Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

My CPU has got 8 cores (it is Intel Core i7-2600 3.40 GHz). When I try to solve a linear matrix equation using LinearSolve for large matrices, Mathematica just uses 4 cores to solve the problem (CPU usage will be 50%). So it means that there is a problem in the Parallel computation options. When I try to Minimize a huge function with several variables, it is even worse and Mathematica just uses one core (CPU usage is about 12%)!!

I am not very familiar with Parallel computation, so will be very appreciative if you can help me to solve my problem. How can I use all capacity of the CPU when I run LinearSolve for very large matrices and Minimize for huge functions and make CPU usage 100% to make the computation time as short as possible on my machine??

Thank you very much.

**

Edit 1:

**

My computer for this analysis:

t = AbsoluteTime[];
primelist = Table[Prime[k], {k, 1, 20000000}];
time2 = AbsoluteTime[] - t   

Yields a load of 12% on my CPU and time2=43.37 and by breaking this analysis into 8 cores:

t = AbsoluteTime[];
job1 = ParallelSubmit[Table[Prime[k], {k, 1, 2500000}]];
job2 = ParallelSubmit[Table[Prime[k], {k, 2500001, 5000000}]];
job3 = ParallelSubmit[Table[Prime[k], {k, 5000001, 7500000}]];
job4 = ParallelSubmit[Table[Prime[k], {k, 7500001, 10000000}]];
job5 = ParallelSubmit[Table[Prime[k], {k, 10000001, 12500000}]];
job6 = ParallelSubmit[Table[Prime[k], {k, 12500001, 15000000}]];
job7 = ParallelSubmit[Table[Prime[k], {k, 15000001, 17500000}]];
job8 = ParallelSubmit[Table[Prime[k], {k, 17500001, 20000000}]];
{a1, a2, a3, a4, a5, a6, a7, a8} = 
 WaitAll[{job1, job2, job3, job4, job5, job6, job7, job8}];
time2 = AbsoluteTime[] - t

Yields 100% load on CPU and time2=17.16

**

Edit 2:

**

To make it completely clear what is happening on my computer and what is my problem, please have a look at the following examples:

First if I want to check number of processors and kernels on my machine:

$ProcessorCount
$KernelCount

The results are 4 and 8 respectively on my machine.

Now if I want to see MKL conditions on my machine and also how much "CPU usage" reported by the system monitor correspond to actual performance, I can run this in Mathematica:

Clear["Global`*"]; a = RandomReal[{1, 2}, {20000, 20000}]; b = RandomReal[{1}, {20000}];
Table[SetSystemOptions["MKLThreads" -> i]; Print["Case=", i]; Print[SystemOptions["MKLThreads"]]; t = AbsoluteTime[]; LinearSolve[a, b]; time2 = AbsoluteTime[] - t; Print["t(", i, ")=", time2]; Print["******"], {i, 4}];

You can see the result including number of MKL threads and computation time for each case below:

Case=1
{MKLThreads->1}
t(1)=202.9560000
******
Case=2
{MKLThreads->2}
t(2)=120.3696000
******
Case=3
{MKLThreads->3}
t(3)=93.5532000
******
Case=4
{MKLThreads->4}
t(4)=88.5300000
******

While the CPU usage for Case1=12%, Case2=25%, Case3=37% and Case4=50%, reported by the system monitor. You can see that in this case "CPU usage" reported by the system monitor correspond to actual performance and the more CPU usage we observe, the less computation time we have.

Now if I increase the number of MKLThreads in SetSystemOptions["MKLThreads" -> ?] to values more than 4 (I mean 5 to 8), I can see that it doesn't have any effect on compuation time and CPU usage. The same thing happens if I change the number of ParallelThreadNumber in SetSystemOptions["ParallelOptions" -> {"ParallelThreadNumber" -> ?}], means that the computation time and CPU usage in this case do not depend on the ParallelThreadNumber. You can see the cases below:

