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inputs/lectures/lec_01.tex

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+\section{Introduction}
+
+\begin{question}
+  Which problems can be modelled as flow problems?
+\end{question}
+
+\begin{example}
+  One interesting application are \emph{Traffic Flows}.
+  \begin{itemize}
+    \item For example, a central authority might coordinate how people evacuate in case
+      of a nearing flood.
+      This is an example of a \vocab{flow over time} 
+    \item In a typical real-life situation, however, there is no such central authority
+      that dictates peoples' routes.
+      Thus, the setting resembles much rather a \emph{game theoretical problem},
+      since each participant will individually decide which route to take,
+      potentially leading to solutions that are not globally optimal.
+
+      This can lead to unexpected outcomes:
+  \end{itemize}
+\end{example}
+
+\begin{example}[Breass's Paradox]
+  Consider the following traffic network, where we want to route a total flow of $1$:
+  \missingfigure{tikz}
+  Here, an edge of cost $x$ will have cost equal to the amount of flow routed along it.
+
+  The optimum solution (in terms of maximum travel time)
+  is of course to route flow $\frac{1}{2}$ along each of the paths.
+
+  It can be shown that this is also the solution that people will actually choose,
+  since it is always better to use the currently less-congested path,
+  leading to an equal spread.
+
+  However, now considering adding an additional
+  \todo{finish remark}.
+\end{example}
+
+
+\section{Max Flows over time}
+
+\begin{definition}[Flow over time]
+  Consider a graph $G = (V,E)$ with \vocab{transit times} $\set{ τ_e } _{e\in E}$
+  and capacities $\set{ u_e } _{e\in E}$.
+
+  A \vocab{flow over time} $f$ time horizon $T$ is a
+  Lebesgue integrable function $f_e\colon [0, T) \to \mathbb{R}_{\geq 0}$
+  for $e\in E$ such that $f_{e}(\theta) = 0$ for $\theta \geq T - τ_e$.
+\end{definition}
+
+\begin{oral}
+  The assumption on the Lebesgue integrability is just here for technical reasons,
+  we will not deal much with it.
+  In practice, most functions that we will encounter are piecewise constant.
+\end{oral}
+
+\begin{remark}+
+  We can interpret this definition as edges signaling pipes,
+  and the flow $f_e$ denoting the inflow rate to edge $e$ over a certain time.
+  Then, it takes $τ_e$ time for the flow to traverse the pipes.
+  Thus, $f_e(\theta - τ_e)$ denotes the excess flow of edge $e$ at time $\theta$.
+\end{remark}
+
+\setcounter{toplevel}{2}
+\begin{definition}
+  Let $f$ be a flow over time with horizon $T$.
+  \begin{enumerate}[a)]
+    \item The \vocab{capacity constraint} is that $f_e(\theta) \leq u_e$ for all $e,\theta$.
+    \item The \vocab{excess for a vertex} $v\in V$ at $\theta$ is defined as
+      \[
+        \ex_f(v, \theta) \coloneq
+        \sum_{e\in δ^-(v)} \int_0^{\zeta - τ_e} f_e(\zeta) d \zeta
+        -
+        \sum_{e\in δ^+(v)} \int_0^{\zeta} f_e(\zeta) d \zeta
+      \]
+    \item We say that $f$ has \vocab{weak flow conservation} if
+      \begin{IEEEeqnarray*}{rCLl}
+        \ex_f(v, \theta) &\geq& 0 & \forall v \in V \setminus \set{ s } \forall \theta \in [0,T)
+        \\
+        \ex_f(v, T) = 0 & \forall v \in V \setminus \set{ s,t } 
+      \end{IEEEeqnarray*}
+    \item We say that $f$ has \vocab{strict flow conservation} if
+      \[
+        \ex_f(v, \theta) = 0 \forall v\in V \\ \set{ s , t}  \forall \theta \in  T
+      \]
+    \item The \vocab{value of f} is $\abs{f} \coloneqq \ex_f(t,T)$.
+    \item A flow is called \vocab{feasible} if it satisfies the capacity constraints 
+      and has weak flow conservation.
+  \end{enumerate}
+\end{definition}
+
+
+Consider a graph $G = (V,E)$ with capacities $c\colon E \to  \mathbb{R}_{\geq 0}$,
+transit times $τ\colon E\to \mathbb{R}_{\geq 0}$,
+a time horizon $T$ and source/sink vertices $s$, $t$.
+
+The \textsc{Maximum Flow over Time Problem} asks for a feasible $s$-$t$-flow
+over time with time horizon $T$ and maximum value of $\abs{f}$.
+
+\begin{definition}[Temporally repeated flow with time horizon $T$]
+  Let $x$ be a static\footnotemark $s$-$t$ (feasible) flow and consider a flow decomposition
+  $(X_p)_{p \in \mathcal{P} \cup \mathcal{C}}$
+  with $\abs{\mathcal{P} \cup \mathcal{C}} \leq \abs{E}$.
+
+  Now, define a flow over time by
+  \[
+    f_e(\theta) \coloneqq \sum_{P \in P_e(\theta) } x_p \qquad
+    \forall  e = (v,w) \in E \; \theta \in [0, T)
+    ,
+  \]
+  where
+  \[
+    P_e(\theta) \coloneqq \set{
+      p\in \mathcal{P} \suchthat
+      p \in \mathcal{P},
+      τ(P_{s,v)} \leq \theta,
+      τ(P_{v, t}) \leq T - \theta
+    }
+    .
