DEA-R: ratio-based comparative efficiency model, its mathematical relation to DEA and its use in applications

Abstract

We propose a new mathematical model for efficiency analysis, which combines DEA methodology with an old idea—Ratio Analysis. Our model, called DEA-R, treats all possible ratios “output/input” as outputs within the standard DEA model. Although DEA and DEA-R generate different summary measures for efficiency, the two measures are comparable. Our mathematical and empirical comparisons establish the validity of DEA-R model in its own right. The key advantage of DEA-R over DEA is that it allows effective integration of the model with experts’ opinions via flexible restrictive conditions on individual “output/input” pairs.

Appendix

Proof of Theorem 1

Let vectors ${\mathbf{a}}$, ${\mathbf{b}}$, ${\mathbf{X}_{k}}$, ${\mathbf{Y}_{k}}$, and matrices ${\mathbf{X}}$, ${\mathbf{Y}}$ and ${\left[ {\mathbf{Y} : \mathbf{X}} \right]}$ be as defined in Sect. 1. Vector ${\left[{\mathbf{bY} : \mathbf{aX}} \right]}$ is defined for any non-negative vector ${\mathbf{b}}$ of the length s and for any non-negative non-trivial vector ${\mathbf{a}}$ of the length m. We will use the symbol ${Q({\mathbf{X},\mathbf{Y}})}$ to denote the set of all such vectors ${[{\mathbf{bY} : \mathbf{aX}}].}$ That is,

$$ Q({\mathbf{X},\mathbf{Y}})=\left\{{[ {\mathbf{bY}:\mathbf{aX}}] \;|\;\mathbf{a}\ge 0, \mathbf{b}\ge 0,\hbox{ L}_\infty (\mathbf{a}) > 0} \right\}. $$

(A.1)

CCR efficiency of a DMU _k is then defined as the maximum value of ${{\mathbf{bY}_k }/{\mathbf{aX}_k }}$ for which the condition ${\mathbf{aX}\ge \mathbf{bY}}$ is satisfied. The term ${{\mathbf{bY}_k }/{\mathbf{aX}_k }}$ is obviously equal to the kth coordinate of the vector ${[{\mathbf{bY} : \mathbf{aX}}]}$, while the condition ${\mathbf{aX}\ge \mathbf{bY}}$ is satisfied if and only if ${\hbox{L}_\infty \left({[{\mathbf{bY} : \mathbf{aX}} ]} \right)\le 1}$. Thus, the original CCR definition of efficiency can also be stated as:

Lemma 1

DEA-CCR efficiency of DMU _k is equal to the maximum value of the kth coordinate among all the vectors from ${Q({\mathbf{X},\mathbf{Y}})}$ whose max norm is not greater than 1.

To establish further equivalent definitions of CCR efficiency, we will need to use the following claims with respect to the set ${Q(\mathbf{X},\mathbf{Y}){:}}$

Claim 1

For any positive scalar K, ${\hbox{K}Q({\mathbf{X},\mathbf{Y}})=Q({\mathbf{X},\mathbf{Y}} ).}$

Proof

For any vector ${\mathbf{u}=\left[{\mathbf{bY} : \mathbf{aX}}\right]}$ from the set ${Q({\mathbf{X},\mathbf{Y}})}$, it is true that ${\hbox{K}\mathbf{u}=\left[ {\left({\hbox{K}\mathbf{b}} \right)\mathbf{Y}:\mathbf{aX}} \right]=\left[ {\mathbf{b}^\prime\mathbf{Y}}:\mathbf{aX}\right]}$, where ${\mathbf{b}^\prime=\hbox{K}\mathbf{b}}$. Since vector ${\mathbf{b}}$ is non-negative and scalar K is positive, vector ${\mathbf{b}^\prime}$ has to be non-negative. This means that ${\hbox{K}\mathbf{u}\in Q\left({\mathbf{X},\mathbf{Y}} \right)}$ and therefore ${\hbox{K}Q\left({\mathbf{X},\mathbf{Y}} \right)\subseteq Q\left( {\mathbf{X},\mathbf{Y}} \right)}$. From the above, it also follows that vector ${\mathbf{v}=\frac{1}{\hbox{K}}\mathbf{u}\in Q\left( {\mathbf{X},\mathbf{Y}} \right)}$. This means that ${\mathbf{u}=\hbox{K}\mathbf{v}}$, and ${\mathbf{u}\in \hbox{K}Q\left( {\mathbf{X},\mathbf{Y}} \right)}$. Hence, ${Q\left( {\mathbf{X},\mathbf{Y}} \right)\subseteq \hbox{K}Q\left( {\mathbf{X},\mathbf{Y}} \right)}$. □

