Reducing COPD readmissions through predictive modeling and incentive-based interventions

Abstract

This paper introduces a case study at a community hospital to develop a predictive model to quantify readmission risks for patients with chronic obstructive pulmonary disease (COPD), and use it to support decision making for appropriate incentive-based interventions. Data collected from the community hospital’s database are analyzed to identify risk factors and a logistic regression model is developed to predict the readmission risk within 30 days post-discharge of an individual COPD patient. By targeting on the high-risk patients, we investigate the implementability of the incentive policy which encourages patients to take interventions and helps them to overcome the compliance barrier. Specifically, the conditions and scenarios are identified for either achieving the desired readmission rate while minimizing the total cost, or reaching the lowest readmission rate under incentive budget constraint. Currently, such models are under consideration for a pilot study at the community hospital.

Appendix: Proofs

Proof

of Proposition 2 In this case, the minimum cost is achieved when the resulting readmission probability equals to the upper bound ${P_{r}^{d}}$. Thus, the equations for α_j will be:

$$\begin{array}{@{}rcl@{}} \alpha_{i}\beta_{i} + \alpha_{j}\beta_{j} &=& {P_{r}^{d}}, \\ \alpha_{i} + \alpha_{j} & =& 1 \end{array} $$

Plugging the second equation into the first one, we can obtain:

$$1 - \alpha_{j} = \frac{{P_{r}^{d}} - \alpha_{j}\beta_{j}}{\beta_{i}}, $$

which leads to

$$\beta_{i} - \alpha_{j}\beta_{i} = {P_{r}^{d}} - \alpha_{j}\beta_{j}. $$

Then itfollows that

$$\alpha_{j} = \frac{\beta_{i} - {P_{r}^{d}}}{\beta_{i} - \beta_{j}}. $$

Using this expression, the minimum cost is obtained by

$$\begin{array}{@{}rcl@{}} &&\min_{j\neq i} \alpha_{i}(C_{r}\beta_{i} + C_{i}) + \alpha_{j}(C_{r}\beta_{j} + C_{j})\\ && = \min_{j\neq i} \left( 1 - \frac{\beta_{i} - {P_{r}^{d}}}{\beta_{i} - \beta_{j}}\right)(C_{r}\beta_{i} + C_{i}) \\ &&\qquad\quad + \frac{\beta_{i} - {P_{r}^{d}}}{\beta_{i} - \beta_{j}}(C_{r}\beta_{j} + C_{j}) \\ && = \min_{j\neq i} \frac{{P_{r}^{d}} - \beta_{j}}{\beta_{i} - \beta_{j}}(C_{r}\beta_{i} + C_{i}) + \frac{\beta_{i} - {P_{r}^{d}}}{\beta_{i} - \beta_{j}}(C_{r}\beta_{j} + C_{j}) \\ && = \min_{j\neq i} \frac{1}{\beta_{i} - \beta_{j}}({P_{r}^{d}}C_{r}\beta_{i} + {P_{r}^{d}}C_{i} - C_{r}\beta_{i}\beta_{j} - C_{i}\beta_{j} \\ &&\qquad\quad + C_{r}\beta_{j}\beta_{i} + C_{j}\beta_{i} - {P_{r}^{d}}C_{r}\beta_{j} - {P_{r}^{d}}C_{j}) \\ && = \min_{j\neq i} \frac{1}{\beta_{i} - \beta_{j}}\left[P_{r}^{d}C_{r}(\beta_{i} - \beta_{j}) + {P_{r}^{d}}C_{i} - C_{i}\beta_{j}\right.\\ &&\qquad\quad \left. + C_{j}\beta_{i} - {P_{r}^{d}}C_{j}\right]\\ && = {P_{r}^{d}}C_{r} + \min_{j\neq i} \frac{1}{\beta_{i} - \beta_{j}}({P_{r}^{d}}(C_{i} - C_{j}) - C_{i}\beta_{j} + C_{j}\beta_{i}). \end{array} $$

Thus, the cost is minimized when the second term of the last equation is minimized. Therefore, selecting j from all k’s such that α_k < α_i to minimize the second term, the optimal solution is obtained.

