Bilevel problem with 3 VI's at lower level
emp
JAMS
GAMS
short
= 1
(u,v) solves VI( [v+y+z-1; u-log(v)], {(u,v) | u >= 0, v >= 0 } )
y solves VI( y+z+3, { y | y free } )
Note that the two VI's (due to the definitional sets) correspond respectively
to a complementarity problem:
0 <= u \perp v + y + z - 1 >= 0
0 <= v \perp u - log(v) >= 0
and a linear equation:
y + z + 3 = 0
The starting value for v is needed to protect the evaluation of log(v).
Contributor: Michael Ferris and Jan-H. Jagla, December 2009
$offtext
positive variable u;
variables v, y, z;
equations f1,f2,f3,h;
f1.. v + y + z =n= 1;
f2.. u =e= log(v);
f3.. y + z =n= -3;
h.. exp(z) + y =e= 2;
v.lo = 0;
v.l = 1;
z.lo = 1;
model mpec /all/;
$onecho > %emp.info%
bilevel
vi f1 u
vi f2 v
vi f3 y
$offecho
solve mpec using emp min z;
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