Dynamic optimization with three functions, where two of them are subfunctions.

Hello,

I am currently working on a excercise sheet on Matlab-functions, and I am having difficulties to code a function the way it is supposed to work. Basically I have 3 functions, where 2 of them are subfunctions.

The first function:

function c = nutzen(S,x,W)

c = 7 * min(S+x,W) - 2*x-3*(x>0)-1*max(0,S+x-W)

end

The second function:

function Sneu = transformationsfunktion(S,x,W)

Sneu = max(0,S+x-W)

end

Then, both of them need to be incorporated into the third:

function V = wertfunktion(S,t)

      if t==0
          V = 0
      end 

      W = 3

      for x = 0:10
          v(x) = nutzen(S,x,W) + wertfunktion(transformationsfunktion(S,x,W),t-1)
      end

      V = max(v(x))

end

To start this code, I have to give values for S and t, but afterwards the code gets into an endless loop...

Actually, it should for x=0:10 calculate the value for v(x) and then after 11 loops give me the max value. Seems to be quite simple.

Does anyone see where my mistake is?

Any help is much appreciated.

[madhan ravi]

S,x,WS ant t ? values

[Amit Kumar]

S=3, t=2, W=3 (is in the code) and x=0:10. (Actual values aren't so important)

[madhan ravi]

I know that but wanted to know whether t was a vector

[madhan ravi]

            v(x) = nutzen(S,x,W) + wertfunktion(transformationsfunktion(S,x,W),t-1)

you have used the wertfunktion function in the function definition instead it should be after

[Amit Kumar]

Thanks for your answer, but I don't really understand what you mean.

The equation :

v(x) = nutzen(S,x,W) + wertfunktion(transformationsfunktion(S,x,W),t-1)

it is given.

[Torsten]

You call "wertfunktion" recursively. Is this intended ?

[Amit Kumar]

Yes, it is.

[Torsten]

Each time you enter "wertfunktion", "wertfunktion" must again be called 11 times. This leads to an infinite recursion.

By the way:

Arrays start at array(1), so setting v(0) will produce an error.

Further, although you set V=0 for t=0, the following loop is also executed. And if you start with t not equal to 0, it will never become 0 because you don't decrease it anywhere in your code.

Further, max(v(x))=v(x). I guess you mean max(v).

Best wishes

Torsten.

[Amit Kumar]

Thanks for you answer, at least I understand now why I am getting into this endless loop.

I am trying to find the error by using the debugger. By the way: Is v(x) the correct expression to assign a value to v with input x? Because while going through the code I noticed it is not calculating any value for v(x)? It simplies decreases t by 1 and then restarts the recursive function.

[Torsten]

As written, v is an array, and you assign a value to array element number x of v.

Note again that arrays start with array element 1, not 0.

[Amit Kumar]

Ok, this I understand now as well. There is no '0th' element in a row. But my exercise sheet asks me to calculate v(x) starting with x=0.

My current code:

function V = wertfunktion(S,t)
    if t==0
        V = 0
    end 

      W = 3

      for x = 1:10
          v(x) = nutzen(S,x,W) + wertfunktion(transformationsfunktion(S,x,W),t-1)
      end
      q = nutzen(S,0,W) + wertfunktion(transformationsfunktion(S,0,W),t-1)
      v = [q v]
      V = max(v)
  end

Now I have to stop this endless loop.