So I'm getting an invalid float error that's causing my code to crash from the following subroutine. The subroutine sums together my distribution function over angle and energy (so I just have a distribution in radius), then I compute my average optical depth, and then I look for the radial zone where that optical depth crosses 2/3. The code is designed to be run over multiple time steps, and in this particular simulation it had run through 49 time steps and then crashed on the 50th, which is confusing me even more.
subroutine SAPass(state, step)
use state_vector_module
use boltztran_memory_module, only: fe, trmfpe
use boltztran_parameter_module
use units_module
implicit none
real, dimension(102) :: Fse
real, dimension(102,8) :: taue
real, dimension(102) :: Te
inetger :: cycsav = -1
integer :: nsze
call opdep(trmfpe, taue)
Fse = 0.
do i=1,102
do j=1,4
do k=1,8
Fse(i) = Fse(i) + fe(i,j,k)
end do
end do
end do
do i=1,102
do j=1,4
do k=1,8
if (Fse(i).eq.0.) then
Te(i) = 0.
else
Te(i) = Te(i) + (fe(i,j,k)*taue(i,k))/Fse(i)
end if
end do
end do
end do
write(*,*) Te !debug check to look for where the error is happening
nsze = 1
do i=1,101
if(Te(i).eq.(2./3.)) then ! This is the line the invalid float error is reported on.
nsze = i
else if((Te(i).gt.(2./3.)).and.(Te(i+1).lt.(2./3.))) then
nsze = i
end if
end do
if(step.eq.cycsav) then
continue
else
cycsav = istep
call tauwrite(taue,step,1)
open(17,name="NSZones.dat",status="unknown",position="append")
write(17,i5) nsze
close(17)
end if
On time step 50 the code crashes and I get the following error:
forrtl: error (65): floating invalid
And it references the specified line in the code. When I look at the print outs for Te during time step 50 the entry for i=4 is a 'NaN' and I can't figure out what in the code could be causing this to become a NaN. The distribution function numbers all look normal for that time step and zone.
You use
Fse(i).eq.0.
Te(i).eq.(2./3.)
It is a very bad idea to compare two floating point numbers for equality. You are probably dividing by an almost-zero in
Te(i) = Te(i) + (fe(i,j,k)*taue(i,k))/Fse(i)
and that results in an error.
It is impossible to say more, because we can't see what fe is and which values it has nor which values taue has.
Related
I am now running a program for a certain iterations. The time step is 0.01. I want to write some information when a specific time is reached. For example:
program abc
implicit none
double precision :: time,step,target
integer :: x
time = 0.d0
step = 0.01
target = 5.d0
do x = 1,6000
time = time + step
"some equations here to calculate the model parameters"
if(time.eq.target)then
write(*,*) "model parameters"
endif
enddo
However, "time" never equals to 1.0 or 2.0 or etc. It shows like "0.999999866" instead of "1.0" and "1.99999845" instead of "2.0".
Although I can use integer "x" to define when to write the information, I prefer to use the time step. Also, I may want to change the time step (0.01/0.02/0.05/etc) or target (5.0/6.0/8.0/etc).
Does anyone knows how to fix this? Thanks ahead.
You have now discovered floating point arithmetic! Just ensure that the time is sufficiently close to the target.
if(abs(time-target) < 0.5d0*step ) then
...
should do the trick.
Floating point arithmetic is not perfect and your variables are always exact up to a certain machine error, depending on your variables' number format (32, 64, 128 bit). The following example illustrates well this characteristic:
PROGRAM main
USE, INTRINSIC :: ISO_FORTRAN_ENV, qp => real128
IMPLICIT NONE
REAL(qp) :: a, b, c
a = 128._qp
b = a/120._qp + 1
c = 120._qp*(b-1)
PRINT*, "a = ", a
PRINT*, "c = ", c
END PROGRAM main
Here is the output to this program with gfortran v.4.6.3:
a = 128.00000000000000000
c = 127.99999999999999999
I was assigned the following problem:
Make a Fortran program which will be able to read a degree[0-360] checking validity range(not type) and it will be able to calculate and print the cos(x) from the following equation, where x is in radians:
cos(x)=1-x^2/2! + x^4/4!-x^6/6!+x^8/8!-...
As a convergence criteria assume 10^(-5) using the absolute error between two successive repeats (I suppose it means do's).
