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Parallelization does not use all cores fully, help :(
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On a new laptop with a 6-core i7-8850H in Mma11.3, I'd like to use its full capabilities. Unicore code
prm = Prime /@ Range[1,1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
Table[f[5^100+i],i,0,10] //AbsoluteTiming
returns 43.9328,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
The multicore code
prm = Prime /@ Range[1, 1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
DistributeDefinitions[prm, f];
ParallelCombine[Table[f[5^100+i],i,#]&, Range[0,10], Join] //AbsoluteTiming
returns 26.1113,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
In Evaluation -> Parallel Kernel Configuration, I have:
Why is the speed-up less than 2x? Why does the multicore code only give 33% use of my CPU and unicore 16%? How can I use all 6 cores fully (or at least 5 cores to the max)?
parallelization
asked 4 hours ago
Leon
315111
 |Â
show 2 more comments
up vote
1
down vote
favorite
On a new laptop with a 6-core i7-8850H in Mma11.3, I'd like to use its full capabilities. Unicore code
prm = Prime /@ Range[1,1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
Table[f[5^100+i],i,0,10] //AbsoluteTiming
returns 43.9328,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
The multicore code
prm = Prime /@ Range[1, 1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
DistributeDefinitions[prm, f];
ParallelCombine[Table[f[5^100+i],i,#]&, Range[0,10], Join] //AbsoluteTiming
returns 26.1113,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
In Evaluation -> Parallel Kernel Configuration, I have:
Why is the speed-up less than 2x? Why does the multicore code only give 33% use of my CPU and unicore 16%? How can I use all 6 cores fully (or at least 5 cores to the max)?
parallelization
asked 4 hours ago
Leon
315111
1
In my experience, Mathematica uses only one thread per core. How does the CPU utilization look (average utilization and time history) with six kernels launched, and with one kernel launched (i.e., not parallel)?
– bbgodfrey
3 hours ago
@bbgodfrey I thought the number of Mma kernels should equal the number of CPU threads. Should it equal the number of CPU cores? Anyway, I tried the same two computations with 6 kernels, and the results are very similar (45sec 15%, 26sec 30%).
– Leon
3 hours ago
1
try addingMethod-> "CoarsestGrained"at the end ofParallelCombine. you can also useLabeledto show which$KernelIdwas used to obtain the results
– Alucard
3 hours ago
CoarsestGrainedandFinestGrainedboth require 26sec.
– Leon
3 hours ago
"The number of kernels available for parallel computation typically corresponds to the number of CPU cores", according to reference.wolfram.com/language/workflow/…. Using more typically just adds overhead. So, with two-thread cores, it is difficult to obtain better than 50% utilization as measured by Task Manager.
– bbgodfrey
3 hours ago
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On a new laptop with a 6-core i7-8850H in Mma11.3, I'd like to use its full capabilities. Unicore code
prm = Prime /@ Range[1,1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
Table[f[5^100+i],i,0,10] //AbsoluteTiming
returns 43.9328,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
The multicore code
prm = Prime /@ Range[1, 1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
DistributeDefinitions[prm, f];
ParallelCombine[Table[f[5^100+i],i,#]&, Range[0,10], Join] //AbsoluteTiming
returns 26.1113,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
In Evaluation -> Parallel Kernel Configuration, I have:
Why is the speed-up less than 2x? Why does the multicore code only give 33% use of my CPU and unicore 16%? How can I use all 6 cores fully (or at least 5 cores to the max)?
parallelization
asked 4 hours ago
Leon
315111
On a new laptop with a 6-core i7-8850H in Mma11.3, I'd like to use its full capabilities. Unicore code
prm = Prime /@ Range[1,1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
Table[f[5^100+i],i,0,10] //AbsoluteTiming
returns 43.9328,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
The multicore code
prm = Prime /@ Range[1, 1000];
f[i_] := Length[(First/@FactorInteger[i]) [Intersection] prm];
DistributeDefinitions[prm, f];
ParallelCombine[Table[f[5^100+i],i,#]&, Range[0,10], Join] //AbsoluteTiming
returns 26.1113,1,4,3,4,3,8,0,2,1,1,2. Task manager measures:
In Evaluation -> Parallel Kernel Configuration, I have:
Why is the speed-up less than 2x? Why does the multicore code only give 33% use of my CPU and unicore 16%? How can I use all 6 cores fully (or at least 5 cores to the max)?
parallelization
parallelization
asked 4 hours ago
Leon
315111
asked 4 hours ago
Leon
315111
asked 4 hours ago
Leon
315111
asked 4 hours ago
Leon
315111
asked 4 hours ago
Leon
315111
315111
1
In my experience, Mathematica uses only one thread per core. How does the CPU utilization look (average utilization and time history) with six kernels launched, and with one kernel launched (i.e., not parallel)?
