Intraday dynamic rescheduling under patient no-shows

Abstract

Patient no-shows are a major source of uncertainty for outpatient clinics. A common approach to hedge against the effect of no-shows is to overbook. The trade-off between patient’s waiting costs and provider idling/overtime costs determines the optimal level of overbooking. Existing work on appointment scheduling assumes that appointment times cannot be updated once they have been assigned. However, advances in communication technology and the adoption of online (as opposed to in-person) appointments make it possible for appointments to be flexible. In this paper, we describe an intraday dynamic rescheduling model that adjusts upcoming appointments based on observed no-shows. We formulate the problem as a Markov Decision Process in order to compute the optimal pre-day schedule and the optimal policy to update the schedule for every scenario of no-shows. We also propose an alternative formulation based on the idea of ‘atomic’ actions that allows us to apply a shortest path algorithm to solve for the optimal policy more efficiently. Based on a numerical study using parameter estimates from existing literature, we find that intraday dynamic rescheduling can reduce expected cost by 15% compared to static scheduling.

Appendices

Appendix A Equivalence of the (DR) and (DRE) formulations

1.1 Proof of Lemma 1

This is a proof by construction. Let $\mathcal {T}$ be an empty sequence and $\mathcal {X}=\{\varvec{x}\}$, i.e., a sequence with just one element. Perform the following steps:

• 1. Let k be the length of the sequence $\mathcal {X}$.

• 2. Define $\tilde{t} \equiv \min j: \tilde{x}_j^k> x_j^\prime ,j>t$.

• 3. If $\tilde{t}=T$, then stop. $(\mathcal {T},\mathcal {X})$ are the sequences we need.

• 4. If $\tilde{t}<T$, let $\tilde{n} = x_{\tilde{t}}^k - x_{\tilde{t}}^\prime$. Append $\tilde{t}$, $\tilde{n}$ times, to the sequence $\mathcal {T}$.

• 5. Define

  $$\begin{aligned} \varvec{\psi }^{n}_{t}(\varvec{x}) = \left\{ \begin{array}{ll} \varvec{\psi }^{n-1}_{t}({\varvec{x}} + \varvec{\Delta }^{t}) &{}\text {if}\ n>1\\ {\varvec{x}} + \varvec{\Delta }^{t} &{}\text {if}\ n=1. \end{array}\right. \end{aligned}$$

• 6. Set $\tilde{\varvec{x}^{k+i}} = \varvec{\psi }^{i}_{t}(\tilde{\varvec{x}}^{k})$, for $1 \le i \le \tilde{n}$. Then, $\tilde{x}_{\tilde{t}}^{k+i} = \tilde{x}_{\tilde{t}}^{k} - i \implies \tilde{x}_{\tilde{t}}^{k+\tilde{n}} = x_{\tilde{t}}^\prime$.

• 7. Go back to step 1. Note that step 2 of the next iteration will choose a higher value of $\tilde{t}$ which guarantees that this sequence of steps (1-7) will only be performed a maximum of $T-t-1$ times.

1.2 Proof of Lemma 2

In the proof of Lemma 1, we identified one sequence of time slot choices that take the schedule from $\varvec{x}$ to $\varvec{x}^\prime$. We now show that any such sequence must have a total update cost of $\sum _{j=t+2}^T \left( {\sigma }_j(\varvec{x}^\prime ) - {\sigma }_j(\varvec{x}) \right) u_{j-t-1}$. First, we note that for any sequence $\left\{ \tilde{t}^k\right\} _{k=1}^{K}$, the number of patients in slot j increases by one when $\tilde{t}^k$ equals $j-1$ and decreases by one when $\tilde{t}^k$ equals j. In the final schedule, $\varvec{x}^\prime$, the number of patients in slot j is given by:

$$\begin{aligned}&x^\prime _j = x_j + \sum _k I(\tilde{t}^k = j-1) - \sum _k I(\tilde{t}^k = j)&\end{aligned}$$

