Wilmott Forums - Mathematica vs Matlab vs Maple, discuss

Forum Navigation:

new topic

FORUMS > The Quantitative Finance FAQs Project

< refresh >

Topic Title: Mathematica vs Matlab vs Maple, discuss
Created On Wed Aug 30, 06 06:38 PM

Topic View:


	Pages: [ << 1 2 3 >> Previous Next ]

ZmeiGorynych
Senior Member

Posts: 874
Joined: Jul 2005

Fri Jul 06, 07 08:37 AM

Oh, and BTW MATLAB's financial toolbox is quite overpriced for what it is. If you consider MATLAB, just get base, optimization and statistics toolboxes (plus any interfaces to excel, datafeeds etc if your interests lie that way).

-------------------------
You give a hungry man a fish, he owes you one fish. You teach him to fish, you lose your monopoly on fishing.

Quote

Top

Bottom

guillaume07
Member

Posts: 37
Joined: Feb 2007

Tue Jul 10, 07 10:03 AM

Scilab is a great tool too and is it free....

Quote

Top

Bottom

Alan
Senior Member

Posts: 3290
Joined: Dec 2001

Tue Jul 10, 07 03:20 PM

Quote
Originally posted by: ZmeiGorynych

Another issue is that a MATLAB license is perpetual (you only get free updates for a year, but you may keep using that version forever), whereas Mathematica, last I heard, had the unspeakable policy of making you pay EVERY YEAR.

That's wrong. You pay for Mathematica once for an individual license which is perpetual;
you pay again (at the upgrade rate) only if you want to upgrade to a new version.

Quote

Top

Bottom

stonexu1984
Member

Posts: 76
Joined: Mar 2007

Sat Jul 14, 07 04:58 AM

garch toolbox is also worth to get. For univariate, I prefer mathwork's garch to UCSD's

Quote

Top

Bottom

MCarreira
Senior Member

Posts: 1256
Joined: Jul 2002

Wed Jul 18, 07 01:23 AM

Quote
Originally posted by: Alan
Quote
Originally posted by: ZmeiGorynych

Another issue is that a MATLAB license is perpetual (you only get free updates for a year, but you may keep using that version forever), whereas Mathematica, last I heard, had the unspeakable policy of making you pay EVERY YEAR.

That's wrong. You pay for Mathematica once for an individual license which is perpetual;
you pay again (at the upgrade rate) only if you want to upgrade to a new version.

If you buy the Premium License (use in 2 computers plus free upgrades plus Premium support and stuff) then you pay every year.

-------------------------
"The best lack all conviction, while the worst
Are full of passionate intensity." - W.B. Yeats

Quote

Top

Bottom

dibble
Senior Member

Posts: 727
Joined: Oct 2006

Wed Jul 18, 07 01:27 AM

-- If you buy the Premium License (use in 2 computers plus free upgrades plus Premium support and stuff) then you pay every year.

But not the full price. You pay a maintenance price.

-- whereas Mathematica, last I heard, had the unspeakable policy of making you pay EVERY YEAR.

The same rules apply for Matlab and Mathematica

-------------------------
"I am not saying that Bush, Cheney and Rumsfeld should be taken outside and shot. What I am saying is that they should be tried on the charge of High treason and then taken outside and shot"

"Religion of course is primarily the mixture of faith and fear. The latter a base human condition, the former a learned human reaction and a mere veil for the latter."

Edited: Wed Jul 18, 07 at 01:29 AM by dibble

Quote

Top

Bottom

jaegerstar2000
Junior Member

Posts: 2
Joined: Jun 2007

Sun Aug 12, 07 11:31 AM

Mathematica has a time series package which has a huge bug in it for determining (G)ARCH.
I want my money back and the sales department is quiet. Technical support are looking
into it. But, no end in sight.......

The bug is that the package does not respect the non-negativity on the coefficients. Which means
sometimes you will receive a complex solution!!!
(see other posts by me on the internet for a more detailed explanation to this problem).

Mathematica version 6.0
Timeseries Package Version: 1.0

Until they fix it, do not purchase the package as you will not get your money back.

