# Proving unequality between two-stage LP and its deterministic problems

 0 Given the following deterministic equivalent definition of two-stage stochastic linear problems: $K = \min_{x} \left\{ c^T{x} + \sum_{k=1}^{M} p_kq_k^Ty_k ~|~ Ax = b; T_kx + W_ky_k = h_k; y_k \geq 0, x \geq 0 \,\,\forall ~k=1,2, \ldots M \right\}$ where $$(T_k, W_k, h_k, y_k)$$ occurs with probability $$p_k > 0 ~\forall ~k = 1,2, \ldots, M$$ Now, I want to prove the following claim: For $$M \geq 2$$, $K \neq p_1 \min_{x} \left\{c^{T}x + q_1^{T} y_1 ~|~ Ax = b; T_1 x + W_1 y_1 = h_1; y_1 \geq 0, x \geq 0 \right\} + \\ \ldots + p_M min_{x} \left \{c^{T}x + q_M^{T} y_M ~|~ Ax = b; T_M x + W_M y_M = h_M; y_M \geq 0, x \geq 0 \right\}$ 1st approach: First, since $$\sum_{k=1}^{M}p_k = 1$$, I planned to establish $min_{x} \left\{c^Tx + \sum_{k=1}^{M} p_kq_k^Ty_k ~|~ Ax = b; T_kx + W_ky_k = h_k; y_k \geq 0, x \geq 0 ~\forall ~k=1,2,\ldots M \right\} \\ > \sum_{k=1}^{M} min_{x} \left\{p_kc^Tx + p_kq_k^Ty_k ~|~ Ax = b; T_kx + W_ky_k = h_k; y_k \geq 0, x \geq 0 \right\}$ However, such inequality seems not to always hold, as we know nothing about the signs of the coefficients of the objective function, besides the fact that the constraints form a polyhedron as a feasible region. The reverse sign is also not correct for every cases, I think. So I abandon this approach. 2nd approach: Define optimal solutions $$x^{*}$$ for K, and $$x_{1}^{*},x_{2}^{*}, \ldots, x_{M}^{*}$$ for $$\min_{x} \left\{p_kc^Tx + p_kq_k^Ty_k ~|~ Ax = b; T_kx + W_ky_k = h_k; y_k \geq 0, x \geq 0 \right\} ~\forall ~k=1,2,\ldots, K$$ By the sake of contradiction, assume $K = \sum_{k=1}^{M} \min_{x} \left\{p_kc^{T}x + p_kq_k^{T} y_k ~|~ Ax = b; T_{k}x + W_{k} y_k = h_k; y_k \geq 0, x \geq 0 \right\}$ This is equivalent to: $$c^Tx^{*} = \sum_{k=1}^{M} p_{k} c^{T} x_k^{*}$$. However, note that $$Ax^{*} = Ax_{1}^{*} = Ax_{2}^{*} = \ldots = Ax_{M}^{*}$$, and $$T_kx^{*} = T_kx_{1}^{*} = \ldots = T_kx_{M}^{*}$$. So, does the first set of equality imply $$x^{*} = x_{1}^{*} = x_{2}^{*} = \ldots = x_{M}^{*}$$, if $$A$$ is an invertible matrix? But if that's the case, shouldn't this means $$\sum_{k=1}^{M} p_kc^Tx_k^{*} = c^Tx^{*} \sum_{k=1}^{M} p_k = c^Tx^{*}$$ (since $$\sum_{k=1}^{M}p_k = 1$$)? But this implies the claim above is wrong... My question: Could someone please help review my attempt above, and let me know how you could arrive at the desired contradiction, or if the claim is actually false? Any other suggestions for other approaches would be really appreciated. asked 28 Feb '18, 03:19 ghjk 63●6 accept rate: 0% fbahr ♦ 4.6k●7●16 FYI -- FAQ: Hey, How do I get that fancy math stuff? > "We have MathJax installed. Use latex between "\$" and "\$" for display math and "\$$" and "\$$" for inline math. Within latex, all backslashes must be doubled." (Sort of paraphrasing here...) -- Also, markdown symbols such as * have to be escaped > "\*". (01 Mar '18, 13:22) fbahr ♦ @fbahr: Thank you so much for your help. I think the problem statement above is false. Could you give it a try? (04 Mar '18, 00:38) ghjk
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