Data Envelopment Analysis
Data Envelopment Analysis (DEA) is an increasingly popular management tool. This write-up isan introduction to Data Envelopment Analysis (DEA) for people unfamiliar with the technique.For a more in-depth discussion of DEA, the interested reader is referred to Seiford and Thrall [1990]or the seminal work by Charnes, Cooper, and Rhodes [1978].
留學生dissertation網DEA is commonly used to evaluate the eciency of a number of producers. A typical statisticalapproach is characterized as a central tendency approach and it evaluates producers relative to anaverage producer In contrast, DEA compares each producer with only the "best" producers. Bythe way, in the DEA literature, a producer is usually referred to as a decision making unit or DMU.DEA is not always the right tool for a problem but is appropriate in certain cases. (See Strengths
and Limitations of DEA.)In DEA, there are a number of producers. The production process for each producer is to takea set of inputs and produce a set of outputs. Each producer has a varying level of inputs and givesa varying level of outputs. For instance, consider a set of banks. Each bank has a certain numberof tellers, a certain square footage of space, and a certain number of managers (the inputs). Thereare a number of measures of the output of a bank, including number of checks cashed, number ofloan applications processed, and so on (the outputs). DEA attempts to determine which of thebanks are most ecient, and to point out speci c ineciencies of the other banks.
A fundamental assumption behind this method is that if a given producer, A, is capable ofproducing Y(A) units of output with X(A) inputs, then other producers should also be able todo the same if they were to operate eciently. Similarly, if producer B is capable of producingY(B) units of output with X(B) inputs, then other producers should also be capable of the sameproduction schedule. Producers A, B, and others can then be combined to form a compositeproducer with composite inputs and composite outputs. Since this composite producer does notnecessarily exist, it is typically called a virtual producer.The heart of the analysis lies in nding the "best" virtual producer for each real producer. Ifthe virtual producer is better than the original producer by either making more output with the
same input or making the same output with less input then the original producer is inecient. Thesubtleties of DEA are introduced in the various ways that producers A and B can be scaled up ordown and combined.
12.1 Numerical Example
To illustrate how DEA works, let's take an example of three banks. Each bank has exactly 10tellers (the only input), and we measure a bank based on two outputs: Checks cashed and Loan
147
148 CHAPTER 12. DATA ENVELOPMENT ANALYSISapplications. The data for these banks is as follows:
 Bank A: 10 tellers, 1000 checks, 20 loan applications
 Bank B: 10 tellers, 400 checks, 50 loan applications#p#分頁標題#e#
 Bank C: 10 tellers, 200 checks, 150 loan applications
Now, the key to DEA is to determine whether we can create a virtual bank that is better thanone or more of the real banks. Any such dominated bank will be an inecient bank.Consider trying to create a virtual bank that is better than Bank A. Such a bank would use nomore inputs than A (10 tellers), and produce at least as much output (1000 checks and 20 loans).Clearly, no combination of banks B and C can possibly do that. Bank A is therefore deemed to beecient. Bank C is in the same situation.
However, consider bank B. If we take half of Bank A and combine it with half of Bank C, thenwe create a bank that processes 600 checks and 85 loan applications with just 10 tellers. Thisdominates B (we would much rather have the virtual bank we created than bank B). Bank B istherefore inecient.
Another way to see this is that we can scale down the inputs to B (the tellers) and still have atleast as much output. If we assume (and we do), that inputs are linearly scalable, then we estimatethat we can get by with 6.3 tellers. We do that by taking .34 times bank A plus .29 times bank B.
The result uses 6.3 tellers and produces at least as much as bank B does. We say that bank B'seciency rating is .63. Banks A and C have an eciency rating of 1.
12.2 Graphical Example
The single input two-output or two input-one output problems are easy to analyze graphically.The previous numerical example is now solved graphically. (An assumption of constant returns toscale is made and explained in detail later.) The analysis of the eciency for bank B looks like thefollowing:
O
20
50
150
200 400 1000 Checks
Loans
C
B
A
V
12.3. USING LINEAR PROGRAMMING 149
If it is assumed that convex combinations of banks are allowed, then the line segment connectingbanks A and C shows the possibilities of virtual outputs that can be formed from these two banks.Similar segments can be drawn between A and B along with B and C. Since the segment AC liesbeyond the segments AB and BC, this means that a convex combination of A and C will createthe most outputs for a given set of inputs.
This line is called the eciency frontier. The eciency frontier de nes the maximum combina-tions of outputs that can be produced for a given set of inputs.
