Hi, Iâ€™m chemical engineer in Korea.
I want to solve some kind of bilevel optimization.
Problem statement.
The problem is that there is 10 m by 10 m square area and there are three sensor. Their location vectors are (x(k),y(k)) where k is number of sensor thus in this model it is 3.
A explosion is occurred on the point (xz, yz) and the probability(P) that sensor can detect this explosion is (1exp(3/d^2)) where d is Euclidean distance between sensor and explosion.
I want to find the optimal sensor location which maximize the product series of detection probability(P1P2P3).
Thus, I have to make model like
Max { min (product series of detection probability(P1P2P3))}
x,y xz,yz
Ã¨ This means that maximizing the (minimum probability at the fixed sensor location)
Questions

I think this is bilevel programming, but obj of leader and follower is same. And I donâ€™t know how to formulate this. I want to know the code bilevel optimization for solving this problem. Alternatively, I make the grid and find all minimum probability under all three sensor location combination varying the explosion vector. Then find the maximum value of probability among this minimum solutions. However, it take too long time. I attach the MatlabGAMS file for this.

I saw the EMP solver and doing for example of bilevel programming. Actually, bilevel programming solution usually like pareto curve. But they give just point. How can I search the set of solution.
My GAMS CODE
set k/1*3/;
variables
xz
yz
minimum;
parameters
x(k)
y(k)
$gdxin tstdat
$load x y
$gdxin
equations
maxi ;
maxi… minimum =e= 1*(1exp(3/((power(xzx(‘1’),2)+power(yzy(‘1’),2))+0.00000000000000000001)))*(1exp(3/((power(xzx(‘2’),2)+power(yzy(‘2’),2))+0.00000000000000000001)))1(1exp(3/((power(xzx(‘3’),2)+power(yzy(‘3’),2))+0.00000000000000000001)));
xz.up=10;
xz.lo=0;
yz.up=10;
yz.lo=0;
Model super/all/;
OPTION nlp=LINDOGLOBAL;
solve super using nlp minimizing minimum;
display x,y, xz.l, yz.l, minimum.l
execute_unload ‘tstsol’, xz,yz,minimum;
My MATLAB CODE
clc;
clear all;
m=0
for xi=4:8
for xj=4:8
for xk=4:8
for yi=4:8
for yj=4:8
for yk=4:8
Xe = [1*(xi1) 1*(xj1) 1*(xk1)]
Ye = [1*(yi1) 1*(yj1) 1*(yk1)]
xe.name = ‘x’;
xe.type = ‘parameter’;
xe.val = Xe’;
xe.form = ‘full’;
xe.dim = 1;
ye.name = ‘y’;
ye.type = ‘parameter’;
ye.val = Ye’;
ye.form = ‘full’;
ye.dim = 1;
wgdx (‘tstdat’, xe,ye);
gams(‘zone’)
rs1.name = ‘xz’;
rs1.form = ‘full’;
r1 = rgdx (‘tstsol’, rs1);
xz = r1.val;
rs2.name = ‘yz’;
rs2.form = ‘full’;
r2 = rgdx (‘tstsol’, rs2);
yz = r2.val;
rs3.name = ‘minimum’;
rs3.form = ‘full’;
r3 = rgdx (‘tstsol’, rs3);
minimum = r3.val;
m=m+1
Rx(m)=xz
Ry(m)=yz
RRx1(m)=Xe(1);
RRx2(m)=Xe(2);
RRx3(m)=Xe(3);
RRy1(m)=Ye(1);
RRy2(m)=Ye(2);
RRy3(m)=Ye(3);
min(m)=minimum
end
end
end
end
end
end
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