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1988-08-09
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* a TIME function for Ryan/McFarland Fortran and Microsoft Version 4.0
* Author: M. Steven Baker
* Date: September 20, 1986
*
real function second()
integer*4 hh,mm,ss,hd
call gettim(hh,mm,ss,hd)
second = float(hh)*3600 + float(mm*60+ss) + float(hd)/100
end
real aa(200,200),a(201,200),b(200),x(200)
real time(8,6),cray,ops,total,norma,normx
real resid,residn,eps,epslon
integer ipvt(200)
lda = 201
ldaa = 200
c
n = 100
cray = .056
write(*,1)
1 format(' Please send the results of this run to:'//
$ ' Jack J. Dongarra'/
$ ' Mathematics and Computer Science Division'/
$ ' Argonne National Laboratory'/
$ ' Argonne, Illinois 60439'//
$ ' Telephone: 312-972-7246'//
$ ' ARPAnet: DONGARRA@ANL-MCS'/)
ops = (2.0e0*n**3)/3.0e0 + 2.0e0*n**2
c
call matgen(a,lda,n,b,norma)
t1 = second()
call sgefa(a,lda,n,ipvt,info)
time(1,1) = second() - t1
t1 = second()
call sgesl(a,lda,n,ipvt,b,0)
time(1,2) = second() - t1
total = time(1,1) + time(1,2)
C
C COMPUTE A RESIDUAL TO VERIFY RESULTS.
C
do 10 i = 1,n
x(i) = b(i)
10 continue
call matgen(a,lda,n,b,norma)
do 20 i = 1,n
b(i) = -b(i)
20 continue
CALL SMXPY(n,b,n,lda,x,a)
RESID = 0.0
NORMX = 0.0
DO 30 I = 1,N
RESID = amax1( RESID, ABS(b(i)) )
NORMX = amax1( NORMX, ABS(X(I)) )
30 CONTINUE
eps = epslon(1.0)
RESIDn = RESID/( N*NORMA*NORMX*EPS )
write(*,40)
40 format(' norm. resid resid machep',
$ ' x(1) x(n)')
write(*,50) residn,resid,eps,x(1),x(n)
50 format(1p5e16.8)
c
write(*,60) n
60 format(//' times are reported for matrices of order ',i5)
write(*,70)
70 format(6x,'sgefa',6x,'sgesl',6x,'total',5x,'mflops',7x,'unit',
$ 6x,'ratio')
c
time(1,3) = total
time(1,4) = ops/(1.0e6*total)
time(1,5) = 2.0e0/time(1,4)
time(1,6) = total/cray
write(*,80) lda
80 format(' times for array with leading dimension of',i4)
write(*,110) (time(1,i),i=1,6)
c
call matgen(a,lda,n,b,norma)
t1 = second()
call sgefa(a,lda,n,ipvt,info)
time(2,1) = second() - t1
t1 = second()
call sgesl(a,lda,n,ipvt,b,0)
time(2,2) = second() - t1
total = time(2,1) + time(2,2)
time(2,3) = total
time(2,4) = ops/(1.0e6*total)
time(2,5) = 2.0e0/time(2,4)
time(2,6) = total/cray
c
call matgen(a,lda,n,b,norma)
t1 = second()
call sgefa(a,lda,n,ipvt,info)
time(3,1) = second() - t1
t1 = second()
call sgesl(a,lda,n,ipvt,b,0)
time(3,2) = second() - t1
total = time(3,1) + time(3,2)
time(3,3) = total
time(3,4) = ops/(1.0e6*total)
time(3,5) = 2.0e0/time(3,4)
time(3,6) = total/cray
c
ntimes = 10
tm2 = 0
t1 = second()
do 90 i = 1,ntimes
tm = second()
call matgen(a,lda,n,b,norma)
tm2 = tm2 + second() - tm
call sgefa(a,lda,n,ipvt,info)
90 continue