For the calculation of the ! the greatest possible kind of integer should be used. Finally the total number of repeats should be printed on screen.
So my code is this:
program ex6_pr2
implicit none
!Variables and Constants
integer::i
real*8::fact,fact2 !fact=factorial
real,parameter::pi=3.14159265
double precision::degree,radiants,cosradiants,s,oldcosradiants,difference !degree,radiants=angle
print*,'This program reads and calculates an angle`s co-sinus'
print*,'Please input the degrees of the angle'
read*,degree
do while(degree<0 .or. degree>360) !number range
read*,degree
print*,'Error input degree'
cycle
end do
radiants=(degree*pi/180)
fact=1
fact2=1
s=0
cosradiants=0
!repeat structure
do i=2,200,1
fact=fact*i
fact2=fact2*(i+2)
oldcosradiants=cosradiants
cosradiants=(-(radiants)**i/fact)+(((radiants)**(i+2))/fact2)
difference=cosradiants-oldcosradiants
s=s+cosradiants
if(abs(difference)<1e-5) exit
end do
!Printing results
print*,s+1.
end program
I get right results for angles such as 45 degrees (or pi/4) and wrong for other for example 90 degrees or 180.
I have checked my factorials where I believe the error is hidden (at least for me).
Well I created another code which seems unable to run due to the following error:FUNCTION name,(RESULT of PROJECT2_EX6~FACT),used where not expected,perhaps missing '()'
program project2_ex6
implicit none
integer(kind=3)::degrees,i,sign
integer::n
double precision::x,err_limit,s_old,s
real,parameter::pi=3.14159265359
print*,'This program calculates the cos(x)'
print*,"Enter the angle's degrees"
read*,degrees
do
if(degrees<0.or.degrees>360) then
print*,'Degrees must be between 0-360'
else
x=pi*degrees/180
exit
end if
end do
sign=1
sign=sign*(-1)
err_limit=1e-5
n=0
s=0
s_old=0
do
do i=1,n
end do
s=(((-1.)**n/(fact(2.*n)))*x**(2.*n))*sign
s=s+s_old
n=n+1
if(abs(s-s_old)<1e-5) then
exit
else
s_old=s
cycle
end if
end do
print*,s,i,n
contains
real function fact(i)
double precision::fact
integer::i
if(i>=1) then
fact=i*fact(i-1)
else
fact=1
end if
return
end function
end program
Although it is your homework, I will help you here. The first thing which is wrong is ýour factorial which you need to replace with
fact = 1
do j = 1,i
fact = fact*j
enddo
second it is easier if you let your do loop do the job so run it as
do i=4,200,2
and predefine cosradians outside the do loob with
cosradiants = 1-radiants**2/2
additionally you need to take into account the changing sign which you can do in the loop using
sign = sign*(-1)
and starting it off with sign = 1 before the loop
in the loop its then
cosradiants= cosradiants+sign*radiants**i/fact
If you have included these things it should work (at least with my code it does)
I am currently writing a code to simulate particle collisions. I am trying to open as much files as there are particles (N) and then put the data for positions and velocities in each of these files for each step of the time integration (using Euler's method, but that is not relevant). For that, I tried using a do loop so it will open all the files I need - then I put all the data in them with a different do loop later - and then close them all.
I first tried just doing a do loop to open the files - but it gave errors of the type "file already open in another unit", so I did the following:
module parameters
implicit none
character :: posvel
integer :: i, j, N
real :: tmax
real, parameter :: tmin=0.0, pi=3.14159265, k=500.0*10E3, dt=10.0E-5, dx=10.0E-3, g=9.806, ro=1.5*10E3
real, dimension(:), allocatable :: xold, xnew, vold, vnew, m, F, r
end module parameters
PROGRAM Collision
use parameters
implicit none
write(*,*) 'Enter total number of particles (integer number):'
read(*,*) N
allocate(xold(N))
allocate(vold(N))
allocate(xnew(N))
allocate(vnew(N))
allocate(m(N))
allocate(F(N))
allocate(r(N))
xold(1) = 0.0
vold(1) = 0.0
m(1) = 6.283*10E-9
r(1) = 10E-4
xold(2) = 5.0
vold(2) = 0.0
m(2) = 6.283*10E-9
r(2) = 10E-4
write(*,*) 'Type total time elapsed for the simulation(real number):'
read(*,*) tmax
do i = 1, N
write(posvel,"(a,i3.3,a)") "posveldata",i,".txt"
open(unit=i,file=posvel, status="unknown")
end do
do i = 1, N
close(unit=i)
end do
END PROGRAM Collision
The last ten lines are the ones that regard to my problem.
