Stata to Mata : The case of Optimize functions

[Youshida Koki]

I am just trying to learn the optimize() function, but cannot find a way to get things going. Let us assume that have
a = ?
b = ?

I have 10 observations and want to find the values of a and b for each observation. Presently, the code uses the equation 1 value of 174 and equation 2 of 22 manually inserted into the Mata code, which produce a = 10, b = 12; but how can I automate the process so that the Stata data is read into the Mata code and get back the results of a and b from the Mata code into Stata.

clear
input float(eq1 eq2)
174 22
 60 14
252 24
  .  .
  .  .
  .  .
  .  .
  .  .
  .  .
  .  .
end

So the optimize function that I have is given below:

 clear mata
 mata

 void mysolver(todo, p, lnf, S, H)
   {
     a   = p[1]
     b   = p[2]
     lnf = ((b^2)+(3*a)-174)^2\   
            (a+b-22)^2   
    }

 S = optimize_init()

 optimize_init_evaluator(S, &mysolver())

 optimize_init_evaluatortype(S, "v0")

 optimize_init_params(S, (1,1)) 

 optimize_init_which(S,  "min" )

 optimize_init_tracelevel(S,"none")

 optimize_init_conv_ptol(S, 1e-16)

 optimize_init_conv_vtol(S, 1e-16)

 p = optimize(S)

p
          1    2
    +-----------+
  1 |  10   12  |
    +-----------+         

 end

[Attaullah Shah]

For now, the immediate solution is given below. I shall write a blog post to provide more details later on

clear
input float(eq1 eq2)
174 22
 60 14
252 24
end

clear mata
 mata

 void mysolver(todo, p, real scalar eq1, real scalar eq2, lnf, S, H)
   {
     a   = p[1]
     b   = p[2]
     
     lnf = ((b^2)+(3*a)-eq1)^2 \   
            (a+b-eq2)^2   
    }

EQ1 = st_data(., "eq1")
EQ2 = st_data(., "eq2")

for (i = 1; i<= 3; ++i){

   eq1 = EQ1[1]

   eq2 = EQ2[1]

  S = optimize_init()

  optimize_init_argument(S, 1, eq1)

  optimize_init_argument(S, 2, eq2)

 optimize_init_evaluator(S, &mysolver())

 optimize_init_evaluatortype(S, "v0")

 optimize_init_params(S, J(1,2,1)) 

 optimize_init_which(S,  "min" )

 optimize_init_tracelevel(S,"none")

 optimize_init_conv_ptol(S, 1e-16)

 optimize_init_conv_vtol(S, 1e-16)

 p = optimize(S)

 p
}
       1    2
    +-----------+
  1 |  10   12  |
    +-----------+
       1   2
    +---------+
  1 |  8   6  |
    +---------+
        1    2
    +-----------+
  1 |   9   15  |
    +-----------+



 end

[Attaullah Shah]

I forgot to change the constants to variables, so you need to change this

eq1 = EQ1[1]
eq2 = EQ2[1]

to this

eq1 = EQ1[i]
eq2 = EQ2[i]

[Jesper Huijs]

I have the following function: Y=R(R-X), the Y and X are known and stated as variables in my dataset with one observation per year. I need to calculate the R for every year, and R has to be non-negative. I am trying to get the implied cost of capital defined by R Easton(2004) method. I know how to do it manually but it will take a lot of time, so i am wondering whether there is a stata function to solve this easily?

[Attaullah Shah]

I have developed such codes details here https://fintechprofessor.com/complet...1-stata-excel/, however, the code is a paid resource.