Revenue management for outpatient appointments: joint capacity control and overbooking with class-dependent no-shows

Abstract

Outpatient appointment scheduling balances efficiency with access to healthcare services, yet appointment no-shows, cancellations, and delay are significant barriers to effective healthcare delivery. Patients with longer appointment delay often waste appointments more frequently, prompting a need for greater flexibility in appointment allocation. We present a joint capacity control and overbooking model where a clinic maximizes profits by controlling bookings from two sequential patient classes with different no-show rates. When booking advance requests, the clinic must balance high no-show probability with the probability of subsequent requests at lower waste rates. We show the optimal policy is computationally intensive to derive; therefore, we develop bounds and approximations which we compare via numerical study with the optimal policy as well as policies from practice and previous literature. We find the optimal policy increases profits 17.8% over first-come-first-serve allocation. We develop a simple policy which performs 0.3% below optimal on average. While pure open access can achieve optimality, it performs 23.0% below optimal on average.

Appendix

Proof

Proposition 1. Assume x ₂ Class 2 patients patients have already been booked. Assume D ₁ ≥ x ₁ and the clinic is considering whether or not to accept the x ₁th patient. The marginal expected profit, $\Updelta_{x_{1}}V_{0}\left(x_{2},x_{1}\right),$ of accepting an additional patient can be written as:

$$ \Updelta_{x_{1}}V_{0}\left(x_{2},x_{1}\right)=V_{0}\left(x_{2},x_{1}\right)-V_{0}\left(x_{2},x_{1}-1\right)=p-hP\left(Z_{2}\left(x_{2}\right)>k-x_{1}\right) $$

(16)

Since the marginal expected profit profit is decreasing in x ₁, the clinic accepts Class 1 patients so long as $\Updelta_{x_{1}}V_{0}\left(x_{2},x_{1}\right)\geq0,$ i.e. $P\left(Z_{2}\left(x_{2}\right)>k-x_{1}\right)\leq\frac{p}{h},$ or until all requests have been booked. $\square$

Proof

Proposition 2. Let $b_{1}\left(x_{2}\right)=b_{1}$ for all values of x ₂. We can write the expected overbooking cost as $L=hE\left[Z_{2}\left(x_{2}\right)+{\hbox{min}}\left(D_{1},b_{1}\left(x_{2}\right)\right)-k\right]^{+}.$ Note that $h\left(z+{\hbox{min}}\left(D_{1},b_{1}\right)-k\right)^{+}$ is an increasing, convex function of z. From Example 8.B.3 on page 368 of Shaked and Shanthikumar (2007) we know $\left\{ Z_{2}\left(x_{2}\right),x_{2}=0,1,2...\right\}$ is stochastically increasing convex. Therefore, $E\phi\left[Z_{2}\left(x_{2}\right)\right]$ is increasing and convex in x ₂ for any convex function including $hE\left[Z_{2}\left(x_{2}\right)+{\hbox{min}}\left(D_{1},b_{1}\right)-k\right]^{+}.$ Therefore, $V_{0}\left(x_{2},b_{1}\left(x_{2}\right)\right)$ is concave in x ₂ since the sum of concave functions is concave. This implies $V_{1}\left(x_{2},0\right)$ must be concave in x ₂ since the component-wise maximum of concave functions is concave. $\square$

Proof

Proposition 3. By definition, for any two random variables, R ₁ and R ₂, and increasing function ϕ

$$ R_{1}\leq_{st}R_{2}\Longleftrightarrow P\left(R_{1}>t\right)\leq P\left(R_{2}>t\right)\forall t\Longleftrightarrow E\phi\left(R_{1}\right)\leq E\phi\left(R_{2}\right) $$

Since $h\left[x-k\right]^{+}$ is an increasing, convex function of x, we can use stochastic ordering between $Z_{2}\left(x_{2}\right)+{\hbox{min}}\left(D_{1},s-x_{2}\right)$ and $Z_{2}\left(x_{2}-1\right)+{\hbox{min}}\left(D_{1},s-x_{2}+1\right)$ to determine whether the expected overbooking cost is increasing or decreasing.

$$ \begin{aligned} Z_{2}\left(x\right)+{\hbox{min}}\left(D_{1},s-x_{2}\right) \geq_{st} & Z_{2}\left(x_{2}-1\right)+{\hbox{min}}\left(D_{1},s-x_{2}+1\right) \\ Z_{2}\left(x_{2}\right)-Z_{2}\left(x_{2}-1\right) \geq_{st} & {\hbox{min}}\left(D_{1},s-x_{2}+1\right)-{\hbox{min}}\left(D_{1},s-x_{2}\right) \end{aligned} $$

(17)

If $P\left(Z_{2}\left(x_{2}\right)-Z_{2}\left(x_{2}-1\right)>t\right)\geq P\left({\hbox{min}}\left(D_{1}s-x_{2}+1\right)-{\hbox{min}}\left(D_{1},s-x_{2}\right)>t\right)$ ∀ t then $Z_{2}\left(x_{2}\right)$ is stochastically increasing in x ₂. If the reverse is true, then $Z_{2}\left(x_{2}\right)$ is stochastically decreasing in x ₂. (Note: each difference is either 0 or 1.)