That worked in codeblocks - it opened just the number of files I needed, but I'm actually using gfortran and it gives me and "end of record" error in the write statement.
How can I make it to execute properly and give me the N different files that I need?
P.S.: It is not a problem of compilation, but after I execute the program.
Your character string in the parameter module has only 1 character length, so it cannot contain the full file name. So please use a longer string, for example
character(100) :: posvel
Then you can open each file as
do i = 1, N
write(posvel,"(a,i0,a)") "posveldata",i,".txt"
open(unit=i,file=trim(posvel), status="unknown")
end do
Here, I have used the format i0 to automatically determine a proper width for integer, and trim() for removing unnecessary blanks in the file name (though they may not be necessary). The write statement can also be written more compactly as
write(posvel,"('posveldata',i0,'.txt')") i
by embedding character literals into the format specification.
The error message "End of record" comes from the above issue. This can be confirmed by the following code
character c
write(c,"(a)") "1"
print *, "c = ", c
write(c,"(a)") "23" !! line 8 in test.f90
print *, "c = ", c
for which gfortran gives
c = 1
At line 8 of file test.f90
Fortran runtime error: End of record
This means that while c is used as an internal file, this "file" does not have enough space to accommodate two characters (here "23"). For comparison, ifort14 gives
c = 1
forrtl: severe (66): output statement overflows record, unit -5, file Internal Formatted Write
while Oracle Fortran12 gives
c = 1
****** FORTRAN RUN-TIME SYSTEM ******
Error 1010: record too long
Location: the WRITE statement at line 8 of "test.f90"
Aborted
(It is interesting that Oracle Fortran reports the record to be "too long", which may refer to the input string.)
I have a code in Fortran IV that I need to run. I was told to try to compile it in Fortran 77 and fix the error. So I named the file with a .f extension and tried to compile it with gfortran. I got the next error referring to the Fortran IV function copied below:
abel.f:432.24:
REAL FUNCTION DGDT*8(IX,NV,XNG,FNG,GNG,X)
1
Error: Expected formal argument list in function definition at (1)
Since I'm not too familiar with Fortran I'd appreciate if someone can tell me how to fix this problem .
REAL FUNCTION DGDT*8(IX,NV,XNG,FNG,GNG,X) AAOK0429
C AAOK0430
C THIS SUBROUTINE COMPUTES THE VALUE OF THE DERIVATIVE OF THE AAOK0431
C G-FUNCTION FOR A SLIT TRANSMISSION FUNCTION GIVEN BY A AAOK0432
C PIECE-WISE CUBIC SPLINE , WHOSE PARAMETERS ARE AAOK0433
C CONTAINED IN XNG,FNG AND GNG. AAOK0434
C AAOK0435
IMPLICIT REAL*8(A-H,O-Z) AAOK0436
C AAOK0437
C ALLOWABLE ROUNDING ERROR ON POINTS AT EXTREAMS OF KNOT RANGE AAOK0438
C IS 2**IEPS*MAX(!XNG(1)!,!XNG(NV)!). AAOK0439
INTEGER*4 IFLG/0/,IEPS/-50/ AAOK0440
DIMENSION XNG(1),FNG(1),GNG(1) AAOK0441
C AAOK0442
C TEST WETHER POINT IN RANGE. AAOK0443
IF(X.LT.XNG(1)) GO TO 990 AAOK0444
IF(X.GT.XNG(NV)) GO TO 991 AAOK0445
C AAOK0446
C ESTIMATE KNOT INTERVAL BY ASSUMING EQUALLY SPACED KNOTS. AAOK0447
