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How can I use Stata to solve a system of nonlinear equations?

Title   Using Stata to solve a system of nonlinear equations
Authors Alan H. Feiveson, NASA
Kerry Kammire, StataCorp
Isabel Cañette, StataCorp
Date June 2001; updated July 2009

We will show you two ways of solving a system of nonlinear equations in Stata. First, we will provide a detailed explanation using nl. Then we will show you the equivalent in Mata.

Suppose you want to solve

    f1(a1,...,an) = 0
    f2(a1,...,an) = 0
    .
    .
    .
    fn(a1,...,an) = 0

(n nonlinear equations in n unknowns a1,..,an).

First, rewrite the first equation so that its right-hand side is 1:

    f1(a1,...,an) + 1 = 1
    f2(a1,...,an) = 0
    .
    .
    .
    fn(a1,...,an) = 0

Then set up a fake dataset with n observations as follows:

  1. The dependent variable y takes on the value 1 for the first observation and 0 for all the others. Stata’s nl estimation will not work if y is a constant, so you need to write the first equation so that the "right-hand sides" are not all the same; that is why I reformulated the problem above. In this example, I used one for the first observation and zero for the others.
  2. Write an nl program that fills in the dependent variable passed to it with the values of the functions. Suppose n=3. Then your program should look something like this:
 program nlfaq

         syntax varlist (min=1 max=1) [if], at(name)

         tempname a1 a2 a3
         scalar `a1' = `at'[1, 1]
         scalar `a2' = `at'[1, 2]
         scalar `a3' = `at'[1, 3]

         tempvar yh
         generate double `yh' = f1(`a1', `a2', `a3') + 1 in 1
         replace `yh' = f2(`a1', `a2', `a3') in 2
         replace `yh' = f3(`a1', `a2', `a3') in 3

         replace `varlist' = `yh'

 end
nl requires that our program accept an if clause, though we can ignore it in our program because we do not have missing data and will not be restricting the estimation sample when calling nl.
  1. Call nl with y as the dependent variable, specifying initial values for a1, a2, ..., and at which the functions can be evaluated.

Here is an example. Suppose I want to solve the following system for A, B, and C:

  exp(A) + B*C = 3
  A/B + C^2    = log(B)
  A/(A+B+C)    = sin(C)

Here is my nl program:

 program nlfaq
 
        syntax varlist(min=1 max=1) [if], at(name)

        tempname A B C
        scalar `A' = `at'[1, 1]
        scalar `B' = `at'[1, 2]
        scalar `C' = `at'[1, 3]

        tempvar yh
        gen double `yh' = exp(`A') + `B'*`C' - 2 in 1
        replace `yh' = `A'/`B' + `C'^2 - log(`B') in 2
        replace `yh' = `A'/(`A'+`B'+`C') - sin(`C') in 3

        replace `varlist' = `yh'

 end

Now I generate the dataset nl requires:

 . clear
 . set obs 3
 obs was 0, now 3

 . generate y = 0

 . replace y = 1 in 1
 (1 real change made)

I estimate using nl:

 . nl faq @ y, parameters(A B C) initial(A 1 B 1 C 1)
 (obs = 3)
       
 Iteration 0:  residual SS =  .5792985
 Iteration 1:  residual SS =  .0364809
 Iteration 2:  residual SS =  .0001378
 Iteration 3:  residual SS =  2.92e-09
 Iteration 4:  residual SS =  1.43e-18
 Iteration 5:  residual SS =  2.25e-31
 
       Source |       SS       df       MS
 -------------+------------------------------         Number of obs =         3
        Model |           1     3  .333333333         R-squared     =    1.0000
     Residual |  2.2495e-31     0           .         Adj R-squared =         .
 -------------+------------------------------         Root MSE      =         .
        Total |           1     3  .333333333         Res. dev.     = -206.4905

 ------------------------------------------------------------------------------
            y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
 -------------+----------------------------------------------------------------
           /A |   .8973072          .        .       .            .           .
           /B |   1.803287          .        .       .            .           .
           /C |   .3033412          .        .       .            .           .
 ------------------------------------------------------------------------------

Finally, I verify the solution:

 . scalar A = [A]_b[_cons]
 . scalar B = [B]_b[_cons]
 . scalar C = [C]_b[_cons]
 . di exp(A) + B*C
 3
 . di A/B + C^2 "   " log(B)
 .58961117   .58961117
 
 . di A/(A+B+C) "   " sin(C) 
 .29871053   .29871053

Now let’s see how to do this with optimize():

. clear mata

. mata:
------------------------------------------------- mata (type end to exit) ------
: 
: void mysolver(todo, p, lnf, S, H)
>         {
>                 a   = p[1]
>                 b   = p[2]
>                 c   = p[3]
>                 lnf = (exp(a) + b*c - 3 )^2 \   
>                       (a/b + c^2 - log(b))^2 \
>                       (a/(a+b+c) - sin(c))^2    
>         }
note: argument todo unused
note: argument S unused
note: argument H unused

: S = optimize_init()

: optimize_init_evaluator(S, &mysolver())

: optimize_init_evaluatortype(S, "v0")

: optimize_init_params(S, (1,1,1))

: 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             3
    +-------------------------------------------+
  1 |  .8973071614   1.803287112   .3033412094  |
    +-------------------------------------------+

: end
--------------------------------------------------------------------------------

Notice that when using optimize, we don’t need to set up an independent variable y. We just define an evaluator function that, for each observation, evaluates to the square of one of the three expressions that need to be equal to zero.

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