SetSystemOptions["ParallelOptions" -> {"ParallelThreadNumber" -> 1}];
Print[SystemOptions["ParallelOptions" -> "ParallelThreadNumber"]];
SetSystemOptions["MKLThreads" -> 8]; Print[SystemOptions["MKLThreads"]];
t=AbsoluteTime[];
LinearSolve[a, b];
time2=AbsoluteTime[] - t;
Print["t=", time2];

The result is (CPU usage=50% during analysis):

{ParallelOptions->{ParallelThreadNumber->1}}
{MKLThreads->4}
t=85.3008000

And for other case:

SetSystemOptions["ParallelOptions" -> {"ParallelThreadNumber" -> 8}];
Print[SystemOptions["ParallelOptions" -> "ParallelThreadNumber"]];
SetSystemOptions["MKLThreads" -> 8]; Print[SystemOptions["MKLThreads"]];
t=AbsoluteTime[];
LinearSolve[a, b];
time2=AbsoluteTime[] - t;
Print["t=", time2];

The result is (Again CPU usage=50% during analysis):

{ParallelOptions->{ParallelThreadNumber->8}}
{MKLThreads->4}
t=85.3476000

As you can see, the CPU usage and computation time do not change when I increase MKLThreads to more than 4 (e.g 5 to 8) and they are also independent of the ParallelThreadNumber.

Another interesting example is about the case I mentioned in edit 1. Please have a look at these examples and results and CPU usage for each case:

1)

Clear["Global`*"];
t = AbsoluteTime[];
primelist = Table[Prime[k], {k, 1, 20000000}];
time2 = AbsoluteTime[] - t

Result: time2=43.37 and CPU usage=12%

2)

Clear["Global`*"];
t = AbsoluteTime[];
job1 = ParallelSubmit[Table[Prime[k], {k, 1, 10000000}]];
job2 = ParallelSubmit[Table[Prime[k], {k, 10000001, 20000000}]];
{a1, a2} = WaitAll[{job1, job2}];
time2 = AbsoluteTime[] - t

Result: time2=30.01 and CPU usage=25%

3)

Clear["Global`*"];
t = AbsoluteTime[];
job1 = ParallelSubmit[Table[Prime[k], {k, 1, 6666666}]];
job2 = ParallelSubmit[Table[Prime[k], {k, 6666667, 13333332}]];
job3 = ParallelSubmit[Table[Prime[k], {k, 13333333, 20000000}]];
{a1, a2, a3} = WaitAll[{job1, job2, job3}];
time2 = AbsoluteTime[] - t

Result: time2=23.46 and CPU usage=37%

4)

Clear["Global`*"];
t = AbsoluteTime[];
job1 = ParallelSubmit[Table[Prime[k], {k, 1, 5000000}]];
job2 = ParallelSubmit[Table[Prime[k], {k, 5000001, 10000000}]];
job3 = ParallelSubmit[Table[Prime[k], {k, 10000000, 15000000}]];
job4 = ParallelSubmit[Table[Prime[k], {k, 15000001, 20000000}]];
{a1, a2, a3, a4} = WaitAll[{job1, job2, job3, job4}];
time2 = AbsoluteTime[] - t 

Result: time2=21.52 and CPU usage=50%

5)

Clear["Global`*"];
t = AbsoluteTime[];
job1 = ParallelSubmit[Table[Prime[k], {k, 1, 3333333}]];
job2 = ParallelSubmit[Table[Prime[k], {k, 3333334, 6666666}]];
job3 = ParallelSubmit[Table[Prime[k], {k, 6666667, 9999999}]];
job4 = ParallelSubmit[Table[Prime[k], {k, 10000000, 13333333}]];
job5 = ParallelSubmit[Table[Prime[k], {k, 13333334, 16666666}]];
job6 = ParallelSubmit[Table[Prime[k], {k, 16666667, 20000000}]];
{a1, a2, a3, a4, a5, a6} = WaitAll[{job1, job2, job3, job4, job5, job6}];
time2 = AbsoluteTime[] - t

Result: time2=18.28 and CPU usage=75%

6)

