How can I get a calculated function value???

I want to calculate Lagrangian fuction.

L(x_wt11, x_wt21, x_wt12, x_wt22) = Vt1 * ( (1-p_wt11)^x_wt11 * (1-p_wt21)^x_wt21 )...
    + Vt2 * ( (1-p_wt12)^x_wt12 * (1-p_wt22)^x_wt22 )...
    - lambda1 * ( 1-(x_wt11 + x_wt12))...
    - lambda2 * ( 1-(x_wt21 + x_wt22))...
    - lambda3 * ( Vw1 * (1-s_wt11/u)^x_wt11 * (1-s_wt12/v)^x_wt12...
                  + Vw2 * (1-s_wt21/u)^x_wt21 * (1-s_wt22/v)^x_wt22 - 2 * beta);

When I input L(0,0,0,0) (or L(1,0,0,0), L(0,1,1,0)...etc)

the answer is not calculated value, although I declare all parameters.

In this equation, "x_wt11, x_wt21, x_wt12, x_wt22" are binary variable,

and anothers are parameters(constant).

I declare parameters before this equation except 'u, v'

because 'u,v' are dependent on "x_wt11, x_wt21, x_wt12, x_wt22" value.

In summary,

1. declare parameters. (except 'u,v')

2. declare Lagrangian function.

3. declare parameters 'u, v'

* The reason why I declaring 'u,v' afterwards, I want to prevent deviding by zero.

* u = x_wt11 + x_wt21 , v = x_wt12 + x_wt22

* So, if "x_wt11=0", "x_wt21=0", and then u is zero.

* To solve this problem, I use "if" statement.

if (x_wt11+x_wt21) == 0
    u=1;
elseif (x_wt11+x_wt21) == 1
    u=1;
elseif (x_wt11+x_wt21) == 2
    u=2;
else 
    u=0;
end

And then, if I input "L(0,0,0,0)"

the answer is like this way, "52/(22*u) + 69/(22*v) + 74/56"

I declare parameters 'u,v' but, Why the answer is not a calculated??? and How I can solve this problem???

Answer 1

[Steven Lord]

Since you're mentioning that you evaluate the function and then define a value for u, I assume you're using symbolic variables from Symbolic Math Toolbox. Let's take a slightly simpler example.

syms x a
f(x) = a*x;
y = f(2)

y = 

When I create y, MATLAB substitutes 2 in place of x in the symfun body. Giving a value to the variable a after f(x) has been created doesn't change f(x).

a = 3;
z = f(2)

z = 

Using subs to substitute the numeric value of a into z will perform the substitution.

z2 = subs(z) % or subs(z, 'a', a)

z2 = 

6

Alternately you could make your function a function of multiple symbolic variables:

syms w b
g(w, b) = b*w;
evaluateWithValueForw = g(4, b)

evaluateWithValueForw = 

evaluateWithValueForb = g(w, 5)

evaluateWithValueForb = 

evaluateWithValueForBoth = g(4, 5)

evaluateWithValueForBoth = 

20

[연승 김]

Thank you very much!!!!

As you advise, it works!!!!!

However, another trouble occur.

The result using subs as you advised, is calculated value but I cannot get the value I wanted.

The result using subs is like "921162842338651293/562949953421312".

921162842338651293/562949953421312 = 1.6363e+03

So, I want to get the value like "1.6363e+03" at once.

I try like below

z2 = subs(z);
z3 = subs(z2);

but I cannot get the value like "1.6363e+03"

Could you give me one more advice???

Thank you!

[Steven Lord]

If the symbolic expression does not contain any symbolic variables, use double to convert it to a double array. If it does contain symbolic variables, use vpa.

[연승 김]

Wow!!!!! Thank you!!

"double" works!!!

I had to spend too much time on this trouble. uu...

Thank you again!

Happy new year and have a nice day~! smile^^

Answer 2

[SaiDileep Kola]

I see no reason why it is not calculated after declaring all values, check if values "u" and "v" are present in the workspace and try to run the equation once again if you see those values are present.

Answer 3

[연승 김]

Thank you for your advice.

I checked it before, but the result is not changed...

I find a trick for this, but it is not a ultimate solution.

The trick is that,

after I receive an answer "52/(22*u) + 69/(22*v) + 74/56",

and I input "52/(22*u) + 69/(22*v) + 74/56".

And then, the anwser is calculated value.

(I don't know why...)

But I want to receive a calculated value at once.

[SaiDileep Kola]

If possible can you share your code snipped and commands being executed, so that I can reproduce the issue at my end

[연승 김]

Thank you for your kind.

Now I'm trying to some tricks and if I solve the trouble, I'll contact to you.

Thank you!!