Programming Logic - Splitting up Tasks Between Threads - math

Question

Lets say you want 5 threads to process data simultaneous. Also assume, you have 89 tasks to process.
Off the bat you know 89 / 5 = 17 with a remainder of 4. The best way to split up tasks would be to have 4 (the remainder) threads process 18 (17+1) tasks each and then have 1 (# threads - remainder) thread to process 17.
This will eliminate the remainder. Just to verify:
Thread 1: Tasks 1-18 (18 tasks)
Thread 2: Tasks 19-36 (18 tasks)
Thread 3: Tasks 37-54 (18 tasks)
Thread 4: Tasks 55-72 (18 tasks)
Thread 5: Tasks 73-89 (17 tasks)
Giving you a total of 89 tasks completed.
I need a way of getting the start and ending range of each thread mathematically/programmability; where the following should print the exact thing I have listed above:
$NumTasks = 89
$NumThreads = 5
$Remainder = $NumTasks % $NumThreads
$DefaultNumTasksAssigned = floor($NumTasks / $NumThreads)
For $i = 1 To $NumThreads
if $i <= $Remainder Then
$NumTasksAssigned = $DefaultNumTasksAssigned + 1
else
$NumTasksAssigned = $DefaultNumTasksAssigned
endif
$Start = ??????????
$End = ??????????
print Thread $i: Tasks $Start-$End ($NumTasksAssigned tasks)
Next
This should also work for any number of $NumTasks.
Note: Please stick to answering the math at hand and avoid suggesting or assuming the situation.

Answer 1

Why? Rather then predetermining the scheduling order, stick all of the tasks on a queue, and then have each thread pull them off one by one when they're ready. Then your tasks will basically run "as fast as possible".
If you pre-allocated, then one thread may be doing a particularly long bit of processing and blocking the running of all the tasks stuck behind it. Using the queue, as each task finishes and a thread frees up, it grabs the next task and keeps going.
Think of it like a bank with 1 line per teller vs one line and a lot of tellers. In the former, you might get stuck behind the person depositing coins and counting it out one by one, the latter you get to the next available teller, while Mr. PocketChange counts away.

Answer 2

I second Will Hartung 's remark. You may just feed them one task at a time (or a few tasks at a at a time, depending if there's much overhead, i.e. if individual tasks get typically completed very fast, relative to the cost of starting/recycling threads). Your subsequent comments effectively explain that your "threads" carry a heavy creation cost, and hence your desire to feed them once with as much work as possible, rather than wasting time creating new "thread" each fed a small amount of work.
Anyway... going to the math question...
If you'd like to assign tasks just once, the following formula, plugged in lieu of the the ????????? in your logic, should do the trick:
$Start = 1
+ (($i -1) * ($DefaultNumTasksAssigned + 1)
- (floor($i / ($Remainder + 1)) * ($i - $Remainder))
$End = $Start + $NumTasksAssigned -1
The formula is explained as follow:
1 is for the fact that your display / logic is one-based not zero-based
The second term is because we generally add ($DefaultNumTasksAssigned + 1) with each iteration.
The third term provides a correction for the last few iterations.
Its first part, (floor($i / ($Remainder + 1)) provides 0 until $i reaches the first thread
that doesn't receive one extra task, and 1 thereafter.
The second part express by how much we need to correct.
The formula for $End is easier, the only trick is the minus 1, it is because the Start and End values are inclusive (so for example between 1 an 19 there are 19 tasks not 18)
The following slightly modified piece of logic should also work, it avoids the "fancy" formula by keeping a running tab of the $Start variable, rather than recomputing it each time..
$NumTasks = 89
$NumThreads = 5
$Remainder = $NumTasks % $NumThreads
$DefaultNumTasksAssigned = floor($NumTasks / $NumThreads)
$Start = 1
For $i = 1 To $NumThreads
if $i <= $Remainder Then // fixed here! need <= because $i is one-based
$NumTasksAssigned = $DefaultNumTasksAssigned + 1
else
$NumTasksAssigned = $DefaultNumTasksAssigned
endif
$End = $Start + $NumTasksAssigned -1
print Thread $i: Tasks $Start-$End ($NumTasksAssigned tasks)
$Start = $Start + $NumTasksAssigned
Next
Here's a Python transcription of the above
>>> def ShowWorkAllocation(NumTasks, NumThreads):
... Remainder = NumTasks % NumThreads
... DefaultNumTasksAssigned = math.floor(NumTasks / NumThreads)
... Start = 1
... for i in range(1, NumThreads + 1):
... if i <= Remainder:
... NumTasksAssigned = DefaultNumTasksAssigned + 1
... else:
... NumTasksAssigned = DefaultNumTasksAssigned
... End = Start + NumTasksAssigned - 1
... print("Thread ", i, ": Tasks ", Start, "-", End, "(", NumTasksAssigned,")")
... Start = Start + NumTasksAssigned
...
>>>
>>> ShowWorkAllocation(89, 5)
Thread 1 : Tasks 1 - 18 ( 18 )
Thread 2 : Tasks 19 - 36 ( 18 )
Thread 3 : Tasks 37 - 54 ( 18 )
Thread 4 : Tasks 55 - 72 ( 18 )
Thread 5 : Tasks 73 - 89 ( 17 )
>>> ShowWorkAllocation(11, 5)
Thread 1 : Tasks 1 - 3 ( 3 )
Thread 2 : Tasks 4 - 5 ( 2 )
Thread 3 : Tasks 6 - 7 ( 2 )
Thread 4 : Tasks 8 - 9 ( 2 )
Thread 5 : Tasks 10 - 11 ( 2 )
>>>
>>> ShowWorkAllocation(89, 11)
Thread 1 : Tasks 1 - 9 ( 9 )
Thread 2 : Tasks 10 - 17 ( 8 )
Thread 3 : Tasks 18 - 25 ( 8 )
Thread 4 : Tasks 26 - 33 ( 8 )
Thread 5 : Tasks 34 - 41 ( 8 )
Thread 6 : Tasks 42 - 49 ( 8 )
Thread 7 : Tasks 50 - 57 ( 8 )
Thread 8 : Tasks 58 - 65 ( 8 )
Thread 9 : Tasks 66 - 73 ( 8 )
Thread 10 : Tasks 74 - 81 ( 8 )
Thread 11 : Tasks 82 - 89 ( 8 )
>>>

Answer 3

I think you've solved the wrong half of your problem.
It's going to be virtually impossible to precisely determine the time it will take to complete all your tasks, unless all of the following are true:
your tasks are 100% CPU-bound: that is, they use 100% CPU while running and don't need to do any I/O
none of your tasks have to synchronize with any of your other tasks in any way
you have exactly as many threads as you have CPUs
the computer that is running these tasks is not performing any other interesting tasks at the same time
In practice, most of the time, your tasks are I/O-bound rather than CPU-bound: that is, you are waiting for some external resource such as reading from a file, fetching from a database, or communicating with a remote computer. In that case, you only make things worse by adding more threads, because they're all contending for the same scarce resource.
Finally, unless you have some really weird hardware, it's unlikely you can actually have exactly five threads running simultaneously. (Usually processor configurations come in multiples of at least two.) Usually the sweet spot is at about 1 thread per CPU if your tasks are very CPU-bound, about 2 threads per CPU if the tasks spend half their time being CPU-bound and half their time doing IO, etc.
tl;dr: We need to know a lot more about what your tasks and hardware look like before we can advise you on this question.