Online Appendix to Accompany AComparisonof Traditional and Open-Access Appointment Scheduling Policies

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1 Online Aendix to Accomany AComarisonof Traditional and Oen-Access Aointment Scheduling Policies Lawrence W. Robinson Johnson Graduate School of Management Cornell University Ithaca, NY Phone: (607) Fax: (607) Rachel R. Chen Graduate School of Management University of California Davis Davis, CA June 2, 2009

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3 Proosition 1: Consider the otimal scheduling olicy {x t } situation where x τ 1 forsometimeslotτ. Then x t 1 for t =1,..., τ. Proof (by contradiction): If τ =1, the roosition holds trivially. For τ 2, we will assume in contradiction that there exists some time slot ˆt <τsuch that x =0; if this is true for multile time slots, we consider the first: ˆt ˆt. =min{t x t =0}. We first examine the case where x t =1 t<ˆt; this includes the case ˆt =1. Because there is no rior overbooking, the doctor must alwaysbeidleduringtimeslotˆt. In this case, we candoatleastaswellbyreassigningallsubsequent atients to the revious time slot, as ½ ¾ x. x t = t for t<ˆt. for t ˆt x t+1 This reassignment will shorten the doctor s day by eliminating the idle time in time slot ˆt, while reducing (or keeing it at zero, when T is large) the doctor s overtime. Since the later atients are just moved ahead one time slot, their waiting times will not change. Thus olicy {x t } is strictly dominated by olicy { x t }, and so could not have been otimal. We now examine the alternative case, where t. =max t x t 2,t<ˆt ª ; this imlies that x t =1 for any intervening time slots t + 1,..., ˆt 1 (if they exist). In this case, we will show that we will do at least as well by reassigning one of the over-booked atients from time slot t to time slot ˆt,as x x. t 1 for t = t t = 1 for t = ˆt x. t otherwise In order to show that olicy {x t } is more exensive than olicy { x t }, we focus on the incremental costs incurred by the atient who had been assigned to time slot t before being reassigned to time slot ˆt. Because all atients are identical, we can without loss of generality require that this atient is always last in line. In other words, this atient will be seen by the doctor only if he or she is the only one in the system. This atient must be seen either (a) strictly before time slot ˆt, or (b) in or after time slot ˆt. If the atient haens to be seen rior to time slot ˆt, the doctor will necessarily be idle in time slot ˆt (since there are no intervening eriods with overbooking), and so this final atient could have been assigned to time 1

4 slot ˆt instead, reducing (or keeing the same) the doctor s idle time while eliminating any waiting time for this atient. On the other hand, in those situations where this final atient cannot be seen until time slot ˆt or later, we could reduce his waiting time by ˆt t by assigning him to the latter time slot ˆt. Thus in every case, we would be the same or better off by assigning this atient to time slot ˆt rather than t. Proosition 2: It is otimal to assign each atient to an individual time slot (t max = N) if and only if 1+β if N T +2 α (1 ) 1+β(1 ) if N = T if N T. (1) Proof: We will break this roosition into three arts. In the first (Proosition 2A), we will show that (1) is a necessary condition for the otimality of the single-booking olicy. In the second (Proosition 2B), we develo conditions under which any olicy with overbooking will be imroved by ostoning a atient to the following eriod. And finally, in the third (Proosition 2C), we will use Proosition 2B to show that (1)isasufficient condition for the otimality of the single-booking olicy. Proosition 2A (Necessity): If single-booking is otimal (t max = N),then(1) is satisfied. Proof: If single-booking is otimal (t max = N), then its cost must be less than the cost of any other olicy. In articular, we can comare its cost to that of a olicy where there is recisely one time slot (identify it as τ; 1 τ t max = N 1) that is double-booked. We start by calculating the state robabilities, exected idle time, and exected waiting time for this new olicy. For t<τ,wehave π t (0) = π t (1) = 1, while for the time eriod that is double-booked (τ) π τ (0) = 2 π τ (1) = 2(1 ) π τ (2) = (1 ) 2. The state robabilities for time slots t>τare recursively defined as 2