$$\begin{array}{@{}rcl@{}} \alpha_{j} &=& \frac{\beta_{i} - {P_{r}^{d}}}{\beta_{i} - \beta_{j}} \text{ for} j = \operatorname*{arg\,min} \limits_{k \neq i \text{ s.t} \beta_{k} < \beta_{i}} \: \\ &&\qquad\left\{\frac{{P_{r}^{d}}(C_{i} - C_{k}) - C_{i}\beta_{k} + C_{k}\beta_{i}}{\beta_{i} - \beta_{k}}\right\}, \end{array} $$

(22)

$$\begin{array}{@{}rcl@{}} \alpha_{i} &=& 1 - \alpha_{j} \text{ for~} i = \operatorname*{arg\,min} \limits_{k} \: \{C_{r}\beta_{k} + C_{k}\}. \end{array} $$

(23)

Expression (10) follows by plugging the results from Eq. 22 into Eqs. 1 and 2.

Proof

of Proposition 4 Rewrite the problem as a standard form of linear programming by adding a non-negative slack variableδ:

$$\begin{array}{@{}rcl@{}} \min & \quad \beta_{b}\alpha_{b} + \beta_{h}\alpha_{h} + \beta_{c}\alpha_{c} + \beta_{n}\alpha_{n} \\ s.t. & \quad \delta + C_{b}\alpha_{b} + C_{h}\alpha_{h} + C_{c}\alpha_{c} + C_{n}\alpha_{n} = \widetilde{C}_{d}, \\ & \quad \alpha_{b} + \alpha_{h} + \alpha_{c} + \alpha_{n} = 1, \\ & \quad \delta \geq 0, \quad \alpha_{i} \geq 0 \quad \text{ for } i = b,h,c,n. \end{array} $$

(24)

Then the solution in Eq. 15 is a basic feasible solution of this problem, since it satisfies all constraints and only two variables are nonzero. To prove that such a solution is optimal, we need to check if it satisfies the optimality condition. Specifically, the basis matrix of the solution and its inverse are:

$$\begin{array}{@{}rcl@{}} B &=& \left( \begin{array}{cc} C_{i} & C_{j} \\ 1 & 1 \end{array}\right), \quad\quad B^{-1} = \frac{1}{C_{i} - C_{j}} \left( \begin{array}{cc} 1 & - C_{j} \\ -1 & C_{i} \end{array}\right). \end{array} $$

The reduced cost for other variable α_k, k ≠ i, j, is

$$\begin{array}{@{}rcl@{}} & \beta_{k}-\left( \beta_{i}\;\beta_{j}\right) \dfrac{1}{C_{i} - C_{j}} \left( \begin{array}{cc} 1 & - C_{j} \\ -1 & C_{i} \end{array}\right) \left( \begin{array}{ll} {C_{k}}\\ {1}\\ \end{array}\right)\\ &=\beta_{k} - \dfrac{1}{C_{i} - C_{j}}(\beta_{i}C_{k} - \beta_{j}C_{k} - \beta_{i}C_{j} + \beta_{j}C_{i} ), \end{array} $$

and the reduced cost for δ is

$$\begin{array}{@{}rcl@{}} && 0 -\left( \beta_{i}\;\beta_{j}\right) \frac{1}{C_{i} - C_{j}} \left( \begin{array}{cc} 1 & - C_{j} \\ -1 & C_{i} \end{array}\right) \left( \begin{array}{c} {1}\\ {0}\\ \end{array}\right)= - \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}. \end{array} $$

The optimality condition implies that the reduced costs are non-negative. In other words, we need to check the signs of these reduced costs.

• For α_k with k s.t $C_{k} \leq \widetilde {C}_{d}$, k ≠ j, the reduced cost for variable α_k can be rewritten as

