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## [O] Using allowframebreaks in org-beamer

 From: Stephen J. Barr Subject: [O] Using allowframebreaks in org-beamer Date: Mon, 11 Nov 2013 13:34:14 -0800 User-agent: mu4e 0.9.9.5; emacs 24.3.1

Greetings,

I am trying to get allowframebreaks to work in an org-mode
presentation. I have the following header + slide.

In the slide that is produced, it seems to drop off the slide after the
8th item, and there is no slide with anything about 9. Is there anything
else that I need to add?

Thanks,
-Stephen

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#+OPTIONS: reveal_center:t reveal_progress:t reveal_history:nil
reveal_control:t reveal_mathjax:t num:nil toc:nil
#+REVEAL_TRANS: linear
#+REVEAL_THEME: night
#+REVEAL_HLEVEL: 2
#+ATTR_REVEAL: :frag highlight-red
#+BEAMER_FRAME_LEVEL: 1
#+LaTeX_CLASS_OPTIONS: [mathserif]

* (8) Optimization of the Average Cost Function - Summary
:PROPERTIES:
:BEAMER_env: frame
:BEAMER_envargs: [allowframebreaks]
:END:

*Equations 18-25*
Summary:
1. Write Total Cost $$TC(Q_{1},Q_{2},R_{1},R_{2})$$  and Average Cost
$$AC(Q_{1},Q_{2},R_{1},R_{2}) = TC/T$$.
2. For fixed $$R_{1}$$ and $$R_{2}$$, we need partials of $$AC$$ wrt $$Q_{1}, Q_{2} = 0$$.
3. Derive conditions for an optimal $$Q_{2}$$.
4. There is no proof, but through experience there is a unique $$Q_{2}$$ which
satisfies opt. $$Q_{2}$$ conditions (Eq. (22)).
5. In some cases, Eq. (22) was positive, so sometimes $$Q_{2} = 0$$.
6. Computational experiments for finding $$Q_{2}$$ from (22) in Table 1
7. *Given optimal $$Q_{2}$$, can find optimal $$Q_{1}$$ from  Eq. (21a)*
1. Note that it has not been proved that $$AC(Q_{1},Q_{2} | R_{1},R_{2}$$ is
convex in $$Q_{1},Q_{2}$$, but by computational experience is convex
8. Use heuristics to search for $$R_{2}$$
1. $$Q_{2}^{*}$$ decreasing in $$R_{2}$$
2. Let $$R_{2}'$$ be the smallest value of $$R_{2}$$ which results in
$$Q_{2}* = 0$$.
3. Then, any value of $$R_{2} > R_{2}'$$ is not optimal.
4. Eq. (25) can calculate an upper bound of $$R_{2}$$, called $$R_{2}'$$.
5. Search for $$R_{2} \in [0, R_{2}']$$
9. From experience, $$R_{1}^{*}$$ is always below the optimal $$R$$ in the
standard problem with no emergency ordering.
1. This creates an upper bound for $$R_{1}$$.
2. Search $$R_{1} \in [0,R]$$.

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