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-   -   Arithmetic overflow in Fortran 95 (http://www.physicsforums.com/showthread.php?t=582285)

edge333 Feb28-12 03:29 PM

Arithmetic overflow in Fortran 95
 
1. The problem statement, all variables and given/known data

Wrote a code using Fortran 95 to solve for an advection-dispersion equation but at the spatial steps specified at dx = 20 m over a total length, L of 20000 m, I keep getting an arithmetic overflow error.

I have run this same program at smaller spatial intervals (dx = 200 m and dx = 2000 m) and they worked perfectly. There must be something about the total number of iterations that Fortran is trying to perform that makes it incapable of computing it (I'm not actually sure).

I'm a novice at using Fortran but curious if this has anything to do with the size of the concentration values that are being computed being too small (i.e. smaller than 10^-38). Does this have anything to do with double precision?

2. Relevant equations

The scheme used to solve the numerical solution to the ADE is an finite differenece upwind scheme.

The first equation in the code under the two "do" statements defines this upwind scheme within the internal bounds of the system.

The second equation in the code outside of the two "do" statements defines the boundary condition at X = L.

3. The attempt at a solution

module global ! Module is used to define arrays and use in main program blocks

implicit none
real len, dx, k, Dl
real, allocatable, dimension (:, :) :: conc ! I used allocatable arrays so that the size of
! the array can be changed, not necessary for this
! simple of a problem
real, allocatable, dimension (:) :: tim
integer ndx, i, n

end module global

program main
use global

real dt , tottime, u

This is where you write out what you're going to be doing. You can just link to where a
! source file is located, but make sure you do it.

dt = 30. ! unit is seconds
len = 20020. ! unit is meters
tottime = 44000. ! unit is seconds (43200 seconds in half a day)
ndx = 1001 ! number of grid points - we want 10 cells so we start at 1.
u = 0.25 ! unit is m/s
Dl = 50. ! unit is m^2/sec
k = 0.1 / 86400. ! unit is sec^-1
nts = (tottime) / dt ! number of time steps
dx = len / real(ndx) ! grid spacing

! Allocate the arrays, conc(entration), and tim(e).
allocate (conc ( ndx , nts ), tim ( nts ))

conc = 0.0 ! set initial concentration elements to 0 mg/L
tim = 0.0 ! initial time is zero
conc (1, :) = 100. ! set BC on LHS to 100 mg/l for all time

! Run the internal finite difference equation and populate the array conc ()

do n = 1, nts - 1
tim (n) = dt * (n - 1)
do i = 2, ndx - 1
conc (i, n + 1) = conc (i, n) - u * (dt / dx) * (conc (i, n) - conc (i - 1, n)) &

+ Dl * (dt / (dx**2)) * (conc (i + 1, n) - 2 * conc (i, n) + conc (i - 1, n)) - k * dt * conc (i, n)
end do
conc (ndx, n + 1) = conc (ndx, n) - u * (dt / dx) * (conc (ndx, n) - conc (ndx - 1, n)) &
+ Dl * (dt / dx**2) * (2 * conc (ndx - 1, n) - 2 * conc (ndx, n)) - k * dt * conc (ndx, n)
end do

! Set BC on RHS where c (ndx + 1, nts) = c (ndx - 1, nts)
! Write the array to a file

open (10, file = 'C:\g95\fortran\CEE 573\Problem Set 5\upwind20.csv', status = 'unknown')

! Write the column headings
write (10, 100) (tim (n) / 86400., n = 1, nts, 10)
100 format ('X, ',2x, 10000 ('t=', f8.4, ','))

! Write the columns
do i = 1, ndx
x = dx * (i - 1)
write (10, 110) x, (conc (i, n), n = 1, nts, 10)
110 format (f10.1, ',', 10000 (f8.4, ','))
end do

end program main

jedishrfu Feb28-12 03:53 PM

Re: Arithmetic overflow in Fortran 95
 
does the computer say what line is getting the overflow error?

edge333 Feb28-12 04:16 PM

Re: Arithmetic overflow in Fortran 95
 
Yeah, it said the error was on line 47 which is the second line of this equation (internal finite difference equation):

conc (i, n + 1) = conc (i, n) - u * (dt / dx) * (conc (i, n) - conc (i - 1, n)) &
+ Dl * (dt / (dx**2)) * (conc (i + 1, n) - 2 * conc (i, n) + conc (i - 1, n)) - k * dt * conc (i, n)

Mark44 Feb28-12 04:39 PM

Re: Arithmetic overflow in Fortran 95
 
To debug this, break up the expression on the right side, so that you do the computation in at least three assignment statements. That should help you determine which term is causing problems.

edge333 Feb28-12 04:52 PM

Re: Arithmetic overflow in Fortran 95
 
Okay, I did that. And so I put the k term on line 48 and the error occurs on line 48. So it must be something with the k term. Why would it not do this if I set the spatial steps higher at dx = 200 m and dx = 2000 m instead of dx = 20 m?

jedishrfu Feb29-12 09:17 AM

Re: Arithmetic overflow in Fortran 95
 
so now you print out the value of k, dt,i,n and conc(i,n) separately
and see which var is causing the issue.

LawrenceC Feb29-12 10:24 AM

Re: Arithmetic overflow in Fortran 95
 
If you have downstream boundary conditions instabilites may exist unless the Peclet number is kept small by making the spatial discritization small. The grid spacing must be sufficiently small so the Laplacian dominates the equation. This can be accomplished by grid refinement. For additional information google 'grid peclet number'.


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