A spatiotemporal Data Envelopment Analysis (S-T DEA) approach: the need to assess evolving units

Abstract

One of the major challenges in measuring efficiency in terms of resources and outcomes is the assessment of the evolution of units over time. Although Data Envelopment Analysis (DEA) has been applied for time series datasets, DEA models, by construction, form the reference set for inefficient units (lambda values) based on their distance from the efficient frontier, that is, in a spatial manner. However, when dealing with temporal datasets, the proximity in time between units should also be taken into account, since it reflects the structural resemblance among time periods of a unit that evolves. In this paper, we propose a two-stage spatiotemporal DEA (S-T DEA) approach, which captures both the spatial and temporal dimension through a multi-objective programming model. In the first stage, DEA is solved iteratively extracting for each unit only previous DMUs as peers in its reference set. In the second stage, the lambda values derived from the first stage are fed to a Multiobjective Mixed Integer Linear Programming model, which filters peers in the reference set based on weights assigned to the spatial and temporal dimension. The approach is demonstrated on a real-world example drawn from software development.

Appendices

Appendix 1: Proof of Proposition 1

Let us assume that a DMU(3) is inefficient and $\lambda _1$, $\lambda _2$ are two non-zero lambdas of its reference set corresponding to DMU(1) and DMU(2) respectively, and assume that $\lambda _1 >\lambda _2$, such that:

$$\begin{aligned}&\lambda _1 +\lambda _2 =1 \end{aligned}$$

(24)

$$\begin{aligned}&0\le \lambda _1 ,\lambda _2 \le 1 \end{aligned}$$

(25)

According to the temporal dimension, DMU(3) is closer to DMU(2), while according to the spatial dimension DMU(3) has a higher resemblance to DMU(1).

The corresponding efficiency from (3) will be $\varphi \le \frac{1}{y_3}\cdot \left( {\lambda _1\cdot y_1+\lambda _2\cdot y_2}\right)$ (where $y_3$ is the output of the DMU under study) and due to (24), the above inequality will be reformulated as follows:

$$\begin{aligned} \begin{array}{l} \varphi \le \frac{1}{y_3 }\cdot \left( {\lambda _1 \cdot y_1 +\lambda _2 \cdot y_2 } \right) \Leftrightarrow \\ \varphi \le \frac{1}{y_3 }\cdot \left[ {\lambda _1 \cdot y_1 +\left( {1-\lambda _1 } \right) \cdot y_2 } \right] \Leftrightarrow \\ \varphi \le \frac{1}{y_3 }\cdot \left[ {y_2 +\left( {y_1 -y_2 } \right) \cdot \lambda _1 } \right] \\ \end{array} \end{aligned}$$

(26)

The efficiency provided by the S-T DEA model will be the following:

$$\begin{aligned} {\hat{\varphi }}\le & {} \frac{y_1 }{y_3 },\,\hbox {if}\,w_t <w_{sp} \end{aligned}$$

(27)

$$\begin{aligned} {\hat{\varphi }}\le & {} \frac{y_2 }{y_3 },\,\hbox {if}\,w_t >w_{sp} \end{aligned}$$

(28)

Therefore, the efficiency ${\hat{\varphi }}$ is calculated from (9) by selecting the combination of $y_1$, $y_2$ that maximizes the value of ${\hat{\varphi }}$ satisfying the constraint. In the right hand side of inequalities (27) and (28), lambda values are omitted due to the constraints (9) and (11) of the S-T DEA model.

In order to prove that $\varphi \ge {\hat{\varphi }}\forall y$ the relative position of $\varphi$ and ${\hat{\varphi }}$ by means of order relationships should be investigated. For this reason, two scenarios about the arrangement of $y_1$ and $y_2$ are examined:

• $y_1<y_2$

In this case, $y_2+\left( {y_1-y_2}\right) \cdot \lambda _1<y_2$ as $\left( {y_1-y_2}\right) \cdot \lambda _1<0$ (for the sake of simplicity denominator $y_3$ is dropped from the analysis since it is equal to all instances). To investigate the order relation of $y_2+\left( {y_1-y_2}\right) \cdot \lambda _1$ with $y_1$, we reformulate the right hand side of inequality (26) with respect to $y_1$ as follows:

$$\begin{aligned} y_1 +\left( {y_2 -y_1 } \right) \cdot \lambda _2 \end{aligned}$$

(29)

