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Activity Diagram + Loops
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Join Date: Dec 2006
Posts: 3
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Hello.
I need to draw an activity digarm for a program code called Finite Differene Time Domain (FDTD). I have attached the code in a document where there are all the for loops. The task is given below
In the piece of code mentioned, you could look at parallelizing the lines of code, as I said
think as if you are using threads. Parallelized grains of code must be independent.
If you think about using a cluster (rather than a parallel machine) then you may want higher
granularity (ie you split the task into bigger chunks (since the communication times are greater).
If anybody could help me out by drwaing an activity digaram for the code below or point me towards tutorials for understanding more.
tHANKS
I need to draw an activity digarm for a program code called Finite Differene Time Domain (FDTD). I have attached the code in a document where there are all the for loops. The task is given below
In the piece of code mentioned, you could look at parallelizing the lines of code, as I said
think as if you are using threads. Parallelized grains of code must be independent.
If you think about using a cluster (rather than a parallel machine) then you may want higher
granularity (ie you split the task into bigger chunks (since the communication times are greater).
If anybody could help me out by drwaing an activity digaram for the code below or point me towards tutorials for understanding more.
tHANKS
It is like matrix multiplication can be distributed among many cpu each of them can compute a row /column independently,
Nice homework I would say.
Nice homework I would say.
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Join Date: Dec 2006
Posts: 3
Reputation:
Solved Threads: 0
Any ideas as to what is happenign in the loops below. I understand ther are three loops. diffiuculy on the array and hwo to reperesent this in a dependency graph.
for (Jx = 1; Jx <= (Nx - 1); Jx++) {
for (Jy = 1; Jy <= (Ny - 1); Jy++) {
for (Jz = 1; Jz <= (Nz - 1); Jz++) {
Eold[1] = E.x.grid[Jx][Jy][Jz];
Eold[2] = E.y.grid[Jx][Jy][Jz];
Eold[3] = E.z.grid[Jx][Jy][Jz];
for (Jx = 1; Jx <= (Nx - 1); Jx++) {
for (Jy = 1; Jy <= (Ny - 1); Jy++) {
for (Jz = 1; Jz <= (Nz - 1); Jz++) {
Eold[1] = E.x.grid[Jx][Jy][Jz];
Eold[2] = E.y.grid[Jx][Jy][Jz];
Eold[3] = E.z.grid[Jx][Jy][Jz];
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