l elaped =0] := et := and := l puh R =0or + d t day l end R if l puh R R l puh R ele if l puh R R ( l length ) ele := l off-duty d + l off-duty d > 0,l puh R (l puh R > 0, or >l length ) [ele] l length S := S { =0} {.(BREAK, ) > 0, l length } Figure 4: Heuritic method addoffduty(s,, ) Otherwie, it inert a break period of duration after the lat ret if l puh R. If l puh R < it i not poible to inert the required amount of off-duty time. The method inert a break period of duration l puh R duration l length after the lat ret if l puh R l length. Otherwie, it inert a break period of The method update to the remaining amount of off-duty time that need to be added. Note that inerting off-duty time after the lat ret may not alway increae the amount of off-duty time on the current day, e.g. if the lat ret end on the previou day and the lat activity on the previou day i a driving period, then inerting a break after the lat ret might imply puh the driving period into 21

22 the current day. If > 0 further off-duty time mut be added on the current day. If l puh R > or >l length and l puh R > 0, the method repeat the previou tep in order to increae the amount of off-duty time on the current day. Otherwie, the method inert to the et S if no further off-duty time i required, or it inert.(break, ) to the et S if a poitive amount of off-duty time mut till be added and l length. S := S {.(BREAK, max{,l length })} [ele] [l puh R > 0] [l elaped =0] := R min{,l puh R := min{,l puh R } } S := S { =0} {.(BREAK, ) > 0, l length } {.(IDLE, ) > 0, <l length } Figure 5: Heuritic method wait(s,, ) Figure 5 illutrate a imilar heuritic method for wait(s,, ). The method firt generate a new chedule by appending a break activity of duration max{,l length } to chedule and include the new chedule in the et S. If l elaped =0, there i no better way to increae the completion time of the chedule and the method terminate. Otherwie, it inert a break period of duration min{,l puh R } after the lat ret in the chedule and update the remaining amount by which the completion time mut be increaed. The method inert to the et S if the completion time i ufficiently increaed, inert.(break, ) to the et S if the completion time mut till be increaed by a poitive amount of l length, or inert.(idle, ) to the et S if the completion time mut till be increaed by a poitive amount of <l length. Thereafter, the method terminate. The heuritic approache preented above only generate chedule by appending off-duty time to a chedule or by trying to inert a much off-duty time after the lat ret a poible. In order to 22

23 olve the CAN-TDSP, however, every alternative of modifying a chedule which doe not reult in a chedule dominated by other need to be generated. Figure 6 illutrate an enumerative method for addoffduty(s,, ) which generate all alternative non-dominated chedule in which the amount of off-duty time on the current day i higher than in chedule. The enumerative method tart with the ame tep a the heuritic method illutrated in Figure 4. After etting := and := it iteratively inert break period of 1 4 hour after the lat ret or break in order to generate all alternative non-dominated chedule. If <l length and l puh B, the method generate a new chedule := l off-duty d +. If l off-duty d break period of duration l length If l puh R < 1 4 <l off-duty d B and add thi chedule to S if l off-duty d = + then thi new chedule i dicarded becaue appending a would be the better alternative. the method terminate becaue all non-dominated alternative have been generated. Otherwie, the method et := be added to. If l off-duty d = l off-duty d R 1 4 and update to the remaining off-duty time that mut + then i added to S. If =0or l end + >d t day the method terminate, becaue no other non-dominated chedule can be generated. Otherwie, the method generate a new chedule by appending a break activity of duration to chedule and include the new chedule in the et S if l length. Then, the method continue with the next iteration. Figure 7 illutrate an enumerative method for wait(s,, ) which generate all non-dominated alternative chedule which can be generated in order to increae the completion time of chedule by a poitive value. The method firt generate a new chedule by appending a break activity of duration max{,l length } to chedule and include the new chedule in the et S. If l elaped =0, there i no better way to increae the completion time of the chedule and the method terminate. Otherwie, it et := and := and iteratively inert break period of 1 4 hour until all non-dominated alternative chedule are determined. If 0 < <l length then a 1 4 hour mut till be added to the completion time. A the minimum duration of a break period to be appended to the chedule i 1 2 hour, it could be beneficial to inert ome break time after the lat break in the chedule. If furthermore l puh B the method generate a new chedule by inerting a break period of 1 4 hour 23

