The examination of pseudo-allocative and pseudo-overall efficiencies in DEA using shadow prices

Abstract

The original Data Envelopment Analysis (DEA) models developed by Charnes et al. (Eur J Oper Res 2:429–444, 1978), Banker et al. (Manag Sci 30:1078–1092, 1984) were both radial models. These models and their varied extensions have remained the most popular DEA models in terms of utilization. The benchmark targets they determined for inefficient units are primarily based on the notion of maintaining the same input and output mixes originally employed by the evaluated unit (i.e. disregarding allocative considerations). This paper presents a methodology to investigate allocative and overall efficiency in the absence of defined input and output prices. The benchmarks determined from models based on this methodology will consider all possible input and/or output mixes. Application of this methodology is illustrated on a model of the financial intermediary function of a bank branch network.

Appendix

Proposition

For any proposed target production ($\alpha \bar{X}$, αy _o), one of the sets of weights corresponding to a fully-defined hyperplane will be of minimal per unit cost.

Proof

First we need to define what is meant by a fully-defined hyperplane. Characterize any enveloping (i.e. efficient) hyperplane by q, the number of (linearly independent) efficient DMUs it contains, and f, the total number of zero input or output weights that characterize this hyperplane. For a production possibility space characterized by m inputs and s outputs, a fully-defined enveloping hyperplane will have $q + f = m + s.$

Begin with any arbitrary set of optimal weights $(v^{*} ,u^{*} ,\omega_{o}^{*} )$ for DMU_o. These weights correspond to an efficient hyperplane upon which DMU_o lies, but not necessarily an actual facet. Let the augmented weight vector ${\mathbf{w}}^{*} = ( - v^{*} ,u^{*} ,\omega_{\text{o}}^{*} ),$ and represent DMU_o by the augmented vector $\mathbf{z}_\mathbf{o} = (x_{o} ,y_{o} ,1).$ Therefore,

$$ \mathbf{w}^{*} \cdot {\mathbf{z}}_{{\mathbf{o}}} = 0\quad ({\text{i}} . {\text{e}} .-v^{*} \cdot x_{o} + u^{*} \cdot y_{o} + \omega_{o}^{*} = 0) $$

This efficient hyperplane can be “rotated” through a direction ${\mathbf{d}} = ( - v,u,\omega_{o} )$ by a degree γ, subject to ${\mathbf{d}} = {\text{k}}{\mathbf{w}}^{*},$ where k is any scalar. The resulting hyperplane will be characterized by the augmented weight vector ${\mathbf{w^{\prime}}} = {\mathbf{w}}^{*} + \gamma {\mathbf{d}},$ where γ is limited such that u′, v′ ≥ 0. For DMU_o to remain on this hyperplane (i.e. maintain its technical efficiency),

$$ \begin{aligned} & {\mathbf{w^{\prime}}} \cdot {\mathbf{z}}_\mathbf{o} = \, 0 \\ & ( - v^{*} - \gamma v) \cdot x_{o} + (u^{*} + \gamma u) \cdot y_{o} + (\omega_{o}^{*} + \gamma \omega_{o} ) = 0 \\ & \gamma (-v \cdot x_{o} + \, u \cdot y_{o} + \omega_{o} ) \, = \, 0 \\ & {\mathbf{d}} \cdot {\mathbf{z}}_\mathbf{o} = \, 0 \\ \end{aligned} $$

The new weight vector needs to be rescaled to maintain normalization of inputs (i.e. $\bar{v} \cdot x_{0} = 1$).

$$ (v^{*} + \gamma v) \cdot x_{o} = \, 1 \, + \gamma (v \cdot x_{o} ) $$

$\therefore\,{\text{let}}\;\bar{w} = ( - \bar{v},\bar{u},\bar{\omega }_{\text{o}} ) = \frac{w\prime }{{1 + \gamma v \cdot X_{\text{o}} }}$

For the production possibility ($\bar{X}$, y _o ), $\bar{u} \cdot y_{o} + \bar{\omega }_{o} = 1$ (since DMU_o is efficient). Thus the unit cost for producing output mix y _o is $\bar{v} \cdot \bar{X}$. The change in cost from “rotating” the hyperplane through d by a degree γ is:

$$ \bar{v} \cdot \bar{X} - v^{*} \cdot \bar{X} = \frac{{(v^{*} + \gamma v) \cdot \bar{X}}}{{1 + (\gamma v \cdot x_{o} )}} - v^{*} \cdot \bar{X} = \frac{{\gamma [v \cdot \bar{X} - (v \cdot x_{o} ) \cdot (v^{*} \cdot \bar{X})]}}{{1 + (\gamma v \cdot x_{o} )}} $$

