Difference between revisions of "BetaT"
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S = 380 - ( 1 - .8918 ) * ( 380 - 1 ) = 339.031 | S = 380 - ( 1 - .8918 ) * ( 380 - 1 ) = 339.031 | ||
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S = 1024 - ( 1 - .8918 ) * ( 1024 - 1 ) = 913.3114 | S = 1024 - ( 1 - .8918 ) * ( 1024 - 1 ) = 913.3114 | ||
Revision as of 18:08, 15 February 2017
BetaT
Assignment 1
Profile Assessment
Naiver Strokes equation for Flow Velocity.
There are a lot of different waves and equations, this one is based off the naiver-stokes equation.
All this program does is calculate the deviation of the wave. It will calculate the velocity and dip. The wave is only going in one direction and is going to drop but at what degree and velocity.
The wave equation takes the distance between wave trophs, so two waves and the distance between them. The height of the wave, and the amount of time it takes each wave to reach its destination. It will perform a calculation to give us the speed per second.
Navier–Stokes equations are useful because they describe the physics of many phenomena of scientific and engineering interest. They may be used to model the weather, ocean currents, water flow in a pipe and air flow around a wing. The Navier–Stokes equations in their full and simplified forms help with the design of aircraft and cars, the study of blood flow, the design of power stations, the analysis of pollution, and many other things. Coupled with Maxwell's equations they can be used to model and study magnetohydrodynamics. courtesy of wikipedia ("https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations")
The problem with this application comes in the main function trying to calculate the finite-difference
// Finite-difference loop: for (int it=1; it<=nt-1; it++) { for (int k=0; k<=nx-1; k++) { un[k][it-1] = u[k][it-1]; } for (int i=1; i<=nx-1; i++) { u[0][it] = un[1][it-1]; u[i][it] = un[i][it-1] - c*dt/dx*(un[i][it-1]-un[i-1][it-1]); } }
The user inputs 2 values which will be used as a reference for the loop.
Testing the application
Contents
Tests ran with no optimization
n | Time in Milliseconds | |
---|---|---|
100 x 100 | 24 | |
500 x 500 | 352 | |
1000 x 1000 | 1090 | |
2000 x 2000 | 3936 | |
5000 x 5000 | 37799 | |
5000 x 10000 | 65955 | |
10000 x 10000 | 118682 | |
12500 x 12500 | 220198 |
gprof
it gets a bit messy down there, but basically 89.19% of the program is spent in the main() calculating those for loops shown above. The additional time is spent allocating the memory which might cause some slowdown when transferring it to the GPU across the bus int he future.
But the main thing to take away here is that main() is 89.19% and takes 97 seconds.
Each sample counts as 0.01 seconds.
% cumulative self self total time seconds seconds calls s/call s/call name 89.19 97.08 97.08 main 4.73 102.22 5.14 1406087506 0.00 0.00 std::vector<std::vector<double, std::allocator<double> >, std::allocator<std::vector<double, std::allocator<double> > > >::operator[](unsigned int) 4.49 107.11 4.88 1406087506 0.00 0.00 std::vector<double, std::allocator<double> >::operator[](unsigned int)
Potential Speed Increase with Amdahls Law
Using Amdahls Law ---- > Sn = 1 / ( 1 - P + P/n )
We can examine how fast out program is capable of increasing its speed.
P = is the part of the program we want to optimize which from above is 89.17% n = the amount of processors we will use. One GPU card has 384 processors or CUDA cores and another GPU we will use has 1020 processor or CUDA cores.
Applying the algorithm gives us.
Amdahls Law for GPU with 384 Cores---- > Sn = 1 / ( 1 - 0.8919 + 0.8919/384 )
Sn = 9.0561125222
Amdahls Law for GPU with 1024 Cores---- > Sn = 1 / ( 1 - 0.8919 + 0.8919/1024 )
Sn = 9.176753777
Therefor According to Amdahls law we can expect a 9x increase in speed. 97 seconds to execute main / 9 amdahls law = 10.7777 seconds to execute after using GPU Interestingly according to the law the difference in GPU cores does not significantly increase speed. Future tests will confirm or deny these results.
Potential Speed Increase with Gustafsons Law
Gustafsons Law S(n) = n - ( 1 - P ) ∙ ( n - 1 )
S = 380 - ( 1 - .8918 ) * ( 380 - 1 ) = 339.031
S = 1024 - ( 1 - .8918 ) * ( 1024 - 1 ) = 913.3114
Using Gustafsons law we see drastic changes in the amount speed increase, this time the additional Cores made a big difference and applying these speed ups we get:
(Quadro K2000 GPU) 97 seconds to execute / 339.031 = 0.29 (GeForce GTX960 GPU) 97 seconds to execute / 913.3114 = 0.11