A joint chance-constrained data envelopment analysis model with random output data

Abstract

Data envelopment analysis (DEA) is a mathematical programming approach for evaluating the technical efficiency performances of a set of comparable decision-making units that transform multiple inputs into multiple outputs. The conventional DEA models are based on crisp input and output data, but real-world problems often involve random output data. The main purpose of the paper is to propose a joint chance-constrained DEA model for analyzing a real-world situation characterized by random outputs and crisp inputs. After developing the model, we carry out the following: First, we obtain an upper bound of this stochastic non-linear model deterministically by applying a piecewise linear approximation algorithm based on second-order cone programming; Second, we obtain a lower bound with use of a piecewise tangent approximation algorithm, which is also based on second-order cone programming; and then we use a numerical example to demonstrate the applicability of the proposed joint chance-constrained DEA framework.

Appendices

Appendix A. Proofs of theorems

Proof of Theorem 1

With the use of the standardized normal distribution (see Charnes and Cooper 1963) the chance constraints are transformed into a deterministic form as follows:

$$ Pr\left[ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rk} } \ge f} \right] \ge \alpha \Leftrightarrow Pr\left[ { - \sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rk} } \le - f} \right] \ge \alpha \Leftrightarrow Pr\left[ {z \le \frac{{ - \sum\nolimits_{r = 1}^{s} {u_{r} y_{rk} } - f}}{{\sqrt {\text{var} \left[ {\sum\nolimits_{r = 1}^{s} {u_{r} \tilde{y}_{rk} } } \right]} }}} \right] \ge \alpha , $$

where the variable z has a normal standard distribution (with zero mean and unit variance). Then, we have:

$$ Pr\left[ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rk} } \ge f} \right] \ge \alpha \Leftrightarrow \sum\limits_{r = 1}^{s} {u_{r} y_{rk} } - \varPhi^{ - 1} \left( \alpha \right)\sqrt {\sum\limits_{r = 1}^{s} {u_{r}^{2} var\left( {\tilde{y}_{rk} } \right)} } \ge f, $$

where $\varPhi^{ - 1} \left( \alpha \right)$ is the inverse cumulative distribution function of the standard normal distribution $\varPhi (\alpha )$. Since $\tilde{y}_{rj} ,\,r = 1, \ldots ,s; \,\,j = 1, \ldots ,n$ are independent normal random variables, these joint constraints can be expressed as:

$$ \prod\limits_{j = 1}^{n} {Pr\left\{ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} } - \sum\limits_{r = 1}^{s} {v_{i} x_{ij} } \le 0} \right\}} = Pr\left[ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} } - \sum\limits_{r = 1}^{s} {v_{i} x_{ij} } \le 0,j = 1, \ldots ,n,} \right] \ge \alpha . $$

Therefore, we have

$$ \prod\limits_{j = 1}^{n} {Pr\left\{ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} } - \sum\limits_{r = 1}^{s} {v_{i} x_{ij} } \le 0} \right\}} \ge \alpha^{1} = \alpha^{{\sum\nolimits_{j = 1}^{n} {\lambda_{j} } }} ,\,\lambda_{j} \ge 0,\sum\limits_{j = 1}^{n} {\lambda_{j} } = 1. $$

which yields

$$ \begin{aligned} & \prod\limits_{i = 1}^{m} {Pr\left\{ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} } - \sum\limits_{r = 1}^{s} {v_{i} x_{ij} } \le 0} \right\}} \ge \alpha^{{\sum\nolimits_{j = 1}^{n} {\lambda_{j} } }} = \prod\limits_{j = 1}^{n} {\alpha^{{\lambda_{j} }} } ,\,\,\sum\limits_{j = 1}^{n} {\lambda_{j} } = 1,\,\lambda_{j} \ge 0, \\ & \Leftrightarrow Pr\left\{ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} } - \sum\limits_{r = 1}^{s} {v_{i} x_{ij} } \le 0} \right\} \ge \alpha^{{\lambda_{j} }} ,\,\,j = 1, \ldots ,n,\,\,\sum\limits_{j = 1}^{n} {\lambda_{j} } = 1,\,\lambda_{j} \ge 0. \\ \end{aligned} $$