– bbgodfrey
3 hours ago
@bbgodfrey I thought the number of Mma kernels should equal the number of CPU threads. Should it equal the number of CPU cores? Anyway, I tried the same two computations with 6 kernels, and the results are very similar (45sec 15%, 26sec 30%).
– Leon
3 hours ago
1
try addingMethod-> "CoarsestGrained"at the end ofParallelCombine. you can also useLabeledto show which$KernelIdwas used to obtain the results
– Alucard
3 hours ago
CoarsestGrainedandFinestGrainedboth require 26sec.
– Leon
3 hours ago
"The number of kernels available for parallel computation typically corresponds to the number of CPU cores", according to reference.wolfram.com/language/workflow/…. Using more typically just adds overhead. So, with two-thread cores, it is difficult to obtain better than 50% utilization as measured by Task Manager.
– bbgodfrey
3 hours ago
 |Â
show 2 more comments
1
In my experience, Mathematica uses only one thread per core. How does the CPU utilization look (average utilization and time history) with six kernels launched, and with one kernel launched (i.e., not parallel)?
– bbgodfrey
3 hours ago
@bbgodfrey I thought the number of Mma kernels should equal the number of CPU threads. Should it equal the number of CPU cores? Anyway, I tried the same two computations with 6 kernels, and the results are very similar (45sec 15%, 26sec 30%).
– Leon
3 hours ago
1
try addingMethod-> "CoarsestGrained"at the end ofParallelCombine. you can also useLabeledto show which$KernelIdwas used to obtain the results
– Alucard
3 hours ago
CoarsestGrainedandFinestGrainedboth require 26sec.
– Leon
3 hours ago
"The number of kernels available for parallel computation typically corresponds to the number of CPU cores", according to reference.wolfram.com/language/workflow/…. Using more typically just adds overhead. So, with two-thread cores, it is difficult to obtain better than 50% utilization as measured by Task Manager.
– bbgodfrey
3 hours ago
1
1
In my experience, Mathematica uses only one thread per core. How does the CPU utilization look (average utilization and time history) with six kernels launched, and with one kernel launched (i.e., not parallel)?
– bbgodfrey
3 hours ago
In my experience, Mathematica uses only one thread per core. How does the CPU utilization look (average utilization and time history) with six kernels launched, and with one kernel launched (i.e., not parallel)?
– bbgodfrey
3 hours ago
@bbgodfrey I thought the number of Mma kernels should equal the number of CPU threads. Should it equal the number of CPU cores? Anyway, I tried the same two computations with 6 kernels, and the results are very similar (45sec 15%, 26sec 30%).
– Leon
3 hours ago
@bbgodfrey I thought the number of Mma kernels should equal the number of CPU threads. Should it equal the number of CPU cores? Anyway, I tried the same two computations with 6 kernels, and the results are very similar (45sec 15%, 26sec 30%).
– Leon
3 hours ago
1
1
try adding
Method-> "CoarsestGrained" at the end of ParallelCombine. you can also use Labeled to show which $KernelId was used to obtain the results– Alucard
3 hours ago
try adding
Method-> "CoarsestGrained" at the end of ParallelCombine. you can also use Labeled to show which $KernelId was used to obtain the results– Alucard
3 hours ago
CoarsestGrained and FinestGrained both require 26sec.– Leon
3 hours ago
CoarsestGrained and FinestGrained both require 26sec.– Leon
3 hours ago
"The number of kernels available for parallel computation typically corresponds to the number of CPU cores", according to reference.wolfram.com/language/workflow/…. Using more typically just adds overhead. So, with two-thread cores, it is difficult to obtain better than 50% utilization as measured by Task Manager.
– bbgodfrey
3 hours ago
"The number of kernels available for parallel computation typically corresponds to the number of CPU cores", according to reference.wolfram.com/language/workflow/…. Using more typically just adds overhead. So, with two-thread cores, it is difficult to obtain better than 50% utilization as measured by Task Manager.
– bbgodfrey
3 hours ago
 |Â
show 2 more comments
1 Answer
1
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up vote
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The following was perfomed on a 4-processor, 8-thread PC running
$Version
(* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" *)
To begin,
ParallelCombine[Table[f[5^100 + i], i, #] &, Range[0, 10], Join] // AbsoluteTiming
(* 37.2452, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
This is equivalent to
ParallelTable[f[5^100 + i], i, 0, 10] // AbsoluteTiming
(* 34.8027, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
which is a bit easier to understand and manipulate, in my view. I then tried both Methods and also tried different orders of i in ParallelTable Nothing mattered. In all cases, Task Manager showed that all four kernels on my computer each ran at about 18% of total CPU capacity. However, two finished almost immediately, while the other two continued for some time, after which another finished and the last continued a bit longer. The reason for this behavior, it turns out, is easily determined.