+  \]
+  Here, the transit times along a fixed path $P$ are defined canonically as
+  \[
+    τ(P_{v,w}) \coloneqq \sum_{e\in P_{v,w}} τ_e
+    .
+  \]
+
+  The resulting flow is called the \vocab{temporally repeated flow with time horizon $T$}
+  associated to the static flow $x$.
+\end{definition}
+\footnotetext{Here, \enquote{static} means just \enquote{not temporal}}
+
+\begin{observation}
+  \label{obs:temporally-repeated-flow-construction}
+  The Temporally repeated flow can be obtained as follows:
+  For each $P \in \mathcal{P}$, we send $x_p$ into $P$
+  from $s$ during $[0, T - τ(P))$ and then pass the flow along $P$.
+
+  Note that the temporally repeated flow even fulfills strict flow conservation.
+\end{observation}
+
+
+\setcounter{toplevel}{6}
+\begin{lemma}
+  Let $x$ be a static, feasible $s$-$t$ flow with
+  flow decomposition $(x_P)_{P \in \mathcal{P} \cup \mathcal{C}}$
+  and $x_P = 0$ for all $P \in \mathcal{P}$ with $τ(P) > T$
+  and for all cycles.
+  Denote by $f$ the corresponding temporally repeated flow $f$.
+
+  Then,
+  \[
+    \abs{f} = T \cdot \abs{x}  - \sum_{e\in E} τ_e \cdot x_e
+    .
+  \]
+\end{lemma}
+
+\begin{proof}
+  Motivated by \autoref{obs:temporally-repeated-flow-construction},
+  we deduce:
+  \begin{IEEEeqnarray*}{rCl}
+    \abs{f}
+    &=& \sum_{P\in \mathcal{P}} (T - τ(P)) \cdot x_P \\
+    &=& T \cdot \sum_{P\in \mathcal{P}} x_P - \sum_{P\in \mathcal{P}} τ(P) \cdot x_P \\
+    &=& T \cdot \abs{x} - \sum_{e\in E} τ_e \underbrace{\sum_{\substack{P\in \mathcal{P} \\ e \in P} } x_p}_{= x_e}
+  \end{IEEEeqnarray*}
+\end{proof}
+
+\begin{corollary}
+  Let $x$ be a static, feasible $s$-$t$ flow and $(x_P)_{P\in \mathcal{P} \cup \mathcal{C}}$
+  a flow decomposition.
+  Then,
+  \[
+    \abs{f} \geq T \cdot \abs{x} - \sum_{e\in E}  τ_e x_e
+    .
+  \]
+\end{corollary}
+
+\begin{proof}
+  Define a new flow by
+  \[
+    \tilde{x}_P \coloneqq \begin{cases}
+      x_P & \text{if $P\in \mathcal{P}$ and $τ(P) \leq T$} \\
+      0 & \text{otherwise}
+    \end{cases}
+  \]
+  for all $P \in \mathcal{P}\cup \mathcal{C}$.
+  Then, the associated temporally repeated flows satisfy
+  \begin{IEEEeqnarray*}{rCl}
+    \abs{f} &=& \abs{\tilde{f}} \\
+            &=& T \abs{\tilde{x}} - \sum_{e \in  E} τ_e \tilde{x}_e \\
+            &\geq & T \abs{x} - \sum_{e\in E} τ_e x_e
+  \end{IEEEeqnarray*}
+  Note that to check the last inequality,
+  it suffices to see that cycles only contribute to the negative sum
+  and do not change the flow value of $\tilde{x}$ compared to $x$.
+  For paths $P$ with $τ(P) > T$, these contribute to $\abs{x} $
+  but not to $\tilde{x}$, however, the effect cancels out by summing
+  the $τ_e$ along this path in the negative summand.
+\end{proof}
+
+\begin{algorithm}[H]
+  \SetKwInput{KwInput}{Input}
+  \SetKwInput{KwOutput}{Output}
+  \SetKw{KwGoTo}{go to}
+  \SetKwProg{Fn}{Def}{:}{}
+  \DontPrintSemicolon
+
+  \caption{Ford-Furkerson Algorithm for maximum flows over time}
+  \KwInput{$G = (V, E)$ with capacities $c$, transit times $τ$, time horizon $T$ and source/sink vertices $s$, $t$.}
+  \KwOutput{A feasible temporally repeated flow}
+  \;
+
+  Compute a feasible static $s$-$t$-flow $x$ maximizing
+  \[
+    T \cdot \abs{x} - \sum_{e \in E} τ_e x_e 
+  \]
+  \;
+
+  Compute a flow decomposition $(x_P)_{P\in \mathcal{P} \cup \mathcal{C}}$.
+  \;
+
+  Output corresponding temporally repeated flow.
+  
+\end{algorithm}
+
+
+\begin{remark}
+  To implement step 1, consider the auxiliary graph $G'$ obtained from $G$
+  by adding an edge $(t,s)$ with infinite capacity.
+  For each edge, set its cost to $τ_e$.
+
+  Now, for each static $s$-$t$ flow $x$ in $G$ gives rise to a circulation
+  in $G'$ by sending $\abs{x}$ along $(t,s)$, whose cost will be precisely
+  \[
+    \sum_{e\in E} x_e \cdot τ_e + \abs{x} \cdot (-T) 
+    .
+  \]
+  Thus, solving the minimum cost flow problem in $G'$ yields a desired flow $f$.
+\end{remark}