  $$\begin{array}{@{}rcl@{}} && \beta_{k} - \frac{1}{C_{i} - C_{j}}(\beta_{i}C_{k} - \beta_{j}C_{k} - \beta_{i}C_{j} + \beta_{j}C_{i} ) \\ && = \beta_{k} + \frac{1}{C_{i} - C_{j}}(- \beta_{i}C_{k} + \beta_{j}C_{k} + \beta_{i}C_{j} - \beta_{j}C_{i} \\ &&\quad -\beta_{i}C_{i} + \beta_{i}C_{i})\\ && =\beta_{k} - \beta_{i} + \frac{1}{C_{i} - C_{j}}(- \beta_{i}C_{k} + \beta_{j}C_{k} - \beta_{j}C_{i} \\ &&\quad + \beta_{i}C_{i})\\ && =\beta_{k} - \beta_{i} + \frac{(C_{i}-C_{k})(\beta_{i} - \beta_{j})}{C_{i} - C_{j}}\\ && =-(C_{i}-C_{k})\left\{\frac{\beta_{i} - \beta_{k}}{C_{i}-C_{k}} - \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}\right\}\\ && =-\frac{C_{i}-C_{k}}{\widetilde{C}_{d}-C_{i}}\left\{\frac{\beta_{i} - \beta_{k}}{C_{i}-C_{k}}(\widetilde{C}_{d}-C_{i})\right.\\ &&\quad \left. - \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}(\widetilde{C}_{d}-C_{i})\right\}\\ && =-\frac{C_{i}-C_{k}}{\widetilde{C}_{d}-C_{i}}\left[\left\{\frac{\beta_{i} - \beta_{k}}{C_{i}-C_{k}}(\widetilde{C}_{d}-C_{i}) + \beta_{i}\right\}\right.\\ &&\quad \left. - \left\{ \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}(\widetilde{C}_{d}-C_{i}) + \beta_{i}\right\}\right]. \end{array} $$

  Due to conditions of i and k, we have C_i − C_k ≥ 0 and $\widetilde {C}_{d}-C_{i} \leq 0$. In addition, the expression inside the square bracket is non-negative from the definitions of i and j. Therefore, the reduced cost of α_k is non-negative.

• For $\alpha _{k^{\prime }}$ with k^′s.t $C_{k^{\prime }} > \widetilde {C}_{d}$,k^′≠i, first, with some algebraic operations, we obtain

  $$\frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}(\widetilde{C}_{d} - C_{i}) + \beta_{i} = \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}(\widetilde{C}_{d} - C_{j}) + \beta_{j}. $$

  Then, the reduce cost for $\alpha _{k^{\prime }}$is:

  $$\begin{array}{@{}rcl@{}} && \beta_{k^{\prime}} - \frac{1}{C_{i} - C_{j}}(\beta_{i}C_{k^{\prime}} - \beta_{j}C_{k^{\prime}} - \beta_{i}C_{j} + \beta_{j}C_{i} ) \\ && = \beta_{k^{\prime}} + \frac{1}{C_{i} - C_{j}}(- \beta_{i}C_{k^{\prime}} + \beta_{j}C_{k^{\prime}} + \beta_{i}C_{j} - \beta_{j}C_{i} \\ &&\quad + \beta_{j}C_{j} +- \beta_{j}C_{j})\\ && = \beta_{k^{\prime}} - \beta_{j} + \frac{1}{C_{i} - C_{j}}(- \beta_{i}C_{k^{\prime}} + \beta_{j}C_{k^{\prime}} + \beta_{i}C_{j}\\ &&\quad - \beta_{j}C_{j})\\ && = \beta_{k^{\prime}} - \beta_{j} + \frac{(C_{j}-C_{k^{\prime}})(\beta_{i} - \beta_{j})}{C_{i} - C_{j}}\\ && = (C_{k^{\prime}}-C_{j})\left\{\frac{\beta_{k^{\prime}} - \beta_{j}}{C_{k^{\prime}}-C_{j}} - \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}\right\}\\ && = \frac{C_{k^{\prime}}-C_{j}}{\widetilde{C}_{d}-C_{j}}\left\{\frac{\beta_{k^{\prime}} - \beta_{j}}{C_{k^{\prime}}-C_{j}}(\widetilde{C}_{d}-C_{j})\right.\\ &&\quad \left. - \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}(\widetilde{C}_{d}-C_{j})\right\}\\ && = \frac{C_{k^{\prime}}-C_{j}}{\widetilde{C}_{d}-C_{j}}\left[\left\{\frac{\beta_{k^{\prime}} - \beta_{j}}{C_{k^{\prime}}-C_{j}}(\widetilde{C}_{d}-C_{j}) + \beta_{j}\right\}\right.\\ &&\quad \left. - \left\{ \frac{\beta_{i} - \beta_{j}}{C_{i} - C_{j}}(\widetilde{C}_{d}-C_{j}) + \beta_{j}\right\}\right]. \end{array} $$

  Again, due to conditions of k^′ and j, we have $C_{k^{\prime }}-C_{j} \geq 0$ and $\widetilde {C}_{d}-C_{j} \geq 0$. Moreover, the expression inside the square bracket is non-negative by definitions of i and j. Therefore, the reduced cost of $\alpha _{k^{\prime }}$ is non-negative.

