Health Care Manage Sc (27) 1:217 229 DOI 1.17/s1729-7-915- Optmal outpatent appontment schedulng Gudo C. Kaandorp Ger Koole Receved: 15 March 26 / Accepted: 28 February 27 / Publshed onlne: 23 May 27 Sprnger Scence + Busness Meda, LLC 27 Abstract In ths paper optmal outpatent appontment schedulng s studed. A local search procedure s derved that converges to the optmal schedule wth a weghted average of epected watng tmes of patents, dle tme of the doctor and tardness (lateness) as obectve. No-shows are allowed to happen. For certan combnatons of parameters the well-nown Baley- Welch rule s found to be the optmal appontment schedule. Keywords Patent schedulng Health care Local search Multmodularty 1 Introducton Outpatent appontment schedulng has been the subect of scentfc nvestgaton snce the begnnng of the fftes of the prevous century when Baley and Welch wrote [1]. he obectve of appontment schedulng s tradng off the nterests of physcans and patents: the patents prefer to have a short watng tme, the G. C. Kaandorp Department of Epdemology & Bostatstcs, Erasmus Medcal Center, Rotterdam, he Netherlands G. Koole Department of Mathematcs, VU Unversty Amsterdam, Amsterdam, he Netherlands URL: http://www.math.vu.nl/ oole G. Koole (B) Department of Mathematcs, Vre Unverstet, De Boelelaan 181a, 181 HV Amsterdam, Holland e-mal: oole@few.vu.nl physcan les to have as lttle dle tme as possble, and to fnsh on tme. Baley and Welch [1] ntroduced the frst advanced schedulng rule and tested t through smulaton. Snce then many papers have appeared that analyzed appontment schedulng n varous settngs (see Cayrl and Veral [2] for an overvew). Most of them use smulaton to analyze the performance of dfferent appontment schedulng rules. A new method s ntroduced to determne optmal schedulng rules for arbtrary numbers of patents. Servce tme duratons are eponentally dstrbuted and patents arrve on tme. No-shows are allowed to happen. he settng s dscrete tme,.e., there s a fnte number of (equally spaced) potental arrval moments. A local search method s descrbed that, startng from an arbtrary appontment schedule, tres to fnd neghborng appontment schedules that are better. From Koole and Van der Slus [7] t follows that when the obectve has a certan property related to convety (called multmodularty) then a locally optmal schedule s globally optmal. he man techncal result of ths paper s the proof that our obectve s ndeed multmodular. hs obectve s a weghted sum of the average epected patent watng tme, the dleness of the doctor durng the sesson length, and the tardness. he tardness s the probablty that the sesson eceeds the planned fnshng tme multpled by the average ecess. he local search method s also mplemented and avalable for publc use on the world wde web at http://obp.math.vu.nl/healthcare/software/ges. For bg nstances (many ntervals) the computaton tmes can be qute long. A faster local search method wth a smaller neghborhood s also mplemented. It s not guaranteed that t termnates wth a global optmum
218 Health Care Manage Sc (27) 1:217 229 soluton, but t gves very good results, also for bg nstances, wthn a reasonable amount of tme. We gve a short lterature overvew. he semnal paper on outpatent schedulng s Baley and Welch [1]. For an overvew of the results obtaned snce then, see Cayrl and Veral [2]. Roughly speang we can classfy the papers as follows: there are those that evaluate schedules (often usng smulaton) and those that desgn algorthms to fnd good schedules. A recent eample of the former, not ncluded n [2], s Hutzschenreuter [6]. In addton to no-shows she consders patents not arrvng on tme, and noneponental servce tmes. hose papers that present algorthms to desgn schedules can also be dvded n two: those that focus on contnuous tme, whch deal wth fndng the optmal nterarrval ntervals, and those that focus on dscrete tme, where the queston s how many arrvals should be scheduled at each potental arrval moment. Pegden and Rosenshne [1] consder a contnuous-tme model. her algorthm fnds the optmal arrval moments, assumng convety of the obectve n the nterarrval tmes. Also Lau and Lau [8]gve a procedure for fndng optmal arrval nstants, agan assumng convety. Hassn and Mendel [5] etend ths wor to no-shows. Wang [16, 17] proves optmalty, for phase-type servce-tme dstrbutons, but for a lmted number of patents. Denton and Gupta [3] formulate the problem as a two-stage stochastc lnear program. her algorthm s a good approach for qucly appromatng large-scale systems. Also Robnson and Chen [12] consder a stochastc lnear program. hey derve a fast heurstc for fndng good and robust nterarrval tmes, usng the fact that nterarrval tmes are dome-shaped, meanng that they are shorter at the begnnng and near the end of the sesson, and longer n the mddle. Let us now consder papers that are most relevant to the current wor as they are dealng wth dscrete tme,.e., a fnte number of potental arrval moments. In Lao et al. [9] a branch-and-bound method s used to fnd the optmal schedule. hs wors only for