86
Chapter 3
Example 3.12 Calculation of the test function
k
With the data in Example 3.10, the vector of residuals is given by Eq. (5) of Example 3.10 and the test
function is found to be
h = eT
P-*f
1 0
l
(2.0
U )
0.117
0.201
0.201
1.887
2.0
0.364
(1)
2
Since rank(R) = 2, the number of degrees of freedom for the^T
distribution is 2, and from Table 3.5 it is
seen that
h
is far too small to raise any suspicion about the quality of the measurements._________________
The test of the data quality is easily carried out, but one cannot from the result that
h > x 1
at, say,
the 97.5% level conclude whether the unsatisfactorily large errors are due to a systematic defect of
the data or to large random errors. However, in many cases one may suspect that one of the
measurements has a systematic error, and if by leaving this measurement out of the analysis the
errors become non-significant it is reasonable to assume that this particular measurement contains a
systematic error. Thus the approach illustrated above can be used for error diagnosis, but only when
the system is over-determined by at least two measurements, i.e., rank(R) ^ 2. Otherwise, when
one measurement is left out the system is no longer over-determined, and the analysis cannot be
carried out. The concept of eliminating one measurement at a time is very simple, as illustrated in
Example 3.13, but it may be time-consuming if no information about the possible source of error is
available; i.e., one has to repeat the calculations with each of the measured rates left out of the
analysis. A more systematic approach for error diagnosis is found in Heiiden
et al
(1994a,b).
Example 3.13 Error diagnosis of yeast fermentation
We now consider the data of von Meyenberg (1969) for high dilution rates. In Example 3.5 it was
concluded that some ethanol was probably missing. As in Example 3.10 we have two non-measured
rates, ammonia utilization and formation of water, whereas there are five measured rates: consumption of
glucose and oxygen and production of carbon dioxide, ethanol and biomass. Thus
' I
0
1
1
1 ^
^0
o '
2
0
0
3
1.83
E c =
3
2
1
2
2
0.5
0.56
0
1
0
0
0
0.17;
,1
0,
and the redundancy matrix is found to be
1
0
1
1
1
0
-0 .2 8 6
-0 .2 8 6
0.143
0.014
0
0.571
0.571
-0 .2 8 6
-0 .0 2 9
0
0.857
0.857
-0 .4 2 9
-0 .0 4 3
(1)
(
2
)