问题
I am not sure this question is on topic here or elsewhere (or not on topic at all anywhere). I have inherited Fortran 90 code that does Newton Raphson interpolation where logarithm of temperature is interpolated against logarithm of pressure.
The interpolation is of the type
t = a ln(p) + b
and where a, b are defined as
a = ln(tup/tdwn)/(alogpu - alogpd)
and
b = ln T - a * ln P
Here is the test program. It is shown only for a single iteration. But the actual program runs over three FOR loops over k,j and i. In reality pthta is a 3D array(k,j,i) and thta is a 1D array (k)
program test
implicit none
integer,parameter :: dp = SELECTED_REAL_KIND(12,307)
real(kind=dp) kappa,interc,pres,dltdlp,tup,tdwn
real(kind=dp) pthta,alogp,alogpd,alogpu,thta,f,dfdp,p1
real(kind=dp) t1,resid,potdwn,potup,pdwn,pup,epsln,thta1
integer i,j,kout,n,maxit,nmax,resmax
kappa = 2./7.
epsln = 1.
potdwn = 259.39996337890625
potup = 268.41687198359159
pdwn = 100000.00000000000
pup = 92500.000000000000
alogpu = 11.43496392350051
alogpd = 11.512925464970229
thta = 260.00000000000000
alogp = 11.512925464970229
! known temperature at lower level
tdwn = potdwn * (pdwn/100000.)** kappa
! known temperature at upper level
tup = potup *(pup/100000.)** kappa
! linear change of temperature wrt lnP between different levels
dltdlp = dlog(tup/tdwn)/(alogpu-alogpd)
! ln(T) value(intercept) where Pressure is 1 Pa and assume a linear
! relationship between P and T
interc = dlog(tup) - dltdlp*alogpu
! Initial guess value for pressure
pthta = exp((dlog(thta)-interc-kappa*alogp)/(dltdlp-kappa))
n=0
1900 continue
!First guess of temperature at intermediate level
t1 = exp(dltdlp * dlog(pthta)+interc)
!Residual error when calculating Newton Raphson iteration(Pascal)
resid = pthta - 100000.*(t1/thta)**(1./kappa)
print *, dltdlp,interc,t1,resid,pthta
if (abs(resid) .gt. epsln) then
n=n+1
if (n .le. nmax) then
! First guess of potential temperature given T1 and
! pressure level guess
thta1 = t1 * (100000./pthta)**kappa
f= thta - thta1
dfdp = (kappa-dltdlp)*(100000./pthta)**kappa*exp(interc + (dltdlp -1.)*dlog(pthta))
p1 = pthta - f/dfdp
if (p1 .le. pdwn) then
if (p1 .ge. pup) then
pthta = p1
goto 1900
else
n = nmax
end if
end if
else
if (resid .gt. resmax) resmax = resid
maxit = maxit+1
goto 2100
end if
end if
2100 continue
end program test
When you run this program with real data from a data file the value of resid is the following
2.7648638933897018E-010
and it does not differ much for the entire execution. Most of the values are in the range
1E-10 and 1E-12
So given these values the following IF condition
IF (abs(resid) .gt. epsln)
never gets called and the Newton Raphson iteration never gets executed. So I looked at two ways to get this to work. One is to remove the exponential call in these two steps
pthta = exp((dlog(thta)-interc-kappa*alogp)/(dltdlp-kappa))
t1 = exp(dltdlp * dlog(pthta)+interc)
i.e. keep everything in the logarithmic space and take the exponent after the Newton Raphson iteration completes. That part does converge without a problem.
The other way I tried to make this work is to truncate
t1 = exp(dltdlp * dlog(pthta)+interc)
When I truncate it to an integer the value of resid changes dramatically from 1E-10 to 813. I do not understand how truncating that function call leads to such a large value change. Truncating that result does result to a successful completion. So I am not sure which is the better way to proceed further.
How can I decide which would be the better way to approach this ?
回答1:
From a research perspective, I'd say your first solution is likely the more appropriate approach. In a physical simulation, one should always work with the logarithm of the properties that are by-definition always positive. In the above code, these would be temperature and pressure. Strictly positive-definite physical variables often result in overflow and underflow in computation, whether you use Fortran or any other programming language, or any possible variable kind. If something can happen, it will happen.
