Divide N floats into M groups so that the sum of each group is as equal as possible?
$begingroup$
Given a list of N non-negative real numbers:
t = [2.99, 7.9, 24.58, ..., 3.1415, 40.4]
I want to partition the elements of $t$ into $M$ groups, $g_1$, $g_2$, ..., $g_M$, so that the sum of each group is as close as possible to $sum(t)/M$, i.e. minimizing the root mean square error:
Minimize: $sqrt{Sigma_i^M (sum(g_i) - sum(t)/M)^2 }$.
Do anyone know a close-to-optimal algorithm of doing this generally? Or do anyone know the name of this problem? (My google-fu failed me).
Background: $t$ represents time in seconds to run $N$ different tasks, and I want to schedule the $N$ tasks on $M$ different computers in such a way that all computers finish their tasks at the same time (or as close as possible).
optimization
asked Dec 9 '18 at 17:43
PenlectPenlect
334
$endgroup$
add a comment |
$begingroup$
Given a list of N non-negative real numbers:
t = [2.99, 7.9, 24.58, ..., 3.1415, 40.4]
I want to partition the elements of $t$ into $M$ groups, $g_1$, $g_2$, ..., $g_M$, so that the sum of each group is as close as possible to $sum(t)/M$, i.e. minimizing the root mean square error:
Minimize: $sqrt{Sigma_i^M (sum(g_i) - sum(t)/M)^2 }$.
Do anyone know a close-to-optimal algorithm of doing this generally? Or do anyone know the name of this problem? (My google-fu failed me).
Background: $t$ represents time in seconds to run $N$ different tasks, and I want to schedule the $N$ tasks on $M$ different computers in such a way that all computers finish their tasks at the same time (or as close as possible).
optimization
asked Dec 9 '18 at 17:43
PenlectPenlect
334
$endgroup$
1
$begingroup$
Wouldn't your intended application suggest to minimize $max_isum(g_i)$ instead?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 17:53
$begingroup$
This looks like a variant on the bin packing problem to me. It is known to be NP-hard in the general case, but there are approaches that are pretty good.
$endgroup$
– Ross Millikan
Dec 9 '18 at 17:57
$begingroup$
Can you confirm that $M$ is imposed ?
$endgroup$
– Jean Marie
Dec 9 '18 at 18:07
$begingroup$
@JeanMarie Yes, M is fix, for example "5", and not subject to change.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
$begingroup$
@HagenvonEitzen Good point, yes, minimizing your expression is also interesting. And now I'm curious if the optimal solution to that will automatically also be the optimal solution for the RMSE.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
add a comment |
$begingroup$
Given a list of N non-negative real numbers:
t = [2.99, 7.9, 24.58, ..., 3.1415, 40.4]
I want to partition the elements of $t$ into $M$ groups, $g_1$, $g_2$, ..., $g_M$, so that the sum of each group is as close as possible to $sum(t)/M$, i.e. minimizing the root mean square error:
Minimize: $sqrt{Sigma_i^M (sum(g_i) - sum(t)/M)^2 }$.
Do anyone know a close-to-optimal algorithm of doing this generally? Or do anyone know the name of this problem? (My google-fu failed me).
Background: $t$ represents time in seconds to run $N$ different tasks, and I want to schedule the $N$ tasks on $M$ different computers in such a way that all computers finish their tasks at the same time (or as close as possible).
optimization
asked Dec 9 '18 at 17:43
PenlectPenlect
334
$endgroup$
Given a list of N non-negative real numbers:
t = [2.99, 7.9, 24.58, ..., 3.1415, 40.4]
I want to partition the elements of $t$ into $M$ groups, $g_1$, $g_2$, ..., $g_M$, so that the sum of each group is as close as possible to $sum(t)/M$, i.e. minimizing the root mean square error:
Minimize: $sqrt{Sigma_i^M (sum(g_i) - sum(t)/M)^2 }$.
Do anyone know a close-to-optimal algorithm of doing this generally? Or do anyone know the name of this problem? (My google-fu failed me).
