Assign same queue item to multiple workers - algorithm

We have a logical queue of tasks, where each task has to be assigned to multiple workers.
The numbers of workers to be assigned are based on a configuration of Minimum and Maximum workers.
A worker should not see the same task that they have already completed. It is not necessary that all workers will see all the tasks.
Total number of workers can change dynamically. Each worker can become online or offline anytime.
Each worker can choose to either complete the task or let it expire.
On expiry the task should be assigned to any worker who has not already completed the task.
Is there a good algorithm to solve this scenario ?

The easy solution:
Greedily assign tasks:
For every task that is ready.
Find the Minimum <= N <= Maximum workers that
didn't see the task yet and assign them.
Repeat until you either run
out of workers or finish all the tasks.
If a worker comes online or finishes task, re-check for all the tasks.
If a new task comes, re-check for the available workers.
This solution might be enough if there are not that many tasks, as it is computation heavy and recomputes everything.
The possible optimizations:
If the greedy solution fails(and it probably will), there are ways to improve on it. I will try to list those that come to my mind but it won't be an exhaustive list.
First, my personal favorite: Network flows. Unfortunately I don't see an easy way to resolve the minimal number of workers requirement, however it would be fast and it would result in a maximum possible workers being assigned at any given moment.
Create the network Source - Workers - Tasks - Sink. Edges workers to tasks will be linked and unlinked as needed:
when a worker is available for a task, create the edge with weight 1, otherwise don't create the edge.
From source link an edge with weight one to each online worker.
From every task link an edge with weight equal to its maximal worker capacity.
You could even differentiate between different kinds of workers, network flows are awesome. The algorithms are fast which makes them suitable even for large graphs. Also they are available in many libraries so you won't have to implement them yourself. Unfortunately, there is no easy way to enforce the minimal workers rule. At least I don't see one right now, there might be some way. Or maybe at least a heuristic
Second, be smart while being greedy:
Create a queue for every task.
When a worker is available, register him for every task he can do into its queue.
When a worker is not available, remove him from the queues.
When a task has enough workers, start progress and disable these workers.
This is still the brute force approach however since you keep the queues, you limit the amount of the necessary computation to a reasonable level. The potential downside is that large tasks(with a large minimal number of workers) might be stalled by the small tasks that will be easier to start - and will eat up the workers. So there will probably be some further checking/balancing and prioritizing needed.
There is definitely more to be tried & done for your task at hand, however the information you provided is rather limited so this advice is not that specific.

Related

A job allocating proble

I got n people, need to do N tasks, each task takes (x man-day), tasks have dependence, such as task A must start after task B finished. How should I arrangement it? Thank you so much!
I set up 4 rules for this question:
one Task could set most num of people to do it (such as 5)
there are logical dependence between tasks
one Task will cost determinated man-days
everyday concurrent working people can not be above current people num(n)
the rules are above, but i don't know how to calculate a minimum total time. Please tell me a method to solve it or some inspiration, thank you so much!
Make topological sort of jobs.
Now you have sequence of events (time; job start/job end)
On start of the job assign a man, if there are free workers, otherwise wait.
On job end free a worker, assign him to the waiting job if exists.
This problem is well-known to be NP-complete. Even the simple version with no dependencies, and only one worker per job, is NP-complete.
So you likely need to accept a reasonable heuristic. Here is one.
With a breadth-first search starting from the jobs which nothing else depends on, figure out for each job the minimum wall-clock time (max out workers on the job, and everything that depends on it, even if that is not actually possible) from starting that job to finishing.
And now you start at the top. Always assign workers to jobs by the following rules:
Longest wallclock time to finish first.
Break ties in favor of longest job.
Break ties in favor of job with fewest max workers.
In other words you're prioritizing putting people to work on the critical path and any slow jobs.

