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 ######################################## #+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 \(R_{1} \in [0,R]\). ########################################