time(4,1) = (second() - t1 - tm2)/ntimes
t1 = second()
do 100 i = 1,ntimes
call sgesl(a,lda,n,ipvt,b,0)
100 continue
time(4,2) = (second() - t1)/ntimes
total = time(4,1) + time(4,2)
time(4,3) = total
time(4,4) = ops/(1.0e6*total)
time(4,5) = 2.0e0/time(4,4)
time(4,6) = total/cray
c
write(*,110) (time(2,i),i=1,6)
write(*,110) (time(3,i),i=1,6)
write(*,110) (time(4,i),i=1,6)
110 format(6(1pe11.3))
c
call matgen(aa,ldaa,n,b,norma)
t1 = second()
call sgefa(aa,ldaa,n,ipvt,info)
time(5,1) = second() - t1
t1 = second()
call sgesl(aa,ldaa,n,ipvt,b,0)
time(5,2) = second() - t1
total = time(5,1) + time(5,2)
time(5,3) = total
time(5,4) = ops/(1.0e6*total)
time(5,5) = 2.0e0/time(5,4)
time(5,6) = total/cray
c
call matgen(aa,ldaa,n,b,norma)
t1 = second()
call sgefa(aa,ldaa,n,ipvt,info)
time(6,1) = second() - t1
t1 = second()
call sgesl(aa,ldaa,n,ipvt,b,0)
time(6,2) = second() - t1
total = time(6,1) + time(6,2)
time(6,3) = total
time(6,4) = ops/(1.0e6*total)
time(6,5) = 2.0e0/time(6,4)
time(6,6) = total/cray
c
call matgen(aa,ldaa,n,b,norma)
t1 = second()
call sgefa(aa,ldaa,n,ipvt,info)
time(7,1) = second() - t1
t1 = second()
call sgesl(aa,ldaa,n,ipvt,b,0)
time(7,2) = second() - t1
total = time(7,1) + time(7,2)
time(7,3) = total
time(7,4) = ops/(1.0e6*total)
time(7,5) = 2.0e0/time(7,4)
time(7,6) = total/cray
c
ntimes = 10
tm2 = 0
t1 = second()
do 120 i = 1,ntimes
tm = second()
call matgen(aa,ldaa,n,b,norma)
tm2 = tm2 + second() - tm
call sgefa(aa,ldaa,n,ipvt,info)
120 continue
time(8,1) = (second() - t1 - tm2)/ntimes
t1 = second()
do 130 i = 1,ntimes
call sgesl(aa,ldaa,n,ipvt,b,0)
130 continue
time(8,2) = (second() - t1)/ntimes
total = time(8,1) + time(8,2)
time(8,3) = total
time(8,4) = ops/(1.0e6*total)
time(8,5) = 2.0e0/time(8,4)
time(8,6) = total/cray
c
write(*,140) ldaa
140 format(/' times for array with leading dimension of',i4)
write(*,110) (time(5,i),i=1,6)
write(*,110) (time(6,i),i=1,6)
write(*,110) (time(7,i),i=1,6)
write(*,110) (time(8,i),i=1,6)
stop
end
subroutine matgen(a,lda,n,b,norma)
real a(lda,1),b(1),norma
c
init = 1325
norma = 0.0
do 30 j = 1,n
do 20 i = 1,n
init = mod(3125*init,65536)
a(i,j) = (init - 32768.0)/16384.0
norma = amax1(a(i,j), norma)
20 continue
30 continue
do 35 i = 1,n
b(i) = 0.0
35 continue
do 50 j = 1,n
do 40 i = 1,n
b(i) = b(i) + a(i,j)
40 continue
50 continue
return
end
subroutine sgefa(a,lda,n,ipvt,info)
integer lda,n,ipvt(1),info
real a(lda,1)
c
c sgefa factors a real matrix by gaussian elimination.
c
c sgefa is usually called by dgeco, but it can be called
c directly with a saving in time if rcond is not needed.
c (time for dgeco) = (1 + 9/n)*(time for sgefa) .
c
c on entry
c
c a real(lda, n)
c the matrix to be factored.