This completes the proof.$\square$

Proof of Theorem 2

Under the assumption of the standardized normal distribution, the chance constraints in (3) can be converted into a deterministic form as follows:

$$ Pr\left\{ {\sum\limits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} } - \sum\limits_{r = 1}^{s} {v_{i} x_{ij} } \le 0} \right\} \ge \alpha^{{\lambda_{j} }} \Leftrightarrow Pr\left[ {z \le \frac{{ - \sum\nolimits_{r = 1}^{s} {u_{r} y_{rj} } + \sum\nolimits_{r = 1}^{s} {v_{i} x_{ij} } }}{{\sqrt {\text{var} \left[ {\sum\nolimits_{r = 1}^{s} {u_{r} \tilde{y}_{rj} } } \right]} }}} \right] \ge \alpha^{{\lambda_{j} }} $$

where z is a normally distributed random variable with zero mean and unit variance. Then, we have:

$$ \sum\limits_{r = 1}^{s} {u_{r} y_{rj} } + \varPhi^{ - 1} \left( {\alpha^{{\lambda_{j} }} } \right)\sqrt {\sum\limits_{r = 1}^{s} {u_{r}^{2} var\left( {\tilde{y}_{rj} } \right)} } \le \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } ,\,\,j = 1,2, \ldots ,n. $$

The proof is complete.$\square$

Proof of Theorem 3

Let $\mathop {max}\nolimits_{t = 1, \ldots ,T} \left\{ {a_{t}^{{}} \lambda + b_{t}^{{}} } \right\} = z_{t}$ which yields $\,z_{t} \ge a_{t}^{{}} \lambda_{j} + b_{t}^{{}} ,\,j = 1, \ldots ,N,\,t = 1, \ldots ,T.$ The constraint

$$ \sum\limits_{r = 1}^{s} {u_{r} y_{rj} } + \mathop {max}\limits_{t = 1, \ldots ,T} \left\{ {a_{t}^{{}} \lambda_{j} + b_{t}^{{}} } \right\}\sqrt {\sum\limits_{r = 1}^{s} {u_{r}^{2} var\left( {\tilde{y}_{rj} } \right)} } \le \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } ,\,j = 1, \ldots ,N $$

can be converted to the following constraints:

$$ \begin{aligned} & \sum\limits_{r = 1}^{s} {u_{r} y_{rj} } + \sqrt {\sum\limits_{r = 1}^{s} {\left( {z_{tj} u_{r}^{{}} } \right)^{2} var\left( {\tilde{y}_{rj} } \right)} } \le \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } ,\,j = 1, \ldots ,N, \\ & z_{tj} \ge a_{t}^{{}} \lambda_{j} + b_{t}^{{}} ,\,j = 1, \ldots ,N,t = 1, \ldots ,T, \\ & \sum\limits_{j = 1}^{N} {\lambda_{j} } = 1. \\ \end{aligned} $$

Let define new variables $z_{t} u_{r}^{{}} = \bar{z}_{tr} ,\lambda_{j} u_{r} = \bar{u}_{rj}$. Then, we have:

$$ \begin{aligned} & z_{tj} \ge a_{t}^{{}} \lambda_{j} + b_{t}^{{}} \Rightarrow z_{t} u_{r} \ge a_{t}^{{}} \lambda_{j} u_{r} + b_{t}^{{}} u_{r} ,\,j = 1, \ldots ,N,\,r = 1, \ldots ,s \\ & \Rightarrow \bar{z}_{tr} \ge a_{t}^{{}} \bar{u}_{rj} + b_{t}^{{}} u_{r} ,,\,r = 1, \ldots ,s,\,t = 1, \ldots ,T, \\ & \sum\limits_{j = 1}^{N} {\lambda_{j} } = 1 \Rightarrow \sum\limits_{j = 1}^{n} {\bar{u}_{rj} } = u_{r} ,\,\,r = 1, \ldots ,s, \\ \end{aligned} $$

We can now conclude that the second set of constraints in Eq. (4) can be written as:

$$ \sum\limits_{r = 1}^{s} {u_{r} y_{rj} } + \mathop {max}\limits_{t = 1, \ldots ,T} \left\{ {a_{t}^{{}} \lambda_{j} + b_{t}^{{}} } \right\}\sqrt {\sum\limits_{r = 1}^{s} {u_{r}^{2} var\left( {\tilde{y}_{rj} } \right)} } \le \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } ,\,j = 1, \ldots ,N. $$

