[O] Using allowframebreaks in org-beamer

[Stephen J. Barr]

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
#+BEAMER_THEME: Madrid
#+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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