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Inside Collection (Textbook):

Textbook by: Charles Severance, Kevin Dowd. E-mail the authors

# A Real Example

Module by: Charles Severance, Kevin Dowd. E-mail the authors

In all of the above examples, we have focused on the mechanics of shared memory, thread creation, and thread termination. We have used the sleep( ) routine to slow things down sufficiently to see interactions between processes. But we want to go very fast, not just learn threading for threading’s sake.

The example code below uses the multithreading techniques described in this chapter to speed up a sum of a large array. The hpcwall routine is from (Reference).

This code allocates a four-million-element double-precision array and fills it with random numbers between 0 and 1. Then using one, two, three, and four threads, it sums up the elements in the array:


#define _REENTRANT        /* basic 3-lines for threads */
#include <stdio.h>
#include <stdlib.h>

void *SumFunc(void *);
double GlobSum;                     /* A global variable */
pthread_mutex_t my_mutex;           /* MUTEX data structure */

#define MAX_SIZE 4000000
double array[MAX_SIZE];             /* What we are summing... */
void hpcwall(double *);

main() {
int i,retval;
double single,multi,begtime,endtime;

/* Initialize things */
for (i=0; i<MAX_SIZE; i++) array[i] = drand48();
pthread_attr_init(&attr);     /* Initialize attr with defaults */

GlobSum = 0;
hpcwall(&begtime);
for(i=0; i<MAX_SIZE;i++) GlobSum = GlobSum + array[i];
hpcwall(&endtime);
single = endtime - begtime;
printf("Single sum=%lf time=%lf\n",GlobSum,single);

/* Use different numbers of threads to accomplish the same thing */
GlobSum = 0;
hpcwall(&begtime);
index[i] = i;
}
hpcwall(&endtime);
multi = endtime - begtime;
printf("Sum=%lf time=%lf\n",GlobSum,multi);
} /* End of the ThreadCount loop */
}

void *SumFunc(void *parm){
int i,me,chunk,start,end;
double LocSum;

/* Decide which iterations belong to me */
me = (int) parm;
start = me * chunk;
end = start + chunk; /* C-Style - actual element + 1 */
if ( me == (ThreadCount-1) ) end = MAX_SIZE;
printf("SumFunc me=%d start=%d end=%d\n",me,start,end);

/* Compute sum of our subset*/
LocSum = 0;
for(i=start;i<end;i++ ) LocSum = LocSum + array[i];

/* Update the global sum and return to the waiting join */
GlobSum = GlobSum + LocSum;
}

First, the code performs the sum using a single thread using a for-loop. Then for each of the parallel sums, it creates the appropriate number of threads that call SumFunc( ). Each thread starts in SumFunc( ) and initially chooses an area to operation in the shared array. The “strip” is chosen by dividing the overall array up evenly among the threads with the last thread getting a few extra if the division has a remainder.

Then, each thread independently performs the sum on its area. When a thread has finished its computation, it uses a mutex to update the global sum variable with its contribution to the global sum:


Single sum=7999998000000.000000 time=0.256624
SumFunc me=0 start=0 end=2000000
SumFunc me=1 start=2000000 end=4000000
Sum=7999998000000.000000 time=0.133530
Efficiency = 0.960923
SumFunc me=0 start=0 end=1333333
SumFunc me=1 start=1333333 end=2666666
SumFunc me=2 start=2666666 end=4000000
Sum=7999998000000.000000 time=0.091018
Efficiency = 0.939829
SumFunc me=0 start=0 end=1000000
SumFunc me=1 start=1000000 end=2000000
SumFunc me=2 start=2000000 end=3000000
SumFunc me=3 start=3000000 end=4000000
Sum=7999998000000.000000 time=0.107473
Efficiency = 0.596950
recs %


There are some interesting patterns. Before you interpret the patterns, you must know that this system is a three-processor Sun Enterprise 3000. Note that as we go from one to two threads, the time is reduced to one-half. That is a good result given how much it costs for that extra CPU. We characterize how well the additional resources have been used by computing an efficiency factor that should be 1.0. This is computed by multiplying the wall time by the number of threads. Then the time it takes on a single processor is divided by this number. If you are using the extra processors well, this evaluates to 1.0. If the extra processors are used pretty well, this would be about 0.9. If you had two threads, and the computation did not speed up at all, you would get 0.5.

At two and three threads, wall time is dropping, and the efficiency is well over 0.9. However, at four threads, the wall time increases, and our efficiency drops very dramatically. This is because we now have more threads than processors. Even though we have four threads that could execute, they must be time-sliced between three processors.1 This is even worse that it might seem. As threads are switched, they move from processor to processor and their caches must also move from processor to processor, further slowing performance. This cache-thrashing effect is not too apparent in this example because the data structure is so large, most memory references are not to values previously in cache.

It’s important to note that because of the nature of floating-point (see (Reference)), the parallel sum may not be the same as the serial sum. To perform a summation in parallel, you must be willing to tolerate these slight variations in your results.

## Footnotes

1. It is important to match the number of runnable threads to the available resources. In compute code, when there are more threads than available processors, the threads compete among themselves, causing unnecessary overhead and reducing the efficiency of your computation.

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