2.7 Extension of the Base Model: Rejection is Allowed
2.8.1 General Framework and A Simple Case
Consider the appointment scheduling problem with T-day horizon. We have K ≤ 2T
types of patients who arrive on day n, where type-k patient prefers days in the set Ak for
k = 1,2, ..., K. We know Ak ⊂ {1,2, ..., T} for k = 1,2, ..., K. We assume that the
number of arrivals on each day follows a Geometric distribution, same as in Section 2.3. With probabilitypk, the arrival belongs to type-kfork = 1,2, ..., K. We havePKk=1pk= 1.
Letx = (x1, ..., xT)denote the system state just before an event on dayn, wherextdenotes
the number of appointments scheduled on dayn+tfort= 1, ..., T. If the event is an arrival (this occurs with probabilityα), the decision must be made such that the arrival is scheduled
on the day i ∈ Ak and the system state transits tox+ei. If the event is a change-of-day
(this occurs with probability 1−α), the system state will transit to y = (x2, ..., xT,0). By
the standard theory of the dynamic programming, the optimality equation is given by:
h(x) +g =α K X k=1 pkmin i∈Ak {h(x+ei)}+ (1−α)[c(x1) +h(y)]
wherex = (x1, x2, ..., xT) andy = (x2, ..., xT,0). Here g is the optimal long run average
cost andh(·,·)is the bias under the optimal policy.
To develop the index policy, we start from a randomized policyγ. The policyγ assigns type-kpatient to dayn+j with probabilityθkj wherek = 1,2, ..., K andj ∈Ak. We have P
j∈Akθkj = 1andθkj = 0forj /∈Ak. We useAnto denote the number of patients arriving
on day n. Among those patients, we assign An,t patients on dayn +t. From the proof of
Theorem 8, we know An,t∼G αρt 1−α(1−ρt) , t= 1, ..., T (2.17) where ρt = K X k=1 pkθkt.
A natural question is how to chooseθkj. We will revisit this question later.
we know the system state is
(x1+An,1, x2+An,2, x3+An,3, ..., xT +An,T).
At the beginning of dayn+ 1, we incur the cost ofc(x1+An,1)and the system state becomes
(x2 +An,2, x3+An,3, ..., xT +An,T,0).
By the end of dayn+ 1, the system state is
(x2+An,2+An+1,1, x3 +An,3+An+1,2, ..., xT +An,T +An+1,T−1, An+1,T).
At the beginning of dayn + 2, we incur the cost ofc(x2+An,2 +An+1,1) and the system state becomes
(x3 +An,3+An+1,2, ..., xT +An,T +An+1,T−1, An+1,T,0).
Similarly, at the beginning of dayn+T, we incur the cost ofc(xT+An,T+An+1,T−1+· · ·+ An+T−1,1)and the system state becomes
Therefore we can write
hγ(x) =E[c(x1+An,1)] + E[c(x2+An,2+An+1,1)] +· · ·
+ E[c(xT +An,T +An+1,T−1+· · ·+An+T−1,1)]
+hγ(An+1,T +· · ·+An+T−1,2, ..., An+T−1,T,0).
Hence the index for daykis
Ik(xk) =E[c(xk+ 1 +An,k+An+1,k−1 +· · ·+An+k−1,1)] −E[c(xk+An,k+An+1,k−1+· · ·+An+k−1,1)]
SinceAn,k has same distribution for alln, so we write
Bk =An,k+An,k−1 +· · ·+An,1.
Thus
Ik(xk) = E[c(xk+ 1 +Bk)]−E[c(xk+Bk)].
This general framework works for all T andK. Next we use an example to study the performance of the index policy numerically.
appointment on day 2 or 3; a type-3 patient requests an appointment on day 3. That is to say, A1 ={1,2,3},A2 ={2,3},A3 ={3}. In this setting, the index function under the general framework reduces to
Ik(xk) = E[c(xk+ 1 +Bk)]−E[c(xk+Bk)], k = 1,2,3
whereB1 =An,1, B2 =An,1+An,2, B3 =An,1+An,2+An,3 =An. We know the distribution
An,t from Equation (2.17). To implement the index policy, we need to know the values of
ρ1, ρ2, ρ3 to compute the distribution ofBk’s. We use the same method to find these values
as in Section 2.5.2. That is,ρ1, ρ2, ρ3 are the solutions to the optimization problem
min (ρ1− 1 3) 2+ (ρ 2− 1 3) 2 + (ρ 3− 1 3) 2 s.t. 0≤ρ1 ≤p1, 0≤ρ2 ≤p1+p2, p3 ≤ρ3 ≤1, ρ1+ρ2+ρ3 = 1.
We solve this quadratic program to obtain the values ofρ1, ρ2, ρ3 given the inputsp1, p2, p3. Table 2.5 includes that the values ofρ1, ρ2under differentp1, p2. Note thatp3 = 1−p1−p2 andρ3 = 1−ρ1−ρ2. Hence the values ofp3 andρ3are known given the values ofp1, p2and ρ1, ρ2.