Therefore, we have the following constraints:

$$ \begin{aligned} & \sum\limits_{r = 1}^{s} {u_{r} y_{rj} } + \sqrt {\sum\limits_{r = 1}^{s} {\bar{z}_{tr}^{2} var\left( {\tilde{y}_{rj} } \right)} } \le \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } ,\,j = 1, \ldots ,N,t = 1, \ldots ,T, \\ & \bar{z}_{tr} \ge a_{t}^{{}} \bar{u}_{rj} + b_{t}^{{}} u_{r} ,\,r = 1, \ldots ,s,\,t = 1, \ldots ,T,\,\,j = 1, \ldots ,N, \\ & \sum\limits_{j = 1}^{n} {\bar{u}_{rj} } = u_{r} ,\,\,r = 1, \ldots ,s, \\ & \bar{z}_{tr} \ge 0,u_{r} \ge 0,\bar{u}_{rj} ,\,r = 1, \ldots ,s,t = 1, \ldots ,T,\,j = 1, \ldots ,n.\, \\ \end{aligned} $$

This completes the proof.$\square$

Proof of Theorem 4

To prove this theorem, consider the following three sets of constraints.

• (I) Since $\alpha \ge 0.5$, we have $\varPhi^{ - 1} \left( \alpha \right) \ge 0$, the expression $- \sum\nolimits_{r = 1}^{s} {u_{r} y_{rk} } + \varPhi^{ - 1} \left( \alpha \right)\sqrt {\sum\nolimits_{r = 1}^{s} {u_{r}^{2} var\left( {\tilde{y}_{rk} } \right)} }$, is a convex function by Lemmas 1, 2 and 3.

• (II) $\sum\nolimits_{r = 1}^{s} {u_{r} y_{rj} } - \sum\nolimits_{i = 1}^{m} {v_{i} x_{ij} } + \sqrt {\sum\nolimits_{r = 1}^{s} {\bar{z}_{tr}^{2} var\left( {\tilde{y}_{rj} } \right)} } \le 0,\,j = 1, \ldots ,N,t = 1, \ldots ,T,$ is a convex function.

• (III) The linear functions $\bar{z}_{tr} \ge a_{t}^{{}} \bar{u}_{rj} + b_{t}^{{}} u_{r} ,$ $\sum\nolimits_{j = 1}^{n} {\bar{u}_{rj} } = u_{r} ,\,r = 1, \ldots ,s,\,t = 1, \ldots ,T,\,\,j = 1, \ldots ,N,$ are convex functions because linear functions are convex.

Therefore, Model (7) is a convex optimization problem and has a global optimal solution.$\square$

Proof of Theorem 5

The proof is given in Beck (2014).$\square$

Proof of Theorem 6

The proof is like that of Theorem 3 and so it is omitted.$\square$

Proof of Theorem 7

Let $X_{P}$ and $X_{T}$ be the feasible region of the constraints of Model (4) for the piecewise Linear approximation and the piecewise tangent approximation of $\varPhi^{ - 1} \left( {\alpha^{{\lambda_{j} }} } \right)$, respectively.

$$ \begin{aligned} & X_{P} = \left\{ {\left( {u,v,\lambda } \right):\sum\limits_{i = 1}^{m} {v_{i} x_{ik} } = 1,\sum\limits_{r = 1}^{s} {u_{r} y_{rj} } + \varPhi^{ - 1} \left( {\alpha^{{\lambda_{j} }} } \right)\sqrt {\sum\limits_{r = 1}^{s} {u_{r}^{2} var\left( {\tilde{y}_{rj} } \right)} } \le \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } ,\,\,\sum\limits_{j = 1}^{n} {\lambda_{j} } = 1,v_{i} \ge 0,u_{r} \ge 0,\lambda_{j} \ge 0} \right\} \\ & \varOmega_{j} = \mathop {max}\limits_{t = 1, \ldots ,T} \left\{ {\varPhi^{ - 1} \left( {\alpha^{{\lambda_{t} }} } \right) + \frac{1}{{\varPhi^{\prime}\left( {\varPhi^{ - 1} \left( {\alpha^{{\lambda_{t} }} } \right)} \right)}}\alpha^{{\lambda_{t} }} ln\left( \alpha \right)\left( {\lambda_{j} - \lambda_{t} } \right)} \right\} \\ & X_{T} = \left\{ {\left( {u,v,\lambda } \right):\sum\limits_{i = 1}^{m} {v_{i} x_{ik} } = 1,\sum\limits_{r = 1}^{s} {u_{r} y_{rj} } + \varOmega_{j} \sqrt {\sum\limits_{r = 1}^{s} {u_{r}^{2} var\left( {\tilde{y}_{rj} } \right)} } \le \sum\limits_{i = 1}^{m} {v_{i} x_{ij} } ,\,\,\sum\limits_{j = 1}^{n} {\lambda_{j} } = 1,v_{i} \ge 0,u_{r} \ge 0,\lambda_{j} \ge 0} \right\} \\ \end{aligned} $$