Since bank B lies below the eciency frontier, it is inecient. Its eciency can be determinedby comparing it to a virtual bank formed from bank A and bank C. The virtual player, called V,is approximately 54% of bank A and 46% of bank C. (This can be determined by an applicationof the lever law. Pull out a ruler and measure the lengths of AV, CV, and AC. The percentage ofbank C is then AV/AC and the percentage of bank A is CV/AC.)The eciency of bank B is then calculated by nding the fraction of inputs that bank V wouldneed to produce as many outputs as bank B. This is easily calculated by looking at the line fromthe origin, O, to V. The eciency of player B is OB/OV which is approximately 63%. This gurealso shows that banks A and C are ecient since they lie on the eciency frontier. In other words,any virtual bank formed for analyzing banks A and C will lie on banks A and C respectively.#p#分頁標題#e#
Therefore since the eciency is calculated as the ratio of OA/OV or OA/OV, banks A and C willhave eciency scores equal to 1.0.
The graphical method is useful in this simple two dimensional example but gets much harder in
higher dimensions. The normal method of evaluating the eciency of bank B is by using an linear
programming formulation of DEA.
Since this problem uses a constant input value of 10 for all of the banks, it avoids the com-
plications caused by allowing di erent returns to scale. Returns to scale refers to increasing or
decreasing eciency based on size. For example, a manufacturer can achieve certain economies of
scale by producing a thousand circuit boards at a time rather than one at a time - it might be only
100 times as hard as producing one at a time. This is an example of increasing returns to scale
(IRS.)
On the other hand, the manufacturer might nd it more than a trillion times as dicult to
produce a trillion circuit boards at a time though because of storage problems and limits on the
worldwide copper supply. This range of production illustrates decreasing returns to scale (DRS.)
Combining the two extreme ranges would necessitate variable returns to scale (VRS.)
Constant Returns to Scale (CRS) means that the producers are able to linearly scale the inputs
and outputs without increasing or decreasing eciency. This is a signi cant assumption. The
assumption of CRS may be valid over limited ranges but its use must be justi ed. As an aside,
CRS tends to lower the eciency scores while VRS tends to raise eciency scores.
12.3 Using Linear Programming
Data Envelopment Analysis, is a linear programming procedure for a frontier analysis of inputs and
outputs. DEA assigns a score of 1 to a unit only when comparisons with other relevant units do not
provide evidence of ineciency in the use of any input or output. DEA assigns an eciency score
less than one to (relatively) inecient units. A score less than one means that a linear combination
of other units from the sample could produce the same vector of outputs using a smaller vector of
inputs. The score reects the radial distance from the estimated production frontier to the DMU
under consideration.
150 CHAPTER 12. DATA ENVELOPMENT ANALYSIS
There are a number of equivalent formulations for DEA. The most direct formulation of the
exposition I gave above is as follows:
Let Xi be the vector of inputs into DMU i. Let Yi be the corresponding vector of outputs. Let
X0 be the inputs into a DMU for which we want to determine its eciency and Y0 be the outputs.
So the X's and the Y 's are the data. The measure of eciency for DMU0 is given by the following
linear program:
Min 
s:t: PiXi  X0
PiYi  Y0
  0
http://www.400190.com/Thesis_Writing/#p#分頁標題#e#where i is the weight given to DMU i in its e orts to dominate DMU 0 and  is the eciency
of DMU 0. So the 's and  are the variables. Since DMU 0 appears on the left hand side of
the equations as well, the optimal  cannot possibly be more than 1. When we solve this linear
program, we get a number of things:
1. The eciency of DMU 0 (), with  = 1 meaning that the unit is ecient.
2. The unit's \comparables" (those DMU with nonzero ).
3. The \goal" inputs (the di erence between X0 and PiXi)
4. Alternatively, we can keep inputs xed and get goal outputs ( 1
 Pi Yi)
DEA assumes that the inputs and outputs have been correctly identi ed. Usually, as the number
of inputs and outputs increase, more DMUs tend to get an eciency rating of 1 as they become
too specialized to be evaluated with respect to other units. On the other hand, if there are too few
inputs and outputs, more DMUs tend to be comparable. In any study, it is important to focus on
correctly specifying inputs and outputs.