Table[f[5^100 + i] // AbsoluteTiming, i, 0, 10]
(* 0.0000608395, 1, 0.0108113, 4, 0.0120675, 3, 0.255642, 4,
2.88463, 3, 0.00794627, 8, 0.0868866, 0, 31.8475, 2,
3.3773, 1, 24.3363, 1, 0.0163816, 2 *)
i = 7 and i = 9 take most of the time, with two kernels computing the others quickly and then finishing. Given that i = 7 takes almost 32 seconds by itself, it is no longer is surprising that ParallelTable takes 35 seconds to handle all i.
Addendum
In my experience, a computation that lends itself well to parallelization will launch four kernels, each using at 16% - 19% of total CPU capacity as measured by Task Manager. Thus, the best I have ever seen is total CPU utilization of about 75%.
answered 2 hours ago
bbgodfrey
43.2k857104
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The following was perfomed on a 4-processor, 8-thread PC running
$Version
(* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" *)
To begin,
ParallelCombine[Table[f[5^100 + i], i, #] &, Range[0, 10], Join] // AbsoluteTiming
(* 37.2452, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
This is equivalent to
ParallelTable[f[5^100 + i], i, 0, 10] // AbsoluteTiming
(* 34.8027, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
which is a bit easier to understand and manipulate, in my view. I then tried both Methods and also tried different orders of i in ParallelTable Nothing mattered. In all cases, Task Manager showed that all four kernels on my computer each ran at about 18% of total CPU capacity. However, two finished almost immediately, while the other two continued for some time, after which another finished and the last continued a bit longer. The reason for this behavior, it turns out, is easily determined.
Table[f[5^100 + i] // AbsoluteTiming, i, 0, 10]
(* 0.0000608395, 1, 0.0108113, 4, 0.0120675, 3, 0.255642, 4,
2.88463, 3, 0.00794627, 8, 0.0868866, 0, 31.8475, 2,
3.3773, 1, 24.3363, 1, 0.0163816, 2 *)
i = 7 and i = 9 take most of the time, with two kernels computing the others quickly and then finishing. Given that i = 7 takes almost 32 seconds by itself, it is no longer is surprising that ParallelTable takes 35 seconds to handle all i.
Addendum
In my experience, a computation that lends itself well to parallelization will launch four kernels, each using at 16% - 19% of total CPU capacity as measured by Task Manager. Thus, the best I have ever seen is total CPU utilization of about 75%.
answered 2 hours ago
bbgodfrey
43.2k857104
add a comment |Â
up vote
3
down vote
The following was perfomed on a 4-processor, 8-thread PC running
$Version
(* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" *)
To begin,
ParallelCombine[Table[f[5^100 + i], i, #] &, Range[0, 10], Join] // AbsoluteTiming
(* 37.2452, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
This is equivalent to
ParallelTable[f[5^100 + i], i, 0, 10] // AbsoluteTiming
(* 34.8027, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
which is a bit easier to understand and manipulate, in my view. I then tried both Methods and also tried different orders of i in ParallelTable Nothing mattered. In all cases, Task Manager showed that all four kernels on my computer each ran at about 18% of total CPU capacity. However, two finished almost immediately, while the other two continued for some time, after which another finished and the last continued a bit longer. The reason for this behavior, it turns out, is easily determined.
Table[f[5^100 + i] // AbsoluteTiming, i, 0, 10]
(* 0.0000608395, 1, 0.0108113, 4, 0.0120675, 3, 0.255642, 4,
2.88463, 3, 0.00794627, 8, 0.0868866, 0, 31.8475, 2,
3.3773, 1, 24.3363, 1, 0.0163816, 2 *)
i = 7 and i = 9 take most of the time, with two kernels computing the others quickly and then finishing. Given that i = 7 takes almost 32 seconds by itself, it is no longer is surprising that ParallelTable takes 35 seconds to handle all i.