c
c lda integer
c the leading dimension of the array a .
c
c n integer
c the order of the matrix a .
c
c on return
c
c a an upper triangular matrix and the multipliers
c which were used to obtain it.
c the factorization can be written a = l*u where
c l is a product of permutation and unit lower
c triangular matrices and u is upper triangular.
c
c ipvt integer(n)
c an integer vector of pivot indices.
c
c info integer
c = 0 normal value.
c = k if u(k,k) .eq. 0.0 . this is not an error
c condition for this subroutine, but it does
c indicate that sgesl or dgedi will divide by zero
c if called. use rcond in dgeco for a reliable
c indication of singularity.
c
c linpack. this version dated 08/14/78 .
c cleve moler, university of new mexico, argonne national lab.
c
c subroutines and functions
c
c blas saxpy,sscal,isamax
c
c internal variables
c
real t
integer isamax,j,k,kp1,l,nm1
c
c
c gaussian elimination with partial pivoting
c
info = 0
nm1 = n - 1
if (nm1 .lt. 1) go to 70
do 60 k = 1, nm1
kp1 = k + 1
c
c find l = pivot index
c
l = isamax(n-k+1,a(k,k),1) + k - 1
ipvt(k) = l
c
c zero pivot implies this column already triangularized
c
if (a(l,k) .eq. 0.0e0) go to 40
c
c interchange if necessary
c
if (l .eq. k) go to 10
t = a(l,k)
a(l,k) = a(k,k)
a(k,k) = t
10 continue
c
c compute multipliers
c
t = -1.0e0/a(k,k)
call sscal(n-k,t,a(k+1,k),1)
c
c row elimination with column indexing
c
do 30 j = kp1, n
t = a(l,j)
if (l .eq. k) go to 20
a(l,j) = a(k,j)
a(k,j) = t
20 continue
call saxpy(n-k,t,a(k+1,k),1,a(k+1,j),1)
30 continue
go to 50
40 continue
info = k
50 continue
60 continue
70 continue
ipvt(n) = n
if (a(n,n) .eq. 0.0e0) info = n
return
end
subroutine sgesl(a,lda,n,ipvt,b,job)
integer lda,n,ipvt(1),job
real a(lda,1),b(1)
c
c sgesl solves the real system
c a * x = b or trans(a) * x = b
c using the factors computed by dgeco or sgefa.
c
c on entry
c
c a real(lda, n)
c the output from dgeco or sgefa.
c
c lda integer
c the leading dimension of the array a .
c
c n integer
c the order of the matrix a .
c
c ipvt integer(n)
c the pivot vector from dgeco or sgefa.
c
c b real(n)
c the right hand side vector.
c
c job integer
c = 0 to solve a*x = b ,
c = nonzero to solve trans(a)*x = b where
c trans(a) is the transpose.
c
c on return
c
c b the solution vector x .
c
c error condition
c
c a division by zero will occur if the input factor contains a
c zero on the diagonal. technically this indicates singularity
c but it is often caused by improper arguments or improper
c setting of lda . it will not occur if the subroutines are
c called correctly and if dgeco has set rcond .gt. 0.0
c or sgefa has set info .eq. 0 .
c
c to compute inverse(a) * c where c is a matrix
c with p columns
c call dgeco(a,lda,n,ipvt,rcond,z)
c if (rcond is too small) go to ...
c do 10 j = 1, p
c call sgesl(a,lda,n,ipvt,c(1,j),0)
c 10 continue
c
c linpack. this version dated 08/14/78 .
c cleve moler, university of new mexico, argonne national lab.