Clear["Global`*"];
t = AbsoluteTime[];
job1 = ParallelSubmit[Table[Prime[k], {k, 1, 2500000}]];
job2 = ParallelSubmit[Table[Prime[k], {k, 2500001, 5000000}]];
job3 = ParallelSubmit[Table[Prime[k], {k, 5000001, 7500000}]];
job4 = ParallelSubmit[Table[Prime[k], {k, 7500001, 10000000}]];
job5 = ParallelSubmit[Table[Prime[k], {k, 10000001, 12500000}]];
job6 = ParallelSubmit[Table[Prime[k], {k, 12500001, 15000000}]];
job7 = ParallelSubmit[Table[Prime[k], {k, 15000001, 17500000}]];
job8 = ParallelSubmit[Table[Prime[k], {k, 17500001, 20000000}]];
{a1, a2, a3, a4, a5, a6, a7, a8} = 
 WaitAll[{job1, job2, job3, job4, job5, job6, job7, job8}];
time2 = AbsoluteTime[] - t

Result: time2=17.16 and CPU usage=100%

But if I make this analysis using ParallelTable, interestingly the CPU usage is 100%, but computation time is 45.81!!! It means that computation time is quite the same with number 1 when I do this analysis with Table on one core (CPU usage=12%)!!

t = AbsoluteTime[];
primelist = ParallelTable[Prime[k], {k, 1, 20000000}];
time2 = AbsoluteTime[] - t

Result: time2=45.81 and CPU usage=100%

I also checked NMinimize for my big function with 75 (or more variables) using all methods available in Mathematica including Automatic, DifferentialEvolution, NelderMead, RandomSearch, and SimulatedAnnealing. The computation time for all of them is quite the same and CPU usage for all methods is only 12%. So it looks that minimization method I use in NMinimize cannot change parallelization conditions.

Now I think my conditions and problems are completely clear, so I would be very appreciative if someone can help me to use all capacity of my CPU in LinearSolve and NMinimize (or Minimize). I still wonder how I can make CPU usage in these cases 100%. In this way we can check whether CPU usage corresponds to actual perfomanse (like what we could see in the examples mentioned above) for LinearSolve and NMinimize or not??

Thank you very much.

**

Edit 3

**

The function I am trying to minimize is a large function including many variables. The general format of the function is sth like this:

(15 (-2.14286*10^-8 Log[E^(-1.*10^6 phi2[2]) + E^(1.*10^6 phi2[2])] + uu1[2]))^2 + 225 (2.14286*10^-8 Log[E^(-1.*10^6 phi2[2]) + E^(1.*10^6 phi2[2])] -2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[2] + phi2[3])) + E^(1.*10^6 (-phi2[2] + phi2[3]))] -2 uu1[2] + uu1[3])^2 + 225 (2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[2] + phi2[3])) + E^(1.*10^6 (-phi2[2] + phi2[3]))] -2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[3] + phi2[4])) + E^(1.*10^6 (-phi2[3] + phi2[4]))] + uu1[2] - 2 uu1[3] + uu1[4])^2 + 225 (2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[3] + phi2[4])) + E^(1.*10^6 (-phi2[3] + phi2[4]))] - 2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[4] + phi2[5])) + E^(1.*10^6 (-phi2[4] + phi2[5]))] + uu1[3] - 2 uu1[4] + uu1[5])^2 + 225 (-2.14286*10^-8 Log[E^(-1.*10^6 phi2[6]) + E^(1.*10^6 phi2[6])]+ 2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[5] + phi2[6])) + E^(1.*10^6 (-phi2[5] + phi2[6]))] + uu1[5] - 2 uu1[6])^2 + 225 (2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[4] + phi2[5])) + E^(1.*10^6 (-phi2[4] + phi2[5]))] - 2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[5] + phi2[6])) + E^(1.*10^6 (-phi2[5] + phi2[6]))] + uu1[4] - 2 uu1[5] + uu1[6])^2 + ((15 (2.14286*10^-8 Log[E^(-1.*10^6 phi2[6]) + E^(1.*10^6 phi2[6])] + uu1[6]))^2) + (0.00918367 phi2[2] + (-0.00175179 - 11/112 (0.007848)) (1 - (1.*10^12 (phi2[2])^2)/(Log[E^(-1.*10^6 phi2[2]) + E^(1.*10^6 phi2[2])])^2) + ...

