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author | Peter Wu <peter@lekensteyn.nl> | 2016-12-13 01:30:36 +0100 |
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committer | Peter Wu <peter@lekensteyn.nl> | 2016-12-13 01:30:36 +0100 |
commit | 7269acd9fbc6c20d4961ce6a5851bfee3a41cb4a (patch) | |
tree | 134873587f6ba3f78e45f338a9d6b73ab5b2e0bb | |
parent | ef9860fec9077d0be59cf0cf2c687b0ef51ffefe (diff) | |
download | 2IMF25-AR-7269acd9fbc6c20d4961ce6a5851bfee3a41cb4a.tar.gz |
JobScheduling: formatting
-rw-r--r-- | part1.tex | 9 |
1 files changed, 5 insertions, 4 deletions
@@ -352,10 +352,11 @@ We can see the list of jobs as an array containing integers that represent the t We first created a rule for every job $j(i)$ that states that $j$ has to end before a decision variable $m$. This can be written as $j(i) + i + 5 < m$. We initialized this variable as $m \gets 50$. From there on, we worked towards the smallest value for which the formula was satisfiable. -We found this value to be $m = 59$. For $m = 58$, z3 returned \"unsat\". -We also reformed all previously mentioned requirements into formula's that can be used in z3: +We found this value to be $m = 59$. For $m = 58$, z3 returned \texttt{unsat}. +We also reformed all previously mentioned requirements into formulae that can be +used in z3 as shown in Listing~\ref{lst:jobScheduling}. -\begin{lstlisting} +\begin{lstlisting}[caption={SMT for Job Scheduling},label={lst:jobScheduling},language=lisp,basicstyle=\scriptsize] (and (>= j3 (+ j1 6)) (>= j3 (+ j2 7))) (and (>= j5 (+ j3 8)) (>= j5 (+ j4 9))) (and (>= j7 (+ j3 8)) (>= j7 (+ j4 9)) (>= j7 (+ j6 11))) @@ -393,7 +394,7 @@ Note that $42 + 12 + 5$ is indeed equal to $59$. This solution was retrieved in the same way as answer a. This answer is $m = 65$. 2 extra lines were added to ensure that job 6 lies within job 12: -\begin{lstlisting} +\begin{lstlisting}[language=lisp,basicstyle=\scriptsize] (>= j6 j12) (<= (+ j6 11) (+ j12 17)) \end{lstlisting} |