• For the slack variable δ, from the condition that $C_{m} > \widetilde {C}_{d}$, where ${m} = \underset {k}{\text {arg\,min}}\;\beta _{k}$, and from definitions of i and j, we obtain that $\frac {\beta _{i} - \beta _{j}}{C_{i} - C_{j}}\ge 0$. Therefore, thereduced cost for δ is also non-negative.

In conclusion, based on the above arguments and the fact that the reduced costs for basic variables α_i and α_j are zero, all the reduced costs are non-negative. Therefore, the optimality condition is satisfied, which implies that Eq. 15 is the optimal condition. Then plugging (15) into Eqs. 1 and 2, the optimal solutions of the problem areobtained.

Proof

of Corollary 1 From Proposition 4, and conditions $C_{h} \leq \widetilde {C}_{d} < C_{b}$and β_b = min_kβ_k, the optimal solution can be derived directly as:

$$\begin{array}{@{}rcl@{}} \alpha_{b} &=& \frac{\widetilde{C}_{d} - C_{j}}{C_{b} - C_{j}}, \;\; \alpha_{j} = \frac{C_{b} - \widetilde{C}_{d}}{C_{b} - C_{j}}, \\ && \text{where } j = \underset{ j\;\in\;\{h,c,n\}} {\text{ arg\,min }} \frac{\beta_{b} - \beta_{j}}{C_{b} - C_{j}}(\widetilde{C}_{d} - C_{b}) + \beta_{b}. \end{array} $$

(25)

In the definition of index j, only the values of $\frac {\beta _{b} - \beta _{j}}{C_{b} - C_{j}}$ are different. Since $\frac {\beta _{b} - \beta _{j}}{C_{b} - C_{j}}$ is negative, we have

$$\begin{array}{@{}rcl@{}} j &=& \underset{ j\;\in\;\{h,c,n\}} {\text{arg\,min}} \frac{\beta_{b} - \beta_{j}}{C_{b} - C_{j}}(\widetilde{C}_{d} - C_{b}) + \beta_{b}\\ &=& \underset{ j\;\in\;\{h,c,n\}} {\text{ arg\,min }} \left|\frac{\beta_{b} - \beta_{j}}{C_{b} - C_{j}}\right|. \end{array} $$

Expression (18) follows by plugging the results from Eq. 25into Eqs. 1and 2.

Proof

of Corollary 3: From Proposition 4, and conditions $C_{n} \leq \widetilde {C}_{d} < C_{c}$and β_b = min_kβ_k, the optimal solution can be derived directly as:

$$\begin{array}{@{}rcl@{}} \alpha_{n} &=& \frac{\widetilde{C}_{d} - C_{i}}{C_{n} - C_{i}}, \;\; \alpha_{i} = \frac{C_{n} - \widetilde{C}_{d}}{C_{n} - C_{i}}, \\ &&\quad \text{ where } i = \underset{ i\;\in\;\{b,h,c\}} {\text{ arg\,min }} \frac{\beta_{n} - \beta_{i}}{C_{n} - C_{i}}(\widetilde{C}_{d} - C_{n}) + \beta_{n}. \end{array} $$

In the definition of index j,only the values of $\frac {\beta _{n} - \beta _{j}}{C_{n} - C_{j}}$are different. Therefore,

$$\begin{array}{@{}rcl@{}} i &=& \underset{ i\;\in\;\{b,h,c\}} {\mathrm{arg\,min}} \frac{\beta_{n} - \beta_{i}}{C_{n} - C_{i}}(\widetilde{C}_{d} - C_{n}) + \beta_{n}\\ &=& \underset{ i\;\in\;\{b,h,c\}} {\text{ arg\,min }} \frac{\beta_{n} - \beta_{i}}{C_{n} - C_{i}}. \end{array} $$

Note that C_n = 0.Then expression (18) follows by plugging the results of Eq. 25into Eqs. 1and 2.