12 J=DABS(X-XNG(1))/(XNG(NV)-XNG(1))*(NV-1)+1 AAOK0448
C ENSURE CASE X=XNG(NV) GIVES J=NV-1 AAOK0449
J=MIN0(J,NV-1) AAOK0450
C INDICATE THAT KNOT INTERVAL INSIDE RANGE HAS BEEN USED. AAOK0451
IFLG=1 AAOK0452
C SEARCH FOR KNOT INTERVAL CONTAINING X. AAOK0453
IF(X.LT.XNG(J)) GO TO 2 AAOK0454
C LOOP TILL INTERVAL FOUND. AAOK0455
1 J=J+1 AAOK0456
11 IF(X.GT.XNG(J+1)) GO TO 1 AAOK0457
GO TO 7 AAOK0458
2 J=J-1 AAOK0459
IF(X.LT.XNG(J)) GO TO 2 AAOK0460
C AAOK0461
C CALCULATE SPLINE PARAMETERS FOR JTH INTERVAL. AAOK0462
7 H=XNG(J+1)-XNG(J) AAOK0463
Q1=H*GNG(J) AAOK0464
Q2=H*GNG(J+1) AAOK0465
SS=FNG(J+1)-FNG(J) AAOK0466
B=3D0*SS-2D0*Q1-Q2 AAOK0467
A=Q1+Q2-2D0*SS AAOK0468
C AAOK0469
C CALCULATE SPLINE VALUE. AAOK0470
8 Z=(X-XNG(J))/H AAOK0471
C TF=((A*Z+B)*Z+Q1)*Z+FNG(J) AAOK0472
C TG=((3.*A*Z+2.*B)*Z+Q1)/H AAOK0473
C DGDT=(TG-TF/X)/X AAOK0474
DGDT=(3.*A*Z*Z+2.*B*Z+Q1)/H AAOK0475
RETURN AAOK0476
C TEST IF X WITHIN ROUNDING ERROR OF XNG(1). AAOK0477
990 IF(X.LE.XNG(1)-2D0**IEPS*DMAX1(DABS(XNG(1)),DABS(XNG(NV)))) GO AAOK0478
1 TO 99 AAOK0479
J=1 AAOK0480
GO TO 7 AAOK0481
C TEST IF X WITHIN ROUNDING ERROR OF XNG(NV). AAOK0482
991 IF(X.GE.XNG(NV)+2D0**IEPS*DMAX1(DABS(XNG(1)),DABS(XNG(NV)))) GO AAOK0483
1 TO 99 AAOK0484
J=NV-1 AAOK0485
GO TO 7 AAOK0486
99 IFLG=0 AAOK0487
C FUNCTION VALUE SET TO ZERO FOR POINTS OUTSIDE THE RANGE. AAOK0488
DGDT=0D0 AAOK0489
RETURN AAOK0490
END AAOK0491
This doesn't look so bad. Modern compilers still accept the real*8 syntax although it isn't standard. So you should (as mentioned) replace the line
REAL FUNCTION DGDT*8(IX,NV,XNG,FNG,GNG,X) AAOK0429
with
REAL*8 FUNCTION DGDT(IX,NV,XNG,FNG,GNG,X) AAOK0429
which compiled successfully for me using gfortran 4.6.2 using gfortran -c DGDT.f.
Good luck, and be on the lookout for other problems. Just because the code compiles does not mean it is running the same way it was designed!
Not really an answer, see the one from Ross. But I just can't stand the requirement for fixed form. Here is how this code probably would look like in F90 with free form:
function DGDT(IX, NV, XNG, FNG, GNG, X)
! THIS FUNCTION COMPUTES THE VALUE OF THE DERIVATIVE OF THE
! G-FUNCTION FOR A SLIT TRANSMISSION FUNCTION GIVEN BY A
! PIECE-WISE CUBIC SPLINE, WHOSE PARAMETERS ARE
! CONTAINED IN XNG,FNG AND GNG.
implicit none
integer, parameter :: rk = selected_real_kind(15)
integer :: ix, nv
real(kind=rk) :: dgdt
real(kind=rk) :: xng(nv)
real(kind=rk) :: fng(nv)
real(kind=rk) :: gng(nv)
real(kind=rk) :: x
! ALLOWABLE ROUNDING ERROR ON POINTS AT EXTREAMS OF KNOT RANGE
! IS 2**IEPS*MAX(!XNG(1)!,!XNG(NV)!).
integer, parameter :: ieps = -50
integer, save :: iflg = 0
integer :: j
real(kind=rk) :: tolerance
real(kind=rk) :: H
real(kind=rk) :: A, B
real(kind=rk) :: Q1, Q2
real(kind=rk) :: SS
real(kind=rk) :: Z
tolerance = 2.0_rk**IEPS * MAXVAL(ABS(XNG([1,NV])))
! TEST WETHER POINT IN RANGE.
if ((X < XNG(1) - tolerance) .or. (X > XNG(NV) + tolerance)) then
! FUNCTION VALUE SET TO ZERO FOR POINTS OUTSIDE THE RANGE.
iflg = 0
DGDT = 0.0_rk
return
end if
! ESTIMATE KNOT INTERVAL BY ASSUMING EQUALLY SPACED KNOTS.