5 π t (0) = [π t 1 (0) + π t 1 (1)] π t (1) = (1 )[π t 1 (0) + π t 1 (1)] + π t 1 (2) π t (2) = (1 )π t 1 (2). Recognizing that π t (2) = (1 ) t+2 τ and relacing π t 1 (0) + π t 1 (1) with 1 π t 1 (2) yields the following state robabilities for t τ: π t (0) = 1 (1 ) t+1 τ π t (1) = (1 ) 1 (1 ) t+1 τ + (1 ) t+1 τ = +(1 2) 1 (1 ) t+1 τ π t (2) = (1 ) t+2 τ. Theexectedidletimewillbe N 1 X N 2 X Ī = π t (0) = (τ 1) + 1 (1 ) t+1 τ +0=(N 1) 1 (1 ) N τ, (2) t=1 t=τ and the exected waiting time will be N 1 X N 1 X W = π t (2) = 0 + (1 ) t+2 τ = 1 (1 ) N τ (1 ) 2 /. (3) t=1 t=τ The exected overtime will deend on the relationshi between T and N. There are three different cases, which we will analyze in turn. Case 1: N T +2: In this case, both olicies will always have days that are at least T time slots long, so that a non-negative amount of overtime will always be incurred. For the olicy where time slot τ is double-booked, the exected overtime will be Ō = N 2+π N 1 (1) + 2π N 1 (2) T = N T 1 (1 ) N τ. (4) Combining (2) through (4), the total cost will be: (N 1) 1 (1 ) N τ + α 1 (1 ) N τ (1 ) 2 / +β N T 1 (1 ) N τ = (N 1 β)+β (N T )+ α(1 ) 2 / 1 β 1 (1 ) N τ. (5) 3

6 The doctor s exeected day length is N. Since the otimal single-booking (t max = N) olicy will also always incur overtime, its exected cost will be N n + β(n T )= (N 1 β)+β (N T ). (6) Comaring (6) with (5), if single-booking is less exensive, then it must be true that (N 1 β)+β (N T ) (N 1 β)+β (N T )+ α(1 ) 2 / 1 β 1 (1 ) N τ α (1 + β). (1 ) 2 Note that this inequality does not deend on time slot τ that is double-booked. Case 2: N = T +1: The only change from Case 1 is to the exected overtime under the t max = N 1 olicy. Positive overtime will be incurred only when there are two atients resent during the final time slot, which occurs with robability π N 1 (2) = (1 ) N+1 τ, yielding costs of (N 1) 1 (1 ) N τ + α 1 (1 ) N τ (1 ) 2 / + β(1 ) N+1 τ = (N 1 β)+β + α(1 ) 2 / 1 β + β 1 (1 ) N τ. (7) When single booking (t max = N) is otimal, then (6) is less exensive than (7). (N 1 β)+β (N T ) (N 1 β)+β + α(1 ) 2 / 1 β + β 1 (1 ) N τ α [1 + β(1 )]. (1 ) 2 Again, this inequality does not deend on which time slot τ is double-booked. Case 3: T N: In this case there will never be overtime for either olicy. When t max = N 1,thecosts will be (N 1) + 1 (1 ) N τ α(1 ) 2 / 1, (8) while for t max = N, the only costs will be the exected idle cost of (N 1). Thus if single-booking is less exensive it must be the case that (N 1) (N 1) + 1 (1 ) N τ α(1 ) 2 / 1 α (1 ), 2 once more regardless of the otimal value of τ. Combining the results of these three cases, we have shown 4

