Introductory Parallel Applications
Courtesy: Dr. David Walker, Cardiff University
Example1 – wave equationCourtesy: David Walker, Cardiff University
Problem – Vibrating string of length L fixed at both ends given an initial displacement.
To determine Ψ(x, t)
displacement
x=0x=L
Vibrating String Problem
To solve ∂2Ψ/c2∂t2 - ∂2Ψ/∂x2 = 0 Subject to Ψ(0, t) = Ψ(L, t) = 0; Ψ(x,
0) = u(x) Numerical approximation: Ψi(t+∆t) = 2Ψi(t) – Ψi(t-∆t) + tau2(Ψi-1(t) -2Ψi(t) + Ψ i+1(t)) where
tau = c ∆t/ ∆x and u(x) = sinx
Vibrating String Problem – Sequential Problemfor(i=0; i<npoints; i++){ x = 2*pi*i/(npoints-1); x = sin(x); ps[i] = old_ps[i] = x;}
for(j=0; j<nsteps; j++){ for(i=1; i<npoints-1; i++){ new_ps[i] = 2*ps[i] – old_ps[i] + tau*tau*(ps[i-1] + 2ps[i] + ps[i+1]); }
for(i=0; i<npoints; i++){ old_ps[i] = ps[i]; ps[i] = new_ps[i]; }}
Vibrating String – Parallel Formulation
The points are divided across processes
Communication between neighboring processors
Data structure – an array for every processor with sizes local_size+2
Vibrating String – Parallel Code/* Set up 1D topology */ MPI_Cart_create(MPI_COMM_WORLD,1,&nprocs,&periods,
reorder,&new_comm); MPI_Cart_coords (new_comm, rank, 1, &mypos) MPI_Cart_shift (new_comm, 0, -1, &right, &left);
/* Initialise array */ local_npts = npoints/nprocs; lbound = mypos*local_npts;
for(i=0;i<local_npts;i++){ x = 2.0*PI*(double)(lbound+i)/(double)(npoints-1); x = sin(x); psi[i+1] = old_psi[i+1] = x; }
Vibrating String – Parallel Code/* Do the updates */ istart = (mypos==0) ? 2 : 1; iend = (mypos==nprocs-1) ? local_npts-1 : local_npts; for(j=0;j<NSTEPS;j++){ MPI_Sendrecv (&psi[1], 1, MPI_DOUBLE, left, 111,
&psi[local_npts+1], 1, MPI_DOUBLE, right, 111, new_comm, &status);
MPI_Sendrecv (&psi[local_npts], 1, MPI_DOUBLE, right, 111, &psi[0], 1, MPI_DOUBLE, left, 111, new_comm, &status);
for(i=istart;i<=iend;i++){ new_psi[i] = 2.0*psi[i] - old_psi[i] + tau*tau*(psi[i-1]-
2.0*psi[i]+psi[i+1]); } for(i=1;i<=local_npts;i++){ old_psi[i] = psi[i]; psi[i] = new_psi[i]; } }
Example 2 – Laplace equation problem
Courtesy: Dr. David Walker, Cardiff University
Electric field around a conducting object in a box
To solve ∂2Ψ/∂x2 + ∂2Ψ/∂y2 = 0 Subject to Ψ(0, y) = Ψ(L1, y) = 0;
Ψ(x, 0) = Ψ(x, L2) = 0; Ψ(x, y) =1 on S(x, y)
S(x, y)
x=0 x=L1
y=0
y=L2
Laplace equation – Numerical approximation
We consider a square problem, L1 = L2 = L
Ψi,jk = ¼(Ψi-1,j
k-1 + Ψi+1,jk-1 + Ψi,j-1
k-1 + Ψi,j+1k-1)
1
0
Laplace Equation – Sequential codedouble phi[NXMAX][NYMAX];double old_phi[NXMAX][NYMAX];int mask[NXMAX][NYMAX];
setup_grid(phi, nptsx, nptsy, mask);
for(k=0; k<nsteps; k++){
for(i=0; i<nptsx; i++){ for(j=0; j<nptsy; j++){ old_phi[i][j] = phi[i, j]; } }
for(i=0; i<nptsx; i++){ for(j=0; j<nptsy; j++){ if(mask[i][j]) phi[i][j] = ¼*(old_phi[i][j-1] + old_phi[i][j+1] +
old_phi[i-1][[j] + old_phi[i][[j-1]); } }
}
Laplace Equation – Sequential codesetup_grid(phi, nptsx, nptsy, mask){
for(i=0; i<nptsx; i++){ for(j=0; j<nptsy; j++){ phi[i, j] = 0.0; mask[i, j] = 1; } }
mask[0…nptsx-1][0] = mask[0..nptsx-1][nptsy-1] = mask[0][0..nptsy-1] = mask[nptsx-1][0..nptsy-1] = 0;
mask[nptsx/2, nptsx/2-1][nptsy/2, nptsy/2-1] = 0; phi[nptsx/2, nptsx/2-1][nptsy/2, nptsy/2-1] = 1;
Laplace Equation - Parallelization
A 2-D topology of processes are considered The updates of boundary values for a given
processor may need communication from its neighboring processors.
Following are main data structures double phi[NYMAX+2][NXMAX+2]; double oldphi[NYMAX+2][NXMAX+2]; int mask[NYMAX+2][NXMAX+2];
LAPLACE Equation - Parallelization/* Work out number of processes in each direction of the process mesh */ dims[0] = dims[1] = 0; MPI_Dims_create (nprocs, 2, dims); npy = dims[0]; npx = dims[1];
/* Set up 2D topology */ periods[0] = periods[1] = 0; MPI_Cart_create (MPI_COMM_WORLD, 2, dims, periods, 1, &new_comm); MPI_Cart_coords (new_comm, rank, 2, coords); myposy = coords[0]; myposx = coords[1];
/* Determine neighbouring processes for communication shifts */ MPI_Cart_shift (new_comm, 0, -1, &down, &up); MPI_Cart_shift (new_comm, 1, -1, &right, &left);
/* Initialise arrays */ setup_grid (phi, npx, npy, myposx, myposy, nptsx, nptsy, &local_nptsx,
&local_nptsy, mask);
Laplace Equation - Parallelization/* Iterate to find solution */for(k=1;k<=nsteps;k++){ for(i=1;i<=local_nptsy;i++) for(j=1;j<=local_nptsx;j++) oldphi[i][j] = phi[i][j];
MPI_Sendrecv (&oldphi[1][1], local_nptsy, MPI_DOUBLE, up, 111, &oldphi[local_nptsx+1][1], local_nptsy, MPI_DOUBLE,down, 111, new_comm, &status);
MPI_Sendrecv (&oldphi[local_nptsx][1],local_nptsy,MPI_DOUBLE,down,112, &oldphi[0][1], local_nptsy, MPI_DOUBLE, up, 112, new_comm, &status);
Laplace Equation - Parallelizationfor(i=1;i<=local_nptsx;i++) sbuf[i-1] = oldphi[i][1]; MPI_Sendrecv (sbuf, local_nptsx, MPI_DOUBLE, left, 113, rbuf, local_nptsx,
MPI_DOUBLE, right, 113, new_comm, &status); for(j=1;j<=local_nptsy;j++) oldphi[i][local_nptsy+1] = rbuf[i-1];
for(i=1;i<=local_nptsx;i++) sbuf[i-1] = oldphi[i][local_nptsy]; MPI_Sendrecv (sbuf, local_nptsx, MPI_DOUBLE, right, 114, rbuf, local_nptsx,
MPI_DOUBLE, left, 114, new_comm, &status); for(i=1;i<=local_nptsx;i++) oldphi[i][0] = rbuf[i-1];
for(i=1;i<=local_nptsx;i++) for(j=1;j<=local_nptsy;j++) if (mask[i][j]) phi[i][j] = 0.25*(oldphi[i][j-1] + oldphi[i][j+1] + oldphi[i-1][j]
+ oldphi[i+1][j]); }
Laplace Equation - Parallelizationsetup_grid (phi, npx, npy, myposx, myposy, nptsx, nptsy, local_ptrx,
local_ptry, mask){
local_nptsx = nptsx/npx; local_nptsy = nptsy/npy; for(j=0;j<=local_nptsy+1;j++) for(i=0;i<=local_nptsx+1;i++){ phi[i][j] = 0.0; mask[i][j] = 1;}
if (myposx == 0) mask[1][1..local_nptsy] = 0;if (myposx == npx-1) mask[local_nptsx][1..local_nptsy] = 0;if (myposy == 0) mask[1..local_nptsx][1] =0;if (myposy == npy-1) mask[1..local_nptsx][local_nptsy] = 0;
for(i=1;i<=local_nptsx;i++){ global_x = local_nptsx*myposx + i - 1; if (global_x == nptsx/2 || global_x == nptsx/2-1){ for(j=1;j<=local_nptsx;j++){ global_y = local_nptsy*myposy + j - 1; if (global_y == nptsy/2 || global_y == nptsy/2-1){ mask[i][j] = 0; phi[i][j] = 1.0; } } }}
*local_ptrx = local_nptsx; *local_ptry = local_nptsy;
Laplace Equation – Performance Analysis
For n2 grid size, N = PxQ processor mesh, tcalc time for a flop, tshift time for communication per grid point
Tseq = T(N) =
4n2tcalc
(4n2/N)tcalc + (2n/P)tshift + (2n/Q)tshift
A more irregular problem – Molecular Dynamics Simulation
Courtesy: Dr. David Walker, Cardiff University
In the previous problems, communication requirements are known in advance
In the current Molecular Dynamics simulation problem, the amount of data that are communicated between processors are not known in advance
The communication is slightly irregular
The Problem A domain consisting of number of particles
(molecules) Each molecule, i is exerted a force, fij by
another molecule, j The sum of all the forces, Fi = ∑jfij makes the
particles assume a new position and velocity
Particles that are r distance apart do not influence each other
Given initial velocities and positions of particles, their movements are followed for discrete time steps
MD - Solution
The cutoff distance, r is used to reduce the time for summation from O(n2)
r
r
Domain decomposed into cells of size rxr
Particles in one cell interact with particles in the neighbouring 8 cells and particles in the same cell
MD - SolutionData structures:An array of all particles. Each element holds <position,
velocity>A 2D array of linked lists, one for each cell. Each element of a
linked list contains pointers to particles.struct particle{ double position[2]; double velocity[2];} Particles[MAX_PARTICLES];
struct list{ particle* part; struct list* next;}*List[MAX_CELLSX][MAX_CELLSY];
Linked ListParticles
MD – Sequential LogicInitialize Particles and Lists;
for each time step for each particle i Let cell(k, l) hold i F[i] = 0; for each particle j in this cell and neighboring 8 cells, and are
r distance from i{ F[i]+= f[i, j]; } update particle[i].{position, velocity} due to F[i]; if new position in new cell (m,n) update Lists[k,l] and
Lists[m,n]
MD – Parallel Version
r
r
A 2D array of processors similar to Laplace
Each processor holds a set of cells
Differences:
•A processor can communicate with the diagonal neighbors
•Amount of data communicated varies over time steps
•Receiver does not know the amount of data
MDS – parallel solution
Steps1. Communication – Each processor communicates parameters of the particles on the boundary cells to its 8 neighboring cells
Challenges – to communicate diagonal cells
2. Update – Each processor calculates new particle velocities and positions
3. Migration – Particles may migrate to cells in other processors
Other challenges:
1. Appropriate packing of data.
2. Particles may have to go through several hops during migration
Assumptions:
1. For simplicity, let us assume that particles are transported to only neighboring cells during migration
MDS – parallel solution – 1st step
Communication of boundary dataA
A
A
A
a a aaaa
a a a
aaa
B
B
B
B
b b bbbb
b b b
bbb
C
C
C
C
c c cccc
c c c
ccc
D
D
D
d
d d dddd
d d d
ddd
MDS – parallel solution – 1st step
Communication of boundary dataA
A
A
A
a a aaaa
a a a
aaa
B
B
B
B
b b bbbb
b b b
bbb
C
C
C
C
c c cccc
c c c
ccc
D
D
D
d
d d dddd
d d d
ddd
A
A
A
A
a a aaaa
a a a
aaa
C Cc c c
C
C
C
C
c c c
c
c c c
c
c c
c c
B
B
B
B
b b b
b
b b b
bb
b
b
b
D
D
D
d
d d d
d
d d d
dd
d
d
d
A Aa a a B Bb b b
D Dd d d
B
B
bb
b
A
A
a
a
a
C
C
c
cc
D
D
d
d
d
MDS – parallel solution – 1st step
Communication of boundary dataA
A
A
A
a a aaaa
a a a
aaa
B
B
B
B
b b bbbb
b b b
bbb
C
C
C
C
c c cccc
c c c
ccc
D
D
D
d
d d dddd
d d d
ddd
A
A
A
A
a a aaaa
a a a
aaa
C Cc c c
C
C
C
C
c c c
c
c c c
c
c c
c c
B
B
B
B
b b b
b
b b b
bb
b
b
b
D
D
D
d
d d d
d
d d d
dd
d
d
d
A Aa a a B Bb b b
D Dd d d
B
B
bb
b
A
A
a
a
a
C
C
c
cc
D
D
d
d
d
D C
B A
Can be achieved by ?
Shift left, shift right, shift up, shift down
MDS – parallel solution – 1st stepLeft shiftnsend = 0;for(i=0; i<local_cellsx; i++){ for each particle p in cell (i, 1){ pack position of p in sbuf nsend += 2 }}
MPI_Sendrecv(sbuf, nsend, …, left,..rbuf, max_particles*2, …, right, &status);
MPI_Getcount(status, MPI_DOUBLE, &nrecv);particles = nrecv/2;for(i=0; i<particles; i++){ read (x,y) from next 2 positions in rbuf; add (x,y) to particles[local_particle_count+i]; determine cell k, l for the particle Add it to list (k, l);}
MDS – parallel solution – 2nd step
Update: Similar to sequential program. A processor has all the required information
for calculating Fi for all its particles Thus new position and velocity determined. If new position belongs to the same cell in
the same processor, do nothing If new position belongs to the different cell
in the same processor, update link lists for old and new cells.
MDS – parallel solution – 3rd step If new position belongs to the different cell in a
different processor – particle migrationfor each particle p update {position, velocity} determine new cell if new cell # old cell delete p from list of old cell if(different processor) pack p into appropriate communication buffer remove p from particle array
Shift leftShift rightShift upShift down
MDS – parallel solution – 3rd step This shifting is a bit different from the previous shifting A processor may just act as a transit point for a particle Hence particles have to be packed with careShift left:MPI_Sendrecv(leftbuf, nsend_left, …, left
rbuf, max_size*4, .., right, &status);MPI_Getcount(status, MPI_DOUBLE, &nrecv);particles = nrecv/4;
for(i=0; i<particles; i++){ read next 4 numbers in {x, y vx, vy} if(particle in this process) add particle to particle array determine cell add particle to list for the cell else put data in the appropriate comm. buffer for the next up or down
shifts}
MDS – comments Generic solution
A particle can move to any cell Force can be from any distance
Load balancing
A Dynamical System- WaTor Courtesy: Dr. David Walker, Cardiff
A 2-D ocean in which sharks and fish survive
2 important featuresa. Need for dynamic load distributionb. Potential conflicts due to updates by
different processors Features shared by other advanced
parallel applications
WaTor – The problem
Ocean divided into grids Each grid cell can be empty or have a
fish or a shark Grid initially populated with fishes and
sharks in a random manner Population evolves over discrete time
steps according to certain rules
WaTor - Rules
Fish: At each time step, a fish tries to move
to a neighboring empty cell. If not empty, it stays
If a fish reaches a breeding age, when it moves, it breeds, leaving behind a fish of age 0. Fish cannot breed if it doesn’t move.
Fish never starves
WaTor - RulesShark: At each time step, if one of the neighboring
cells has a fish, the shark moves to that cell eating the fish. If not and if one of the neighboring cells is empty, the shark moves there. Otherwise, it stays.
If a shark reaches a breeding age, when it moves, it breeds, leaving behind a shark of age 0. shark cannot breed if it doesn’t move.
Sharks eat only fish. If a shark reaches a startvation age (time steps since last eaten), it dies.
Inputs and Data StructuresInputs: Size of the grid Distribution of sharks and fishes Shark and fish breeding ages Shark starvation ageData structures:A 2-D grid of cells struct ocean{ int type /* shark or fish or empty */ struct swimmer* occupier; }ocean[MAXX][MAXY]A linked list of swimmers struct swimmer{ int type; int x,y; int age; int last_ate; int iteration; swimmer* prev; swimmer* next; } *List;
Sequential Code Logic
•Initialize ocean array and swimmers list
•In each time step, go through the swimmers in the order in which they are stored and perform updates
Towards a Parallel Code 2-D data distribution similar to
Laplace and molecular dynamics is used. Each processor holds a grid of ocean cells.
For communication, each processor needs data from 4 neighboring processors.
2 new challenges – potential for conflicts, load balancing
1st Challenge – Potential for Conflicts Unlike previous problems,
border cells may change during updates due to fish or shark movement
Border cells need to be communicated back to the original processor. Hence update step involves communication
In the meantime, the original processor may have updated the border cell. Hence potential conflicts
Time T
Time T+1
S F SF
F F
F
F
FF
S
F
F
S
F
F
F
F
F
S
2 Techniques Rollback updates for those particles
(fish or shark) that have crossed processor boundary and are in potential conflicts.
May lead to several rollbacks until a free space is found.
2nd technique is synchronization during updates to avoid conflicts in the first place.
2 Techniques During update, a processor
x sends its data first to processor y, allows y to perform its updates, get the updates from y, and then performs its own updates.
Synchronization can be done by sub-partitioning.
Divide a grid owned by a processor into sub-grids.
1 2
3 4
1 2
3 4
update
Load Imbalance
The workload distribution changes over time
2-D block distribution is not optimal
Techniques: Dynamic load balancer Static load balancing by a different data
distribution
Dynamic load balancing
Performed at each time step Orthogonal recursive bisection
Problems: Complexity in finding the neighbors and data for communication
Static Data Distribution
Cyclic0,0 0,1 0,2 0,3 0,0 0,1 0,2 0,3
1,0 1,1 1,2 1,3 1,0 1,1 1,2 1,3 2,0 2,1 2,2 2,3 2,0 2,1 2,2 2,3 3,0 3,1 3,2 3,3 3,0 3,1 3,2 3,3
Problems: Increase in boundary data; increase in communication
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