c
c subroutines and functions
c
c blas saxpy,sdot
c
c internal variables
c
real sdot,t
integer k,kb,l,nm1
c
nm1 = n - 1
if (job .ne. 0) go to 50
c
c job = 0 , solve a * x = b
c first solve l*y = b
c
if (nm1 .lt. 1) go to 30
do 20 k = 1, nm1
l = ipvt(k)
t = b(l)
if (l .eq. k) go to 10
b(l) = b(k)
b(k) = t
10 continue
call saxpy(n-k,t,a(k+1,k),1,b(k+1),1)
20 continue
30 continue
c
c now solve u*x = y
c
do 40 kb = 1, n
k = n + 1 - kb
b(k) = b(k)/a(k,k)
t = -b(k)
call saxpy(k-1,t,a(1,k),1,b(1),1)
40 continue
go to 100
50 continue
c
c job = nonzero, solve trans(a) * x = b
c first solve trans(u)*y = b
c
do 60 k = 1, n
t = sdot(k-1,a(1,k),1,b(1),1)
b(k) = (b(k) - t)/a(k,k)
60 continue
c
c now solve trans(l)*x = y
c
if (nm1 .lt. 1) go to 90
do 80 kb = 1, nm1
k = n - kb
b(k) = b(k) + sdot(n-k,a(k+1,k),1,b(k+1),1)
l = ipvt(k)
if (l .eq. k) go to 70
t = b(l)
b(l) = b(k)
b(k) = t
70 continue
80 continue
90 continue
100 continue
return
end
subroutine saxpy(n,da,dx,incx,dy,incy)
c
c constant times a vector plus a vector.
c jack dongarra, linpack, 3/11/78.
c
real dx(1),dy(1),da
integer i,incx,incy,ix,iy,m,mp1,n
c
if(n.le.0)return
if (da .eq. 0.0e0) return
if(incx.eq.1.and.incy.eq.1)go to 20
c
c code for unequal increments or equal increments
c not equal to 1
c
ix = 1
iy = 1
if(incx.lt.0)ix = (-n+1)*incx + 1
if(incy.lt.0)iy = (-n+1)*incy + 1
do 10 i = 1,n
dy(iy) = dy(iy) + da*dx(ix)
ix = ix + incx
iy = iy + incy
10 continue
return
c
c code for both increments equal to 1
c
20 continue
do 30 i = 1,n
dy(i) = dy(i) + da*dx(i)
30 continue
return
end
real function sdot(n,dx,incx,dy,incy)
c
c forms the dot product of two vectors.
c jack dongarra, linpack, 3/11/78.
c
real dx(1),dy(1),dtemp
integer i,incx,incy,ix,iy,m,mp1,n
c
sdot = 0.0e0
dtemp = 0.0e0
if(n.le.0)return
if(incx.eq.1.and.incy.eq.1)go to 20
c
c code for unequal increments or equal increments
c not equal to 1
c
ix = 1
iy = 1
if(incx.lt.0)ix = (-n+1)*incx + 1
if(incy.lt.0)iy = (-n+1)*incy + 1
do 10 i = 1,n
dtemp = dtemp + dx(ix)*dy(iy)
ix = ix + incx
iy = iy + incy
10 continue
sdot = dtemp
return
c
c code for both increments equal to 1
c
20 continue
do 30 i = 1,n
dtemp = dtemp + dx(i)*dy(i)
30 continue
sdot = dtemp
return
end
subroutine sscal(n,da,dx,incx)
c
c scales a vector by a constant.
c jack dongarra, linpack, 3/11/78.
c
real da,dx(1)
integer i,incx,m,mp1,n,nincx
c
if(n.le.0)return
if(incx.eq.1)go to 20
c
c code for increment not equal to 1
c
nincx = n*incx
do 10 i = 1,nincx,incx
dx(i) = da*dx(i)
10 continue
return
c
c code for increment equal to 1
c
20 continue
do 30 i = 1,n
dx(i) = da*dx(i)
30 continue
return
end
integer function isamax(n,dx,incx)
c
c finds the index of element having max. absolute value.
c jack dongarra, linpack, 3/11/78.
