Geoscience Reference
In-Depth Information
u -values algebraically by forward substitution, for example with Equation (12.11) as
follows
y 1
x 1
u 1 =
(12.14)
y 2
x 2 u 1
x 1
u 2 =
and so on for u 3 ,
u 4 ,...,
u n . This could also be done by backward substitution, starting
with u n ,
as follows
y m
x p
u n =
(12.15)
y m 1
x p 1 u n
x p
u n 1 =
and so on for the remaining values u n 2 ,
u n 3 ,...,
u 1 . This procedure would be fine if the
data were accurate and if the real system in nature were indeed to perform as formulated
in these equations. Unfortunately, hydrologic data are invariably subject to considerable
error and natural catchments tend to exhibit some nonlinear and non-stationary features
in their response characteristics. Therefore, an optimal solution must be sought, which
makes use of all m available equations. Among the more common solution methods of
a set of equations like (12.9) and (12.11) are those based on the least squares criterion
(see Snyder, 1955) and on other mathematical programming techniques (see Deininger,
1969; Diskin and Boneh, 1973, Box et al. , 1994).
The method of least squares
The underlying criterion in this method consists of the minimization of the sum of
the squares of the differences between the measured data y i and the calculated values
x k u i k + 1 in Equation (12.9). These differences are called the residuals, say
ε i .A
simple example will illustrate how their squares can be minimized.
Example 12.4. Application of least squares method
Consider again the simple case described by Equation (12.13) and illustrated in Figure
12.8. The sum of the squares of the residuals is
i ε
i
x 1 u 1 ) 2
x 2 u 1 ) 2
x 3 u 1 ) 2
=
( y 1
+
( y 2
x 1 u 2
+
( y 3
x 1 u 3
x 2 u 2
x 3 u 2 ) 2
x 3 u 3 ) 2
x 3 u 4 ) 2
(12.16)
+
( y 4
x 1 u 4
x 2 u 3
+
( y 5
x 2 u 4
+
( y 6
ε
i
This sum can be minimized by putting
/∂
u i =
0 for each value of i . This yields
=
respectively for i
1 and 2
( y 1
x 1 u 1 ) x 1 +
( y 2
x 1 u 2
x 2 u 1 ) x 2 +
( y 3
x 1 u 3
x 2 u 2
x 3 u 1 ) x 3 =
0
and
( y 2
x 1 u 2
x 2 u 1 ) x 1 +
( y 3
x 1 u 3
x 2 u 2
x 3 u 1 ) x 2
+
( y 4
x 1 u 4
x 2 u 3
x 3 u 2 ) x 3 =
0
(12.17)
Search WWH ::




Custom Search