small nstances. Vanden Bosche, Detz and Smeon [15] use a method that resembles the method of ths paper n a number of ways. hey derve upper and lower bounds for the optmal appontment schedule. o show these bounds they use what they call submodularty (Lemma 1 of [15]), whch s n fact multmodularty on a subset of the equatons that we use (see the append). Usng the results of [15] upper and lower bound schedules (whch often concde) can be made startng from specfc schedules. Our results gve convergence to the optmal schedule startng from any schedule. he results of [15] are etended to dfferent types of patents n Vanden Bosche and Detz [13], and also to no-shows n Vanden Bosche and Detz [14]. he ncluson of dfferent types of patents s relatvely straghtforward, the sequence s optmzed usng local search, and for each sequence the optmal schedule s determned usng the method of [15]. Also our proofs nowhere use the fact that servce tmes are equally dstrbuted. Summarzng, compared to the wor of Vanden Bosche and coauthors, our stronger sub/multmodularty results allow us to formulate an algorthm that converges from any ntal schedule to the optmal one. he paper s structured as follows. In Secton 2 a model s defned, n whch we can compute for an arbtrary appontment schedule the obectve. In Secton 3 the local search algorthm s descrbed. Secton 4 s devoted to numercal results. he proof that our obectve s multmodular s gven s Append A. 2 Model For the schedulng problem we have to ntroduce some varables. A treatment/operaton room s operatonal durng ntervals wth length d (for eample a day from 8: am tll 4: pm splt n ntervals wth length 1 mn gves = 48 and d = 1). Wthn these ntervals a total of N patents should be scheduled. Patent servce tmes are assumed to be eponentally dstrbuted wth rate (and epectaton 1 ). Let t {,...,N} be the number of patents scheduled at the start of nterval t. A schedule s a vector ( 1,..., ) wth t=1 t = N. So we have: β = 1 : average servce tme : number of ntervals d: length of nterval N: total number of patents t : number of patents scheduled at the start of nterval t, t = 1,..., In the model we mae the followng assumptons: he servce tmes of patents are ndependent and eponentally dstrbuted. Patents always come on tme (no-shows are modeled later on). In the followng sectons we wll gve the formulas for calculatng for a gven schedule the mean watng tme, dle tme and tardness (lateness), whch we call W(), I() and L(), respectvely. o compare schedules we gve weghts α W, α I,andα L to the three man factors to obtan the overall obectve functon C() = α W W() +
Health Care Manage Sc (27) 1:217 229 219 α I I() + α L L(). Our problem can now be stated as follows: { mn α W W() + α I I() + α L L() t t = N t N } (1) For a gven schedule ( = ( 1,..., )) the probabltes of havng patents n the queue ust before new arrval(s) and ust after arrval(s) can be calculated. hs can be used to calculate the mean watng tme, dle tme and tardness. We ntroduce the followng notaton: p t ()=P ( patents n queue ust before the arrval(s) at nterval t ) and p t +()=P( patents n queue ust after the arrval(s) at nterval t ). We start empty, thus p 1 () = 1. Iteratvely the other probabltes can be calculated as follows: p 1 () 1, p t +( ) =, < t, p t +( ) = p t ( t ), t, p (t+1) () = N = p t +()b, p (t+1) ( ) = N = p t +()a,. where a = P(# potental departures = ) = (d) e d! and 1 b = P(# potental departures ) = 1 a. Because of the eponental servce tme dstrbutons the potental number of departures n any nterval has a Posson dstrbuton. 2.1 Mean watng tme of a patent If a patent arrves and fnds patents n the queue (ncludng the patent who s currently beng treated), then the mean watng tme of that patent wll be /. In our model patents arrve ust before a new nterval alone or n groups. he th one of that group has a mean watng tme of N = p t ( ) ( + 1) 1.hss ust the mean watng tme of one patent, so we must sum them all over the groups and ntervals an dvde = that through all N patents. hus we fnd the followng formula for the mean watng tme: W() = 1 N t N p t ( ) ( + 1) 1 t=1 =1 = 2.2 Mean dle tme of a doctor (2) For calculatng the mean dle tme of a doctor, we calculate frst the mean maespan M(), whch s the tme the last patent fnshes. hen t s easy to fnd the mean dle tme I(), because I() = M() N/. Set t {ma t t > }. Now we now for sure that the maespan s greater than ( t 1)d. he dstrbuton of the number of patents n the queue at tme t s nown. So the average maespan s M() = ( t 1)d + N p t +( ). =1 So now we obtan the followng formula for the mean dle tme: I() = ( ( t 1)d + 2.3 Mean tardness N ) p t +( ) N =1 (3) For the mean tardness of the day we loo at the end of the last nterval. Now f there are patents n queue, then the etenson s on average 1. We now the patent dstrbuton ust after the last nterval, sothe tardness functon s as follows: L() = N p (+1) ( ) =1 2.4 Includng no-shows (4) We can add no-shows to our model. hs s an mportant generalzaton as no-shows occur frequently n practce. Every patent now has a probablty ρ of not showng up. We assume that ρ s the same for all patents and that the patents are ndependent. hus the number of arrvals at tme t has a Bnomal( t,ρ) dstrbuton. hs changes the formulas used n the model as follows. p t () remans the same. p t +() s somewhat dfferent, because t s not nown how many patents are eactly comng. We must sum over the dstrbuton
