# How can i minimize

3 views (last 30 days)
Jeffrey Eiyike on 23 Jun 2016
Answered: Titus Edelhofer on 24 Jun 2016
min h
45<= x1 + u1 -u3 -u4 <= 45 + h100
40<= x2 + u2 -u3 +u4 <= 40 + h100
80<= x3 + u3 <= 80 + h100
u1,u2,u3,u4 >= 0
u1<= 170 ,u2<= 50 , u3<= 100, u4<= 70

#### 1 Comment

Torsten on 24 Jun 2016
What does h100 mean ? How is it related to h ?
Best wishes
Torsten.

Titus Edelhofer on 23 Jun 2016
Hi Jeffrey,
there is probably some missing information. I guess it's clear that h>=0. Since there are no restrictions on x1, x2, x3, you might choose
u1 = 0; u2 = 0; u3 = 0; u4 = 0;
x1 = 45; x2 = 40; x3 = 80;
=>
h = 0;
is the optimal solution... ? Or am I missing something?
Titus

#### 1 Comment

Jeffrey Eiyike on 23 Jun 2016
I am suppose to minimise h .

Jeffrey Eiyike on 23 Jun 2016
That's the question on the jpg

Titus Edelhofer on 24 Jun 2016
Hi Jeffrey,
I'm not sure I fully understand. If u1, u2, u3, u4 and x (=x1,x2,x3) are given, the computing lambda is trivial. There is still some information missing.
Titus

#### 1 Comment

Jeffrey Eiyike on 24 Jun 2016
u1<= 170 ,u2<= 50 , u3<= 100, u4<= 70 and x1= 130 x2= 120 x3= 150

Titus Edelhofer on 24 Jun 2016
Hi Jeffrey,
if I'm not mistaken, this should work:
% let the vector of unknowns be [u1 u2 u3 u4 lambda], then we
% have the following inequalities:
% u1 -u3 -u4 -lambda*100 <= 45-x1
% -u1 +u3 +u4 <= -45+x1
% and likewise for the other two
% in matrix form A*x<=b we have:
A = [ ...
1 0 -1 -1 -100;
0 1 -1 1 -100;
0 0 1 0 -100;
-1 0 1 1 0;
0 -1 1 -1 0;
0 0 -1 0 0];
% where b is:
b = [45-130; 40-120; 80-150; -45+130; -40+120; -80+150];
% now solve for the unknowns: we don't care about u1, u2, u3, u4 but lambda should be small:
uLambda = linprog([0;0;0;0;1], A, b, [], [], [0;0;0;0;0], [170;50;100;70;inf])
% extract lambda
lambda = uLambda(5);
Titus