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I encountered this lemma in a research paper related to End-to-End inventory management model.

Please note that $d_{[t_1,t_2]} = \sum_{t=t_1}^{t_2} d_i$, where $d_t$ denotes demand at time instance t. $I_{t_1}$ denotes the inventory level at time $t_1$.

I could understand why $f(a)$ is a convex function. But I didn't understand why is it a piece-wise function. Subsequenly, I am not able to visualize how the function has extreme points at $d_{[t_1,s]}-I_{t_1}$?

Can you please help me understand how $f(a)$ is a piece-wise function and how it has extreme points at $d_{[t_1,s]}-I_{t_1}$?

Proof of Lemma

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For a linear function $g(a)$, the function $g(a)^+=\max(g(a),0)$ is piecewise linear (with two pieces). Also, finite combinations (weighted sums) of piecewise linear functions are piecewise linear. So $f(a)$ is piecewise linear. Because $h\ge 0$ and $b\ge 0$, $f(a)$ is also convex. For a convex piecewise linear function, the claim that “the optimal solution should be one of the extreme points” is a bit too strong because one of the slopes could be $0$, yielding an infinite number of optimal solutions. But it is true that at least one extreme point is optimal.

To find the extreme points, note that for $\max(g(a),0)$, the extreme point occurs where $g(a)=0$.

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  • $\begingroup$ Can you please help out with how the extreme points came out to be $d_{[t_1,s]}−I_{t_1}$? I cant figure it out even though I have been trying to understand it for hours now. $\endgroup$ Commented Jun 2 at 15:04

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