Where uu1[i], uu3[i] and phi2[i] are variables. The issue is that the number of variables can increase to a large number (for example 5000 or even more) which make the function hugely big!! So if I cannot use all capacity of the CPU it takes maybe days to minimize such a function, even though one computer with full capacity of the CPU is not enough to solve such a problem too, but the first step is to learn how to configure parallelization for NMinimize (or FindRoot) on a single machine to be able to extend it to parallelization on several remote machines.

Edit 4:

An example of the complete form of the function with 75 variables is:

Pastebin link

Where variables (unknown parameters) are:

{uu1[2], uu3[2], phi2[2], uu1[3], uu3[3], phi2[3], uu1[4], uu3[4], phi2[4], uu1[5], uu3[5], phi2[5], uu1[6], uu3[6], phi2[6], uu1[7], uu3[7], phi2[7], uu1[8], uu3[8], phi2[8], uu1[9], uu3[9], phi2[9], uu1[10], uu3[10], phi2[10], uu1[11], uu3[11], phi2[11], uu1[12], uu3[12], phi2[12], uu1[13], uu3[13], phi2[13], uu1[14], uu3[14], phi2[14], uu1[15], uu3[15], phi2[15], uu1[16], uu3[16], phi2[16], uu1[17], uu3[17], phi2[17], uu1[18], uu3[18], phi2[18], uu1[19], uu3[19], phi2[19], uu1[20], uu3[20], phi2[20], uu1[21], uu3[21], phi2[21], uu1[22], uu3[22], phi2[22], uu1[23], uu3[23], phi2[23], uu1[24], uu3[24], phi2[24], P1, F1, M1, PN, FN, MN}

I know this function is extremely instable, but the optimum point of the function is also known which is equal to zero, so I am trying to find the values of the parameters which make the function minimum (zero). The parameters which can make the whole function as small as possible are the best answers.

**

Edit 5

**

Thank you all guys for your helpful comments. Based on what KAI and Oleksandr R. mentioned, and as far as I could understand, LinearSolve uses all capacity of the CPU cores to solve the equation involving large matrices. Consequently, it seems that if I want to solve some linear matrix equations for a few times, the best method is to solve each of them one by one in a LOOP to make it most efficient. In this way Mathemathica is able to use all capacity of CPU cores in each step and solve the equation in the most efficient way (in each step) and goes to the next step. But if you have a look at these 2 examples, apparently it is not like this and Parallelization forces Mathematica to solve the problem involving LinearSolve in a way that is likely more efficient and faster. You can check these examples on your computer. Based on the comments we had here, I am wondering how we can explain these examples.

Example 1:

Clear["Global`*"];
t = AbsoluteTime[];
NN = 8;
CC = Array[cc, NN];
For[i = 1, i < (NN + 1), i++,
  Clear[a, b];
  a = RandomReal[{i, i + 1}, {6000, 6000}];
  b = RandomReal[{i}, {6000}];
  CC[[i]] = LinearSolve[a, b];
  ];
time2 = AbsoluteTime[] - t

For example 1 CPU usage is 50% and time2=23.4

Example 2:

