Fix code to calculate minimizer value in non linear optimization problem

[rohailk]

hi,

I am trying to solve the following problem

The initial point is =(0 0 0)

My code is the following

proc optmodel;
var x{1..3};
min f = x[1]^2  + 2*x[2]^2 + 5*x[3]^2 - 4*x[1] - 20*x[2] - 20*x[3] ;

/* starting point */
   x[1]= 0;
   x[2]= 0;
   x[3]=0;
solve; 
print x x.dual;
quit;

This code works fine

Now if P is the descent direction of f at Xo then

I need to find the minimizer Aplha for this problem

I mean to say that I am trying to solve this question:

I have tried the following code to calculate Alpha

proc optmodel;
   num n = 3;
   var x {1..n};
   min f = x[1]^2  + 2*x[2]^2 + 5*x[3]^2 - 4*x[1] - 20*x[2] - 20*x[3] ;
   
/* starting point */
   x[1]= 0;
   x[2]= 0;
   x[3]=0;
   solve; 
   print x x.dual;

   num p {1..n};
   for {j in 1..n} p[j] = -x[j].dual;
   var alpha >= 0;
   min g  = (x[1]+alpha*p[1])^2 + 2*(x[2]+alpha*p[1])^2 + 5*(x[3]+alpha*p[1])^2-4*(x[1]+alpha*p[1]) -20*(x[2]+alpha*p[1]) -20*(x[3]+alpha*p[1])
   solve;
   print alpha;
quit;

But this code is giving me wrong value of alpha...its showing alpha=0

Please help me to fix this code

[RobPratt]

There are three errors:

1. You need to FIX x for the first solve, as in the previous examples.

2. You need a semicolon at the end of the declaration of g.

3. Some of the appearances of p[1] in g should instead be p[2] or p[3].

The corrected statement is:

min g = (x[1]+alpha*p[1])^2 + 2*(x[2]+alpha*p[2])^2 + 5*(x[3]+alpha*p[3])^2-4*(x[1]+alpha*p[1]) -20*(x[2]+alpha*p[2]) -20*(x[3]+alpha*p[3]);

[rohailk]

Thanks for the quick response.

I fixed the code now like this:

proc optmodel;
   num n = 3;
   var x {1..n};
   min f = x[1]^2  + 2*x[2]^2 + 5*x[3]^2 - 4*x[1] - 20*x[2] - 20*x[3] ;
   
/* starting point */
   x[1]= 0;
   x[2]= 0;
   x[3]=0;
   solve; 
   print x x.dual;

   num p {1..n};
   for {j in 1..n} p[j] = -x[j].dual;
   var alpha >= 0;
   min g = (x[1]+alpha*p[1])^2 + 2*(x[2]+alpha*p[2])^2 + 5*(x[3]+alpha*p[3])^2-4*(x[1]+alpha*p[1]) -20*(x[2]+alpha*p[2]) -20*(x[3]+alpha*p[3]);

   solve;
   print alpha;
quit;

but when I run it. I get the value of alpha=0. Is it ok or need some adjustment in the code please?

thanks

[RobPratt · Accepted solution]

You still need to correct the first error by adding a FIX statement.

[rohailk]

thank you so much..i fixed it like this

proc optmodel;
   num n = 3;
   var x {1..n};
   min f = x[1]^2  + 2*x[2]^2 + 5*x[3]^2 - 4*x[1] - 20*x[2] - 20*x[3] ;
   
/* starting point */
   fix x[1] = 0;
   fix x[2] = 0;
   fix x[3] = 0;

   solve; 
   print x x.dual;

   num p {1..n};
   for {j in 1..n} p[j] = -x[j].dual;
   var alpha >= 0;
   min g = (x[1]+alpha*p[1])^2 + 2*(x[2]+alpha*p[2])^2 + 5*(x[3]+alpha*p[3])^2-4*(x[1]+alpha*p[1]) -20*(x[2]+alpha*p[2]) -20*(x[3]+alpha*p[3]);

   solve;
   print alpha;
quit;