This is true about other physical quantities as well, for example, energy (the typical energy of a Gamma-Ray-Burst is ~10^54 ergs), volume of objects in arbitrary dimensions (the volume of a 100-dimensional sphere of radius 10meters is ~ 10^100), or even probability (the likelihood function in many statistical problems can take values of ~10^{-1000} or less). Working with log-transform of positive-definite variables would enable your code to handle numbers as big as ~10^10^307 (for a double precision variable).
A few notes also regarding the Fortran syntax used in your code:
The variable
RESMAXis used in your code without initialization.When assigning values to variables, it is important to specify the kind of the literal constants appropriately, otherwise, the program results might be affected. For example, here is the output of your original code compiled with Intel Fortran Compiler 2018 in debug mode:
-0.152581477302743 7.31503025786548 259.608693509165 -3.152934473473579E-002 99474.1999921620And here is the same code's output, but with all literal constants suffixed with the kind parameter
_dp(see the revised version of your code below):-0.152580456940175 7.31501855886952 259.608692604963 -8.731149137020111E-011 99474.2302854451The output from the revised code in this answer is slightly different from the output of the original code in the above question.
There is no need to use
.gt.,.ge.,.le.,.lt., ..., for comparison. These are legacy FORTRAN syntax, as far as I am aware. Use instead the more attractive symbols (<,>,<=,>=,==) for comparison.There is no necessity to use a
GOTOstatement in a Fortran program. This is again legacy FORTRAN. Frequently, simple elegant do-loops and if-blocks can replaceGOTOstatements, just as in the revised code below.There is no need to use kind-specific intrinsic functions in Fortran anymore (such as
dexp,dlog, ... for double precision). Almost all (and perhaps all) of Fortran intrinsic functions have generic names (exp,log, ...) in the current Fortran standard.
The following is a revision of the program in this question, that resolves all of the above obsolete syntax, as well as the problem of dealing with extremely large or small positive-definite variables (I probably went too far in log-transforming some variables that would never cause overflow or underflow, but my purpose here was to just show the logic behind log-transformation of positive-definite variables and how to deal with their arithmetics without potentially causing overflow/underflow/error_in_results).
program test
implicit none
integer,parameter :: dp = SELECTED_REAL_KIND(12,307)
real(kind=dp) kappa,interc,pres,dltdlp,tup,tdwn
real(kind=dp) pthta,alogp,alogpd,alogpu,thta,f,dfdp,p1
real(kind=dp) t1,resid,potdwn,potup,pdwn,pup,epsln,thta1
integer i,j,kout,n,maxit,nmax,resmax
real(kind=dp) :: log_resmax, log_pthta, log_t1, log_dummy, log_residAbsolute, sign_of_f
real(kind=dp) :: log_epsln, log_pdwn, log_pup, log_thta, log_thta1, log_p1, log_dfdp, log_f
logical :: residIsPositive, resmaxIsPositive, residIsBigger
log_resmax = log(log_resmax)
resmaxIsPositive = .true.
kappa = 2._dp/7._dp
epsln = 1._dp
potdwn = 259.39996337890625_dp
potup = 268.41687198359159_dp
pdwn = 100000.00000000000_dp
pup = 92500.000000000000_dp
alogpu = 11.43496392350051_dp
alogpd = 11.512925464970229_dp
thta = 260.00000000000000_dp
alogp = 11.512925464970229_dp
log_epsln = log(epsln)
log_pup = log(pup)
log_pdwn = log(pdwn)
log_thta = log(thta)
! known temperature at lower level
tdwn = potdwn * (pdwn/1.e5_dp)**kappa
! known temperature at upper level
tup = potup *(pup/1.e5_dp)** kappa
! linear change of temperature wrt lnP between different levels
dltdlp = log(tup/tdwn)/(alogpu-alogpd)
! ln(T) value(intercept) where Pressure is 1 Pa and assume a linear
! relationship between P and T
interc = log(tup) - dltdlp*alogpu
! Initial guess value for pressure
!pthta = exp( (log(thta)-interc-kappa*alogp) / (dltdlp-kappa) )
log_pthta = ( log_thta - interc - kappa*alogp ) / ( dltdlp - kappa )
n=0
MyDoLoop: do
!First guess of temperature at intermediate level
!t1 = exp(dltdlp * log(pthta)+interc)
log_t1 = dltdlp * log_pthta + interc
!Residual error when calculating Newton Raphson iteration(Pascal)
!resid = pthta - 1.e5_dp*(t1/thta)**(1._dp/kappa)
log_dummy = log(1.e5_dp) + ( log_t1 - log_thta ) / kappa
if (log_pthta>=log_dummy) then
residIsPositive = .true.