Example 12.3.1 Consider analyzing the eciencies of 3 DMUs where 2 inputs and 3 outputs are
used. The data is as follows:
DMU Inputs Outputs
1 5 14 9 4 16
2 8 15 5 7 10
3 7 12 4 9 13
The linear programs for evaluating the 3 DMUs are given by:
 LP for evaluating DMU 1:
min THETA
st
5L1+8L2+7L3 - 5THETA <= 0
14L1+15L2+12L3 - 14THETA <= 0
9L1+5L2+4L3 >= 9
4L1+7L2+9L3 >= 4
16L1+10L2+13L3 >= 16
L1, L2, L3 >= 0
12.3. USING LINEAR PROGRAMMING 151
 LP for evaluating DMU 2:
min THETA
st
5L1+8L2+7L3 - 8THETA <= 0
14L1+15L2+12L3 - 15THETA <= 0
9L1+5L2+4L3 >= 5
4L1+7L2+9L3 >= 7
16L1+10L2+13L3 >= 10
L1, L2, L3 >= 0
 LP for evaluating DMU 3:
min THETA
st
5L1+8L2+7L3 - 7THETA <= 0
14L1+15L2+12L3 - 12THETA <= 0
9L1+5L2+4L3 >= 4
4L1+7L2+9L3 >= 9
16L1+10L2+13L3 >= 13
L1, L2, L3 >= 0
The solution to each of these is as follows:
 DMU 1.
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficien Increase Decrease
$B$10 theta 1 0 1 1E+30 1
$B$11 L1 1 0 0 0.92857142 0.619047619
$B$12 L2 0 0.24285714 0 1E+30 0.242857143
$B$13 L3 0 0 0 0.36710963 0.412698413
Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$B$16 IN1 -0.103473 0 0 1E+30 0
$B$17 IN2 -0.289724 -0.07142857 0 0 1E+30
$B$18 OUT1 9 0.085714286 9 0 0
$B$19 OUT2 4 0.057142857 4 0 0
$B$20 OUT3 16 0 16 0 1E+30
152 CHAPTER 12. DATA ENVELOPMENT ANALYSIS
 DMU 2.
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficien Increase Decrease#p#分頁標題#e#
$B$10 theta 0.773333333 0 1 1E+30 1
$B$11 L1 0.261538462 0 0 0.866666667 0.577777778
$B$12 L2 0 0.22666667 0 1E+30 0.226666667
$B$13 L3 0.661538462 0 0 0.342635659 0.385185185
Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$B$16 IN1 -0.24820512 0 0 1E+30 0.248205128
$B$17 IN2 -0.452651 -0.0666667 0 0.46538461 1E+30
$B$18 OUT1 5 0.08 5 10.75 0.655826558
$B$19 OUT2 7 0.0533333 7 1.05676855 3.41509434
$B$20 OUT3 12.78461538 0 10 2.78461538 1E+30
 DMU 3.
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficien Increase Decrease
$B$10 theta 1 0 1 1E+30 1
$B$11 L1 0 0 0 1.08333333 0.722222222
$B$12 L2 0 0.283333333 0 1E+30 0.283333333
$B$13 L3 1 0 0 0.42829457 0.481481481
Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$B$16 IN1 -0.559375 0 0 1E+30 0
$B$17 IN2 -0.741096 -0.08333333 0 0 1E+30
$B$18 OUT1 4 0.1 4 16.25 0
$B$19 OUT2 9 0.066666667 9 0 0
$B$20 OUT3 13 0 13 0 1E+30
Note that DMUs 1 and 3 are overall ecient and DMU 2 is inecient with an eciency rating
of 0.773333.
Hence the ecient levels of inputs and outputs for DMU 2 are given by:
 Ecient levels of Inputs:
0:261538 " 5
14 #+ 0:661538 " 7
12 # = " 5:938
11:6 #
12.3. USING LINEAR PROGRAMMING 153
 Ecient levels of Outputs:
0:261538 2
64
9
4
16
375
+ 0:661538 2
64
4
9
13
375
=2
64
5
7
12:785
375
Note that the outputs are at least as much as the outputs currently produced by DMU 2 and
inputs are at most as big as the 0.773333 times the inputs of DMU 2. This can be used in two
di erent ways: The inecient DMU should target to cut down inputs to equal at most the ecient
levels. Alternatively, an equivalent statement can be made by nding a set of ecient levels of
inputs and outputs by dividing the levels obtained by the eciency of DMU 2. This focus can then
be used to set targets primarily for outputs rather than reduction of inputs.
Alternate Formulation
There is another, probably more common formulation, that provides the same information. We
can think of DEA as providing a price on each of the inputs and a value for each of the outputs.
The eciency of a DMU is simply the ratio of the inputs to the outputs, and is constrained to be
no more than 1. The prices and values have nothing to do with real prices and values: they are an
arti cial construct. The goal is to nd a set of prices and values that puts the target DMU in the
best possible light. The goal, then is to
Max uT Y0
vTX0
s:t:
uT Yj#p#分頁標題#e#
vTXj
 1; j = 0; :::; n;
uT  0;
vT  0:
Here, the variables are the u's and the v's. They are vectors of prices and values respectively.
This fractional program can be equivalently stated as the following linear programming problem
(where Y and X are matrices with columns Yj and Xj respectively).
Max uT Y0
s:t: vTX0 = 1;
uT Y