24 et d to current day of [ele] [l end + >d t day ] S := S {.(BREAK, max{,l length })} [ele] [l elaped =0] et := and := [ele] [ <l length, l puh B ] := B S := S { l off-duty d = l off-duty d + } [ele] [l puh R < 1 4 ] := R 1 4, := l off-duty d + l off-duty d S := S { l off-duty d [ele] = l off-duty d + } l end =0or + d t day S := S {.(BREAK, ) l length } Figure 6: Enumerative method addoffduty(s,, ) 24

25 S := S {.(BREAK, max{,l length })} [ele] [l elaped =0] et := and := [ele] [0 < <l length ] S := S { B 1 4 lpuh B 1 4 } {.(IDLE, 1 4 ) lon-duty + l idle <t day t off-duty } [ele] [l puh R < 1 4 ] := R 1 4, := 1 4 S := S { l end = lend + } [ele] [ =0] S := S {.(BREAK, ) l length } Figure 7: Enumerative method wait(s,, ) after the lat break in and add the new chedule to S. If l on-duty + l idle <t day t off-duty the method generate a new chedule by appending an idle period of 1 4 hour to and add the new chedule to S. If l puh R < 1 4 the method terminate becaue all non-dominated alternative have been generated. Otherwie, the method et := R 1 4 and decreae by 1 4. If lend = l end + then the updated chedule i added to S. If =0then the completion time ha been ufficiently increaed and the method terminate. If l length, the method generate a new chedule by appending a break activity of duration to chedule and include the new chedule in the et S. Then, the method continue with the next iteration. 25

26 8 Computational Experiment In order to evaluate the cheduling method preented in thi paper we generated three et of benchmark intance for a planning horizon tarting on Monday 0.00 AM and ending on Friday PM. In all benchmark et one hour of work time hall be conducted at each location in the tour. The driving time between two ubequent location i randomly et to a value between 2 and 10 hour for the firt et, between 10 and 20 hour for the econd et, and between 2 and 20 hour for the third et. Auming an average peed of 75 km/h, thi implie that the ditance between two ubequent location range from 150 km to 1500 km. The duration of the time window of the location are randomly et to a value between 1 and 12 hour. The tart time of the time window are randomly et to a time between 15 minute and 6 hour after the earliet departure at the previou location increaed by the pure driving time multiplied by 1.5. All intance generated have an accumulated on-duty time of 70 hour or le. Table 2 how the reult of our computational experiment. In the table CAN refer to the exact approach uing the enumerative method illutrated in Figure 6 and 7. CAN1 refer to the approach uing the heuritic which only append break period, and CAN2 refer to the approach uing the heuritic illutrated in Figure 4 and 5. In order to compare the reult of thee method with other exiting approache we adapted the exact approach for the U.S. Truck Driver Scheduling Problem (US-TDSP) preented by Goel and Kok (2010) to the Canadian cae. For thi, however, we had to relax condition (9) to (13) of the CAN-TDSP. In order to tackle the remaining condition only mall change to the approach had to be made. CAN-R8 refer to the approach for the relaxed problem uing the relevant parameter of Table 1, and CAN-R10 refer to the approach for the relaxed problem uing the ame parameter, except for t ret which i et to 10 hour. A CAN-R8 and CAN- R10 do not conider all contraint, ome of the truck driver chedule found by thee method may not be feaible for the CAN-TDSP. In the table we preent the number of feaible chedule for the relaxed problem. For further comparion, we alo ran experiment with the method preented by Goel and Kok (2010) in order to determine for how many intance a feaible chedule for the US-TDSP exit. We can ee in Table 2 that the heuritic approache CAN1 and CAN2 are very fat and have imilar running time a the CAN-R8 and CAN-R10 approache which are known to have a wort 26

27 Algorithm Driving Time Intance Feaible Total CPU Time CAN CAN CAN CAN-R CAN-R US CAN CAN CAN CAN-R CAN-R US CAN CAN CAN CAN-R CAN-R US Table 2: Reult cae complexity of O(λ 2 ). The CAN2 heuritic find a feaible chedule, if one exit, for almot 99 percent of the intance. For the et of benchmark intance with driving time between 10 and 20 hour, the CAN2 heuritic even ucceed in finding a feaible chedule for all intance for which a feaible chedule exit. The exact approach CAN require much more time, which i not urpriing a the number of alternative explored by the method i ignificantly higher. It mut be noted that the CAN approach make ue of ome additional dominance criteria which are not reported in thi paper. Thee criteria can be eaily derived from the dominance criteria preented in thi paper, but 27