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Program caused arithmetic error: Floating underflow [message #88771]

Wed, 18 June 2014 02:37

g.nacarts
Messages: 148
Registered: November 2013

Senior Member

Hi

I have the following expression:

value = 0

for i=0, 283 do begin
for j=0, 283 do begin
value = (value + (A[i,j]-B[i,j])^2.)/17000.
endfor
endfor

when I type, print value I got the following error:
% Program caused arithmetic error: Floating underflow

Does anyone know why this happens?

Regards,
Gina

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Re: Program caused arithmetic error: Floating underflow [message #88772 is a reply to message #88771]

Wed, 18 June 2014 02:52

Fabzi
Messages: 305
Registered: July 2010

Senior Member

Hi,

On 18.06.2014 11:37, g.nacarts@gmail.com wrote:
> % Program caused arithmetic error: Floating underflow

typing the exact same error message in search engines gives very helpful
and pertinent links, the first result in g**g** being:

http://www.idlcoyote.com/math_tips/underflow.html

Cheers,

Fabien

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Re: Program caused arithmetic error: Floating underflow [message #88773 is a reply to message #88771]

Wed, 18 June 2014 03:10

g.nacarts
Messages: 148
Registered: November 2013

Senior Member

Thanks for your response. Actually I already had a look on this link but I couldn't manage to solve my problem. I used the L to make it LONG but I didn't see any number in the outputs.

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Re: Program caused arithmetic error: Floating underflow [message #88775 is a reply to message #88773]

Wed, 18 June 2014 04:37

Fabzi
Messages: 305
Registered: July 2010

Senior Member

On 18.06.2014 12:10, g.nacarts@gmail.com wrote:
> I couldn't manage to solve my problem.

Well it's mostly not a problem and could be ignored. It depends on your
numbers A and B:

IDL> print, 1e-30 / 17000
5.88235e-35
IDL> print, 1e-50 / 17000
0.00000
% Program caused arithmetic error: Floating underflow

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Re: Program caused arithmetic error: Floating underflow [message #88776 is a reply to message #88771]

Wed, 18 June 2014 04:40

Matthew Argall
Messages: 286
Registered: October 2011

Senior Member

Note that in the last iteration, your result will essentially be divided by ((17000^283)^283).

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Re: Program caused arithmetic error: Floating underflow [message #88777 is a reply to message #88771]

Wed, 18 June 2014 05:11

g.nacarts
Messages: 148
Registered: November 2013

Senior Member

Well, I can't ignore it because I need the values of the parameter "value" as shown in the expression. But since this cause an overflow I didn't get any result. There is any way to overcome this issue?

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Re: Program caused arithmetic error: Floating underflow [message #88778 is a reply to message #88777]

Wed, 18 June 2014 05:28

Fabzi
Messages: 305
Registered: July 2010

Senior Member

I don't think that normal computers are made to represent the number
((17000^283)^283) as Matthew pointed out. Maybe you need to explain why
you want to divide your value by the size of the universe ... twice.

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Re: Program caused arithmetic error: Floating underflow [message #88779 is a reply to message #88777]

Wed, 18 June 2014 05:59

Fabzi
Messages: 305
Registered: July 2010

Senior Member

... if you sure that the formulation of your loop is correct you could
try using double precision numbers. This will just allow value to become
even closer to 0 than it was already:

IDL> print, 1e-50 / 17000
0.00000
% Program caused arithmetic error: Floating underflow
IDL> print, 1d-50 / 17000d
5.8823529e-55

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Re: Program caused arithmetic error: Floating underflow [message #88780 is a reply to message #88771]

Wed, 18 June 2014 08:38

g.nacarts
Messages: 148
Registered: November 2013

Senior Member

Why the last iteration will be divided by (17000^283)^283? I thought that for each iteration the value be divided by 17000.

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Re: Program caused arithmetic error: Floating underflow [message #88781 is a reply to message #88780]

Wed, 18 June 2014 09:05

Mats Löfdahl
Messages: 263
Registered: January 2012

Senior Member

Den onsdagen den 18:e juni 2014 kl. 17:38:45 UTC+2 skrev g.na...@gmail.com:
> Why the last iteration will be divided by (17000^283)^283? I thought that for each iteration the value be divided by 17000.

The value from the first iteration will be divided by that number. The contribution from each iteration will be divided by one factor 17000 less than the previous one. The value from the last iteration will be divided by 17000 only.

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Re: Program caused arithmetic error: Floating underflow [message #88782 is a reply to message #88780]

Wed, 18 June 2014 09:10

Chip Helms
Messages: 24
Registered: November 2012

Junior Member

On Wednesday, June 18, 2014 3:38:45 PM UTC, g.na...@gmail.com wrote:
> Why the last iteration will be divided by (17000^283)^283? I thought that for each iteration the value be divided by 17000.

If you still find yourself having issues locating the problem, you might double check that the issue isn't occurring somewhere else nearby in the code (I believe the default behavior is for IDL to issue math error warnings only when it returns to interactive prompt). If you set !EXCEPT=2, IDL will issue warnings (complete with line numbers) at the time the error occurs.

http://www.exelisvis.com/docs/error_handling_system_va.html# sysvars_272074529_1002623

Also, currently, your code is dividing the current running total by 17000 at each step (so the division occurs 284*284 or 80656 times, granted it's not the same as finding the total and dividing by 17000d^80656). I'm guessing this is to converge on the parameter value iteratively?

Cheers,
Chip

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