Instead, Engle recommends Eviews. I recommend either: MATLAB, Ox, Eviews, SAS, Gauss and maybe
R-Project: tSeries for fitting GARCH.

Ian Gregory.

Quote

Top

Bottom

ZmeiGorynych
Senior Member

Posts: 874
Joined: Jul 2005

Sun Aug 19, 07 10:41 PM

Quote
Originally posted by: guillaume07
Scilab is a great tool too and is it free....

We played with it a year or so ago and found it was incredibly slow on datasets of any size at all (considering it's MATLAB we're benchmarking against, that's quite an achievement

) as well as not supporting the full set of matlab commands, nor the same quality of GUI.

Why bother with clones - I'd either get the real thing, or use another language such as python.

-------------------------
You give a hungry man a fish, he owes you one fish. You teach him to fish, you lose your monopoly on fishing.

Quote

Top

Bottom

Hamilton
Senior Member

Posts: 5551
Joined: Jul 2001

Sun Aug 26, 07 03:19 AM

I would like to know if any actuaries are using Mathematica and, if so, to what extent does it replace APL in their applications.

-------------------------
Despotism has so often been established in the name of liberty that experience should warn us to judge parties by their practices rather than their preachings.

�Raymond Aron, The Opium of the Intellectuals

Quote

Top

Bottom

Alan
Senior Member

Posts: 3290
Joined: Dec 2001

Sun Aug 26, 07 04:38 PM

Can anyone match or beat either of these running times?
If so what's your setup? (Matlab/Mathematica Version #, CPU, Brand, Model, Op. Sys., Memory)

Matlab
---------------------------
M = rand(300,300);
t=0.25; T=2;
tic; inv(M) * (expm(-t*M) - expm(-T*M)); toc;
Elapsed time is 0.288171 seconds

Mathematica
---------------------------
M = Table[Random[], {i, 1, 300}, {j, 1, 300}];
t = 0.25; T= 2;
Timing[Inverse[M].(MatrixExp[-M t] - MatrixExp[-M T]);]
{0.375 Second, Null}

Edited: Sun Aug 26, 07 at 10:57 PM by Alan

Quote

Top

Bottom

AVt
Senior Member

Posts: 750
Joined: Dec 2001

Sun Aug 26, 07 07:31 PM

with a bit guessing ( random = uniform in [0,1] and timing = execution time [not CPU time] ):

on my medium sized PC Maple needs about 10 times more, 90% seems to be spent on
computing the exponentials (working with hardware precision, no symbolics intended)

edited: forgot to say it needs ~ 2 seconds

Edited: Sun Aug 26, 07 at 07:35 PM by AVt

Quote

Top

Bottom

Alan
Senior Member

Posts: 3290
Joined: Dec 2001

Sun Aug 26, 07 11:05 PM

Yes, you guessed right on the random numbers.
The timings are simply the standard timings for each software system respectively.
According to Mathematica: Timing[ ] "includes only CPU time spent in the Mathematica kernel"
I don't have Matlab, so not sure there.

My newest desktop is a couple years old, and I get 2.5 secs for the Mathematica problem.
But user:tonyd reports these times I have posted on a 2.66GHz Intel PC, but we don't know the rest of his setup.
My deskotp is a 2.99GHZ Intel PC running Mathematica Ver. 5.0 (Ver. 6.0.1 is current as of this post)
I would like to learn what it takes to improve my setup to get tonyd's timings -- simply V6 or new hardware or ? --
just made a similar query to the MM newsgroup.

Edited: Mon Aug 27, 07 at 12:44 AM by Alan

Quote

Top

Bottom

AVt
Senior Member

Posts: 750
Joined: Dec 2001

Mon Aug 27, 07 09:08 PM

My timing was for concurrent Maple 11 on Win XP using an Athlon 2.2 GHz & 1 GB memory (so 64 Bit is not supported).

So in MMA does not give the time for symbolics outside the kernel - same as Maple. But it should be small, it is almost
pure floating point computation.

Almost: looking into the help and the code there should be some symbolics involved, it uses a general MatrixFunction(A, F)
where F is analytic (so works for F=sin as well as for exp).