22 Health Care Manage Sc (27) 1:217 229 Fg. 1 Base-case scenaro wth dfferent weghts αw=.5 2 αw=1 αw=2 αw=1 8: 8:1 8:2 8:3 8:4 8:5 9: 9:1 9:2 9:3 9:4 9:5 1: 1:1 1:2 1:3 1:4 1:5 11: 11:1 11:2 11:3 11:4 11:5 of how many patents wll be arrve. hs gves for p t () and p t +(): p 1 () 1, p t +( ) = t = ( t ) ρ t (1 ρ) p t ( ),, p (t+1) () = N p t +()b, = p (t+1) ( ) = N p t +()a,. = 2.4.1 Mean watng tme of a patent For the mean watng tme, for all ntervals we must addtonally sum the watng tme over the dstrbuton of the number of arrvng patents. hs gves the followng equaton: W() = 1 N(1 ρ) ( =1 t t=1 =1 N = ( t p t ( ) + 1 2.4.2 Mean dle tme of a doctor ) ρ t (1 ρ) ) (5) Agan we frst calculate the maespan. Now we do not now when the last patent s comng. But we can calculate the probablty that the last patent s comng at nterval t. hs probablty s P(last patent s comng at nterval t) = P(all patents after nterval t are no-shows) P(# arrvals at tme t 1) = ρ N t =1 (1 ρ t ). If the last patent s comng at nterval t, we now for sure the maespan s greater than (t 1)d.ocalculate the ecess after nterval t, we sum over the dstrbuton of the number of patents that come (havng n mnd that at least one patent comes) tmes the mean ecess. What we fnd s then: M()= P(last patent s comng at tme t) t: t > E(mean maespan last patent s comng at tme t) = ρ N t =1 (1 ρ t ) t: t > ( t (t 1)d + N = =1 p t () + ( t ) ρ t (1 ρ) 1 ρ t he mean dle tme s then gven by: I() = M() N(1 ρ) 1 (6) he queston can be ased how mportant ths mean dle tme s, because now the tme between the real last patent and the last planned patent s not added as dle tme. So n the case of no-shows the dle tme s less relevant as obectve and should have a relatvely low weght. 2.4.3 Mean tardness he formula of the mean tardness s the same (of course wth the new probabltes p t () and p t +()). ) able 1 Outcome values for dfferent schedules α W =.5 α W = 1 α W = 2 α W = 1 Indvdual Baley Welch Mean watng tme 26.46 19.9 15.35 9.85 12.37 16.75 Mean dle tme 21.86 36.69 54.2 88.58 72.14 5.7 Mean tardness 7.99 9.6 12.61 29.79 19.62 11.42 Obect value (α W =.5) 25.59 4.23 29.81 Obect value (α W = 1) 36.83 46.41 38.18 Obect value (α W = 2) 54.12 58.78 54.94 Obect value (α W = 1) 146. 157.72 188.95
Health Care Manage Sc (27) 1:217 229 221 Fg. 2 a Optmal schedules (ρ aganst β) b Optmal schedules (N aganst β) c Optmal schedules (N aganst ρ) a ρ=, β=18 ρ=.1, β=2 ρ=.25, β=24 ρ=.5, β=36 2 b N=8, β=25 N=1, β=2 N=16, β=12.5 N=2, β=1 8: 8:1 8:2 8:3 8:4 8:5 9: 9:1 9:2 9:3 9:4 9:5 1: 1:1 1:2 1:3 1:4 1:5 11: 11:1 11:2 11:3 11:4 11:5 c N=9, ρ= N=1, ρ=.1 N=12, ρ=.25 N=18, ρ=.5 2 8: 8:1 8:2 8:3 8: 8:4 8:1 8:2 8:3 8:4 8:5 9: 9:1 9:2 9:3 9:4 9:5 1: 1:1 1:2 1:3 1:4 1:5 11: 11:1 11:2 11:3 11:4 11:5 8:5 9: 9:1 9:2 9:3 9:4 9:5 1: 1:1 1:2 1:3 1:4 1:5 11: 11:1 11:2 11:3 11:4 11:5 3 Local search o compute the schedule wth the lowest obectve value we could try all possble schedules (the soluton space) and loo whch one has the lowest obectve value. But the number of all possble schedules s huge (t s ( ) N+ 1 N ), so we need a search algorthm to reduce the computaton tme. A local search algorthm starts wth a feasble soluton and tres teratvely to mprove the current soluton by searchng a better soluton n ts neghborhood untl a local mnmum s found. In general the local mnmum s not a global mnmum, but for the current problem and a well-chosen neghborhood t s possble to show that the local search algorthm fnshes n the global mnmum. We ntroduce our neghborhood. Defne the vectors u 1, ( 1,,...,, 1), u 2, (1, 1,,...,), u 3, (, 1, 1,,...,), =,.. u 1, (,...,1, 1, ), u (,...,, 1, 1) and tae V ={u 1,...,u }. As the neghborhood of schedule we tae all vectors of the form + v 1 + + v wth v 1,...,v V such that + v 1 + +v. hen the algorthm s as follows. Algorthm for computng an optmal schedule 1. Start wth some schedule 2. For all U V : for y = + v U v such that y compute C(y); f C(y) <C() then := y and start agan wth step 2 3. s the optmal schedule A vector u t can be nterpreted as movng a patent from tme slot t to tme slot t 1. hus the neghborhood of conssts of all combnaton of sngle-nterval shfts startng from. In Append A we prove that wth ths neghborhood the local search algorthm converges to the global optmal soluton. In the onlne tool we also mplemented a smaller neghborhood that gves much faster results. Under ths opton we smply tae y = + u for all u U n step 2 of the algorthm, thus we only consder U wth U =1. able 2 Outcome values(ρ aganst β) ρ =, β = 18 ρ =.1, β = 2 ρ =.25, β = 24 ρ =.5, β = 36 Mean watng tme 13.43 15.35 18.93 27.29 Mean dle tme 51.67 54.2 56.96 6.66 Mean tardness 1.4 12.61 17.28 28.59 Obect value 47.24 54.12 66.53 95.29