7 that if single booking is otimal, then (1) will be satisfied. Proosition 2B: Consider an arbitrary schedule with overbooking, so that t max <N. Define τ to be the last time slot in which there is overbooking; τ =max{t. x t 2}. If α (1 ) 2 1+β if t max T +1 1+β(1 ) if t max = T 1 if t max T 1 then exected costs will decrease if we ostone one atient from eriod τ to eriod τ +1., (9) Proof: The costs for the first τ 1 eriods are unaffected by this ostonement, and so can be ignored in the comarison. Note that π τ (0)/ is the robability that the system would have been emty without considering the atient to be ostoned, and 1 π τ (0)/ is the robability that the system would have one or more other atients besides this one. We will examine both of these ossibilities in turn. Case 1: Thesystemisnotemtyineriodτ, even without considering this final atient. The robability of this eventuality is 1 π τ (0)/. The only consequence of ostoning the final atient to the following eriod will be that if he arrives he will wait for one fewer eriods. Thus in this case, ostonement will decrease exected system costs by α(1 ). Case 2: The system is emty in eriod τ, rior to the consideration of this final atient. This ossibility will haen with robability π τ (0)/. Because we have just reassigned the atient away from a time eriod that turns out to have been otherwise emty, the costs will increase; define C to be this cost increase. Before we calculate C, we will establish the conditions under which we would like to defer the atient to the following eriod, considering both Case 1 and Case 2. The exected cost increase from deferring a atient from τ to τ +1is non-ositive if: [1 π τ (0)/][ (1 )α]+[π τ (0)/] C 0,or C [/π τ (0) 1] (1 )α. Now since at least two atients arrive in eriod τ, as addition to (ossibly) atients carried forward from earlier eriods, π τ (0) 2,or/π τ (0) 1/. Thus it would be sufficient to satisfy the tighter constraint C α(1 ) 2 /. (10)5

8 So at this oint it is sufficient to show that (9) imlies(10). We now continue ontocalculate C by determining the costs of this arbitrary olicy both without and with rescheduling a atient from eriod τ to eriod τ +1. We do so at the acknowledged exense of a minor abuse of notation: henceforward the state robabilities will be imlicitly redefined to be conditional on being in Case 2. Case2a: (The arbitrary schedule with overbooking, conditional on being in Case 2, with no ostonement.) In this case the state robabilities will be π t (1) = 1 π t (0) = for t = τ,..., t max. Thus over these final t max τ +1eriods, the atients waiting time is zero ( W =0), the exected doctor s idle time will be Ī = (t max τ), and the exected overtime Ō will deend on difference between t max and the number of time slots er day T (caed at τ 1 as we are focusing only on those costs incurred in eriod τ and beyond), as 0 if t max T Ō = t max max {T,τ 1} if t max T +1 so that the total cost of this olicy over the final t max τ +1eriods will be (t max τ) if T t max (11) (t max τ)+β [t max max {T,τ 1}] if T t max 1. (12), Case2b: (The arbitrary schedule with overbooking, conditional on being in Case 2, when a atient is ostoned from eriod τ to τ +1.) In this case, the state robabilities and erformance measures deend on whether τ = t max or τ<t max, each of which we will analyze in turn. Case 2bi (τ = t max ): When we reschedule the final atient to eriod τ +1, the state robabilities are π τ (0) = 1, with π τ+1 (0) = and π τ+1 (1) = 1. 6

9 Because the costs will be identical for the two olicies through eriod τ 1, weneedtoconsider only those costs incurred in eriod τ and τ +1. The doctor will always be idle in eriod τ, but never in the final eriod τ +1, so that the idle time will be Ī =1. (13) There will be no atient waiting time in these two eriods, so W =0. (14) The exected overtime Ō incurred over these two eriods will deend on relationshi between t max and the length of the day T,as Case 2bii (τ <t max ): Ō = 0 if t max T 1 1 if t max = T 2 if t max T +1. (15) In this case there is no longer a atient in eriod τ (π τ (0) = 1), there are two atients scheduled for eriod τ +1, and one atient scheduled for every subsequent eriod, through t max. Because the analyses for the state robabilities and exected idle and waiting costs are identical to those in Proosition 2A (excet that τ and N there are relaced by τ +1and t max here), only the results will be resented here. For t = τ +1,..., t max, the state robabilities are: π t (0) = 1 (1 ) t τ π t (1) = +(1 2) 1 (1 ) t τ π t (2) = (1 ) t+1 τ. Now the exected doctor s idle time will be Ī = t max X 1 t=τ π t (0) = 1 + t max X 1 t=τ+1 1 (1 ) t τ = (t max τ)+(1 ) tmax τ, (16) and the exected atient waiting time will be Xt max Xt max W = π t (2) = 0 + (1 ) t τ+1 = 1 (1 ) t max τ (1 ) 2 /. (17) t=τ t=τ+1 The exected overtime Ō deends on the relationshi between T and t max. Note that the length of the doctor s day will be t max 1+k with robability π tmax (k) for k =0, 1, 2; its exected value is D = t max 1+π tmax (1) + 2π tmax (2) = t max +(1 ) t max τ. 7

