# From deterministic to stochastic LP formulations

I am having a hard time understanding the very first example in "A Tutorial on Stochastic Programming".

More specifically the authors show that one can formulate the stochastic variant of (1.2) with Eq (1.8).

## My question:

How should one interpret (1.8) (i.e. how do the new inequalities allow us to explore two different scenarios in the problem?)

# The deterministic problem

## Stochastic re-formulation

(1.8) is a simple reformulation of the deterministic LP (1.2) It's still a deterministic LP. This may be somewhat confusing since the authors are going back forth between a deterministic LP formulation (assuming that $d$ is known) and a stochastic LP formulation (in which we're minimizing the expected value over random values of $d$.) See (1.9) for the expected value formulation corresponding to the deterministic $d$ formulation in (1.8).
• Thank you Brian! I understand it's still a deterministic LP, what I was/am struggling with is how the new variable $t$ and the new constraints work (i.e. what does $t$ represent, and how do these inequalities help us explore the two scenarios mentioned?) – Josh Jun 28 '16 at 17:43
• This is a standard technique in deterministic LP formulation: You can minimize $\max(a,b)$ by minimizing $t$ subject to the constraints $t \geq a$ and $t \geq b$. Here $t$ is an auxilliary variable whose value ends up being the maximum of $a$ and $b$. – Brian Borchers Jun 28 '16 at 17:46
• Well, either $x \geq d$ (with $[x-d]_{+}=x-d$ and $[d-x]_{+}=0$) or $x<d$ (with $[d-x]_{+}=d-x$ and $[x-d]_{+}=0$.) In the first case where $x \geq d$, the value that you want is $cx+h(x-d)$. In the second where $x<d$, the value that you want is $cx+b(d-x)$. So, you take the maximum of $cx+h(x-d)$ and $cx+b(d-x)$. – Brian Borchers Jun 28 '16 at 20:02
• The value of the expression in (1.1) is exactly $cx+h(x-d)$ if $x\geq d$, and the value is $cx+b(d-x)$ if $x < d$. Furthermore, if $x \geq d$, then $cx+h(x-d) \geq cx+b(d-x)$ (keep in mind that $h$ and $b$ are positive.) On the other hand, $x<d$, then $cx+b(d-x) \geq cx+h(x-d)$. – Brian Borchers Jun 28 '16 at 20:57