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186 CHAPTER 4 LINEAR PROGRAMMING APPLICATIONS

However, a simple examination of Table 4.17 suggests that such comparison is
not going to be easy given that we have multiple inputs and multiple outputs.
County hospital, for example, has a higher output of trained nurses than University
hospital but a lower output of trained paramedics. And to make matters worse
their inputs are different also! To progress, let us introduce a simple expression of
efficiency:
output
Efficiency ¼
input

That is, an organization will be more efficient the more output it produces in relation
to its input. However, if we have multiple outputs and inputs, we must show this as
some sort of average output to average input. Clearly, a simple average of outputs/
inputs will not work given the varied units of measurement, so instead we have:

weighted outputs
Efficiency ¼
weighted inputs
That is, an organization’s efficiency is the ratio of its weighted outputs to its
weighted inputs. Conventionally, efficiency is shown between 0 and 1 (or 0 to
100). However, we are still left with the problem of weights. How do we determine
suitable weights? After all, different hospitals may place differing priorities on
individual outputs and inputs. County hospital, for example, may decide to put more
importance on out-patient treatments rather than paramedics trained while City
hospital may put a higher weight on paramedics trained and less on in-patient days
provided. DEA resolves this problem very simply. The DEA approach is to find the
set of weights for hospital x that maximizes that particular hospital’s efficiency score
(and with the word maximizes we are back in LP territory once more). By definition,
those weights will optimize hospital x’s efficiency score. We then use those same
weights to calculate the efficiency score for the other hospitals in the data set.
Clearly those weights may not be the optimal weights for the other hospitals.
However, by looking at the efficiency scores calculated, we can determine hospital
x’s efficiency relative to the other hospitals. For example, let us suppose that hospital
x’s efficiency score was 0.8 (on a scale from 0 to 1) whilst that for hospital y was 0.9 –
calculated using the optimal weights for hospital x. We know that hospital x’s weights
are optimal for hospital x – this is the best efficiency score possible. And yet another
hospital, using weights that may not be optimal for itself, has a higher efficiency. We
would conclude that hospital x was relatively inefficient compared to hospital y.
Hospital y is achieving more outputs and/or less input than hospital x.
So, we are looking to find a set of weights for hospital x that maximizes its efficiency
score. That efficiency score will be relative to the efficiency scores of the other hospitals
in the data set and all efficiency scores will be constrained to be between 0 and 1. We
shall use as our decision variables, u i and v j ,where u i refer to the output weights and v j
refer to the input weights. In summary we will then have:
weighted outputs of hospital x
Max:
weighted inputs of hospital x
s:t:
weighted outputs of hospital 1
1
weighted inputs of hospital 1
weighted outputs of hospital 2
1
weighted inputs of hospital 2

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