10 Thus the exected overtime will be 0 if t max T 1 Ō = (1 ) tmax τ+1 if t max = T. (18) (1 ) tmax τ + t max max {T,τ 1} if t max T +1 Note that the exressions for Ī, W,and Ō given in (13) (15) are just secial cases of (16) (18) for τ = t max ; thus we need only consider the latter three exressions as we calculate the exected costs for all values of τ as (t max τ)+(1 ) tmax τ + α 1 (1 ) tmax τ (1 ) 2 / if t max T 1 (19) (t max τ)+(1 ) tmax τ + α 1 (1 ) t max τ (1 ) 2 / +β(1 ) t max τ+1 if t max = T (20) (t max τ)+(1 ) tmax τ + α 1 (1 ) tmax τ (1 ) 2 / +β (1 ) t max τ + t max max {T,τ 1} if t max T +1 (21) Case 2bα (t max T 1): In this case we need to comare the increase in cost of this olicy over the cost of the olicy in Case 2a. With C =(19) (11), (10)willholdwhen (t max τ)+(1 ) tmax τ + α 1 (1 ) tmax τ (1 ) 2 / (t max τ) α(1 ) 2 / (1 ) t max τ 1 α(1 ) 2 / 0 α (1 ) 2. Case 2bβ (t max = T ): With C =(20) (11), (10)becomes (t max τ)+(1 ) tmax τ + α 1 (1 ) t max τ (1 ) 2 / +β(1 ) tmax τ+1 (t max τ) α(1 ) 2 / (1 ) tmax τ 1 α(1 ) 2 / + β(1 ) 0 8 α [1 + β(1 )]. (1 ) 2

11 Case 2bγ (t max T +1): With C =(21) (12), (10)becomes (t max τ)+(1 ) t max τ + α 1 (1 ) t max τ (1 ) 2 / +β (1 ) tmax τ + t max T (t max τ) β [t max T ] α(1 ) 2 / (1 ) tmax τ 1 α(1 ) 2 / + β 0 α [1 + β]. (1 ) 2 Thus if (9) holds, then in all three cases C 0, showing that we can always lower the costs by moving one overbooked atient from eriod τ to eriod τ +1. Proosition 2C (Sufficiency): If (1) is satisfied, then single-booking is otimal (t max = N). Proof: We start by showing that if (1) issatisfied, then (9) will be satisfied as well, for all t max N 1. First consider the case of (1)whereN T +2,sothatα (1 + β)/(1 ) 2. Because this is the tightest of the three lower bounds on α,itwillholdforallthreecasesof(9). Next consider the second case of (1) where N = T +1and α (1 + β(1 ))/(1 ) 2. Since t max <Nwith overbooking, t max T here, so that we need to consider only the last two conditions of (9), both of which are satisfied in this case. Finally, in the third case of (1) wheren T, we know α /(1 ) 2. Since t max <N,then t max T 1, and so we are restricted to the third case of (9), which is again satisfied. Wehaveshown that if (1)issatisfied, then (9) will be satisfied as well. Since (9)issatisfied, then from Proosition 2B we know that exected costs will decrease when we reassign one atient from eriod τ to eriod τ +1. Reeated alication of this roosition will eventually move this overbooked atient from eriod τ to eriod t max +1. Reeating this roosition with each overbooked atient eventually converts this arbitrary olicy into the single-booking olicy, decreasing costs at every ste and showing that when (9) holds, the single-booking olicy is less exensive than any arbitrary olicy with overbooking. Corollary 3: If it is otimal to overbook one atient (t max = N 1), thenx 1 =2and x t =1for t =2,..., N 1. 9