(ρ1, ρ2) p1 = 0 p1 = 0.2 p1 = 0.4 p1 = 0.6 p1 = 0.8 p1 = 1 p2 = 0 (0,0) (0.1,0.1) (0.2,0.2) (0.3,0.3) (13,13) (13,13) p2 = 0.2 (0,0.2) (0.2,0.2) (0.3,0.3) (13,31) (13,13) p2 = 0.4 (0,0.4) (0.2,0.4) (13,13) (13,13) p2 = 0.6 (0,0.5) (0.2,0.4) (13,13) p2 = 0.8 (0,0.5) (0.2,0.4) p2 = 1 (0,0.5)
Table 2.5: The values ofρ1, ρ2under differentp1, p2.
the shortest-queue policy. Since the optimal policy is numerically intractable in the general framework, the metrics proposed in 2.6, i.e., gapHP and η(HP,SQ), are no longer well- defined. We introduce a new metricimp(IP,SP). In mathematical terms,
imp(IP,SP) = LRACSP−LRACIP LRACSP
.
This metric can be interpreted as the percentage improvement achieved in using the index policy (IP) instead of the shortest-queue policy (SP). For example, if the long-run average cost under SP is 100 while the long-run average cost under IP is 95, then using IP instead of SP achieves the improvement of5%.
Since it is numerically prohibitive to compute the long-run average cost under both IP and SP, we use simulation to compare their performances. We describe the design of the simulation study below. We use the cost function introduced in Section 2.4.1. The expected number of arrivals per day, τ, is 15, same as where the cost function is minimized. Both IP and SP are operated under the same demand stream. We run 200 replications and each replication consists of 1000 days. At the end of each replication, we calculate theLRACIP,
imp(IP,SP)from 100 replications. We use the formula “mean±2×standard deviation” to obtain the95% confidence interval (CI) of imp(IP,SP). The results are included in Table 2.6. 95% CI p1 = 0 p1 = 0.2 p1 = 0.4 p1 = 0.6 p1 = 0.8 p1 = 1 p2 = 0 0±0 5.1±0.9 8.9±1.4 11.0±2.0 12.5±2.0 14.1±2.1 p2 = 0.2 3.4±0.7 7.8±1.2 10.2±1.6 11.9±2.1 13.5±2.0 p2 = 0.4 5.1±0.9 9.2±1.3 10.8±1.8 12.8±2.2 p2 = 0.6 6.0±1.0 10.1±1.8 11.4±1.9 p2 = 0.8 7.3±1.2 10.8±1.8 p2 = 1 8.2±1.2
Table 2.6: The 95% CI ofimp(IP,SP)(in percentage) whenτ = 15.
As we can see from Table 2.6, the cost improvement increases withp1(decreases withp3) for any fixedp2and increases withp2 (decreases withp3) for any fixedp1. Notep1(p3) is the proportion of the type-1 (type-3) patient, the most (least) flexible type. Hence, we conclude that the cost improvement of using IP over SP is higher when the system faces a more flexible demand.
We now consider the situation where the expected number of arrivals per day is not the same as where the cost functionc(·)is minimized. Table 2.7 and 2.8 include the 95% CI of
imp(IP,SP)when τ is 20 and 10 respectively. As we can see from both tables, the index policy performs better than the shortest-queue policy. However, the cost improvement when the clinic is overstaffed (τ = 10) is not as high as the situation when the clinic is understaffed (τ = 20).
95% CI p1 = 0 p1 = 0.2 p1 = 0.4 p1 = 0.6 p1 = 0.8 p1 = 1 p2 = 0 0±0 5.6±1.0 9.1±1.3 11.4±1.8 12.7±2.0 14.1±2.2 p2 = 0.2 3.5±0.7 8.0±1.4 10.4±1.7 12.2±2.1 13.7±2.2 p2 = 0.4 5.5±0.8 9.4±1.4 10.9±1.8 12.9±2.0 p2 = 0.6 6.7±1.0 10.5±1.7 11.9±1.9 p2 = 0.8 8.1±1.2 11.3±1.8 p2 = 1 9.1±1.3
Table 2.7: The 95% CI ofimp(IP,SP)(in percentage) whenτ = 20.
95% CI p1 = 0 p1 = 0.2 p1 = 0.4 p1 = 0.6 p1 = 0.8 p1 = 1 p2 = 0 0±0 2.7±0.6 4.4±0.9 5.0±1.2 5.1±1.4 5.5±1.4 p2 = 0.2 2.1±0.4 4.1±0.9 4.5±1.0 4.9±1.2 5.4±1.5 p2 = 0.4 3.0±0.6 4.4±0.9 4.5±1.1 5.1±1.3 p2 = 0.6 3.1±0.7 4.6±1.2 4.8±1.2 p2 = 0.8 3.6±0.8 4.9±1.9 p2 = 1 4.0±0.9
Table 2.8: The 95% CI ofimp(IP,SP)(in percentage) whenτ = 10.