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

The name of the test program is

If we use `epsilonf`=0.1 and `epsilon`=0, we get the first solution
with the following number of boxes and computation time:

full bisection | smear bisection | |

direct storage | 71, 31200ms | 38, 20220ms |

reverse storage | 35, 14710ms | 38, 20280ms |

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 33 boxes against 38 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. But it must be noted that the use of the Hessian allows to
reduce this overestimation and consequently the differences in term of
used boxes between the two bisection mode is slightly reduced compared
to the one we have observed when using only the Jacobian (see
section 2.4.3.4).