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authorKoen van der Heijden <koen.vd.heijden1@gmail.com>2016-12-14 12:24:23 +0100
committerKoen van der Heijden <koen.vd.heijden1@gmail.com>2016-12-14 12:24:23 +0100
commit86f397d5a69fe0d8e9c51117f9e5f7b21a2972f2 (patch)
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parentb9a6d7f1a98bc9e0e18879379586bf8fbe7db0de (diff)
download2IMF25-AR-86f397d5a69fe0d8e9c51117f9e5f7b21a2972f2.tar.gz
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@@ -58,9 +58,20 @@ are an explosive combination: they are not allowed to be put in the same truck.
\subsection*{Solution:}
\subsubsection*{(a)}
+We generalized this problem to fitting pallets into $T$ trucks, with capacity $c_T$ and maximum amount of pallets $p_T$. Then we added variables $N$, $P$, $C$, $S$ and $D$ representing Nuzzles, Prittles, Crottles, Skipples and Dupples respectively.
+The size of $P$ is to be determined.
+To put a constraint on the capacity of a truck, we created the following formula:\\
+$\sum_{x \in X} (ite (x \in T) (weight(x)) (0))$, where $X \in {N, P, C, S, D}$.\\ We summed the result for all $X$, and compared it to $c_T$.
+We did exactly the same for the amount of pallets, where we substituted $weight(x)$ with $1$, and $c_T$ with $p_T$.
+After this, we added the static constraints that $s \in S$ must be in truck $1$, $2$ or $3$ as these are our cooling trucks.
+Finally we added a constraint saying all $n \in N$ are distinct.
-\subsubsection*{(b)}
+Since we have $8$ trucks, with maximum amount of pallets $8$, we can carry at most $64$ pallets. This means that we can carry at most $22$ pallets of prittles. We searched for the solution of how many pallets we could actually carry in a binary sense: First we tried $11$ pallets, then $16$, then $19$, $21$ and finally $22$. These tries were all satisfiable, thus we can carry at most $22$ pallets of prittles.
+\subsubsection*{(b)}
+For part b of the assignment, we added a set of constraints saying that whenever a truck contains a pallet of Crottles, it may not contain a pallet of Prittles.
+This was captured by a set of constraints looping over all pallets of Crottles, and then for the Truck containing them, looping over all pallets to check whether they are not Prittles (giving us at most $10 \cdot 8 = 80$ new constraints).
+After that, we again searched for a solution in a binary way. This time, we did not get a satisfiable solution anymore above 19 pallets of Prittles. Thus, we can take at most 19 pallets of Prittles if they may not be carried together with Crottles.
\section*{Problem: Chip design}
Give a chip design containing two power components and ten regular components satisfying the following constraints: