# Parfor with Iterations Having Unequal Execution Times

Asked by Greg on 7 Dec 2011
Latest activity Commented on by Greg on 7 Dec 2011

I've run into an interesting problem with parfor...

I'm trying to use parfor to evaluate a bunch of different input cases of a program simultaneously. I know beforehand that the cases are not going to take the same amount of time to evaluate, and I can tell you with reasonable accuracy which ones are going to take the longest. The number of cases that each worker is running is very low (about 40 total cases being distributed to 12 workers).

Some of these cases will take as much as three or four times longer than the faster ones, and so a lot of user time could be saved if I could predict what order the parfor loop was going to execute in and try to distribute the jobs intelligently.

The simple example would be this: say I have four jobs to run on two workers, three of which will take 10 seconds and one of which will take 30 seconds. Obviously the ideal solution is for one worker to evaluate the long job and the other worker to evaluate the other three in the same amount of time. What tends to happen, though, is that both workers will evaluate one of the short jobs first, and then one of them will wind up with 40 seconds of working time while the other worker sits idle for half that time.

Even more annoying, I think that sometimes there will be an available job sitting in the "queue" with a worker available, but it won't run because that job is supposed to go to a worker that is busy. The example here would be two 10-second jobs and two 30-second jobs, and one worker winds up evaluating both 30-second jobs.

Anyone have any clever ideas of how to work around this problem?

## Products

Answer by Titus Edelhofer on 7 Dec 2011

Hi,

if your cases are input cases you might try to use the (slightly more elaborate way) of distributing the code using a job with 40 tasks. It helps at least for the distribution of the tasks, because they get distributed one by one. You might still end up that the longest run is the last in the row to be computed.

One step further would be to create more then one job where you distribute your tasks more clever.

Titus

## 1 Comment

Greg on 7 Dec 2011

After looking into the job scheduler stuff a bit, I think this is going to be the "right" way to do what I want, though it's going to take a little bit more effort to get there.

Answer by Jaap on 7 Dec 2011

is it possible to make an inner loop that goes through the faster cases?

or: if you know everything beforehand, make two parfor loops. while in the first, you will do all short/fast cases (hoping that rem(12,cases)==0). the next loop, you will do the rest in the same fashion.

## 1 Comment

Greg on 7 Dec 2011

That's not a bad idea, but it might come back to bite me if I'm a little bit wrong about which one is going to take the longest.

Answer by Titus Edelhofer on 7 Dec 2011

Hi,

here a rather ugly way which gives you full control on which worker processes which index:

```function myparfortest
```
```% this is a usual parfor loop
x = zeros(200, 1);
parfor i=1:200
x(i) = rank(magic(i));
end
```
```% use spmd to do the same, although
spmd
y = zeros(200, 1);
% suppose we have 3 workers:
% worker1: process i=1, 4, 7, ...
% worker2: process i=2, 5, 8, ...
% worker3: process i=3, 6, 9, ...
for i=labindex:numlabs:200
y(i) = rank(magic(i));
end
end
```
```% now collect the result
Y = zeros(200, 1);
for i=1:length(y)
yi = y{i};
Y(i:length(y):end) = yi(i:length(y):end);
end
```
```% let's see if we did right:
isequal(x, Y)
```

Titus

## 1 Comment

Greg on 7 Dec 2011

This is a little hack-ish, but it's easy and it has promise if you know the execution order precisely beforehand. It won't dynamically re-evaluate like the job scheduler would, though, if some jobs start finishing earlier than expected.