We know

$$ \varPhi^{ - 1} \left( {\alpha^{{\lambda_{j} }} } \right) \ge \mathop {max}\limits_{t = 1, \ldots ,T} \left\{ {\varPhi^{ - 1} \left( {\alpha^{{\lambda_{t} }} } \right) + \frac{1}{{f\left( {\varPhi^{ - 1} \left( {\alpha^{{\lambda_{t} }} } \right)} \right)}}\alpha^{{\lambda_{t} }} ln\left( \alpha \right)\left( {\lambda_{j} - \lambda_{t} } \right)} \right\} = \varOmega_{j} $$

Then $X_{T} \subseteq X_{P}$, and consequently $\theta_{T}^{*} \le \theta_{P}^{*}$.$\square$

Appendix B. Proofs of lemmas

Proof of Lemma 1

See Beck (2014).$\square$

Proof of Lemma 2

Let $\eta \in \left( {0, 1} \right)$. Then $\varPsi \left( {\eta X_{1} + \left( {1 - \eta } \right)X_{2} } \right) = \sqrt {\eta^{2} X_{1}^{t} VX_{1} + \left( {1 - \eta } \right)^{2} X_{2}^{t} VX_{2}^{{}} + 2\eta \left( {1 - \eta } \right)X_{1}^{t} VX_{2}^{{}} }$, which leads to: $2\eta \left( {1 - \eta } \right)X_{1}^{t} VX_{2}^{{}} \le \eta \left( {1 - \eta } \right)\left[ {X_{1}^{t} VX_{1}^{{}} + X_{2}^{t} VX_{2}^{{}} } \right]$. Clearly,

$$ \begin{aligned} \varPsi \left( {\eta X_{1} + \left( {1 - \eta } \right)X_{2} } \right) & = \sqrt {\eta^{2} X_{1}^{t} VX_{1} + \left( {1 - \eta } \right)^{2} X_{2}^{t} VX_{2}^{{}} + 2\eta \left( {1 - \eta } \right)X_{1}^{t} VX_{2}^{{}} } \\ & \le \eta \sqrt {X_{1}^{t} VX_{1} } + \left( {1 - \eta } \right)\sqrt {X_{2}^{t} VX_{2} } = \eta \varPsi \left( {X_{1} } \right) + \left( {1 - \eta } \right)\varPsi \left( {X_{2} } \right) \\ \end{aligned} $$

Thus, $\varPsi \left( X \right)$ is a convex function.$\square$

Proof of Lemma 3

Clearly, $\alpha^{{\lambda_{j} }}$ is convex and hence $\varPhi^{ - 1} \left( . \right)$ is increasing and convex. It follows that $\alpha^{{t\lambda_{1} + \left( {1 - t} \right)\lambda_{2} }} \le t\alpha^{{\lambda_{1} }} + \left( {1 - t} \right)\alpha^{{\lambda_{2} }}$. Since $\varPhi^{ - 1} \left( . \right)$ is an increasing function, we have $\varPhi^{ - 1} \left( {\alpha^{{t\lambda_{1} + \left( {1 - t} \right)\lambda_{2} }} } \right) \le \varPhi^{ - 1} \left( {t\alpha^{{\lambda_{1} }} + \left( {1 - t} \right)\alpha^{{\lambda_{2} }} } \right) \le t\varPhi^{ - 1} \left( {\alpha^{{\lambda_{1} }} } \right) + \left( {1 - t} \right)\varPhi^{ - 1} \left( {\alpha^{{\lambda_{2} }} } \right)$. Therefore, $\varPhi^{ - 1} \left( {\alpha^{{\lambda_{j} }} } \right)$ is convex.$\square$