J = abs(x-xng(1)) / (xng(nv)-xng(1)) * (nv-1) + 1
! ENSURE CASE X=XNG(NV) GIVES J=NV-1
J = MIN(J,NV-1)
! INDICATE THAT KNOT INTERVAL INSIDE RANGE HAS BEEN USED.
IFLG = 1
! SEARCH FOR KNOT INTERVAL CONTAINING X.
do
if ( (x >= xng(j)) .or. (j==1) ) EXIT
j = j-1
! LOOP TILL INTERVAL FOUND.
end do
do
if ( (x <= xng(j+1)) .or. (j==nv-1) ) EXIT
j = j+1
! LOOP TILL INTERVAL FOUND.
end do
! CALCULATE SPLINE PARAMETERS FOR JTH INTERVAL.
H = XNG(J+1) - XNG(J)
Q1 = H*GNG(J)
Q2 = H*GNG(J+1)
SS = FNG(J+1) - FNG(J)
B = 3.0_rk*SS - 2.0_rk*Q1 - Q2
A = Q1 + Q2 - 2.0_rk*SS
! CALCULATE SPLINE VALUE.
Z = (X-XNG(J))/H
DGDT = ( (3.0_rk*A*Z + 2.0_rk*B)*Z + Q1 ) / H
end function DGDT
Note, I did not test this in any way, also there might be some wrong guesses in there, like that ieps should be a constant. Also, I am not so sure about iflg, and the ix argument does not appear to be used at all. So I might got something wrong. For the tolerance it is better to use a factor instead of a difference and a 2.**-50 will not change the value for a the maxval in a double precision number here. Also note, I am using some other F90 features besides the free form now.
DISCLAIMER: Just mentioning a possible solution here, not recommending it...
As much as all other answers are valid and that supporting some Fortran IV code as is is a nightmare, you still might want / need to avoid touching it as much as possible. And since Fortran IV had some strange behaviours when it comes to loops for example (with loops always cycled at least once IINM), using a "proper" Fortran IV compiler might be a "good" idea.
Anyway, all this to say that the Intel compiler for example, supports Fortran IV natively with the -f66 compiler switch, and I'm sure other compilers do as well. This may be worth checking.
Trying to get a one-parameter least squares minimisation working in fortran77. Here's the code; it compiles and seems to work except....it gets caught in an infinite loop between values of h1= 1.8E-2 and 3.5E-2.
Having a look now but, odds are, I'm not going to have much luck sussing the issue on my own. All help welcome!
PROGRAM assignment
! A program designed to fit experiemental data, using the method
! of least squares to minimise the associated chi-squared and
! obtain the four control parameters A,B,h1 and h2.
!*****************************************************************
IMPLICIT NONE
INTEGER i
DOUBLE PRECISION t(17),Ct(17),eCt(17)
DOUBLE PRECISION h1loop1,h1loop2,deltah,Cs
DOUBLE PRECISION chisqa,chisqb,dchisq
OPEN(21, FILE='data.txt', FORM='FORMATTED', STATUS='OLD')
DO i=1,17
READ(21,*)t(i),Ct(i),eCt(i)
END DO
CLOSE(21)
!Read in data.txt as three one dimensional arrays.
!*****************************************************************
!OPEN(21, FILE='outtest.txt', FORM='FORMATTED', STATUS='NEW')
!DO i=1,17
! WRITE(21,*)t(i),Ct(i),eCt(i)
!END DO
!CLOSE(21)
!
!Just to check input file is being read correctly.
!*****************************************************************
!**********************Minimising Lamda1 (h1)*********************
deltah= 0.0001
h1loop2= 0.001
h1loop1= 0.0 !Use initial value of 0 to calculate start-point chisq
DO 10
chisqa= 0.0
DO 20 i= 1, 17
Cs= exp(-h1loop1*t(i))
chisqa= chisqa + ((Ct(i) - Cs)/eCt(i))**2
20 END DO
chisqb= 0.0
DO 30 i= 1, 17
h1loop2= h1loop2 + deltah
Cs= exp(-h1loop2*t(i))
chisqb= chisqb + ((Ct(i) - Cs)/eCt(i))**2
30 END DO
!Print the two calculated chisq values to screen.