222 Health Care Manage Sc (27) 1:217 229 able 3 Outcome values(n aganst β) N = 8,β = 25 N = 1,β = 2 N = 16,β = 12.5 N = 2,β = 1 Mean watng tme 16.74 15.35 11.83 11.9 Mean dle tme 54.82 54.2 53.53 49.3 Mean tardness 15.56 12.61 8.1 5.6 Obect value 6. 54.12 42.47 37.63 4 Numercal eamples In ths secton we gve some numercal eamples. All these computatons were done wth our webtool whch s avalable for epermentaton at http://obp. math.vu.nl/healthcare/software/ges. Let the followng be the base-case scenaro. A medcal practce s operatonal between 8: am and 12: am. We splt ths nterval up n 48 ntervals of 5 mn. hus = 48 and d = 5. A treatment duraton s on average 2 mn (1/ = 2) and the percentage of noshows s 1% (ρ =.1). We want to plan ten patents (N = 1). o analyze ths model wth the small neghborhood (whch s not guaranteed to gve the optmal soluton) too a few seconds, analyzng the full neghborhood (what we dd for all cases consdered n ths secton) too around 12 h for each nstance. Frst we compute for base-case scenaro the the optmal schedule, for dfferent weghts n our obectve functon. he weght for the tardness s taen 1 (α L = 1), for the dle tme t s taen.2 (α I =.2). he dle tme has a relatvely low weght because of the noshows. We too four dfferent weghts for α W (.5, 1, 2, and 1), and determned the optmal schedules for each of these cases. he schedules are gven n Fg. 1. It s seen that f the watng tme has gven a bgger weght then the patents are more spread out to the end of the schedule, as one would epect. In the optmal schedule wth α W =.5 there are two patents scheduled at the begnnng of the day. Note that the optmal schedule for α W =.5 s close to the Baley Welch rule. In all cases the tmes between consecutve arrvals frst ncreases and then decreases agan. hs s the dome-shaped form that we dscussed n the lterature overvew. o have a better loo on the results we compare the optmal schedules wth two estng schedules: the ndvdual bloc schedule and the Baley Welch rule. Wth the ndvdual bloc schedule the worng day s dvded n the same number of ntervals as there are patents. In each bloc eactly one patent s scheduled. he Baley Welch rule s smlar as the ndvdual bloc schedule, but wth the last patent moved to the begnnng of the day. So n our base-case scenaro the ndvdual bloc schedule and the Baley Welch rule plan every 24 mn a patent, wth the ecepton that the Baley Welch rule schedules two patents at 8: am and none at 11:36 am. he results of the schedules are gven n able 1. he optmal schedules are of course better than the two estng schedules, but t can been seen for α W = 2 that the Baley Welch schedule s almost as good as the optmal one. Now we wll loo what happens wth the optmal schedules f we change some parameters. he changes are chosen such that the total worload does not change. he worload for the base-case scenaro s Nβ(1 ρ)=1 2.9=18 mnutes. We change the parameters two at a tme, ρ and β, N and β,andn and ρ, respectvely. Let α W = 2 and the other parameters fed as n the base-case scenaro. he optmal schedules are gven n Fg. 2. he correspondng outcome values are shown n ables 2, 3, 4. From able 2 we see that f ρ becomes larger (thus β decreases) the mean watng tme, dle tme and tardness all becomes larger, because of the hgher uncertanty. From the results of able 3 t s seen that f β becomes smaller (thus N ncreases) then the mean watng tme, dle tme and tardness all becomes smaller because of reduced uncertanty. he results of able 4 show us that f ρ becomes larger (thus N decreases) the able 4 Outcome values(n aganst ρ) N = 9, ρ = N = 1, ρ =.1 N = 12, ρ =.25 N = 18, ρ =.5 Mean watng tme 14.44 15.35 17.48 21.73 Mean dle tme 5.12 54.2 56.43 58.7 Mean tardness 1.83 12.61 14.63 17.35 Obect value 49.73 54.12 6.89 72.43
Health Care Manage Sc (27) 1:217 229 223 mean watng tme, dle tme and tardness all becomes larger because of the hgher uncertanty. A fnal change n parameters would be changng and d. hs would evdently lead to more smultaneous arrvals. 5 Conclusons In ths paper a method s presented to obtan optmal outpatent schedules n case of a fnte number of possble arrval epochs. he proof of the optmalty reles on showng that the obectve s multmodular, whch s a generalzaton of convety to lattces. Numercal results are presented. he nterarrval tmes have a dome shape, as observed earler n the lterature: the frst nterarrval tmes are short, then they get longer, and become agan short. Note that n certan cases the optmal rule s close to the Baley-Welch rule. For certan parameter values the Baley-Welch rule s ndeed optmal. Acnowledgements he authors would le to than the three anonymous referees for ther valuable suggestons. Append A: Local search method o prove that the local search algorthm converges to the global optmum, we frst show that our obectve functon s multmodular. We start by defnng multmodularty. A.1 Multmodularty Multmodularty (Hae [4]) s a property of functons on Z m. Defne the vectors v,...,v m Z m as follows: v = ( 1,,...,) v 1 = (1, 1,,...,) v 2 = (, 1, 1,,...,). v m 1 = (,..., 1, 1) v m = (,...,, 1) Let V ={v,...,v m }.Now: Defnton A.1 A functon f : Z m R s called multmodular f for all Z m,v,w V,v = w, f ( + v) + f ( + w) f () + f ( + v + w) (7) Central n the theory of multmodular functons s the concept of an atom. Defnton A.2 For some Z m and σ a permutaton of {,...,m}, we defne the atom S(,σ) as the conve set wth etreme ponts + v σ(), + v σ() + v σ(1),..., + v σ() + +v σ(m). It s shown n Hae [4] that each atom s a smple, and each unt cube s parttoned nto m! atoms; all atoms together span R m. In Koole and Van der Slus [7] the followng theorem s shown. It forms the bass of our neghborhood choce. heorem A.3 For f multmodular, a pont Z m s a global mnmum f and only f f () f (y) for all y = such that y Z m s an etreme pont of S(,σ) for some σ. Our problem (1) sa 1 dmensonal problem: gven 1,..., 1 we derve by = N 1 t=1 t. We wll show that t has a multmodular obectve functon. he set of allowable solutons s gven by { Z 1, 1 t=1 t N}. hs doman s not equal to Z 1, so the queston arses f the local search algorthm stll converges to the global mnmum. Accordng Lemma 2 n Koole and Van der Slus [7] heorem A.3 remans vald for ths subset of Z 1. Provng that our obectve functon s multmodular for the 1- dmensonal problem (1) s equvalent to showng that the obectve functon n dmensons satsfes Eq. 7 for v, w V,where V = u 1, u 2, u 3,. u 1, u ( 1,,...,, 1), (1, 1,,...,), (, 1, 1,,...,), =.. (,...,1, 1, ), (,...,, 1, 1) Note that u t s nothng else then movng a patent from tme slot t to tme slot t 1. Now we show that our obectve functon s multmodular and that t can be mnmzed by a local search algorthm that s guaranteed to termnate n the global mnmum. Our neghborhood s the set of all possble combnatons of the vectors u t added to the current schedule. heorem A.4 he watng tme functon W(), the dle tme functon I() and the tardness functon L(), as defned n Eqs. 2, 3, and4, are multmodular for all u, u V for whch =.
224 Health Care Manage Sc (27) 1:217 229 Proof of heorem A.4 It s easy to see that f the maespan s multmodular then also the dle tme s multmodular. hus t s suffcent to show that the maespan, the watng tme and the tardness are multmodular. hus t has to be shown that W( + u ) + W( + u ) W() + W( + u + u ), M( + u ) + M( + u ) M() + M( + u + u ) and ( + u ) + ( + u ) L() + ( + u + u ) for every possble and wth 1 <. Weuse couplng (see Rghter [11]) for ths proof, to compare the dfferent schedules, + u, + u and + u + u. For every possble combnaton of and, all dfferent possbltes of patent flows are dstngushed to detect the dfference between the number of patents n queue for each schedule for each tme nterval. Frst the proof s gven for 2 <. In Fg. 3, dfferent paths are shown for the dfferent schedules. (A) Let us start wth Case A. Schedule (A1) and schedule (A3) are followng the same path untl tme 1. Also Schedule (A2) and schedule (A4) are followng the same path untl that tme. In Case A the queue emptes between tme and tme 1, sofromthattmeonall the paths are the same. hus ust before tme 1, there are say patents n queue. Let = + 1. hen ust after tme 1 there are patents n queue for schedules (A1) and (A2) and + 1 for schedules (A3) and (A4). hus after tme 1 schedule (A1) and schedule (A2) are followng the same path and also Case A he queue emptes somewhere (A1) -1 (A2) -1 +u +1 (A3) -1 +u +1 (A4) -1 Case B he queue emptes not # departures < # departures = # departures > l (B a 1) (B b 1) (B c 1) -1-1 -1 l-1-1 (B a 2) -1 (B b 2) -1 (B c 2) -1 +u +1 l+1 (B a 3) +1 1 (B b 3) +1-1 (B c 3) -1 +u -1 l -1 (B a 4) -1 (B b 4) -1 (B c 4) -1 Fg. 3 Case A & B (2 < )
Health Care Manage Sc (27) 1:217 229 225 (B) schedule (A3) and schedule (A4) are followng the same path. Now say that untl tme 1 schedule (A1) has a total watng tme α 1, then schedule (A3) also has that total watng tme α 1. Say that untl tme 1 schedule (A2) has a total watng tme α 2, then schedule (A4) has the same total watng tme. Just after tme 1 schedules (A1) and (A2) follow the same path, so they have the same total watng tme, say β 1. Schedule (A3) and (A4) also follow the same path, thus they also have the same total watng tme, say β 2. Now t s easy to see that the watng tme satsfes α 2 + β 1 (A2)+α 1 + β 2 (A3)=α 1 + β 1 (A1)+α 2 + β 2 (A4). For the maespan and tardness only the end of a day s mportant, so we want to now what happens at the end of the path of each schedule. Schedules (A1) and (A2) follow after tme 1 the same path, and therefore they have the same maespan and tardness. Schedules (A3) and (A4) follow after tme 1 the same path, therefore they also have the same maespan and tardness. So (A2)+(A3)=(A1)+(A4) for the maespan and the tardness. Now loo at Case B. he queue does not empty between tme and 1, so now ust before tme 1 t can be that for schedules (B2) and (B4) there s one patent less n queue, because one patent more could be treated (because the movement of one patent from tme to tme 1). Otherwse all