How to determine if task is busy or not?

How to determine if task is busy or not?

[Tomek Walkuski]

Is there any way to determine if task is busy or not?

I have pool of worker tasks and I want to choose one of them to give
some work to do.

Re: How to determine if task is busy or not?

[Ludovic Brenta]

Tomek Walkuski wrote:
> Is there any way to determine if task is busy or not?
>
> I have pool of worker tasks and I want to choose one of them to give
> some work to do.

I don't think you should see the problem in these terms.  You don't
have to select a task from the pool; the tasks select themselves as
they become available for a new job, e.g. each task would look like

loop
   Queue.Get_Next (Job);
   Process (Job);
end loop;

where Queue is a protected object and Get_Next removes a job from it
and passes it to the task. Of course, if there are no jobs in the
queue, it blocks until one arrives.

HTH

--
Ludovic Brenta.

Re: How to determine if task is busy or not?

[Dmitry A. Kazakov]

On Sat, 8 Aug 2009 03:56:11 -0700 (PDT), Ludovic Brenta wrote:

> Tomek Walkuski wrote:
>> Is there any way to determine if task is busy or not?
>>
>> I have pool of worker tasks and I want to choose one of them to give
>> some work to do.
> 
> I don't think you should see the problem in these terms.  You don't
> have to select a task from the pool; the tasks select themselves as
> they become available for a new job, e.g. each task would look like
> 
> loop
>    Queue.Get_Next (Job);
>    Process (Job);
> end loop;
> 
> where Queue is a protected object and Get_Next removes a job from it
> and passes it to the task. Of course, if there are no jobs in the
> queue, it blocks until one arrives.

Just to add a pair words...

The scheme described by Ludovic is not only simpler, it is also race
condition free. The reverse scheme has a race condition in it. If you have
asked if a task is idle, you do not know if it still is right now (or not
terminated). To eliminate race condition you need one atomic operation
"take job or fall back" and poll the pool of task with this operation. It
is not only inefficient it is also a potential live lock or at least
unbalanced servicing policy, if you have many job suppliers.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de

Re: How to determine if task is busy or not?

[Maciej Sobczak]

On 8 Sie, 13:27, "Dmitry A. Kazakov" <mail...@dmitry-kazakov.de>
wrote:

> The scheme described by Ludovic is not only simpler, it is also race
> condition free. The reverse scheme has a race condition in it. If you have
> asked if a task is idle, you do not know if it still is right now

Depends. If it is the asking entity that is also providing the job,
then the problem above cannot happen. In other words, if the "manager"
asks the task whether it is busy or idle and the answer is "idle",
then that answer it true until the same manager provides new work
unit, because there is no other way for the task to switch from "idle"
to "busy". Which means that the manager can always safely handle a new
work unit to the "idle" task, no race is possible.

The race can happen the other way round, though: the manager asks the
task for its status, the task answers "busy" and the manager takes
some decision based on this answer (move on to another task, perhaps?)
even though it might not be true anymore (the task in question might
be already "idle" by that time). This kind of race might be
acceptable, though, at most at the price of underutilized on
unbalanced server.

The advantage of such a setup is that the manager knows whether the
work can be handled *immediately* by any one of the tasks in the pool.
False negatives can be harmless, as long as the positives are
accurate.

Using a queue, as suggested by Ludovic, makes sense only when the work
units can *wait* for being processed. This might or might not be the
case.

There are examples of systems where either of these schemes are good
(batch processing vs. real-time?). I would not say that one is better
than another.

To Tomek - the simplest way to deploy your original idea is to use a
flag (Busy/Idle - an enumeration type is perfect here), which is
shared and used by both the worker task and its manager. Protected
object is a proper solution here and you might even use the same
protected object for managing both the flag and the work unit itself.

--
Maciej Sobczak * www.msobczak.com * www.inspirel.com

Database Access Library for Ada: www.inspirel.com/soci-ada

Re: How to determine if task is busy or not?

[Tomek Wałkuski]

On 8 Sie, 16:59, Maciej Sobczak <see.my.homep...@gmail.com> wrote:

> Depends. If it is the asking entity that is also providing the job,
> then the problem above cannot happen. In other words, if the "manager"
> asks the task whether it is busy or idle and the answer is "idle",
> then that answer it true until the same manager provides new work
> unit, because there is no other way for the task to switch from "idle"
> to "busy". Which means that the manager can always safely handle a new
> work unit to the "idle" task, no race is possible.
>
Only manager is providing work to the tasks. So, as you said, no race
is possible.

> To Tomek - the simplest way to deploy your original idea is to use a
> flag (Busy/Idle - an enumeration type is perfect here), which is
> shared and used by both the worker task and its manager. Protected
> object is a proper solution here and you might even use the same
> protected object for managing both the flag and the work unit itself.
>
That was my first thought :)

Re: How to determine if task is busy or not?

[Dmitry A. Kazakov]

On Sat, 8 Aug 2009 07:59:51 -0700 (PDT), Maciej Sobczak wrote:

> On 8 Sie, 13:27, "Dmitry A. Kazakov" <mail...@dmitry-kazakov.de>
> wrote:
> 
>> The scheme described by Ludovic is not only simpler, it is also race
>> condition free. The reverse scheme has a race condition in it. If you have
>> asked if a task is idle, you do not know if it still is right now
> 
> Depends. If it is the asking entity that is also providing the job,
> then the problem above cannot happen. In other words, if the "manager"
> asks the task whether it is busy or idle and the answer is "idle",
> then that answer it true until the same manager provides new work
> unit, because there is no other way for the task to switch from "idle"
> to "busy". Which means that the manager can always safely handle a new
> work unit to the "idle" task, no race is possible.

