Nonlinear-Nonlinear bilevel problem from [Kleniati & Adjiman, 2014]
Optimal solution
| Objective values | Solution point(s) |
|---|---|
| F* = -10.000 | x* = (1.0, -1.0, -1.0, -1.0, -1.0) |
| f* = -3.100 | y* = (-1.0, -1.0, -1.0, -1.0, -1.0) |
Description in the AMPL format
set I:= 1..5;
set J:= 1..5;
set K:= 1..11;
var x{i in I} >= -1, <= 1; # Outer variables
var y{j in J} >= -1, <= 1; # Inner variables
var l{k in K} >= 0, <= 100; # KKT Multipliers
minimize outer_obj: sum {i in I} -x[i]^2 + sum {j in J} -y[j]^2;
subject to
# Outer constraints:
outer_con_1: -x[1] + y[1]*y[2] <= 0;
outer_con_2: x[2]*y[1]^2 <= 0;
outer_con_3: x[1] + y[3] -exp(x[2]) <= 0;
# Inner constraints:
inner_con: x[1]-y[3]^2 <= 0.2;
# Inner objective:
inner_obj: 0.1*y[3] + y[1]^3 + (x[1]+x[2])*y[2]^2 + (y[4]^2 + y[5]^2)*x[3]*x[4]*x[5] = 0;
# KKT conditions:
stationarity_1: 3*y[1]^2 - l[2] + l[3] = 0;
stationarity_2: 2*(x[1] + x[2])*y[2] - l[4] + l[5] = 0;
stationarity_3: -2*l[1]*y[3] - l[6] + l[7] = -0.1;
stationarity_4: 2*x[3]*x[4]*x[5]*y[4] -l[8] + l[9] = 0;
stationarity_5: 2*x[3]*x[4]*x[5]*y[5] - l[10] + l[11] = 0;
complementarity_1: l[1]*(x[1]-y[3]^2-0.2) = 0;
complementarity_2: l[2]*(-1-y[1]) = 0;
complementarity_3: l[3]*(y[1]-1) = 0;
complementarity_4: l[4]*(-1-y[2]) = 0;
complementarity_5: l[5]*(y[2]-1) = 0;
complementarity_6: l[6]*(-1-y[3]) = 0;
complementarity_7: l[7]*(y[3]-1) = 0;
complementarity_8: l[8]*(-1-y[4]) = 0;
complementarity_9: l[9]*(y[4]-1) = 0;
complementarity_10: l[10]*(-1-y[5]) = 0;
complementarity_11: l[11]*(y[5]-1) = 0;
References