log_residAbsolute = log_pthta + log( 1._dp - exp(log_dummy-log_pthta) )
else
residIsPositive = .false.
log_residAbsolute = log_dummy + log( 1._dp - exp(log_pthta-log_dummy) )
end if
print *, "log-transformed values:"
print *, dltdlp,interc,log_t1,log_residAbsolute,log_pthta
print *, "non-log-transformed values:"
if (residIsPositive) print *, dltdlp,interc,exp(log_t1),exp(log_residAbsolute),exp(log_pthta)
if (.not.residIsPositive) print *, dltdlp,interc,exp(log_t1),-exp(log_residAbsolute),exp(log_pthta)
!if (abs(resid) > epsln) then
if ( log_residAbsolute > log_epsln ) then
n=n+1
if (n <= nmax) then
! First guess of potential temperature given T1 and
! pressure level guess
!thta1 = t1 * (1.e5_dp/pthta)**kappa
log_thta1 = log_t1 + ( log(1.e5_dp)-log_pthta ) * kappa
!f = thta - thta1
if ( log_thta>=thta1 ) then
log_f = log_thta + log( 1._dp - exp( log_thta1 - log_thta ) )
sign_of_f = 1._dp
else
log_f = log_thta + log( 1._dp - exp( log_thta - log_thta1 ) )
sign_of_f = 1._dp
end if
!dfdp = (kappa-dltdlp)*(1.e5_dp/pthta)**kappa*exp(interc + (dltdlp -1._dp)*log(pthta))
! assuming kappa-dltdlp>0 is TRUE always:
log_dfdp = log(kappa-dltdlp) + kappa*(log(1.e5_dp)-log_pthta) + interc + (dltdlp -1._dp)*log_pthta
!p1 = pthta - f/dfdp
! p1 should be, by definition, positive. Therefore:
log_dummy = log_f - log_dfdp
if (log_pthta>=log_dummy) then
log_p1 = log_pthta + log( 1._dp - sign_of_f*exp(log_dummy-log_pthta) )
else
log_p1 = log_dummy + log( 1._dp - sign_of_f*exp(log_pthta-log_dummy) )
end if
!if (p1 <= pdwn) then
if (log_p1 <= log_pdwn) then
!if (p1 >= pup) then
if (log_p1 >= log_pup) then
log_pthta = log_p1
cycle MyDoLoop
else
n = nmax
end if
end if
else
!if (resid > resmax) resmax = resid
residIsBigger = ( residIsPositive .and. resmaxIsPositive .and. log_residAbsolute>log_resmax ) .or. &
( .not.residIsPositive .and. .not.resmaxIsPositive .and. log_residAbsolute<log_resmax ) .or. &
( residIsPositive .and. .not. resmaxIsPositive )
if ( residIsBigger ) then
log_resmax = log_residAbsolute
resmaxIsPositive = residIsPositive
end if
maxit = maxit+1
end if
end if
exit MyDoLoop
end do MyDoLoop
end program test
Here is a sample output of this program, which agrees well with the output of the original code:
log-transformed values:
-0.152580456940175 7.31501855886952 5.55917546888014
-22.4565579499410 11.5076538974964
non-log-transformed values:
-0.152580456940175 7.31501855886952 259.608692604963
-1.767017293116268E-010 99474.2302854451
For comparison, here is the output from the original code:
-0.152580456940175 7.31501855886952 259.608692604963
-8.731149137020111E-011 99474.2302854451
来源：https://stackoverflow.com/questions/48990778/newton-raphson-iteration-unable-to-iterate