Background: $t$ represents time in seconds to run $N$ different tasks, and I want to schedule the $N$ tasks on $M$ different computers in such a way that all computers finish their tasks at the same time (or as close as possible).
optimization
optimization
asked Dec 9 '18 at 17:43
PenlectPenlect
334
asked Dec 9 '18 at 17:43
PenlectPenlect
334
asked Dec 9 '18 at 17:43
PenlectPenlect
334
asked Dec 9 '18 at 17:43
PenlectPenlect
334
asked Dec 9 '18 at 17:43
PenlectPenlect
334
334
1
$begingroup$
Wouldn't your intended application suggest to minimize $max_isum(g_i)$ instead?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 17:53
$begingroup$
This looks like a variant on the bin packing problem to me. It is known to be NP-hard in the general case, but there are approaches that are pretty good.
$endgroup$
– Ross Millikan
Dec 9 '18 at 17:57
$begingroup$
Can you confirm that $M$ is imposed ?
$endgroup$
– Jean Marie
Dec 9 '18 at 18:07
$begingroup$
@JeanMarie Yes, M is fix, for example "5", and not subject to change.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
$begingroup$
@HagenvonEitzen Good point, yes, minimizing your expression is also interesting. And now I'm curious if the optimal solution to that will automatically also be the optimal solution for the RMSE.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
add a comment |
1
$begingroup$
Wouldn't your intended application suggest to minimize $max_isum(g_i)$ instead?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 17:53
$begingroup$
This looks like a variant on the bin packing problem to me. It is known to be NP-hard in the general case, but there are approaches that are pretty good.
$endgroup$
– Ross Millikan
Dec 9 '18 at 17:57
$begingroup$
Can you confirm that $M$ is imposed ?
$endgroup$
– Jean Marie
Dec 9 '18 at 18:07
$begingroup$
@JeanMarie Yes, M is fix, for example "5", and not subject to change.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
$begingroup$
@HagenvonEitzen Good point, yes, minimizing your expression is also interesting. And now I'm curious if the optimal solution to that will automatically also be the optimal solution for the RMSE.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
1
1
$begingroup$
Wouldn't your intended application suggest to minimize $max_isum(g_i)$ instead?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 17:53
$begingroup$
Wouldn't your intended application suggest to minimize $max_isum(g_i)$ instead?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 17:53
$begingroup$
This looks like a variant on the bin packing problem to me. It is known to be NP-hard in the general case, but there are approaches that are pretty good.
$endgroup$
– Ross Millikan
Dec 9 '18 at 17:57
$begingroup$
This looks like a variant on the bin packing problem to me. It is known to be NP-hard in the general case, but there are approaches that are pretty good.
$endgroup$
– Ross Millikan
Dec 9 '18 at 17:57
$begingroup$
Can you confirm that $M$ is imposed ?
$endgroup$
– Jean Marie
Dec 9 '18 at 18:07
$begingroup$
Can you confirm that $M$ is imposed ?
$endgroup$
– Jean Marie
Dec 9 '18 at 18:07
$begingroup$
@JeanMarie Yes, M is fix, for example "5", and not subject to change.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
$begingroup$
@JeanMarie Yes, M is fix, for example "5", and not subject to change.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
$begingroup$
@HagenvonEitzen Good point, yes, minimizing your expression is also interesting. And now I'm curious if the optimal solution to that will automatically also be the optimal solution for the RMSE.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
$begingroup$
@HagenvonEitzen Good point, yes, minimizing your expression is also interesting. And now I'm curious if the optimal solution to that will automatically also be the optimal solution for the RMSE.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
add a comment |
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1
$begingroup$
Wouldn't your intended application suggest to minimize $max_isum(g_i)$ instead?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 17:53
$begingroup$
This looks like a variant on the bin packing problem to me. It is known to be NP-hard in the general case, but there are approaches that are pretty good.
$endgroup$
– Ross Millikan
Dec 9 '18 at 17:57
$begingroup$
Can you confirm that $M$ is imposed ?
$endgroup$
– Jean Marie
Dec 9 '18 at 18:07
$begingroup$
@JeanMarie Yes, M is fix, for example "5", and not subject to change.
$endgroup$
– Penlect
Dec 9 '18 at 21:01
$begingroup$
@HagenvonEitzen Good point, yes, minimizing your expression is also interesting. And now I'm curious if the optimal solution to that will automatically also be the optimal solution for the RMSE.
$endgroup$
– Penlect
Dec 9 '18 at 21:01