Claim 2

Let ${\mathbf{v}}$ and ${\mathbf{w}}$ be vectors of the length m and s, respectively. Then ${Q\left({\left[ {\mathbf{X}:\mathbf{v}} \right],\left[ {\mathbf{Y}:\mathbf{w}} \right]} \right)=Q\left({\mathbf{X},\mathbf{Y}} \right)}$, i.e., when the matrices ${\mathbf{X}}$ and ${\mathbf{Y}}$ are transformed by dividing their rows by some positive scalars, then the set ${Q({\mathbf{X},\mathbf{Y}})}$ stays unchanged.

Proof

Let ${\mathbf{u}}$ be a vector from the set ${Q\left({\left[{\mathbf{X} : \mathbf{v}}\right],\left[ {\mathbf{Y} : \mathbf{w}}\right]}\right)}$, that is ${\mathbf{u}=\left[{\mathbf{b}\left[{\mathbf{Y} : \mathbf{w}} \right]:\mathbf{a}\left[ {\mathbf{X}:\mathbf{v}} \right]} \right]}$. It can easily be shown that this is equivalent to ${\mathbf{u}=\left[ {\left[ {\mathbf{b} : \mathbf{w}} \right]\mathbf{Y}:\left[ {\mathbf{a} : \mathbf{v}} \right]\mathbf{X}} \right]=\left[ {\mathbf{b}^\prime\mathbf{Y}} : \mathbf{a}^\prime\mathbf{X}\right]}$, where ${\mathbf{a}^\prime=\left[ {\mathbf{a}:\mathbf{v}} \right]}$ and ${\mathbf{b}^\prime=\left[ {\mathbf{b}:\mathbf{w}} \right]}$. Clearly, ${\mathbf{a}^\prime}$ is a non-negative non-trivial vector of the length m, while ${\mathbf{b}^\prime}$ is a non-negative vector of the length s. It follows then that ${\mathbf{u}\in Q\left( {\mathbf{X},\mathbf{Y}} \right)}$, which further means that ${Q\left( {\left[ {\mathbf{X} : \mathbf{v}} \right],\left[ {\mathbf{Y} : \mathbf{w}} \right]} \right)\subseteq Q\left( {\mathbf{X},\mathbf{Y}} \right)}$. On the other hand, if ${\mathbf{u}=\left[ {\mathbf{bY}:\mathbf{aX}} \right]}$ is any vector from ${Q({\mathbf{X},\mathbf{Y}})}$, then we can define non-negative vectors ${\mathbf{a}^{\prime\prime}}$ and ${\mathbf{b}^{\prime\prime}}$ of the length m and s, respectively, as: ${\mathbf{a}^{\prime\prime}=\left[ {a_1 v_1,a_2 v_2,\ldots,a_m v_m} \right]}$ and ${\mathbf{b}^{\prime\prime}=\left[ {b_1 w_1,b_2 w_2,\ldots,b_s w_s }\right]}$. Then, ${\mathbf{a}=\left[ {\mathbf{a}^{\prime\prime} : \mathbf{v}} \right]}$ and ${\mathbf{b}=\left[{\mathbf{b}^{\prime\prime} : \mathbf{w}}\right]}$, and so ${\mathbf{u}=\left[ {\left[ {\mathbf{b}^{\prime\prime} : \mathbf{w}} \right]\mathbf{Y} : \left[{\mathbf{a}^{\prime\prime} : \mathbf{v}}\right]\mathbf{X}} \right]=\left[{\mathbf{b}^{\prime\prime}\left[{\mathbf{Y} : \mathbf{w}}\right] : \mathbf{a}^{\prime\prime}\left[{\mathbf{X} : \mathbf{v}}\right]}\right]}$. We conclude that ${\mathbf{u}\in Q\left({\left[{\mathbf{X} : \mathbf{v}} \right],\left[{\mathbf{Y} : \mathbf{w}}\right]} \right)}$ and that also ${Q\left({\mathbf{X},\mathbf{Y}}\right)\subseteq Q\left( {\left[{\mathbf{X} : \mathbf{v}}\right],\left[ {\mathbf{Y} : \mathbf{w}}\right]} \right)}$. □