Recommend algorithm of fair distributed resources allocation consensus

There are distributed computation nodes and there are set of computation tasks represented by rows in a database table (a row per task):
A node has no information about other nodes: can't talk other nodes and doesn't even know how many other nodes there are
Nodes can be added and removed, nodes may die and be restarted
A node connected only to a database
There is no limit of tasks per node
Tasks pool is not finite, new tasks always arrive
A node takes a task by marking that row with some timestamp, so that other nodes don't consider it until some timeout is passed after that timestamp (in case of node death and task not done)
The goal is to evenly distribute tasks among nodes. To achieve that I need to define some common algorithm of tasks acquisition: when a node starts, how many tasks to take?
If a node takes all available tasks, when one node is always busy and others idle. So it's not an option.
A good approach would be for each node to take tasks 1 by 1 with some delay. So
each node periodically (once in some time) checks if there are free tasks and takes only 1 task. In this way, shortly after start all nodes acquire all tasks that are more or less equally distributed. However the drawback is that because of the delay, it would take some time to take last task into processing (say there are 10000 tasks, 10 nodes, delay is 1 second: it would take 10000 tasks * 1 second / 10 nodes = 1000 seconds from start until all tasks are taken). Also the distribution is non-deterministic and skew is possible.
Question: what kind/class of algorithms solve such problem, allowing quickly and evenly distribute tasks using some sync point (database in this case), without electing a leader?
For example: nodes use some table to announce what tasks they want to take, then after some coordination steps they achieve consensus and start processing, etc.
So this comes down to a few factors to consider.
How many tasks are currently available overall?
How many tasks are currently accepted overall?
How many tasks has the node accepted in the last X minutes.
How many tasks has the node completed in the last X minutes.
Can the row fields be modified? (A field added).
Can a node request more tasks after it has finished it's current tasks or must all tasks be immediately distributed?
My inclination is do the following:
If practical, add a "node identifier" field (UUID) to the table with the rows. A node when ran generates a UUID node identifier. When it accepts a task it adds a timestamp and it's UUID. This easily let's other nodes be able to determine how many "active" nodes there are.
To determine allocation, the node determines how many tasks are available/accepted. it then notes how many many unique node identifiers (including itself) have accepted tasks. It then uses this formula to accept more tasks (ideally at random to minimize the chance of competition with other nodes). 2 * available_tasks / active_nodes - nodes_accepted_tasks. So if there are 100 available tasks, 10 active nodes, and this node has accepted 5 task already. Then it would accept: 100 / 10 - 5 = 5 tasks. If nodes only look for more tasks when they no longer have any tasks then you can just use available_tasks / active_nodes.
To avoid issues, there should be a max number of tasks that a node will accept at once.
If node identifier is impractical. I would just say that each node should aim to take ceil(sqrt(N)) random tasks, where N is the number of available tasks. If there are 100 tasks. The first node will take 10, the second will take 10, the 3rd will take 9, the 4th will take 9, the 5th will take 8, and so on. This won't evenly distribute all the tasks at once, but it will ensure the nodes get a roughly even number of tasks. The slight staggering of # of tasks means that the nodes will not all finish their tasks at the same time (which admittedly may or may not be desirable). By not fully distributing them (unless there are sqrt(N) nodes), it also reduces the likelihood of conflicts (especially if tasks are randomly selected) are reduced. It also reduces the number of "failed" tasks if a node goes down.
This of course assumes that a node can request more tasks after it has started, if not, it makes it much more tricky.
As for an additional table, you could actually use that to keep track of the current status of the nodes. Each node records how many tasks it has, it's unique UUID and when it last completed a task. Though that may have potential issues with database churn. I think it's probably good enough to just record which node has accepted the task along with when it accepted it. This is again more useful if nodes can request tasks in the future.

A new project management algorithm wanted for finite number of workers

I'm studying task-based parallel computing and got interested in a variation of the old project management problem -- the critical path of an activity-on-vertex (AOV) project network, which can be calculated using the topological sorting algorithm if there's no deadlock cycle. The total time of those activities on a critical path gives the minimum completion time of the project.
But this is assuming we always have enough workers simultaneously finishing the activities with no dependence on each other. If the number of workers (processors/cores) available is finite, certain activities can wait not because some activities they depend on have not yet been finished, but simply because all workers are now busy doing other activities. This is a simplified model for today's multi-core parallel computing. If there's only one worker who has to do all the activities, the project completion time is the total time of all activities. We are back to single-core serial computing that way.
Is there an efficient algorithm that gives the minimum completion time of an AOV network given a finite number of workers available? How should we wisely choose which activities to do first when the doable activities is more than the number of workers so as to minimize the idling time of workers later on? The minimum time should be somewhere in between the critical path time (infinite workers) and the total time of all activities (one worker). It should also be greater than equal to the total time divided by the number of workers (no idling). Is there an algorithm to get that minimum time?
I found a C++ conference video called "work stealing" that almost answers my question. At 18:40, the problem is said on the slide to be NP-hard if activities cannot be paused, further divided, or transferred from worker to worker. Such restrictions make decisions of which workers to finish which jobs (activities) too hard to make. Work stealing is therefore introduced to avoid making such difficult decisions beforehand. Instead, it makes such decions no longer crucial so long as certain apparent greedy rules are followed. The whole project will be always finished as soon as possible under the constraint of either the critical path or the no-idling time of the finite number of workers or both. The video then goes on talking about how to make the procedure of "work stealing" between different workers (processors) more efficient by making the implementation distributed and cache-friendly, etc.
According to the video, future C++ shared-memory parallel coding will be task-based rather than loop-based. To solve a problem, the programmer defines a bunch of tasks to finish and their dependence relations to respect, and then the coding language will automatically schedule the tasks on multiple cores at run time in a flexible way. This "event-driven"-like way of implementing a flexible code by a distributed task queuing system will become very useful in parallel computing.
When an optimization problem is NP-hard, the best way to solve it is to find ways to avoid it.

