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I'm new with the Mathematica softwere, so I'm sorry if my request will be stupid. I have to solve this optimization problem: enter image description here

where ps pg and pgs are the variables, and beta,delta and l are parameters with values between 0 and 1. How can I handle with this?

I tried doing this:

Minimize[{0.5 (x1 + x2 + x3 (b (1 - a) + 1 - b) + l (1 - b) (1 - g)), 
  x1 - x2 >= b (1 - g) x3 + (1 - b) (x3 + (1 - g) l), x1 >= 0, 
  x2 >= 0, x3 >= 0, 
  x1 - x2 <= b (x3 + (1 - g) l) + (1 - b) (1 - g) x3}, {x1, x2, x3}]

but I had no result :( Thanks everyone for helping me

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I think you mean g instead of a in your code. Also the help on Minimize has the conditions separated by && instead of commas. It still had trouble not knowing what b, g and l are. I got it to work with

b = .5;
g = .5;
l = .5;

Minimize[{0.5 (x1 + x2 + x3 (b (1 - g) + 1 - b) + l (1 - b) (1 - g)), 
  x1 - x2 >= b (1 - g) x3 + (1 - b) (x3 + (1 - g) l) && 0 <= x1 <= 1 &&
    0 <= x2 <= 1 && 0 <= x3 <= 1 && 
   x1 - x2 <= b (x3 + (1 - g) l) + (1 - b) (1 - g) x3}, {x1, x2, x3}]

(*{0.125, {x1 -> 0.125, x2 -> 0., x3 -> 0.}}*)

The solution is probably not unique with b, g, and l unknown. You will find playing with those values, that there are combinations that have no minimum.

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  • $\begingroup$ b ang g are probabilities, so 0<=b<=1 and same for g. I know also that g>b. l is another parameter...I only know that l>0. The solution has to be x2=x3=0 and x1=(1-b)(1-g)l. The reason is that (copy and paste from the article): "The problem is one of linear programming. Hence, the solution lies on a vertex of the restricted domain of the program. It can be shown that the restrictions pg>=0,pgs>=0 and ps-pg>=b(1-g)pgs+(1-b)(pgs+(1-g)l) must be binding at the optimum." thank you for helping me, best regards PS: why did you put in the constraint that x1,x2,x3<1? $\endgroup$
    – Paolo Marzani
    Commented Dec 7, 2017 at 12:20
  • $\begingroup$ I assumed they were, but it doesn't change the answer. I added your assumptions for b, g, and l and they didn't change anything either. $\endgroup$
    – Bill Watts
    Commented Dec 7, 2017 at 19:10

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