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Given an optimization problem as follows: $$ \begin{array}{cc} \operatorname{Max} Z=3 x_{1}+9 x_{2}-2 y_{1}-4 y_{2} \\ \text { subject to, } y_{1}+y_{2}=15 \\ 5 x_{1}+2 x_{2} \leq 10 \\ x_{1}, x_{2}, y_{1}, y_{2} \geq 0 \end{array} $$ Assume that variables $x_{1}$ and $x_{2}$ are uncertain and that there are three different scenarios, $(1.3 X, X, 0.6 X)$ for the values of $x_{1}$ and $x_{2}$, each occurring with a probability of $1 / 3$.
(a) Formulate the scenario based stochastic models as a deterministic equivalent model
(b) Reformulate the problem as an Expected Value problem.

(a) Deterministic equivalent model is: $$ \begin{aligned} \operatorname{Max} z=& \frac{1}{3}(1.3)\left(3 x_{11}+9 x_{21}\right)+\frac{1}{3}(1)\left(3 x_{12}+9 x_{22}\right) \\ &+\frac{1}{3}(0.6)\left(3 x_{13}+9 x_{23}\right)-2 y_{1}-4 y_{2} \end{aligned} $$ subject to: $$\quad y_{1}+y_{2}=15$$ $$ \begin{aligned} 1.3\left(5 x_{11}+2 x_{21}\right) \leq 10 \\ 1\left(5 x_{12}+2 x_{22}\right) \leq 10 \\ 0.6\left(5 x_{13}+2 x_{23}\right) \leq 10 \end{aligned} $$ where $x_{i s}, y_{i} \geq 0 ; i=1,2 ; s=1,2,3$

(b) I have confusion with this problem. What I did,

$$\mathbb E[q]=\left( \frac13 (1.3)\left(3,9\right)+\frac13 (1)\left(3,9\right)+\frac13 (0.6)\left(3,9\right)\right)=\left(2.9,8.7\right)$$

$$\mathbb E[W]=\left( \frac13 (1.3)\left(5,2\right)+\frac13 (1)\left(5,2\right)+\frac13 (0.6)\left(5,2\right)\right)=\left(4.83333,1.93333\right)$$ $$ \begin{aligned} \operatorname{Max} z=& \left(2.9 x_{1}+8.7 x_{2}\right)-2 y_{1}-4 y_{2} \end{aligned} $$ subject to: $$\quad y_{1}+y_{2}=15$$ $$ \begin{aligned} 4.83333 x_{1}+1.93333 x_{2} \leq 10 \end{aligned} $$ where $x_{i}, y_{i} \geq 0 ; i=1,2 $;

Is my approach is correct?

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