(A1)

$$\begin{aligned}&\implies \sum _k I(\tilde{t}^k = j-1) = x_j^\prime - x_j + \sum _k I(\tilde{t}^k = j).&\end{aligned}$$

(A2)

The formulation defined by (DRE) only allows for $\tilde{t}^k < T$, therefore, $\sum _k I(\tilde{t}^k = T) = 0$ (we cannot postpone patients who are schedule to arrive at the end of the day). Substituting this into Equation A1 gives us a recursive definition for $\sum _k I(\tilde{t}^k = j-1)$ (for $t<j-1<T$):

$$\begin{aligned} \sum _k I(\tilde{t}^k = j-1)&= x^\prime _j - x_j + \sum _{j^\prime =j+1}^{T} ( x^\prime _{j^\prime } - x_{j^\prime })\\&= {\sigma }_j(\varvec{x}^\prime ) - {\sigma }_j(\varvec{x}). \end{aligned}$$

By definition, the cost of each atomic update corresponding to a choice of $\tilde{t}^k$ when the current time is t is given by ${u}_{\tilde{t}^k - t}$. The total cost over all the updates given by the sequence $\left\{ \tilde{t}^k\right\} _{k=1}^{K}$ would be:

$$\begin{aligned} \sum _j \sum _k {u}_{\tilde{t}^k - t} I(\tilde{t}^k = j)&= \sum _j {u}_{j - t} ({\sigma }_j(\varvec{x}) - {\sigma }_j(\varvec{x}^\prime )). \end{aligned}$$

Proofs for Backward scheduling

1.1 Proof for Lemma 3

Let $\mathcal {S} = \left\{ \tilde{s}\right\} _{k=1}^{K}$ be an optimal sequence of atomic updates. Let $\left\{ \tilde{t}\right\} _{k=1}^{K}$ be the sequence of time slots and on which the updates are applied. Let $\left\{ \tilde{\varvec{x}}\right\} _{k=1}^{K+1}$ be the resulting sequence of schedules, i.e.,

$$\begin{aligned} \tilde{s}^k = \left\{ \begin{array}{ll} {\tilde{\varvec{x}}^{k}} + \varvec{\Delta }^{\tilde{t}^{k}} &{}\text {if the}\ k^{th}\ \text {update is a forward update}\\ {\tilde{\varvec{x}}^{k}} - \varvec{\Delta }^{\tilde{t}^{k}} &{}\text {otherwise.} \end{array}\right. \end{aligned}$$

Let $\tilde{s}^{l}$ be a backward update that is preceded by a forward update $\tilde{s}^{l-1}$. We show that for any such pair, swapping them so that the backward update precedes the forward update also results in a feasible update sequence. Doing these swaps will eventually lead to a sequence where all backward updates precede all forward updates. Through these swaps, we can eventually construct an update sequence where all backward updates precede all forward updates. Note that swapping two updates does not affect the rescheduling cost and therefore, the new sequence is also optimal. To show that it is feasible to swap the two updates, $\tilde{s}^l$ and $\tilde{s}^{l-1}$, we consider three possible cases:

The two updates cancel each other, i.e., $\tilde{\varvec{x}}^{l} = \tilde{\varvec{x}}^{l+1}$. The pair of updates can be dropped from the sequence without affecting the final schedule. This would imply that the original sequence $\left\{ S\right\} _{k=1}^{K}$ was not optimal which is a contradiction. Therefore, this case will never arise.

Both updates remove one patient from slot $\tilde{t}^l$. For feasibility, there must be at least two patient in $\tilde{t}^l$ before the updates are applied, i.e., $\tilde{x}^{l-1}_{\tilde{t}^{l}} \ge 2$. Therefore, the updates can be swapped while still being feasible.

The two updates act on separate slots and do not affect each other’s feasibility. Therefore, they can be swapped.