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Old   March 19, 2002, 05:35
Default I faced difficulty in a fortran error
  #1
J. Kim
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Dear all,

I faced an error in fortran compiler: Divide by zero I found out the casue of that. However, I did not know a way to solve it. Here is a error.

---------------- KmT=475.0e-6 KmD=155.0e-3 CDT=2.36e-10 CDD=2.57e-10

C RENEW ATP CONCENTRATION AT BOTTOM WALL

DO 510 I=1,L1

F_ATP=F(I,1,5)*VmaxT/(KmT*CDT)-S/(CDT)

F(I,1,5)=F(I,2,5)+F_ATP*YDIF(2) 510 CONTINUE C RENEW ADP CONCENTRATION AT BOTTOM WALL

DO 520 I=1,L1

F_ADP=F(I,1,6)*VmaxD/(KmD*CDD)-F(I,1,5)*VmaxT/(KmT*CDD)

F(I,1,6)=F(I,2,6)+F_ADP*YDIF(2) 520 CONTINUE

---------------- As you can see, the values of KmT*CDT, CDT, KmD*CDD, and KmT*CDD is very small like the unit.

How can I change them? I would be appreciated for this if you would help me.
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Old   March 19, 2002, 05:59
Default Re: I faced difficulty in a fortran error
  #2
STN
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Hi Kim....

Could you tell which, exactly, line is error ?

Is it this line " F_ATP=F(I,1,5)*VmaxT/(KmT*CDT)-S/(CDT)" or "F_ADP=F(I,1,6)*VmaxD/(KmD*CDD)-F(I,1,5)*VmaxT/(KmT*CDD)"?

If yes, It may because KmT or KmD is not define as REAL or DOUBLEPRECISION parameter. Normaly, FORTRAN define all parameter which has I,J,K,L,M,N in the first letter is INTEGER.So your KmT or KmD is ZERO.

Hope this help.. STN
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Old   March 19, 2002, 08:36
Default Re: I faced difficulty in a fortran error
  #3
Seb Perron
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1) First to make sure that all you variables are declared with the appropriate type, add

implicit none

at the first line of your routine.

2) Compute both

dtmp0=KmT*CDT

and

dtmp1=KmT*CDD

and before making the division, do a simple check on their value.

Good luck.
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Old   March 19, 2002, 09:21
Default Re: I faced difficulty in a fortran error
  #4
J. Kim
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Thanks really for your warm answer.

Like you gave me comment, I already did so. There is still error: Divied by zero.... Are there any solution? I will post the problem again. --------------------- ..... KmT=475.0e-6 KmD=155.0e-3 CDT=2.36e-10 CDD=2.57e-10 ..... DO 510 I=1,L1

F_ATP=F(I,1,5)*VmaxT/(KmT*CDT)-S/(CDT)

F(I,1,5)=F(I,2,5)+F_ATP*YDIF(2) 510 CONTINUE

DO 520 I=1,L1

F_ADP=F(I,1,6)*VmaxD/(KmD*CDD)-F(I,1,5)*VmaxT/KmT*CDD)

F(I,1,6)=F(I,2,6)+F_ADP*YDIF(2) 520 CONTINUE

----------------

----------------------
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Old   March 19, 2002, 10:20
Default Re: I faced difficulty in a fortran error
  #5
Wen Long
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Hi, Kim:

Maybe your KmT, KmD and CDT are too small, and divition by a small number is apt to make errors. Can you normalize it first?

Say, KmT_1=Kmt/1.0e-6, KmD=KmD/1.0e-3 ...

Then you devide by (KmT_1*KmD) and times the result with 1.0e+9,

hope this helps,

wen
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Old   March 19, 2002, 12:19
Default Re: I faced difficulty in a fortran error
  #6
Francisco
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Dear J.Kim,

Ive pasted your piece of code into my Fortran compiler (F90 for Windows) and with slight modifications (which I hope did not change the divison operation), it seemed to work (no divison by zero). Here is the small code I used:

program kim

implicit none

real*8 KmT, KmD,CDT,CDD,F_ATP,F_ADP,S,VmaxD,VmaxT

KmT=475.0e-6

KmD=155.0e-3

CDT=2.36e-10

CDD=2.57e-10

F_ATP=VmaxT/(KmT*CDT)-S/(CDT)

F_ADP=VmaxD/(KmD*CDD)-F_ATP*VmaxT/(KmT*CDD)

write(*,*) F_ATP,F_ADP,CDT

pause

end

Obviously, I might have removed the source of error, but try it yourself.

Best regards,

Francisco.

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Old   March 19, 2002, 17:36
Default Re: Thanks for your good comment.
  #7
J. Kim
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