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Specified variable and added change to the last review point.

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Janita Willumsen
2023-12-05 15:32:39 +01:00
parent 5df0a2411b
commit 4b711a477e
5 changed files with 22 additions and 8 deletions
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4
latex/sections/abstract.tex
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@@ -6,7 +6,7 @@
temperature, when undergoing phase transition. To generate spin configurations,
we used the Metropolis-Hastings algorithm, which applies a Markov chain Monte
Carlo sampling method. We determined the time of equilibrium to be approximately
3000 Monte Carlo cycles, and used the following samples to find the probability
$5000$ Monte Carlo cycles, and used the following samples to find the probability
distribution at temperature $T_{1} = 1.0 J / k_{B}$, and $T_{2} = 2.4 J / k_{B}$.
For $T_{1}$ the mean energy per spin is $\langle \epsilon \rangle \approx -1.9969 J$,
with a variance $\text{Var} (\epsilon) = 0.0001$. And for $T_{2}$, close to the critical
@@ -16,6 +16,6 @@
susceptibility. We have estimated the critical temperatures of finite lattice sizes,
and used these values to approximate the critical temperature of a lattice of
infinite size. Using linear regression, we estimated the critical temperature
to be $T_{c}(L = \infty) \approx 2.2695 J/k_{B}$.
to be $T_{c}^{*}(L = \infty) \approx 2.2693 J/k_{B}$.
\end{abstract}
\end{document}

8
latex/sections/appendices.tex
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@@ -216,4 +216,12 @@ the magnetic susceptibility.
\label{fig:phase_susceptibility_1M}
\end{figure}
Result of profiling using Score-P in Figure \ref{fig:scorep_assessment}.
\begin{figure}[H]
\centering
\includegraphics[width=\linewidth]{../images/profiling.pdf}
\caption{Score-P assessment of parallel efficiency.}
\label{fig:scorep_assessment}
\end{figure}
\end{document}

4
latex/sections/conclusion.tex
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@@ -30,6 +30,6 @@ heat capacity and susceptibility for lattices of size $L = {20, 40, 60, 80, 100}
We observed a phase transition in the temperature range $T \in [2.1, 2.4] J / k_{B}$.
Using the values from the finite lattices, we approximated the critical temperature
of a lattice of infinite size. Using linear regression, we numerically estimated
$T_{c}(L = \infty) \approx 2.2695 J/k_{B}$ which is close to the analytical solution
$T_{C}(L = \infty) \approx 2.269 J/k_{B}$ Lars Onsager found in 1944.
$T_{c}^{*}(L = \infty) \approx 2.2693 J/k_{B}$ which is close to the analytical solution
$T_{c}(L = \infty) \approx 2.269 J/k_{B}$ Lars Onsager found in 1944.
\end{document}

13
latex/sections/results.tex
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@@ -102,8 +102,13 @@ MPI processes. Each process was set to spawn $10$ threads, resulting in a total
$100$ threads working in parallel. We include results for $1$ million MC cycles
in Appendix \ref{sec:additional_results}
$\boldsymbol{Rewrite}$ We used a profiler to make sure the program was fully optimized which found that the
workload was balanced, the threads was not left idle to long/not a lot of downtime.
To evaluate the performance of the parallelization, we used a profiler. The
assessment output can be found in Appendix \ref{sec:additional_results} in Figure
\ref{fig:scorep_assessment}. The assessment shows a lower score for the MPI load
balance, compared to the OpenMP load balance. The master process gatheres all the
data using blocking communication, resulting in the other processes waiting. This
results in one process, the master, having to work more. The OpenMP load balance
score is very good, suggesting the threads are not left idle for long periods.
In Figure \ref{fig:phase_energy_10M}, for the larger lattices we observe a sharper
increase in $\langle \epsilon \rangle$ in the temperature range $T \in [2.25, 2.35]$. %]
@@ -179,10 +184,10 @@ the lattice size increase toward infinity, $1/L$ approaches zero. %
\end{figure}
We used linear regression to find the intercept $\beta_{0}$, which gives us an estimated value
of the critical temperature for a lattice of infinite size. The estimated critical temperature
is $T_{c \text{num}} \approx 2.2693 J/k_{B}$. We also compared the
is $T_{c}^{*}(L = \infty) \approx 2.2693 J/k_{B}$. We also compared the
estimate with the analytical solution, the relative error of our estimate is
\begin{equation*}
\text{Relative error} = \frac{T_{c \text{ numerical}} - T_{c \text{ analytical}}}{T_{c \text{ analytical}}} \approx 5.05405 \cdot 10^{-5} J/k_{B}
\text{Relative error} = \frac{T_{c}^{*} - T_{c \text{ analytical}}}{T_{c \text{ analytical}}} \approx 5.05405 \cdot 10^{-5} J/k_{B}
\end{equation*}
\end{document}