dfferent schedules wll have the same number n queue and then Case A apples. So for schedule (B2) and (B4) there are then 1 patents n queue and for schedules (B1) and (B3) there are patents n queue. Concernng the watng tme, let us say agan that schedules (B1) and (B3) have a total watng tme of α 1 and that schedules (B2) and (B4) have a total watng tme of α 2. Defne agan = + 1. hen ust after tme 1 there are patents n queue for schedule (B1), 1 for schedule (B2), + 1 for path (B3), one more because of the movement of one patent from tme to tme 1 and for schedule (B4). Now we dstngush between the followng three possbltes for the number of departures between tme 1 an.letl = l +. (a) he number of departures s less than. For schedule (Ba1) there wll be say l( 1) patents left ust before tme and ust after tme t wll be then l. Let the total watng tme between tme 1 and be β and after tme γ 1. For schedule (Ba2) the number of patent s l 1 ust before tme. Soust after tme there are l 1 patents n queue and the total watng tme between tme 1 and s then β d and after tme γ 2. For schedule (Ba3) the number of patent s l + 1, ust before tme. Justafter tme there are l patents n queue (one patent less arrves) and the total watng tme between tme 1 and s then β + d and after tme agan γ 1. For schedule (Ba4) the number of patent s l, ust before tme. Just after tme there are l 1 patents n queue (one patent comes less) and the total watng tme between tme 1 and s agan β and after tme agan γ 2. Now we see that the watng tme satsfes α 2 + β d + γ 2 (Ba2)+α 1 + β + d + γ 1 (Ba3) = α 1 + β + γ 1 (Ba1)+α 2 + β + γ 2 (Ba4) he end of the path (after tme ) ofschedules (Ba1) and (Ba3) s the same. he same holds for (Ba2) and (Ba4). So n ths case (Ba2)+(Ba3)=(Ba1)+(Ba4), for the maespan and tardness. (b) he second possblty s that there are eactly departures between tme 1 and. For schedule (B b 1) there wll be = patents left ust before tme and ust after tme t wll be l. Let the total watng tme between tme 1 and β and after tme γ 1. For schedule (B b 2) the number of patents s also, ust before tme. So ust after tme there are l patents n queue and the total watng tme between tme 1 and s then β d and after tme agan γ 1. For schedule (B b 3) the number of patents s + 1 = 1, ust before tme. So ust after tme there are l patents n queue and the total watng tme between tme 1 and s then β + d and after tme agan γ 1. For schedule (B b 4) the number of patents s =, ust before tme
226 Health Care Manage Sc (27) 1:217 229. So ust after tme there are l 1 patents n queue and the total watng tme between tme 1 and s then β (same as (B b 1))andaftertme γ 2 whch s of course smaller then γ 1. Now we see that the watng tme satsfes α 2 + β d + γ 1 (Ba2)+α 1 + β + d + γ 1 (Ba3) α 1 + β + γ 1 (Ba1)+α 2 + β + γ 2 (Ba4). he end of the path (after tme ) ofschedules (B b 1), (B b 2) and (B b 3) are the same so the maespan and tardness are the same for these schedules. At the end of the path of schedule (B b 4) there s one patent less (or n the worst case the same), so the maespan and tardness s also less or equal than the other schedules. So we can conclude that (B b 2)+(B b 3) (B b 1)+(B b 4), for the maespan and tardness. (c) he last possblty s that there are more than departures between tme 1 and.soforallpaths((bc1), (Bc2), (Bc3) and (Bc4)) there wll be no patents left ust before tme. Just after tme there wll be for schedule (Bc1) and (Bc2) l patents n queue and have a total watng tme of γ 1.(Bc3) and (Bc4) have then l 1 patents n queue and a total watng tme of γ 2. Now the total watng tme between tme 1 and tme, f there are s > departures s m (n 1)d n=1 s = m(m 1)d 2s (the frst patent has a watng tme of, the second d s, the thrd 2d s, etc...), wth m the number of patents ust after tme 1. Because ths s a conve functon t s clear that the watng tme functon satsfes α 2 + ( 1)( 2)d + 2s γ 1 (Bc2) + α 1 + (+1)d + γ 2s 2 (Bc3) α 1 + ( 1)d + γ 2s 1 (Bc1)+α 2 + ( 1)d +γ 2s 2 (Bc4). he ends of the paths of schedule (Bc1) and (Bc2) are the same and the ends of paths of schedule (Bc3) and (Bc4) are the same. herefore s (Bc2)+(Bc3)=(Bc1)+(Bc4), for the maespan and tardness. All cases for 2 < are done. Now for 1 = <. For Case C untl Case E (Fg. 4) counts that before tme 1 the queue somewhere emptes, so after that tme all schedules wll followng the same path and ust before tme 1 there are patents n queue for all schedules. untl tme 1 schedule (1) and (3) have a total watng tme α 1 and schedule (2) and (4) a total watng tme α 2. Just after tme 1 there wll be patents for schedule (1) and (2) and + 1 patents for schedule (3) and (4). Now after tme 1 we can dstngush the followng four possbltes. (C) Now for Case C there are equal or less than departures so ust before tme there are for schedule (C1) and (C2) l patents left and for schedule (C3) and (C4) l + 1 patents left. Between tme 1 and schedule (C1) and (C2) have the same total watng tme, say β 1 and schedule (C3) and (C4) have the same total watng tme, say β 2. Just after tme