Quote

The MatrixFunction(A) command returns the Matrix obtained by interpolating [lambda, F( lambda )] for each of the
eigenvalues lambda of A, including multiplicities.

Here the Matrix polynomial is r(lambda) = F(lambda) - p(lambda)*q(lambda) where p(x) is the characteristic polynomial,
q(lambda) is the quotient, and r(lambda) is the remainder.

The latter only makes sense for me by switching to rationals for the characteristic polynomial.

BTW: the determinant of such a beast is quite large and like 10^146, I have doubts in the reliability of the result,
but forgot the command for numerical stability concerning matrices (i.e. I think there is some).

---

u:=RandomTools:-Generate(float(range=0.0 .. 1.0, digits=14, method=uniform), makeproc=true);
M:=evalf(RandomMatrix(d, generator=u));

st:=time():
MatrixInverse(M).(MatrixExponential(-t*M) - MatrixExponential(-T*M));
time()-st;

results in 2 seconds.

Quote

Top

Bottom

spacemonkey
Senior Member

Posts: 306
Joined: Aug 2002

Mon Sep 03, 07 10:52 PM

Tried this on the free Matlab clone octave:-

octave:26> M = rand(300,300);
octave:27> t = 0.25; T = 2;
octave:28> tic; inv(M) * (expm(-t*M) - expm(-T*M)); toc
ans = 3.2403 (seconds)

Not bad when you consider my computer - 1.7Ghz P4, 256Mb.

I also tried Maxima, but I couldn't get it to calculate the answer at all. Maxima has a lot of potential, but it seems pretty difficult to use. Still it beats the hell out of Mathematica on price.

Posts: 63
Joined: Jun 2007

Thu Sep 06, 07 10:43 AM

Quote
Originally posted by: Alan
Can anyone match or beat either of these running times?
If so what's your setup? (Matlab/Mathematica Version #, CPU, Brand, Model, Op. Sys., Memory)

Matlab
---------------------------
M = rand(300,300);
t=0.25; T=2;
tic; inv(M) * (expm(-t*M) - expm(-T*M)); toc;
Elapsed time is 0.288171 seconds

Mathematica
---------------------------
M = Table[Random[], {i, 1, 300}, {j, 1, 300}];
t = 0.25; T= 2;
Timing[Inverse[M].(MatrixExp[-M t] - MatrixExp[-M T]);]
{0.375 Second, Null}

Hi Alan,

My setup : Windows XP, AMD Opteron P250, 2.39GHz, 4Go RAM
Mathematica v5.0 -> {1.852 Second, Null}
Mathematica v6.0 -> {0.422 Second, Null}

When I ran the Mathematica benchmarks (Help -> About Mathematica -> System Information -> Kernel)
it 'rated me' 1.37, worse being 0.21 and best being 2.84.

My conclusion : simply v6 [ + intel oriented hardware better than AMD (for mathematica at least) ].

Personal comment : I also tried to do my own MatrixExp. I did compute the eigensystem and did
the exponentiation with exp(M) = P.exp(D).P^(-1). This should be less efficient than native MatrixExp
because there are faster algorithm to do matrix exponentiation, as quoted in their help.
Anyway, with v5.0, home made exponentiation is faster, which is NOT normal.
With v6.0, it's no longer the case.

I'll try to get a matlab licence.

-------------------------
"I've never seen so many men wasted so badly". Blondie - The Good, the Bad, the Ugly.

Posts: 63
Joined: Jun 2007

Thu Sep 06, 07 04:29 PM

Quote
Originally posted by: Alan
Can anyone match or beat either of these running times?
If so what's your setup? (Matlab/Mathematica Version #, CPU, Brand, Model, Op. Sys., Memory)

Matlab
---------------------------
M = rand(300,300);
t=0.25; T=2;
tic; inv(M) * (expm(-t*M) - expm(-T*M)); toc;
Elapsed time is 0.288171 seconds

Mathematica
---------------------------
M = Table[Random[], {i, 1, 300}, {j, 1, 300}];
t = 0.25; T= 2;
Timing[Inverse[M].(MatrixExp[-M t] - MatrixExp[-M T]);]
{0.375 Second, Null}