From (29), it can be seen that as $\left( {y_2-y_1}\right) \cdot \lambda _2>0$, then $y_1+\left( {y_2-y_1}\right) \cdot \lambda _2>y_1$. Therefore, the following order relationship is formulated:

$$\begin{aligned} y_1<y_1+\left( {y_2-y_1}\right) \cdot \lambda _2=y_2+\left( {y_1-y_2}\right) \cdot \lambda _1<y_2 \end{aligned}$$

(30)

The initial assumption $y_1<y_2$ is reformulated as $y_2=y_1+k$, where k is a real non-negative number. By substitution in (26), the following is derived:

$$\begin{aligned} \begin{array}{l} \varphi \le \frac{1}{y_3 }\cdot \left[ {\lambda _1 \cdot y_1 +\lambda _2 \cdot \left( {y_1 +k} \right) } \right] \Leftrightarrow \\ \varphi \le \frac{1}{y_3 }\cdot \left[ {\left( {\lambda _1 +\lambda _2 } \right) \cdot y_1 +\lambda _2 \cdot k} \right] \Leftrightarrow \\ \varphi \le \frac{1}{y_3 }\cdot \left( {y_1 +\lambda _2 \cdot k} \right) \\ \end{array} \end{aligned}$$

(31)

Comparing the right hand sides of inequalities (27) and (31), it is obvious that $y_1+\lambda _2\cdot k>y_1$, consequently, $\varphi \ge {\hat{\varphi }}$.

• $y_1 >y_2$

In this case $y_1 +\left( {y_2 -y_1 } \right) \cdot \lambda _2 <y_1$ as $\left( {y_2 -y_1 } \right) \cdot \lambda _2 <0$. To investigate the order relationship with $y_2$, we reformulate (22) with respect to $y_2$ as follows:

$$\begin{aligned} y_2 +\left( {y_1 -y_2 } \right) \cdot \lambda _1 \end{aligned}$$

(32)

From (30), it can be seen that as $\left( {y_2 -y_1 } \right) \cdot \lambda _2 >0$, then $y_2 +\left( {y_1 -y_2 } \right) \cdot \lambda _1>y_2$. Therefore, the following order relationship is formulated:

$$\begin{aligned} y_2 <y_2 +\left( {y_1 -y_2 } \right) \cdot \lambda _1 =y_1 +\left( {y_2 -y_1 } \right) \cdot \lambda _2 <y_1 \end{aligned}$$

(33)

The initial assumption $y_1 <y_2$ is reformulated as $y_1 =y_2 +m$, where m is a real non-negative number. By substitution in (26), the following is derived:

$$\begin{aligned} \begin{array}{l} \varphi \le \frac{1}{y_3 }\cdot \left[ {\lambda _1 \cdot \left( {y_1 +m} \right) +\lambda _2 \cdot y_2 } \right] \Leftrightarrow \\ \varphi \le \frac{1}{y_3 }\cdot \left[ {\left( {\lambda _1 +\lambda _2 } \right) \cdot y_2 +\lambda _1 \cdot m} \right] \Leftrightarrow \\ \varphi \le \frac{1}{y_3 }\cdot \left( {y_2 +\lambda _1 \cdot m} \right) \\ \end{array} \end{aligned}$$

(34)

Comparing the right hand sides of inequalities (28) and (34), it is obvious that $y_2 +\lambda _1\cdot m>y_2$, consequently, $\varphi \ge {\hat{\varphi }}$.

Appendix 2: Proof of Proposition 2

Let $\lambda _i^{DEA}\in {\mathcal {F}}^{DEA}$ be the optimal lambdas of DEA output model described by (1)–(4), $\lambda _i^{S-T\,DEA}\in {\mathcal {F}}^{S-T\,DEA}$ be the optimal lambdas of S-T DEA model, described by (9)–(13), and ${\mathcal {F}}^{\bullet }$ be the efficiency set of model $\bullet$. By construction of tables A and ${\varvec{\Delta }}$, a lambda that appears in ${\mathcal {F}}^{DEA}$ will also appear in ${\mathcal {F}}^{S-T\,DEA}$. From the S-T DEA model, only one weight dependent solution will be selected among the lambdas provided by the initial DEA model (1)–(4) and thus, the feasibility of the model is guaranteed and furthermore it stands that: ${\mathcal {F}}^{S-T\,DEA}\subseteq {\mathcal {F}}^{DEA}$.