there are for all schedules l patents. So after tme follows schedule (C1) and (C3) the same path and have a total watng tme of γ 1 and follows schedule (C2) and (C4) the same path and have a total watng tme of γ 2. Now s easy to see that the watng tme satsfes α 2 + β 1 + γ 2 (C2)+α 1 + β 2 + γ 1 (C3)=α 1 + β 1 + γ 1 (C1)+α 2 + β 2 + γ 2 (C4). he ends of paths of schedule (C1) and (C3) are the same and the ends of paths of schedule (C2) and (C4) are the same. herefore s (C2)+(C3)=(C1)+(C4), for the maespan and tardness. (D,E) Now for Case D and Case E there are more than departures between tme 1 and tme. So ust before tme there are no patents left for all schedules. Between tme 1 and schedule (1) and (2) have the same total watng tme, say β 1 and schedule (3) and (4) have the same total watng tme, say β 2. Just after tme there are for schedule (1) and (2) l patents n queue and for schedule (3) and (4) l 1. Between tme and tme schedule (1) and (2) have a total watng tme of say γ 1 and schedule (3) and (4) have a total watng tme of say γ 2. Between tme and can happen the followng two cases: (D) he queue emptes. So for all schedules there are say m patents left ust before tme ( Case D ). Let m =m+.just after tme there wll be then m patents for schedule (D1) and (D3), wth a total watng tme of δ 1 and m +1 patents for schedule (D2) and (D4), wth a total watng tme of δ 2.Sothewat-
Health Care Manage Sc (27) 1:217 229 227 Case C he queue emptes somewhere # Departures l (C1) l (C2) +u +u +1 l+1 +1 l+1 (C3) (C4) Case D he queue emptes somewhere # Departures > he queue emptes somewhere m m (D1) Case E he queue emptes somewhere # Departures > he queue emptes not m m (E1) m m +1 (D2) m m +1 (E2) +u +1-1 m m (D3) +u +1-1 m-1 m -1 (E3) +u +1-1 m m +1 (D4) +u +1-1 m-1 m (E4) Fg. 4 Case C, D & E (1 = < ) (E) ng tme functon satsfes α 2 +β 1 +γ 2 +δ 2 (D2) + α 1 + β 2 + γ 1 + δ 1 (D3) = α 1 + β 1 +γ 1 +δ 1 (D1)+α 2 + β 2 + γ 2 + δ 2 (D4). he ends of paths of schedule (D1) and (D3) are the same and the ends of paths of schedule (D2) and (D4) are the same. herefore s (D2)+(D3)=(D1)+(D4), for the maespan and tardness. Now emptes the queue not between tme and tme, so now ust before tme there s one patent less (m 1) n queue for schedule (E3) and (E4). Just after tme there wll be for schedule (E1) and (E4) m patents, for schedule (E2) m + 1 and for schedule (E3) m 1 n queue. Now the total watng tme f there s m patents left s gven by m(m 1) 1. Because ths 2 s a conve functon t s clear that the watng tme functon satsfes α 2 + β 1 + γ 2 + (m +1)m 1 2 (D2)+α 1 + β 2 + γ 1 + (m 1)(m 2) 1 2 (D3) α 1 + β 1 + γ 1 + m (m 1) 1 (D1) + α 2 2 + β 2 + γ 2 + m (m 1) 1 2 (D4). Let s=d( 1). he man fnshng tme of the day wll be at s+m 1 for schedule (E1) and (E4), s + (m +1) 1 for schedule (E2) and s+(m 1) 1 for schedule (E3). So s+(m +1) 1 (E2)+s+(m 1) 1 (E3)=s+m 1 (E1)+s + m 1 (E4) for the maespan and tardness. For Case F untl Case I (Fg. 5) counts that before tme 1 the queue does not empty, so ust before tme 1 there are patents n queue for schedule (1) and (3) and for schedule (2) and (4) one less, so 1.Untltme 1 schedule (1) and (3) have
228 Health Care Manage Sc (27) 1:217 229 Case F # Departures < Case G # Departures = he queue emptes not l (F1) he queue emptes not (G1) -1-1 l-1-1 (F2) -1-1 (G2) +u +1 l+1 (F3) +u +1 1 (G3) +u -1 l -1 (F4) +u -1-1 (G4) Case H he queue emptes not # Departures > he queue emptes somewhere m m (H1) Case I he queue emptes not # Departures > he queue emptes not m m (I1) -1-1 m m +1 (H2) -1-1 m m +1 (I2) +u +1-1 m m (H3) +u +1-1 m-1 m -1 (I3) +u -1-1 m m +1 (H4) +u -1-1 m-1 m (I4) Fg. 5 Case F, G, H & I (1 = < ) a total watng tme α 1 and schedule (2) and (4) a total watng tme α 2. Just after tme 1 there wll be patents fore schedule (1) and (4), 1 patents for schedule (2) and + 1 patents for schedule (3). (F) For Case F, after tme 1 we can dstngush the followng four possbltes. For schedule (F1) there are say l patents left ust before tme and ust after tme t s l. Say that the total watng tme between tme 1 and s β and after tme γ 1. For schedule (F2) the number of patent s l 1 ust before tme. So ust after tme there are l 1 patents n queue and the total watng tme between tme 1 and s then β d and after tme γ 2. For schedule (F3) the number of patent s l + 1 ust before tme. Just after tme there are l patents n queue (one patent comes less) and the total watng tme between tme 1 and s then β + d and after tme agan γ 1 (same path as schedule (F1)). For schedule (F4) the number of patent s l ust before tme. Just after tme there are l 1 patents n queue (one patent comes less) and the total watng tme between tme 1 and s agan β and after tme agan γ 2 (same path as schedule (F2)). Now we see that the watng tme satsfes α 2 + β d + γ 2 (F2)+α 1 + β + d + γ 1 (F3)=α 1 + β + γ 1 (F4)+α 2 + β + γ 2 (F4) he end of the path (after tme ) ofschedules (F1) and (F3) are the same, and also (F2) and (F4) are the same. So n ths case (F2)+(F3)=(F1)+(F4), for the maespan and tardness.