Clear["Global`*"];
t = AbsoluteTime[];
job1 = ParallelSubmit[a1 = RandomReal[{1, 2}, {6000, 6000}]; b1 = RandomReal[{1}, {6000}]; c1 = LinearSolve[a1, b1]];
job2 = ParallelSubmit[a2 = RandomReal[{2, 3}, {6000, 6000}]; b2 = RandomReal[{2}, {6000}]; c2 = LinearSolve[a2, b2]];
job3 = ParallelSubmit[a3 = RandomReal[{3, 4}, {6000, 6000}]; b3 = RandomReal[{3}, {6000}]; c3 = LinearSolve[a3, b3]];
job4 = ParallelSubmit[a4 = RandomReal[{4, 5}, {6000, 6000}]; b4 = RandomReal[{4}, {6000}]; c4 = LinearSolve[a4, b4]];
job5 = ParallelSubmit[a5 = RandomReal[{5, 6}, {6000, 6000}]; b5 = RandomReal[{5}, {6000}]; c5 = LinearSolve[a5, b5]];
job6 = ParallelSubmit[a6 = RandomReal[{6, 7}, {6000, 6000}]; b6 = RandomReal[{6}, {6000}]; c6 = LinearSolve[a6, b6]];
job7 = ParallelSubmit[a7 = RandomReal[{7, 8}, {6000, 6000}]; b7 = RandomReal[{7}, {6000}]; c7 = LinearSolve[a7, b7]];
job8 = ParallelSubmit[a8 = RandomReal[{8, 9}, {6000, 6000}]; b8 = RandomReal[{8}, {6000}]; c8 = LinearSolve[a8, b8]];
{R1, R2, R3, R4, R5, R6, R7, R8} = WaitAll[{job1, job2, job3, job4, job5, job6, job7, job8}];
time2 = AbsoluteTime[] - t

For example 2 CPU usage=100% and time2=19.8

share|improve this question
2  
4 cores rather than 8 might be a license limitation. –  b.gatessucks Sep 23 '12 at 7:29
2  
@makmaak, do you have 8 cores or 4 plus hyperthreading? If that is the case the MKL will not only use the true number of cores for performance reasons. –  user21 Sep 23 '12 at 14:26
1  
Can you divide the area over which you want to minimize in 8 parts? If so, you could try ParallelTable to start parallel evaluations in all those areas. Pick the lowest result from the table returned. –  Sjoerd C. de Vries Sep 23 '12 at 15:05
1  
Also, what minimization method are you using? Not all methods are particularly amenable to parallelization, even in principle. In those cases you are basically stuck with domain decomposition as suggested by Sjoerd. –  Oleksandr R. Sep 23 '12 at 18:05
1  
... in that case, one can load up the CPU to 100% of its front end/scheduler capacity and see a performance gain by doing so. To understand this you need to know how and why SMT works, which is quite a technical topic. Jon Stokes has given a very good introduction that I would suggest as a starting point. Focusing on your actual problem, i.e. NMinimize, basically the current implementation is serial only, but the algorithms it uses are parallelizable in principle to some extent, which is why I asked which of the algorithms ... –  Oleksandr R. Sep 24 '12 at 7:17

2 Answers 2

It appears that your chip only has 4 cores according to Intel. It can create 4 additional virtual cores with hyperthreading, but virtual cores and physical cores are not always equivalent. The statement that you have 8 cores is only true in some circumstances.

As your question indicates, LinearSolve uses the Intel MKL library. Intel indicates that hyperthreading will generally not be useful in this case. See the hyperthreading section at this website. This section is especially relevant:

Note: If the requested number of threads exceeds the number of physical cores (perhaps because of hyper-threading), and MKL_DYNAMIC is not changed from its default value of TRUE, Intel MKL will scale down the number of threads to the number of physical cores.

Therefore, Mathematica appears to be solving the problem in the most efficient way, and doesn't seem to want to be forced to solve the problem in a way that will likely be less efficient.

share|improve this answer
    
Thank you for you answer. Could you please have a look at Edit 5 in my question? It is not a contradiction with what you mentioned?? –  mak maak Sep 27 '12 at 4:17
    
I don't get a performance improvement in Example 2 of Edit 5 when I go from 4 cores to 4 physical plus 4 virtual cores. My computer uses more resources (100% vs 50% of the CPU), but the Timing of the problem is basically the same. You might want to test to make sure that you are actually getting the improvement in timing because of the additional cores and not just from using a slightly different method. It looks like the latter might be the reason for the increased speed, but I'm not sure. –  KAI Sep 27 '12 at 18:20

This doesn't address the parallelization, but hopefully gives some indication of how even simple manipulation of the objective function might speed things up.