Addendum
In my experience, a computation that lends itself well to parallelization will launch four kernels, each using at 16% - 19% of total CPU capacity as measured by Task Manager. Thus, the best I have ever seen is total CPU utilization of about 75%.
answered 2 hours ago
bbgodfrey
43.2k857104
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The following was perfomed on a 4-processor, 8-thread PC running
$Version
(* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" *)
To begin,
ParallelCombine[Table[f[5^100 + i], i, #] &, Range[0, 10], Join] // AbsoluteTiming
(* 37.2452, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
This is equivalent to
ParallelTable[f[5^100 + i], i, 0, 10] // AbsoluteTiming
(* 34.8027, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
which is a bit easier to understand and manipulate, in my view. I then tried both Methods and also tried different orders of i in ParallelTable Nothing mattered. In all cases, Task Manager showed that all four kernels on my computer each ran at about 18% of total CPU capacity. However, two finished almost immediately, while the other two continued for some time, after which another finished and the last continued a bit longer. The reason for this behavior, it turns out, is easily determined.
Table[f[5^100 + i] // AbsoluteTiming, i, 0, 10]
(* 0.0000608395, 1, 0.0108113, 4, 0.0120675, 3, 0.255642, 4,
2.88463, 3, 0.00794627, 8, 0.0868866, 0, 31.8475, 2,
3.3773, 1, 24.3363, 1, 0.0163816, 2 *)
i = 7 and i = 9 take most of the time, with two kernels computing the others quickly and then finishing. Given that i = 7 takes almost 32 seconds by itself, it is no longer is surprising that ParallelTable takes 35 seconds to handle all i.
Addendum
In my experience, a computation that lends itself well to parallelization will launch four kernels, each using at 16% - 19% of total CPU capacity as measured by Task Manager. Thus, the best I have ever seen is total CPU utilization of about 75%.
answered 2 hours ago
bbgodfrey
43.2k857104
The following was perfomed on a 4-processor, 8-thread PC running
$Version
(* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" *)
To begin,
ParallelCombine[Table[f[5^100 + i], i, #] &, Range[0, 10], Join] // AbsoluteTiming
(* 37.2452, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
This is equivalent to
ParallelTable[f[5^100 + i], i, 0, 10] // AbsoluteTiming
(* 34.8027, 1, 4, 3, 4, 3, 8, 0, 2, 1, 1, 2 *)
which is a bit easier to understand and manipulate, in my view. I then tried both Methods and also tried different orders of i in ParallelTable Nothing mattered. In all cases, Task Manager showed that all four kernels on my computer each ran at about 18% of total CPU capacity. However, two finished almost immediately, while the other two continued for some time, after which another finished and the last continued a bit longer. The reason for this behavior, it turns out, is easily determined.
Table[f[5^100 + i] // AbsoluteTiming, i, 0, 10]
(* 0.0000608395, 1, 0.0108113, 4, 0.0120675, 3, 0.255642, 4,
2.88463, 3, 0.00794627, 8, 0.0868866, 0, 31.8475, 2,
3.3773, 1, 24.3363, 1, 0.0163816, 2 *)
i = 7 and i = 9 take most of the time, with two kernels computing the others quickly and then finishing. Given that i = 7 takes almost 32 seconds by itself, it is no longer is surprising that ParallelTable takes 35 seconds to handle all i.
Addendum
In my experience, a computation that lends itself well to parallelization will launch four kernels, each using at 16% - 19% of total CPU capacity as measured by Task Manager. Thus, the best I have ever seen is total CPU utilization of about 75%.
answered 2 hours ago
bbgodfrey
43.2k857104
edited 1 hour ago
answered 2 hours ago
bbgodfrey
43.2k857104
answered 2 hours ago
bbgodfrey
43.2k857104
answered 2 hours ago
bbgodfrey
43.2k857104
43.2k857104
add a comment |Â
add a comment |Â
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1
In my experience, Mathematica uses only one thread per core. How does the CPU utilization look (average utilization and time history) with six kernels launched, and with one kernel launched (i.e., not parallel)?
– bbgodfrey
3 hours ago
@bbgodfrey I thought the number of Mma kernels should equal the number of CPU threads. Should it equal the number of CPU cores? Anyway, I tried the same two computations with 6 kernels, and the results are very similar (45sec 15%, 26sec 30%).
– Leon
3 hours ago
1
try adding
Method-> "CoarsestGrained"at the end ofParallelCombine. you can also useLabeledto show which$KernelIdwas used to obtain the results– Alucard
3 hours ago
CoarsestGrainedandFinestGrainedboth require 26sec.– Leon
3 hours ago
"The number of kernels available for parallel computation typically corresponds to the number of CPU cores", according to reference.wolfram.com/language/workflow/…. Using more typically just adds overhead. So, with two-thread cores, it is difficult to obtain better than 50% utilization as measured by Task Manager.
– bbgodfrey
3 hours ago