12 Proof: The cost of this olicy is given by equations (5), (7), and (8), deending on the relationshi between N and T. Note that in every case, the eriod of double-booking aears only in the term 1 (1 ) N τ, which is decreasing in τ. Since single-booking is not otimal, the lower bound (1) on α must not hold, indicating that the multilicand of 1 (1 ) N τ is negative in all three cases. Thus costs are increasing in τ, so that the minimum cost olicy will set τ =1, comleting the roof. Proosition 4: Consider the otimal scheduling olicy {x t } for the alternate assumtion on the doctor s length of day, where he or she must always stay through eriod T. Then (a) Proosition 1 holds for these assumtions as well.. (b) If N T,thenx t =1for t =1,..., N and zero otherwise. (c) If N T,thent max T Proof: Part (a): The analysis and conclusions in the roof of Proosition 1 are valid here as well. Part (b): Under this olicy, there is no atient waiting and no overtime. Since the doctor sees all the atients within his or her shortest ossible day length, his or her exected idle time will be at its minimum ossible value, T n. Thus this olicy is otimal. Part (c): Assume in contradiction that t max <T. Then there must exist at least one time slot τ<t with x τ 2; if more than one such time slot exists, we consider the last: τ =max{t. x t 2}. We will show that we can erform at least as well by reassigning one of the overbooked atients from eriod τ to eriod t max +1 n max,as x x t 1 for t = τ t = 1 for t = t max +1 x t otherwise. In order to show that olicy {x t } is more exensive than olicy { x t }, we focus on the incremental costs incurred by the one atient who had been assigned to times slot τ instead of time slot t max +1. Because all atients are identical, we can without loss of generality require that this atient is always last in line, and be the last one seen that day. This atient must be seen either (a) before the end of time slot t max, or (b) in or after time slot t max +1. In case (a), the doctor will be idle in time slot t max +1,but since t max +1 T, he or she cannot take advantage of this idleness to leave the office early. In this case 10

13 the atient could have been assigned to the later eriod t max +1, eliminating any waiting time he might have, keeing idle time at zero, and keeing the length of the doctor s day unchanged. On the other hand, in case (b), we could have assigned the final atient to time slot t max +1instead, reducing the atient s waiting time by t max +1 τ without increasing any other costs. Thus in either case, erformance is always as least as good when the atient is reassigned to time slot t max +1. Reeated alication of this reassignment leads to t max T. Proosition 5: If the same-day scheduling olicy is referable to the traditional scheduling olicy, then it will also be referable to any hybrid system. Proof: The roof deends on develoing a lower bound on the cost of the hybrid schedule, which we then comare to the cost of the same-day olicy. As before, let the number of atients reviously scheduled er day be N, andlets reresent the number of eole who call in the morning, a Poisson-distributed random variable with mean λ. Let n reresent the total exected daily workload, n = λ +(1 )N. In develoing the lower bound, assume that we know, rior to assigning atients to secific time slots, the realization of s for that future day. Using this additional knowledge can only lead to an otimal olicy whose costs will be less than that of the otimal olicy for the true hybrid model, and so will rovide a lower bound to that roblem. The otimal olicy, given foreknowledge of the realization of s, willbeto reserve the first s time-slots for these future same-day atients, allowing them to be served with neither idle time nor waiting time. The N traditionally-scheduled atients will start their day in time slot s +1, but aart from overtime are otherwise unaffected by the s same-day atients. Define C T ( n, T ) to be the exected cost, under traditional scheduling atients with an exected workload of n when the length of the day is T. Then the lower bound on the hybrid olicy will be X C H:LB ( n, T )= (s λ)c T ( n λ, T s). (22) s=0 Note that for large values of s, the length of the day available to the traditional atients could be negative, which reflects the fact that some of the s atients were seen on overtime. This requires no changes in the analysis of the traditional scheduling model. 11

14 Since the Poisson distribution is closed under addition, we can subdivide the same-day arrivals into two Poisson arrival rocesses: the first with mean λ, and the second with mean n λ. Time slots are still assigned sequentially as atients call in. Define C SD ( n, T ) to be the exected cost of scheduling atients with an exected workload of n under a same-day schedule when the length of the day is T. Then the cost of the same-day olicy can be written as X C SD ( n, T )= (s λ)c SD ( n λ, T s). (23) s=0 Comaring (22) and(23) shows that they are each weighted averages (with the same weights) of the traditional and same-day scheduling olicies, resectively. So to the extent to which traditional olicies are more exensive than same-day olicies (i.e., excet for very small values of α and ), so too will the lower bound on the otimal hybrid olicy be more exensive than the same-day olicy. 12