• For t ≥ 1

  • $P\left(Z_{2}\left(x_{2}\right)-Z_{2}\left(x_{2}-1\right)>t\right)=0=P\left({\hbox{min}}\left(D_{1},s-x_{2}+1\right)-{\hbox{min}}\left(D_{1},s-x_{2}\right)>t\right)$

• For t = 0

  • $P\left(Z_{2}\left(x_{2}\right)-Z_{2}\left(x_{2}-1\right)>t\right)=\alpha_{2}$

  • $P\left({\hbox{min}}\left(D_{1},s-x_{2}+1\right)-{\hbox{min}}\left(D_{1},s-x_{2}\right)>t\right)=P\left(D_{1}\geq s-x_{2}+1\right)$

• For t < 0

  • $P\left(Z_{2}\left(x_{2}\right)-Z_{2}\left(x_{2}-1\right)>t\right)=1=P\left({\hbox{min}}\left(D_{1},s-x_{2}+1\right)-{\hbox{min}}\left(D_{1},s-x_{2}\right)>t\right)$

For any increasing function $\phi, Z_{2}\left(x_{2}\right)$ is stochastically increasing in x ₂ and $E\phi\left(Z_{2}\left(x_{2}\right)\right)$ is increasing in x ₂ if $\alpha_{2}\geq P\left(D_{1}\geq s-x_{2}+1\right).$ $\square$

Proof

Proposition 4. Let superscripts A and B denote the two cases, respectively, where the Class 1 booking limit stays constant or decreases by one.

Case A: $b_{1}\left(x_{2}\right)=b_{1}\left(x_{2}-1\right)$

$$ \triangle V_{1}^{A}\left(x_{2},0\right) = p\alpha_{2}-h\triangle L^{A}\left(x_{2},b_{1}\left(x_{2}\right)\right) $$

If the x ₂th customer does not attend,

$$ \triangle L^{A}\left(x_{2},b_{1}\left(x_{2}\right)\right) =0= E\left[Z_{2}\left(x_{2}-1\right)+\hbox{min}\left(b_{1}\left(x_{2}-1\right),D_{1}\right)-k\right]^{+}-E\left[Z_{2}\left(x_{2}-1\right)+\hbox{min}\left(b_{1}\left(x_{2}-1\right),D_{1}\right)-k\right]^{+} $$

If the x ₂th customer attends and Class 1 demand is at least $b_{1}\left(x_{2}-1\right),$

$$ \begin{aligned} \triangle L^{A}\left(x_{2},b_{1}\left(x_{2}\right)\right) & = E\left[Z_{2}\left(x_{2}-1\right)+1+b_{1}\left(x_{2}-1\right)-k\right]^{+}-E\left[Z_{2}\left(x_{2}-1\right)+b_{1}\left(x_{2}-1\right)-k\right]^{+}\\ &= \sum_{z=k-y\left(x_{2}-1\right)-1}^{\infty}P\left(Z_{2}\left(x_{2}\right)>z\right)-\sum_{z=k-y\left(x_{2}-1\right)}^{\infty}P\left(Z_{2}\left(x_{2}\right)>z\right)\\ &= P\left(Z_{2}\left(x_{2}\right)>k-b_{1}\left(x_{2}-1\right)-1\right) \end{aligned} $$

If the x ₂th customer attends and Class 1 demand is equal to $d<b_{1}\left(x_{2}-1\right),$

$$ \begin{aligned} \triangle L^{A}\left(x_{2},b_{1}\left(x_{2}\right)\right) &= E\left[Z_{2}\left(x_{2}-1\right)+1+d-k\right]^{+}-E\left[Z_{2}\left(x_{2}-1\right)+d-k\right]^{+}\\ &= P\left(Z_{2}\left(x_{2}\right)>k-d-1\right) \end{aligned} $$

Conditioning upon these sub-cases we write the marginal expected profit as

$$ \begin{aligned} \triangle V_{1}^{A}\left(x_{2},0\right) &= p\alpha_{2}-h\alpha_{2}S\left(x_{2},b_{1}\left(x_{2}\right)\right) &-h\alpha_{2}P\left(Z_{2}\left(x_{2}-1\right)>k-b_{1}\left(x_{2}-1\right)-1\right)P\left(D_{1}\geq b_{1}\left(x_{2}-1\right)\right) \end{aligned} $$

(18)

where

$$ S\left(x_{2},b_{1}\left(x_{2}\right)\right)=\sum_{d=k-x_{2}+1}^{b_{1}\left(x_{2}-1\right)-1}P\left(D_{1}=d\right)P\left(Z_{2}\left(x_{2}-1\right)>k-d-1\right) $$

(19)

Case B $b_{1}\left(x\right)=b_{1}\left(x_{2}-1\right)-1$

$$ \triangle V_{1}^{B}\left(x_{2},0\right) = p\alpha_{2}-pP\left(D_{1}\geq b_{1}\left(x_{2}-1\right)\right)-h\triangle L^{B}\left(x_{2},b_{1}\left(x_{2}\right)\right) $$

Conditioning on sub-cases as in Case A, we derive the marginal change in expected overbooks as follows and write the marginal expected profit for Case B as

$$ \begin{aligned} \triangle V_{1}^{B}\left(x_{2},0\right)& = p\alpha_{2}-h\alpha_{2}S\left(x_{2},b_{1}\left(x_{2}\right)\right)\\ & -P\left(D_{1}\geq b_{1}\left(x_{2}-1\right)\right)\left(p-h\left(1-\alpha_{2}\right)P\left(Z_{2}\left(x_{2}-1\right)>k-b_{1}\left(x_{2}-1\right)\right)\right) \end{aligned} $$

(20)

$\square$