What's the best Task scheduling algorithm for some given tasks?

We have a list of tasks with different length, a number of cpu cores and a Context Switch time.
We want to find the best scheduling of tasks among the cores to maximize processor utilization.
How could we find this?
Isn't it like if we choose the biggest available tasks from the list and give them one by one to the current ready cores, it's going to be best or you think we must try all orders to find out which is the best?
I must add that all cores are ready at the time unit 0 and the tasks are supposed to work concurrently.
The idea here is that there's no silver bullet, for what you must consider what are the types of tasks being executed, and try to schedule them as nicely as possible.
CPU-bound tasks don't use much communication (I/O), and thus, need to be continuously executed, and interrupted only when necessary -- according to the policy being used;
I/O-bound tasks may be continuously put aside in the execution, allowing other processes to work, since it will be sleeping for many periods, waiting for data to be retrieved to primary memory;
interative tasks must be continuously executed, but needs not to be executed without interruptions, as it will generate interruptions, waiting for user inputs, but it needs to have a high priority, in order not to let the user notice delays in the execution.
Considering this, and the context switch costs, you must evaluate what types of tasks you have, choosing, thus, one or more policies for your scheduler.
Edit:
I thought this was a simply conceptual question. Considering you have to implement a solution, you must analyze the requirements.
Since you have the length of the tasks, and the context switch times, and you have to maintain the cores busy, this becomes an optimization problem, where you must keep the minimal number of cores idle when it reaches the end of the processes, but you need to maintain the minimum number of context switches, so that your overall execution time does not grow too much.
As pointed by svick, this sounds like a partition problem, which is NP-complete, and in which you need to divide a sequence of numbers into a given number of lists, so that the sum of each list is equal to each other.
In your problem you'd have a relaxation on the objective, so that you no longer need all the cores to execute the same amount of time, but you want the difference between any two cores execution time to be as small as possible.
In the reference given by svick, you can see a dynamic programming approach that you may be able to map onto your problem.

Fair job processing algorithm

I've got a machine that accepts user uploads, performs some processing on them, and then returns the result. It usually takes a few minutes to process each upload received.
The problem is, a few users can upload a lot of jobs that basically deny processing to other users for a long time. I thought of just setting a hard cap and using priority queues, e.g. after 5 uploads in an hour, all new uploads are given a lower processing priority. I basically want to process ALL jobs, but I don't want the user who uploaded 1000 jobs to make everyone wait.
My question is, is there a better way to do this?
My goal is to minimize the time between the upload and the result being returned. It would be ideal if the algorithm could work in a distributed manner as well.
Thanks
Implementation will vary widely depending on what these jobs are and how long they take and how varied the processing times are, as well as how likely there is to be a fatal error during the process.
That being said, an easy way to maintain an even distribution of jobs across users is to maintain a list of all the users who have submitted jobs. When you are ready to get a new job, rather than just taking the next job out of a random queue, cycle through the users taking the top job from each user each time.
Again, this can be accomplished a number of ways, I would recommend a map from users to their respective list of jobs submitted. Cycle through the keys of the map each time you are ready for a new job. then get the list of jobs for whatever key you are on, and do the first job.
This is assuming that each job is "atomic" in that one job is not dependent on being executed next to the jobs it was submitted with.
Hope that helps, of course I could have completely misunderstood what you are asking for.
You don't have to roll-your-own. There is Sun Grid Engine. An open-source tool that is built to do that sort of thing, and if you are willing to pay, there is Platform LSF, which I use at work.
What is the maximum # of jobs a user can submit? Can users submit 1 job a a time OR is it a batch of jobs?
So your algorithm would go something like this
If the User has submitted jobs Then
Check how many jobs per hour
If the jobs per hour > than the average Then
Modify the users profile to a lower priority
Else
Check Users priority level and restore
End If
If the priority = HIGH
process right away
Else If priority = MEDIUM
Check Queue for High Priority
If High Priority Found (rerun this loop)
Else Process
Else If priority = LOW
Check Queue for High Priority
If High Priority Found (rerun this loop)
Else Process
Check Queue for Medium Priority
If Medium Priority Found (rerun this loop)
Else Process
Process Queue
End If
You can use a graph algorithm like Edmond's Blossom V to assign all users and jobs to a process. If a user can upload more then another user it would be more simplier for him to find a process. With the Blossom V algorithm you can define a threshold to not exceed the maximum process the server can handle.

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