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that uses it by default (In native code, you can access the number of processors with the Win32 GetSy stem l n fo API) Next in this function we calculate the number of elements each thread will process, c h u n k, by dividing the iteration count by the number of processors As an illustration, say we had 1 00,000 iterations to perform (Le, ( h i - l ow ) 1 00,000) and a degree of parallelism of 16 (ie, p 1 6); each thread would then execute 6,250 iterations (Le, c h u n k 6,250) It's not a requirement that the iteration count is evenly divisible by p, so we have to take care of some edge conditions With our partitioning strategy, the last partition could end up with more iterations to run than others We immediately create a C o u n t down E v e n t of count p: this is an event abstraction that becomes signaled once p threads have called S i gn a l on it We then queue up p work items in the CLR thread pool (each of which signals the latch upon completion) and wait on the latch Each work item queued to the pool iterates over its iteration space: the p i d is just the loop counter i passed as the second argument to Qu e u e U s e rWo r k Item This is used for a subtle reason: if we used i directly from the C# anonymous delegate passed to the thread pool, it would be hoisted into a closure and shared by all iterations; the result is that the wrong value of i would be used by any given iteration, and, in fact, most threads would probably observe i as p (depending on various race conditions), which is outside of its legal range Each thread iterates from 10 + p i d * c h u n k to 10 + ( p i d + 1 ) * c h u n k o r h i, whichever i s larger, and calls the body function, passing the iteration index as the argument We check for h i because, if the task is the last of the group, it must iterate until h i in case the iteration count was not evenly divisible by p Notice the indices that any given thread processes are adja cent and contiguous; this usually (but not always) helps improve cache locality, particularly when the indices will be used to index into an array After executing the part of the loop for which the thread is responsible, it calls l a t c h S i g n a l to indicate that it has finished Finally, the thread that ran the parallel loop waits for all iterations to finish by calling l a t c h Wa i t This call unblocks once all iterations are done There are a few noteworthy comments First, we could make a slight optimization and initialize the latch with one fewer signal and run one of.

1 1 +2 2 2

To illustrate better why this is necessary, imagine for a moment that the consumer replaced the S i g n a l O b j e c tAndWa it call with two independent calls to R e l e a s eMutex and then Wa it F o rS i ng l eOb j e c t instead

Comparison of Actual to Budgeted Time Report Budgeted Hours 142 129 100 140 292 81 40 Actual Hours 174 120 100 145 305 80 60 Variance 32 +9 0 5 7 +1 20

m_mutex Re l e a seMutex ( ) ; m event Wa itOne ( ) ;

All it takes is three threads, one producer and two consumers, and bad luck to encounter a deadlock due to a missed signal

3 = 3 = E(X)E(Y ) 2 and we see that the quantities in parentheses are E(X) and E(Y ), respectively This is true of the general case, that is, if X and Y are independent, then E(X Y ) = E(X) E(Y ) A proof can be fashioned by generalizing the example above This can be extended to any number of random variables: if X,Y ,Z, are mutually independent in pairs, then E(X Y Z ) = E(X) E(Y ) E(Z) Although the examples given here involve discrete random variables, the results concerning the expected values are true for continuous random variables as well

te ( co n s ume r ) R e l e a seMutex ( g_hMut ex ) ; R e l e a s eMutex ( g_hMut ex ) ; SetEvent ( g_hSy n c E vent ) ; SetEvent ( g_hSyn c Event ) ; Wa i t F o r S i ngleObj e c t ( ) ; Wait F o rS i n g leObj e c t ( ) ; t 1 ( c o n s umer ) t 2 ( p rod u c e r )

Work Center ID Number PL-42 PL-45 RN-28 RN-36 TS-04 ZZ-10 ZZ-12