Going back to Lemma 1, we can now prove that the maximum value of the kth coordinate among all the vectors from ${Q\left({\mathbf{X},\mathbf{Y}}\right)}$ cannot be attained on those vectors whose max norm is less than 1. Namely, if vector ${\mathbf{u}=\left[ {\mathbf{bY} : \mathbf{aX}} \right]}$ is such that ${\hbox{L}_\infty \left(\mathbf{u} \right) < 1}$, then vector ${\mathbf{v}=\frac{1}{\hbox{L}_\infty (\mathbf{u})}\mathbf{u}}$ is L_∞-unit vector, which according to Claim 1 also belongs to ${Q({\mathbf{X},\mathbf{Y}})}$. It follows then that the kth coordinate of vector ${\mathbf{v}}$ is greater than the kth coordinate of vector ${\mathbf{u}}$ and so the latter one cannot be the maximum kth coordinate among all the vectors ${\left[ {\mathbf{bY}:\mathbf{aX}} \right]}$ whose max norm is not greater than 1. Thus, we have

Lemma 2

CCR efficiency of DMU _k is equal to the maximum value of the kth coordinate among all the L _∞ -unit vectors from ${Q\left({\mathbf{X},\mathbf{Y}}\right)}$.

Now, let ${N\left({\mathbf{X},\mathbf{Y}}\right)}$ be the set of all L_∞-unit vectors from ${Q\left({\mathbf{X},\mathbf{Y}}\right)}$. Taking any vector ${\mathbf{u}=\left[{\mathbf{bY} : \mathbf{aX}}\right]}$ from ${N\left({\mathbf{X},\mathbf{Y}}\right)}$ and dividing it by its kth coordinate we obtain the vector

$$ \mathbf{u}^\prime=\frac{1}{u_k }\mathbf{u}, $$

(A.2)

whose max norm is equal to ${\frac{1}{u_k }}$ and whose kth coordinate is equal to 1. According to Claim 1, vector ${\mathbf{u}^\prime}$ belongs to ${Q\left({\mathbf{X},\mathbf{Y}}\right)}$. Therefore, by the transformation (A.2), set ${N\left({\mathbf{X},\mathbf{Y}}\right)}$ is transformed to a set of vectors from ${Q\left({\mathbf{X},\mathbf{Y}}\right)}$ whose kth coordinate is equal to 1. On the other hand, every vector ${\mathbf{u}^\prime}$ from ${Q\left({\mathbf{X},\mathbf{Y}}\right)}$ whose kth coordinate is equal to 1 can be obtained by applying transformation (A.2) on some vector ${\mathbf{u}}$ from ${N\left({\mathbf{X},\mathbf{Y}}\right)}$. This means that by transformation (A.2), set ${N(\mathbf{X},\mathbf{Y})}$ is transformed to the set of all vectors from ${Q\left({\mathbf{X},\mathbf{Y}} \right)}$, whose kth coordinate is equal to 1. For the latter set we will use the symbol ${N_k \left({\mathbf{X},\mathbf{Y}}\right)}$, that is:

$$ N_k \left({\mathbf{X},\mathbf{Y}}\right)=\left\{{\mathbf{u}\in Q_k \left({\mathbf{X},\mathbf{Y}}\right) \;|\; u_k =1} \right\}. $$

(A.3)

