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###

Example 4

In this example (see section 15.1.3) we deal with a complex problem
of three equations in three unknowns
.
We are looking for a solution in the domain:

The system has a solution which is approximatively:

This problem is extremely ill conditioned as for the `TestDomain`
the functions intervals are:

The name of the test program is `Test_Solve_General_Gradient`.
With `espsilonf`=0 and `epsilon`=0.001 and if we stop at the
first solution we find with the maximum equation ordering:

with 73 boxes using the direct storage mode (with the reverse
storage mode only 37 boxes are needed).
With the maximum middle-point equation ordering we find the same intervals
with 67 boxes in the direct storage mode (41 for the
reverse storage mode).
The importance of normalizing the functions has been mentioned in
section 2.3.5.4. But in this example the use of the Jacobian matrix
enable to drastically reduce the computation time.
If we use `epsilonf`=0.1 and `epsilon`=0 and if we stop at the first solution we
find an exact solution using 73 boxes:

in a time which is about 1/100 of the time necessary when we
don't use
the Jacobian.
In the single bisection smear mode (i.e. only one variable is
bisected in the process) the same root is obtained in 21080ms (about
50% less time than when using the full bisection mode) with
only 40 boxes in the direct storage mode.

Note that for `epsilonf`=0.1 and `epsilon`=0 we find the only root with 73
boxes in 39760ms
(41 boxes and 24650ms in the single bisection smear mode).

Note that we may improve the efficiency of the procedure by using
simplification procedures such as the 2B (section 2.17)
and the 3B method. An interesting point in this example is that the
bisection mode 1 (bisecting along the variable whose interval has the
largest diameter) is more effective than using the bisection mode 2
(using the smear function) with 53 boxes against 108 for the mode 2
for `epsilonf`=1e-6 and `epsilon`=1e-6.
This can easily been explained by the complexity of the Jacobian
matrix elements that leads to a large overestimation of their values
when using interval: in that case the smear function is not very
efficient to determine which variable has the most influence on the
equations.

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Jean-Pierre Merlet
2012-12-20