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| author | Thibaut <thibaut.horel@gmail.com> | 2011-12-22 18:19:20 -0800 |
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| committer | Thibaut <thibaut.horel@gmail.com> | 2011-12-22 18:19:20 -0800 |
| commit | 02f0fa6e179f3e3d6d6c4be22f1257c866d831af (patch) | |
| tree | 36e49a83ebc8e2f9a8f51cbff384f5fa531e7beb /notes.tex | |
| parent | 6cd9d223c6beb635b9eec27bf666602551af879e (diff) | |
| download | adscheduling-02f0fa6e179f3e3d6d6c4be22f1257c866d831af.tar.gz | |
Diffstat (limited to 'notes.tex')
| -rw-r--r-- | notes.tex | 50 |
1 files changed, 42 insertions, 8 deletions
@@ -1,6 +1,7 @@ \documentclass{article} \usepackage[utf8]{inputenc} -\usepackage{amsmath} +\usepackage{amsmath,amsthm} +\newtheorem{lemma}{Lemma} \begin{document} \section{Introduction} @@ -137,7 +138,7 @@ solution: \end{enumerate} \subsection{Trivial Case} -For now we will assume \emph{Fractional allocation}. In this case we +From now on we will assume \emph{Fractional allocation}. In this case we can now serve each request exactly, \eqref{eq:constraint} is automatically true, and we only need to find a subset $A$ of advertisers that maximizes: @@ -165,17 +166,50 @@ the knapsack. To make the problem more interesting we add the following two constraints: -\begin{itemize} -\item if one ad appears on more than one screen in a branch of the - tree, the screens must be contiguous -\item no ad must be seen on more than two screens -\end{itemize} +\begin{enumerate} +\item\label{con:cont} if one ad appears on more than one screen in a branch, the screens must be contiguous +\item\label{con:limit} no ad must be seen on more than two screens +\end{enumerate} \subsubsection{Case of the path} +An allocation will be called \emph{feasible} if it satisfies the two +afore-mentioned constraints. + +\begin{lemma} + Reordering the ads by decreasing order of $n_i$ in a feasible + allocation only breaks the constraint \ref{con:limit}. by one screen: + no ad is seen on more than three contiguous screens. +\end{lemma} + +\begin{proof} + Case disjunction and lots of trivial inequalities but it works. The + fact that the capacities of the nodes are decreasing along a branch is + important (this is the first time the topology of the graph is used). +\end{proof} + +This lemma gives a way to find an approximation of the optimal +solution using dynamic programming: consider the ads by increasing +order of $n_i$ and start packing them from the bottom. + +\paragraph{Question:} the $n_i$ are not necessarily integers, we can +do rounding. What is the typical precision needed? + +\paragraph{Price of anarchy:} The greedy dynamic programming algorithm +will have a global welfare greater than the optimal feasible +allocation. Looking at how the algorithm proceeds, for each +advertiser, we exceed the optimal solution by at most the capacity of +the node we are currently filling. Does this show that the price of +anarchy is bounded by 2? It seems that this gives a bound on the Price +of Stability. This bound could potentially be proven tight by a simple +example. + +\subsubsection{Case of the tree} + +Try to find a swapping argument similar to the one used for the path. -Articles to read: \cite{Johnson:PCKP,DBLP:journals/telsys/ShawCC97} +\nocite{Johnson:PCKP,DBLP:journals/telsys/ShawCC97} \bibliographystyle{plain} \bibliography{papers.bib} |