Here is (most of) the function from your "Edit 3", trimmed to remove unbalanced brackets:

f=(15 (-2.14286*10^-8 Log[E^(-1.*10^6 phi2[2])+E^(1.*10^6 phi2[2])]+uu1[2]))^2+225 (2.14286*10^-8 Log[E^(-1.*10^6 phi2[2])+E^(1.*10^6 phi2[2])]-2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[2]+phi2[3]))+E^(1.*10^6 (-phi2[2]+phi2[3]))]-2 uu1[2]+uu1[3])^2+225 (2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[2]+phi2[3]))+E^(1.*10^6 (-phi2[2]+phi2[3]))]-2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[3]+phi2[4]))+E^(1.*10^6 (-phi2[3]+phi2[4]))]+uu1[2]-2 uu1[3]+uu1[4])^2+225 (2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[3]+phi2[4]))+E^(1.*10^6 (-phi2[3]+phi2[4]))]-2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[4]+phi2[5]))+E^(1.*10^6 (-phi2[4]+phi2[5]))]+uu1[3]-2 uu1[4]+uu1[5])^2+225 (-2.14286*10^-8 Log[E^(-1.*10^6 phi2[6])+E^(1.*10^6 phi2[6])]+2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[5]+phi2[6]))+E^(1.*10^6 (-phi2[5]+phi2[6]))]+uu1[5]-2 uu1[6])^2+225 (2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[4]+phi2[5]))+E^(1.*10^6 (-phi2[4]+phi2[5]))]-2.14286*10^-8 Log[E^(-1.*10^6 (-phi2[5]+phi2[6]))+E^(1.*10^6 (-phi2[5]+phi2[6]))]+uu1[4]-2 uu1[5]+uu1[6])^2+((15 (2.14286*10^-8 Log[E^(-1.*10^6 phi2[6])+E^(1.*10^6 phi2[6])]+uu1[6]))^2);

Extract the variables and test the timing of NMinimize:

vars = Cases[f, _[_Integer], -1] // Union;
NMinimize[f, vars] // Timing

(*  {5.039, {6.38213*10^-14, {phi2[2] -> -1.26493*10^-10, phi2[3] -> -1.30904*10^-10, 
    phi2[4] -> 5.5286*10^-11, phi2[5] -> 1.87167*10^-11, phi2[6] -> 6.97739*10^-12, 
    uu1[2] -> 5.3047*10^-9, uu1[3] -> 4.24388*10^-9, uu1[4] -> -1.37845*10^-13, 
    uu1[5] -> -4.24391*10^-9, uu1[6] -> -5.30474*10^-9}}}  *)

Okay, that took 5 seconds. Now try a simple rescaling of the variables:

g = ExpandAll[f /. {p_phi2 :> 10.^-6 p, u_uu1 :> 10.^-5 u}];
NMinimize[g, vars] // Timing

(*  {0.109, {6.39186*10^-14, {phi2[2] -> 0.0438337, phi2[3] -> 0.0778526, 
    phi2[4] -> 0.0688122, phi2[5] -> 0.0663468, phi2[6] -> 0.0552145, 
    uu1[2] -> 0.000531695, uu1[3] -> 0.000425536, uu1[4] -> -2.53421*10^-7, 
    uu1[5] -> -0.000426044, uu1[6] -> -0.000533116}}}  *)

That only took a tenth of a second. There is a factor of 50 speed increase just by rescaling the variables. I don't know how that might scale with a 75 or 5000 dimensional problem, but it does suggest that you would be better off improving the objective function, as Oleksandr said, than throwing more CPU cores at it.

share|improve this answer
    
Thank you for your reply. I agree with what you mentioned, this function is extremely instable and needs serious attention, but I am wondering how to do this. You actually changed the function completely in this example, while I need the original function to be minimized. Moreover, even after rescaling for this function, I think if we can use more CPU cores (when the number of variables is very large) it is better than using one core. If you like to check the complete form of the function for 75 dimensions, please have a look at Edit 4. –  mak maak Sep 26 '12 at 5:17
    
@makmaak, I'm afraid I don't know much about this sort of thing. I had a look at your full function in Edit 4, but couldn't find a good approach to improve the stability. Good luck with it... –  Simon Woods Sep 30 '12 at 14:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.