Health Care Manage Sc (27) 1:217 229 229 (G) (H,I) Now for Case G there are eactly departures between tme 1 and. For schedule (G1) there wll be = patents left ust before tme and ust after tme t wll be l. Say that the total watng tme between tme 1 and s β and after tme γ 1. For schedule (G2) the number of patent shall also be ust before tme.soustafter tme there are l patents n queue and the total watng tme between tme 1 and s then β d and after tme γ 2. For schedule (G3) the number of patent shall be + 1 = 1 ust before tme. So ust after tme there are l patents n queue and the total watng tme between tme 1 and s then β + d and after tme agan γ 1 (same path as schedule (G1)). For schedule (G4) the number of patent shall be = ust before tme. So ust after tme there are l 1 patents n queue and the total watng tme between tme 1 and s then β (same as (G1)) and after tme γ 3 whch s of course smaller than γ 2, because 1 patent s less to do. Now we see that the watng tme satsfes α 2 + β d + γ 2 (G2) + α 1 + β + d + γ 1 (G3) α 1 + β + γ 1 (G1)+α 2 + β + γ 3 (G4). he end of the path (after tme ) of schedules (G1) and (G3) are the same so the maespan and tardness are the same for these schedules. At the end of the path of schedule (G4) there are one patent less (or n the worst case the same) than at the end of path (G2), so the maespan and tardness shall also be less or equal than schedule (G2). So we can conclude that (G2)+(G3) (G1)+(G4), for the maespan and tardness. Now for Case H and I there are more than departures between tme 1 and tme. So ust before tme there are no patents left for all schedules. Between tme 1 and schedules (1) and (4) have the same total watng tme of ( 1)d 2s (4) ( +1) d 2s, schedule (2) ( 1)( 2)d 2s (same as n Case Bc ). and schedule Just after tme the schedules wll follow the same path as n Case D and Case F, whch we already dscussed, so for the maespan and tardness t s mmedately clear that t satsfes the multmodularty. Now for the watng tme t s also clear because before tme t satsfes the multmodularty, and after tme also. Now we dstngushed all possble cases and we proved that n each case the watng tme, maespan and tardness are multmodular. hus the same holds for the dle tme. he proof can easly be etended to nclude noshows. hs s done by condtonng on the no-shows: we get the same model as wthout no-shows but wth less patents planned. References 1. Baley NJ, Welch JD (1952) Appontment systems n hosptal outpatent departments. Lancet 259:115 118 2. Cayrl, Veral E (23) Outpatent schedulng n health care: a revew of lterature. Prod Oper Manag 12:519 549 3. Denton B, Gupta D (23) A sequental boundng approach for optmal appontment schedulng. IIE rans 35:13 116 4. Hae B (1985) Etremal splttng of pont processes. Math Oper Res 22:543 556 5. Hassn R, Mendel S (26) Schedulng arrvals to queues: a model wth no-shows. Worng paper 6. Hutzschenreuter A (25) Queueng models for outpatent appontment schedulng. M.Sc hess, Unversty of Ulm 7. Koole G, van der Slus E (23) Optmal shft schedulng wth a global servce level constrant. IIE rans 35:149 155 8. Lau H, Lau AH (2) A fast procedure for computng the total system cost of an appontment schedule for medcal and ndred facltes. IIE rans 32:833 839 9. Lao C, Pegden CD, Rosenshne M (1993) Plannng tmely arrvals to a stochastc producton or servce system. IIE rans 25:36 73 1. Pegden CD, Rosenshne M (199) Schedulng arrvals to queues. Comput Oper Res 17:343 348 11. Rghter R (1994) Schedulng. In: Shaed M, Shanthumar, JG (eds) Stochastc orders and ther applcatons. Academc, New Yor 12. Robnson LW, Chen RR (23) Schedulng doctors appontments: optmal and emprcally-based heurstc polces. IIE rans 35:295 37 13. Vanden Bosche PM, Detz DC (2) Mnmzng epected watng n a medcal appontment system. IIE rans 32: 841 848 14. Vanden Bosche PM, Detz DC (21) Schedulng and sequencng arrvals to an appontment system. J Serv Res 4: 15 25 15. Vanden Bosche PM, Detz DC, Smeon JR (1999) Schedulng customer arrvals to a stochastc servce system. Nav Res Logst 46:549 559 16. Wang PP (1993) Statc and dynamc schedulng of customer arrvals to a sngle-server system. Nav Res Logst 4:345 36 17. Wang PP (1997) Optmally schedulng N customer arrval tmes for a sngle-server system. Comput Oper Res 24: 73 716