Note that the set of max norms of all the vectors from ${N_k \left({\mathbf{X},\mathbf{Y}}\right)}$ is equal to the set of inverse values of the kth coordinates of all the vectors from ${N\left({\mathbf{X},\mathbf{Y}}\right)}$. Since the maximum of a set of positive numbers is equal to the inverse value of the minimum of inverse values of the elements of that set, we can write:

Lemma 3

CCR efficiency of DMU _k is equal to the inverse value of the smallest possible max norm of all the vectors from ${N_k \left({\mathbf{X},\mathbf{Y}}\right)}$, i.e., ${e_k =\frac{1}{\mathop {\hbox{min} }\limits_{\mathbf{u}\in N_k (\mathbf{X},\mathbf{Y})} L_\infty (\mathbf{u})}}$.

Next, we will transform matrices ${\mathbf{X}}$ and ${\mathbf{Y}}$ using Claim 2 while choosing suitable vectors ${\mathbf{v}}$ and ${\mathbf{w}}$. Let vector ${\mathbf{v}}$ be kth column of matrix ${\mathbf{X}}$ and vector ${\mathbf{w}}$ be kth column of matrix ${\mathbf{Y}}$. For this choice of ${\mathbf{v}}$ and ${\mathbf{w}}$, we will denote matrices ${\left[ {\mathbf{X}:\mathbf{v}} \right]}$ and ${\left[{\mathbf{Y} : \mathbf{w}}\right]}$ as ${\mathbf{X}\left(k\right)}$ and ${\mathbf{Y}\left(k \right)}$, respectively. Clearly, kth columns of matrices ${\mathbf{X}\left(k\right)}$ and ${\mathbf{Y}\left(k \right)}$ are columns of ones. From Claim 2, it follows that ${Q\left({\mathbf{X}\left(k\right),\mathbf{Y}\left(k \right)}\right)=Q\left({\mathbf{X},\mathbf{Y}}\right)}$, and so ${N_k \left({\mathbf{X}\left(k\right),\mathbf{Y}\left(k\right)} \right)=N_k \left({\mathbf{X},\mathbf{Y}}\right)}$. According to ` 3, we can then state that the CCR efficiency of DMU _k is equal to the inverse value of the smallest possible max norm of all the vectors

$$\mathbf{u}=\left[{\mathbf{bY}\left(k \right) : \mathbf{aX}\left(k\right)}\right] $$

(A.4)

from the set ${N_k \left({\mathbf{X}\left(k \right),\mathbf{Y}\left(k \right)}\right)}$.

The above definition of CCR efficiency is useful because the definition of the set ${N_k \left({\mathbf{X}\left(k \right),\mathbf{Y}\left(k \right)}\right)}$ can be simplified. Since the kth columns of matrices ${\mathbf{X}}$ and ${\mathbf{Y}}$ are columns of ones, then the kth coordinate of the vector (A.4) is equal to the quotient of the sums of the coordinates of the vectors ${\mathbf{b}}$ and ${\mathbf{a}}$. Therefore, any vector (A.4) will be an element of the set ${N_k \left({\mathbf{X}\left(k \right),\mathbf{Y}\left(k \right)} \right)}$ if and only if ${\mathbf{a}}$ and ${\mathbf{b}}$ are non-negative vectors whose sums of the coordinates are equal and positive. Further on, if the sum of the coordinates of the vectors ${\mathbf{a}}$ and ${\mathbf{b}}$ is equal to a positive number S ≠ 1, then we can substitute vectors ${\mathbf{a}}$ and ${\mathbf{b}}$ by the vectors ${\mathbf{a}^{\prime}=\frac{1}{\hbox{S}}\mathbf{a}}$ and ${\mathbf{b}^{\prime}=\frac{1}{\hbox{S}}\mathbf{b}}$. This substitution does not change vector (A.4) while the sum of the coordinates of vectors ${\mathbf{a}^{\prime}}$ and ${\mathbf{b}^{\prime}}$ is equal to 1. Consequently, set ${N_k \left({\mathbf{X}\left(k \right),\mathbf{Y}\left(k \right)}\right)}$ is the set of all the vectors (A.4), where ${\mathbf{a}}$ and ${\mathbf{b}}$ are non-negative vectors whose sum of the coordinates is equal to 1, QED.