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From: Patrick Scheibe on 10 Jul 2010 04:00 Hi, just think your calculation of the new P through: The algorithm relies on the order of the Do since it uses not only oldP for the calculation of the "new P". So for calculating the value P[[i]] it is required that the values of P for smaller i are already calculated to evaluate the term Sum[A[[i, j]] P[[j]], {j, 1, i - 1}] on the right side of your iteration scheme. This is bad for data-parallelism where you rely on the fact that all values P[[i]] can be calculated separately from each other. Remember that the order of calculation in a parallel algorith is often not determined. What you should do is to go on a popular "scholar" web search an find "parallel Successive Over Relaxation" publications. Then come back to Mathematica and read more about DistributeDefinitions to make data (in your case the A and B and omega) available to all started parallel kernels. Hope this helps a bit. Cheers Patrick On Thu, 2010-07-08 at 03:14 -0400, pratip wrote: > Dear Group, > > Here is a simple code to solve linear systems. It is motivated from > Fullerton University numerics example. Now if someone can suggest how > to parallelize this code. > > SORmethod[A0_, B0_, P0_, omega_, max_] := > Module[{A = N[A0], B = N[B0], i, j, k = 0, n = Length[P0], P = P0, > oldP = P0}, > While[ k < max, > Do[P[[i]] = (1 - omega) oldP[[i]] + > omega/A[[i, > i]] (B[[i]] - Sum[A[[i, j]] P[[j]], {j, 1, i - 1}] - > Sum[A[[i, j]] oldP[[j]], {j, 1 + i, n}]) , {i, 1, n}]; > oldP = P; > k = k + 1; ]; > Return[P] ] > > To test the solver. > > n = 10; > A = DiagonalMatrix[Table[2., {n}]]; > For[i = 1, i <= n - 1, i++, > A[[i, i + 1]] = A[[i + 1, i]] = 1.;]; > B = Table[5. - Abs[i - 5], {i, 1, n}]; > P = Table[1., {i, n}]; > X3 = SORmethod[A, B, P, 1.5, 200] // Chop > > Check the above result with Mathematica LinearSolve > > LinearSolve[A, B] // N > > Straightforward ParallelDo or ParallelSum is not working. One can see > for large n (abt 400-4000) we will significantly get help if we can > parallize the code. > > Best regards, > > Pratip > |