initial mem.c and related commit

vm-beyondphys/Makefile

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+
+all: mem
+
+mem: mem.c
+	gcc -o mem mem.c -Wall -O
+

vm-beyondphys/README.md

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+
+# Overview
+
+In this homework, you'll be investigating swap performance with a simple
+program found in `mem.c`. The program is really simple: it just allocates an
+array of integers of a certain size, and then proceeds to loop through it
+(repeatedly), incrementing each value in the array. 
+
+Type `make` to build it (and look at the file `Makefile` for details about how
+the build works).
+
+Then, type `./mem` followed by a number to run it. The number is the size (in
+MB) of the array. Thus, to run with a small array (size 1 MB):
+
+```sh
+prompt> ./mem 1
+```
+
+and to run with a larger array (size 1 GB):
+
+```sh
+prompt> ./mem 1024
+```
+
+The program prints out the time it takes to go through each loop as well as
+the bandwidth (in MB/s). Bandwidth is particularly interesting to know as it
+gives you a sense of how fast the system you're using can move through data;
+on modern systems, this is likely in the GB/s range. 
+
+Here is what the output looks like for a typical run:
+
+```sh
+prompt> ./mem 1000
+allocating 1048576000 bytes (1000.00 MB)
+  number of integers in array: 262144000
+loop 0 in 448.11 ms (bandwidth: 2231.61 MB/s)
+loop 1 in 345.38 ms (bandwidth: 2895.38 MB/s)
+loop 2 in 345.18 ms (bandwidth: 2897.07 MB/s)
+loop 3 in 345.23 ms (bandwidth: 2896.61 MB/s)
+^C
+prompt> 
+```
+
+The program first tells you how much memory it allocated (in bytes, MB, and in
+the number of integers), and then starts looping through the array. The first
+loop (in the example above) took 448 milliseconds; because the program
+accessed the 1000 MB in just under half a second, the computed bandwidth is
+(not surprisingly) just over 2000 MB/s. 
+
+The program continues by doing the same thing over and over, for loops 1, 2,
+etc. 
+
+Important: to stop the program, you must kill it. This task is accomplished on
+Linux (and all Unix-based systems) by typing control-C (^C) as shown above.
+
+Note that when you run with small array sizes, each loop's performance numbers
+won't be printed. For example:
+
+```sh
+prompt>  ./mem 1
+allocating 1048576 bytes (1.00 MB)
+  number of integers in array: 262144
+loop 0 in 0.71 ms (bandwidth: 1414.61 MB/s)
+loop 607 in 0.33 ms (bandwidth: 3039.35 MB/s)
+loop 1215 in 0.33 ms (bandwidth: 3030.57 MB/s)
+loop 1823 in 0.33 ms (bandwidth: 3039.35 MB/s)
+^C
+prompt> 
+```
+
+In this case, the program only prints out a sample of outputs, so as not to
+flood the screen with too much output. 
+
+The code itself is simple to understand. The first important part is a memory
+allocation: 
+
+```c
+    // the big memory allocation happens here
+    int *x = malloc(size_in_bytes);
+```
+
+Then, the main loop begins:
+
+```c
+    while (1) {
+	x[i++] += 1; // main work of loop done here.
+```
+
+
+The rest is just timing and printing out information. See `mem.c` for details.
+
+Much of the homework revolves around using the tool vmstat to monitor what is
+happening with the system. Read the vmstat man page (type `man vmstat`) for
+details on how it works, and what each column of output means.
+
+
+

vm-beyondphys/mem.c

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+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <assert.h>
+#include <sys/time.h>
+
+// Simple routine to return absolute time (in seconds).
+double Time_GetSeconds() {
+    struct timeval t;
+    int rc = gettimeofday(&t, NULL);
+    assert(rc == 0);
+    return (double) ((double)t.tv_sec + (double)t.tv_usec / 1e6);
+}
+
+// Program that allocates an array of ints of certain size,
+// and then proceeeds to update each int in a loop, forever.
+int main(int argc, char *argv[]) {
+    if (argc != 2) {
+	fprintf(stderr, "usage: spin <memory (MB)>\n");
+	exit(1);
+    }
+    long long int size = (long long int) atoi(argv[1]);
+    long long int size_in_bytes = size * 1024 * 1024;
+
+    printf("allocating %lld bytes (%.2f MB)\n", 
+	   size_in_bytes, size_in_bytes / (1024 * 1024.0));
+
+    // the big memory allocation happens here
+    int *x = malloc(size_in_bytes);
+    if (x == NULL) {
+	fprintf(stderr, "memory allocation failed\n");
+	exit(1);
+    }
+
+    long long int num_ints = size_in_bytes / sizeof(int);
+    printf("  number of integers in array: %lld\n", num_ints);
+
+    // now the main loop: each time through, touch each integer
+    // (and increment its value by 1).
+    int i = 0;
+    double time_since_last_print = 2.0; 
+    double t = Time_GetSeconds();
+    int loop_count = 0;
+    while (1) {
+	x[i++] += 1; // main work of loop done here.
+
+	// if we've gone through the whole loop, reset a bunch of stuff
+	// and then (perhaps) print out some statistics. 
+	if (i == num_ints) {
+	    double delta_time = Time_GetSeconds() - t;
+	    time_since_last_print += delta_time;
+	    if (time_since_last_print >= 0.2) { // only print every .2 seconds
+		printf("loop %d in %.2f ms (bandwidth: %.2f MB/s)\n", 
+		       loop_count, 1000 * delta_time, 
+		       size_in_bytes / (1024.0*1024.0*delta_time));
+		time_since_last_print = 0;
+	    }
+
+	    i = 0;
+	    t = Time_GetSeconds();
+	    loop_count++;
+	}
+    }
+
+    return 0;
+}
+