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DATA ENVELOPMENT ANALYSIS 187

weighted outputs of hospital 3
1
weighted inputs of hospital 3
weighted outputs of hospital 4
1
weighted inputs of hospital 4
weights 0

To see how this looks mathematically, let us examine County hospital’s efficiency.
We then have:

36:72u 1 þ 45:98u 2 þ 175u 3 þ 23u 4
Max :
275:7v 1 þ 348:5v 2 þ 104:1v 3
s:t:

48:14u 1 þ 43:1u 2 þ 253u 3 þ 41u 4
1
285:2v 1 þ 123:8v 2 þ 106:72v 3
34:62u 1 þ 27:11u 2 þ 148u 3 þ 27u 4
1
162:3v 1 þ 128:7v 2 þ 64:21v 3
36:72u 1 þ 45:98u 2 þ 175u 3 þ 23u 4
1
275:7v 1 þ 348:5v 2 þ 104:1v 3
33:16u 1 þ 56:46u 2 þ 160u 3 þ 84u 4
1
210:4v 1 þ 154:1v 2 þ 104:04v 3
u 1 ; u 2 ; u 3 ; u 4 ; v 1 ; v 2 ; v 3 0

The objective function is set to optimize the efficiency score by finding the most
favourable values for the weights/decision variables. We then have a constraint for
the efficiency score using the optimum weights for County for each of the four
hospitals in the data set. Each constraint must be no greater than 1 given that we are
assessing efficiency at between 0 and 1. Clearly the constraints are not in the
standard linear format, but this is easily remedied. Transforming the first constraint,
for example, we have:

48:14u 1 þ 43:1u 2 þ 253u 3 þ 41u 4
1
285:2v 1 þ 123:8v 2 þ 106:72v 3
48:14u 1 þ 43:1u 2 þ 253u 3 þ 41u 4 285:2v 1 þ 123:8v 2 þ 106:72v 3
48:14u 1 þ 43:1u 2 þ 253u 3 þ 41u 4 285:2v 1 123:8v 2 106:72v 0

We can transform the other three constraints in the same way. Transforming the
objective function, however, is less obvious. To do so we need to observe that when
we are seeking to maximize a ratio, as in this case, it is the relative magnitude
of the numerator and denominator that are of interest and not their individual
values. We can therefore achieve the same by setting the denominator to some
constant value and then maximize the numerator. Here we shall set the denomi-
nator, the weighted inputs, to take a value of 1. Our objective function then
becomes:
Max :
36:72u 1 þ 45:98u 2 þ 175u 3 þ 23u 4

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