WRITE(6,*) 'Chi-squared a=',chisqa,'for Lamda1=',h1loop1
WRITE(6,*) 'Chi-squared b=',chisqb,'for Lamda1=',h1loop2
dchisq= chisqa - chisqb
IF (dchisq.GT.0.0) THEN
h1loop1= h1loop2
ELSE
deltah= deltah - ((deltah*2)/100)
END IF
IF (chisqb.LE.6618.681) EXIT
10 END DO
WRITE(6,*) 'Chi-squared is', chisqb,' for Lamda1 = ', h1loop2
END PROGRAM assignment
EDIT: Having looked at it again I've decided I have no clue what's screwing it up. Should be getting a chi-squared of 6618.681 from this, but it's just stuck between 6921.866 and 6920.031. Help!
do i=1
is not starting a loop, for a loop you need to specify an upper bound as well:
do i=1,ub
that's why you get the error message about the doi not having a type, in fixed format spaces are insignificant...
Edit: If you want to have an infinite loop, just skip the "i=" declaration completely. You can use an exit statement to leave the loop, when a certain criterion has been reached:
do
if (min_reached) EXIT
end do
Edit2: I don't know why you stick to F77 fixed format. Here is your program in free format, with some fixes of places, which looked weird, without digging too much into the details:
PROGRAM assignment
! A program designed to fit experiemental data, using the method
! of least squares to minimise the associated chi-squared and
! obtain the four control parameters A,B,h1 and h2.
!*****************************************************************
IMPLICIT NONE
integer, parameter :: rk = selected_real_kind(15)
integer, parameter :: nd = 17
integer :: i,t0
real(kind=rk) :: t(nd),t2(nd),Ct(nd),eCt(nd),Ctdiff(nd),c(nd)
real(kind=rk) :: Aa,Ab,Ba,Bb,h1a,h1b,h2a,h2b,chisqa,chisqb,dchisq
real(kind=rk) :: deltah,Cs(nd)
OPEN(21, FILE='data.txt', FORM='FORMATTED', STATUS='OLD')
DO i=1,nd
READ(21,*) t(i),Ct(i),eCt(i)
END DO
CLOSE(21)
!Read in data.txt as three one dimensional arrays.
!*****************************************************************
!OPEN(21, FILE='outtest.txt', FORM='FORMATTED', STATUS='NEW')
!DO i=1,17
! WRITE(21,*)t(i),Ct(i),eCt(i)
!END DO
!CLOSE(21)
!
!Just to check input file is being read correctly.
!*****************************************************************
!****************************Parameters***************************
Aa= 0
Ba= 0
h1a= 0
h2a= 0
!**********************Minimising Lamda1 (h1)*********************
deltah= 0.001_rk
h1b= deltah
minloop: DO
chisqa= 0
DO i= 1,nd
Cs(i)= exp(-h1a*t(i))!*Aa !+ Ba*exp(-h2a*t(i))
Ctdiff(i)= Ct(i) - Cs(i)
c(i)= Ctdiff(i)**2/eCt(i)**2
chisqa= chisqa + c(i)
h1a= h1a + deltah
END DO
! Use initial h1 value of 0 to calculate start-point chisq.
chisqb= 0
DO i= 1,nd
h1b= h1b + deltah
Cs(i)= exp(-h1b*t(i))!*Ab !+ Bb*exp(-h2b*t(i))
Ctdiff(i)= Ct(i) - Cs(i)
c(i)= Ctdiff(i)**2/eCt(i)**2
chisqb= chisqb + c(i)
END DO
! First-step h1 used to find competing chisq for comparison.
WRITE(6,*) 'Chi-squared a=', chisqa,'for Lamda1=',h1a
WRITE(6,*) 'Chi-squared b=', chisqb,'for Lamda1=',h1b
! Prints the two calculated chisq values to screen.
dchisq= chisqa - chisqb
IF (dchisq.GT.0) THEN
h1a= h1b
ELSE IF (dchisq.LE.0) THEN
deltah= (-deltah*2)/10
END IF
IF (chisqb.LE.6000) EXIT minloop
END DO minloop
WRITE(6,*) 'Chi-squared is', chisqb,'for Lamda1=',h1b
END PROGRAM assignment