(7)

The denominator will be positive, i.e. $1 + \gamma (v \cdot x_{o} ) = (v^{*} + \gamma v) \cdot x_{o} > 0$, since $v^{\prime} = v^{*} + \gamma v > 0 \, (v^{\prime} \ne 0\quad {\text{since}}\,{\mathbf{d}} \ne k{\mathbf{w}}^{*} )$. Thus, the sign of the resultant change will depend on the sign of $\Updelta_{d} = [v \cdot \bar{X} - (v \cdot X_{o} ) \cdot (v^{*} \cdot \bar{X})]$ and γ. For convenience, limit γ to be positive as a negative value for γ can be represented by “rotation” through the direction −d (which is a valid direction since $- {\mathbf{d}} \cdot {\mathbf{z}}_{{\mathbf{o}}} = 0$). Thus, for a given production possibility, the sign of the change will be independent of γ, but be determined by d (or specifically v′).

If Δ _d  = 0, then the “cost” of (${\bar{\text{X}}}$, y _o ) is indifferent to rotation of the hyperplane through d. If Δ _d  > 0 (<0), then cost increases (decreases) with increasing γ. To see this, re-express Eq. 7 as follows:

$$ \bar{v} \cdot \bar{X} - v^{*} \cdot \bar{X} = \frac{{[v \cdot \bar{X} - (v \cdot x_{o} ) \cdot (v^{*} \cdot \bar{X})]}}{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 \gamma }}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$\gamma $}} + (v \cdot x_{o} )}} $$

Thus, as γ increases, the denominator in Eq. 7 decreases.

For any given target (represented by λ), the goal of the maximization portion of the minimax program is to choose a set of weights (i.e. a hyperplane) to maximize the associated unit “cost”. Assume that an optimal solution w* exists that does not correspond to a fully-defined hyperplane. This plane contains q efficient DMUs (q ≥ 1 since the hyperplane must contain DMU_o) and f zero weights. This hyperplane can be rotated to some positive extent along any direction d that satisfies:

$$ \begin{aligned} {\mathbf{d}} \cdot \user2{z}_{j} & = \, 0\quad {\text{ for each DMU on the hyperplane}}, \\ v_{i} & = \, 0\quad {\text{if}}\,v_{i}^{*} = 0{\text{ and}} \\ u_{r} & = \, 0\quad {\text{if}}\,u_{r}^{*} = 0. \\ \end{aligned} $$

Specifically, rotation will be possible until another constraint in the program becomes binding (i.e. another input or output weight becomes zero, or another efficient DMU joins the q DMUs on the hyperplane). Note that for any d that satisfies the above conditions, −d will also.

Hence, it can be shown that there will exist a fully-defined hyperplane w ^f that will contain each of the q DMUs and have $w_{i}^{f} = 0$ if $w_{i}^{*} = 0$. (This can be arrived at by successive rotations through directions d _i , each introducing another DMU or zero weight into the hyperplane, until a fully-defined hyperplane is achieved). Define ${\mathbf{d}} = {\mathbf{w}}^{f} -{\mathbf{w}}^{*}$, which will satisfy both ${\mathbf{d}} \cdot {\mathbf{z}}_{{\mathbf{o}}} = 0$ and $- {\mathbf{d}} \cdot {\mathbf{z}}_{{\mathbf{o}}} = 0$, since ${\mathbf{w}}^{{\mathbf{f}}} \cdot {\mathbf{z}}_{{\mathbf{o}}} = 0$ and ${\mathbf{w}}^{*} \cdot {\mathbf{z}}_{{\mathbf{o}}} = 0$. If Δ _d  > 0 (<0), then any positive rotation along d (−d) would increase the cost for the target. However, w* is optimal (i.e. maximal cost), thus Δ _d  = 0 and the cost does not change with respect to rotation through d. Specifically, choosing γ = 1 results in the fully-defined hyperplane w ^f.

Thus if any hyperplane that was not fully-defined represented an optimal solution to the minimax problem for a particular target, that will exist a fully-defined hyperplane that is also optimal. Thus, only fully-defined hyperplanes need be considered. Q.E.D.