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-rw-r--r--csci5451/ass1.tex11
-rw-r--r--csci5451/ass1p6.c30
2 files changed, 27 insertions, 14 deletions
diff --git a/csci5451/ass1.tex b/csci5451/ass1.tex
index 6dbdfc1..24f2e0c 100644
--- a/csci5451/ass1.tex
+++ b/csci5451/ass1.tex
@@ -7,10 +7,13 @@
\begin{document}
\maketitle
\section{Question 1}
- \subsection{Overlapping Intervals}
-
- \subsection{Non-Overlapping Intervals}
- \(t_{transfer} = t_s+t_w*d*m/k*k\) or \(\sum_{i=1}^{k} t_s+t+w*d*m/k\)
+ \subsection*{Overlapping Intervals}
+ The time for one message to be sent 1 hop is \(t_s+t_w*m/k\). After the first message is sent forward one hop, we account for the final message \((d-1)*t_w*m/k\), making the final expression
+ \[k(t_s+t_w+m/k)+((d-1)t_w*m/k)\]
+ \subsection*{Non-Overlapping Intervals}
+ \[t_{transfer} = t_s*k+t_w*d*m\]
+ \subsection*{}
+ For both cases, as \(k\) goes to \(m\), the time to transfer will increase greatly. If \(t_s\) is very large, the optimal value of \(k\) is 1. In other words, it is better to transfer the message all at once instead of in \(k\) parts. If \(t_s\) is 0 it has little to no effect on overall transmission time regardless of \(k\) being large or small.
\section{Question 2}
\subsection*{A}
Shared memory has all of the processors access one large pool of memory while distributed memory has each processor have a section of the memory.
diff --git a/csci5451/ass1p6.c b/csci5451/ass1p6.c
index 3b4a1bb..d87eddd 100644
--- a/csci5451/ass1p6.c
+++ b/csci5451/ass1p6.c
@@ -26,7 +26,8 @@
// is run for a fixed number of time steps rather than until
// temperatures reach steady state.
-int main(int argc, char **argv) {
+int main(int argc, char **argv)
+{
int max_time = 50; // Number of time steps to simulate
int width = 20; // Number of cells in the rod
double initial_temp = 50.0; // Initial temp of internal cells
@@ -38,25 +39,30 @@ int main(int argc, char **argv) {
// Allocate memory
H = malloc(sizeof(double *) * max_time);
int t, p;
- for (t = 0; t < max_time; t++) {
+ for (t = 0; t < max_time; t++)
+ {
H[t] = malloc(sizeof(double *) * width);
}
// Initialize constant left/right boundary temperatures
- for (t = 0; t < max_time; t++) {
+ for (t = 0; t < max_time; t++)
+ {
H[t][0] = L_bound_temp;
H[t][width - 1] = R_bound_temp;
}
// Initialize temperatures at time 0
t = 0;
- for (p = 1; p < width - 1; p++) {
+ for (p = 1; p < width - 1; p++)
+ {
H[t][p] = initial_temp;
}
// Simulate the temperature changes for internal cells
- for (t = 0; t < max_time - 1; t++) {
- for (p = 1; p < width - 1; p++) {
+ for (t = 0; t < max_time - 1; t++)
+ {
+ for (p = 1; p < width - 1; p++)
+ {
double left_diff = H[t][p] - H[t][p - 1];
double right_diff = H[t][p] - H[t][p + 1];
double delta = -k * (left_diff + right_diff);
@@ -71,19 +77,23 @@ int main(int argc, char **argv) {
// Column headers
printf("%3s| ", "");
- for (p = 0; p < width; p++) {
+ for (p = 0; p < width; p++)
+ {
printf("%5d ", p);
}
printf("\n");
printf("%3s+-", "---");
- for (p = 0; p < width; p++) {
+ for (p = 0; p < width; p++)
+ {
printf("------");
}
printf("\n");
// Row headers and data
- for (t = 0; t < max_time; t++) {
+ for (t = 0; t < max_time; t++)
+ {
printf("%3d| ", t);
- for (p = 0; p < width; p++) {
+ for (p = 0; p < width; p++)
+ {
printf("%5.1f ", H[t][p]);
}
printf("\n");