As I said, only if single job supplier (manager) is here. In that case it
is still a poor scheme because of polling. A better (I tempted to say
proper) one is when idle tasks queue themselves into the pool of idle tasks
maintained by the manager. But a protected object of tasks would be better
here as well.

> The advantage of such a setup is that the manager knows whether the
> work can be handled *immediately* by any one of the tasks in the pool.
> False negatives can be harmless, as long as the positives are
> accurate.

No, it is not an advantage.

1. Job is done to get a result, it is a lengthy process, so there is a
worker to perform it asynchronously. Until the job is done it is of little
interest whether it has been started or not. (Some interest exists if jobs
are cancelable if not initiated)

2. The manager is only a mediator here. If a job cannot be started
immediately what the manger is supposed to do? If it does not cancel the
job, the only option is to wait. This is what the job queue is for.

> Using a queue, as suggested by Ludovic, makes sense only when the work
> units can *wait* for being processed. This might or might not be the
> case.

See above. If there is some service timeout a queue is serviced by a
monitor task (a special "mister no" worker). Still better.

> There are examples of systems where either of these schemes are good
> (batch processing vs. real-time?). I would not say that one is better
> than another.
> 
> To Tomek - the simplest way to deploy your original idea is to use a
> flag (Busy/Idle - an enumeration type is perfect here), which is
> shared and used by both the worker task and its manager. Protected
> object is a proper solution here and you might even use the same
> protected object for managing both the flag and the work unit itself.

No, you don't need any extra shared data. Do conditional entry call do
determine if the worker is ready:

task body Worker is
begin
   loop
      accept New_Job (Work_Item : in out Job);
      ... -- Service
   end loop;
end Worker;

The manager does a conditional entry call (RM 9.7.3):

   select
      Worker.New_Job (Thing_To_Do);
   else
      -- The chap is busy
      ...
   end select;

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de

Re: How to determine if task is busy or not?

[Ludovic Brenta]

Dmitry A. Kazakov wrote on comp.lang.ada:
> Maciej Sobczak wrote:
>> Dmitry A. Kazakov wrote:
>
>>> The scheme described by Ludovic is not only simpler, it is also race
>>> condition free. The reverse scheme has a race condition in it. If you have
>>> asked if a task is idle, you do not know if it still is right now
>
>> Depends. If it is the asking entity that is also providing the job,
>> then the problem above cannot happen. In other words, if the "manager"
>> asks the task whether it is busy or idle and the answer is "idle",
>> then that answer it true until the same manager provides new work
>> unit, because there is no other way for the task to switch from "idle"
>> to "busy". Which means that the manager can always safely handle a new
>> work unit to the "idle" task, no race is possible.
>
> As I said, only if single job supplier (manager) is here. In that case it
> is still a poor scheme because of polling. A better (I tempted to say
> proper) one is when idle tasks queue themselves into the pool of idle tasks
> maintained by the manager. But a protected object of tasks would be better
> here as well.
>
>> The advantage of such a setup is that the manager knows whether the
>> work can be handled *immediately* by any one of the tasks in the pool.
>> False negatives can be harmless, as long as the positives are
>> accurate.
>
> No, it is not an advantage.
>
> 1. Job is done to get a result, it is a lengthy process, so there is a
> worker to perform it asynchronously. Until the job is done it is of little
> interest whether it has been started or not. (Some interest exists if jobs
> are cancelable if not initiated)
>
> 2. The manager is only a mediator here. If a job cannot be started
> immediately what the manger is supposed to do? If it does not cancel the
> job, the only option is to wait. This is what the job queue is for.
>
>> Using a queue, as suggested by Ludovic, makes sense only when the work
>> units can *wait* for being processed. This might or might not be the
>> case.
>
> See above. If there is some service timeout a queue is serviced by a
> monitor task (a special "mister no" worker). Still better.
>
>> There are examples of systems where either of these schemes are good
>> (batch processing vs. real-time?). I would not say that one is better
>> than another.
>
>> To Tomek - the simplest way to deploy your original idea is to use a
>> flag (Busy/Idle - an enumeration type is perfect here), which is
>> shared and used by both the worker task and its manager. Protected
>> object is a proper solution here and you might even use the same
>> protected object for managing both the flag and the work unit itself.
>
> No, you don't need any extra shared data. Do conditional entry call do
> determine if the worker is ready:
>
> task body Worker is
> begin
>    loop
>       accept New_Job (Work_Item : in out Job);
>       ... -- Service
>    end loop;
> end Worker;
>
> The manager does a conditional entry call (RM 9.7.3):
>
>    select
>       Worker.New_Job (Thing_To_Do);
>    else
>       -- The chap is busy
>       ...
>    end select;

But that becomes unwieldy in the presence of multiple worker tasks,
any one of which needs to be idle when enqueuing a job.  How about:

protected Queue is
   procedure Enqueue (J : in Job);
   entry Get_Next (J : out Job);
private
   -- left as an exercise :)
end Queue;

protected body Queue is
   procedure Enqueue (J : in Job) is
   begin
      if Get_Next_Job'Count < 1 then
         raise No_Worker_Is_Presently_Waiting_For_A_Job;
      else
         -- ...
      end if;
   end Enqueue;

   entry Get_Next (J : out Job) when Jobs_Enqueued is
   begin
      ...
   end Get_Next;
end Queue;

Start all workers before the manager; all workers then start waiting
on the Get_Next entry.  The manager keeps enqueuing new jobs until all
workers are busy, at which point no worker is waiting on Get_Next
anymore and the manager gets an exception to signal that condition.

Then the difficult part is to ensure that all workers collectively
consume jobs at least as fast as the manager can enqueue them.

--
Ludovic Brenta.