c
real dx(1),dmax
integer i,incx,ix,n
c
isamax = 0
if( n .lt. 1 ) return
isamax = 1
if(n.eq.1)return
if(incx.eq.1)go to 20
c
c code for increment not equal to 1
c
ix = 1
dmax = abs(dx(1))
ix = ix + incx
do 10 i = 2,n
if(abs(dx(ix)).le.dmax) go to 5
isamax = i
dmax = abs(dx(ix))
5 ix = ix + incx
10 continue
return
c
c code for increment equal to 1
c
20 dmax = abs(dx(1))
do 30 i = 2,n
if(abs(dx(i)).le.dmax) go to 30
isamax = i
dmax = abs(dx(i))
30 continue
return
end
REAL FUNCTION EPSLON (X)
REAL X
C
C ESTIMATE UNIT ROUNDOFF IN QUANTITIES OF SIZE X.
C
REAL A,B,C,EPS
C
C THIS PROGRAM SHOULD FUNCTION PROPERLY ON ALL SYSTEMS
C SATISFYING THE FOLLOWING TWO ASSUMPTIONS,
C 1. THE BASE USED IN REPRESENTING FLOATING POINT
C NUMBERS IS NOT A POWER OF THREE.
C 2. THE QUANTITY A IN STATEMENT 10 IS REPRESENTED TO
C THE ACCURACY USED IN FLOATING POINT VARIABLES
C THAT ARE STORED IN MEMORY.
C THE STATEMENT NUMBER 10 AND THE GO TO 10 ARE INTENDED TO
C FORCE OPTIMIZING COMPILERS TO GENERATE CODE SATISFYING
C ASSUMPTION 2.
C UNDER THESE ASSUMPTIONS, IT SHOULD BE TRUE THAT,
C A IS NOT EXACTLY EQUAL TO FOUR-THIRDS,
C B HAS A ZERO FOR ITS LAST BIT OR DIGIT,
C C IS NOT EXACTLY EQUAL TO ONE,
C EPS MEASURES THE SEPARATION OF 1.0 FROM
C THE NEXT LARGER FLOATING POINT NUMBER.
C THE DEVELOPERS OF EISPACK WOULD APPRECIATE BEING INFORMED
C ABOUT ANY SYSTEMS WHERE THESE ASSUMPTIONS DO NOT HOLD.
C
C *****************************************************************
C THIS ROUTINE IS ONE OF THE AUXILIARY ROUTINES USED BY EISPACK III
C TO AVOID MACHINE DEPENDENCIES.
C *****************************************************************
C
C THIS VERSION DATED 4/6/83.
C
A = 4.0E0/3.0E0
10 B = A - 1.0E0
C = B + B + B
EPS = ABS(C-1.0E0)
IF (EPS .EQ. 0.0E0) GO TO 10
EPSLON = EPS*ABS(X)
RETURN
END
SUBROUTINE SMXPY (N1, Y, N2, LDM, X, M)
REAL Y(*), X(*), M(LDM,*)
C
C PURPOSE:
C MULTIPLY MATRIX M TIMES VECTOR X AND ADD THE RESULT TO VECTOR Y.
C
C PARAMETERS:
C
C N1 INTEGER, NUMBER OF ELEMENTS IN VECTOR Y, AND NUMBER OF ROWS IN
C MATRIX M
C
C Y REAL(N1), VECTOR OF LENGTH N1 TO WHICH IS ADDED THE PRODUCT M*X
C
C N2 INTEGER, NUMBER OF ELEMENTS IN VECTOR X, AND NUMBER OF COLUMNS
C IN MATRIX M
C
C LDM INTEGER, LEADING DIMENSION OF ARRAY M
C
C X REAL(N2), VECTOR OF LENGTH N2
C
C M REAL(LDM,N2), MATRIX OF N1 ROWS AND N2 COLUMNS
C
C ----------------------------------------------------------------------
C
C CLEANUP ODD VECTOR
C
J = MOD(N2,2)
IF (J .GE. 1) THEN
DO 10 I = 1, N1
Y(I) = (Y(I)) + X(J)*M(I,J)
10 CONTINUE
ENDIF
C
C CLEANUP ODD GROUP OF TWO VECTORS
C
J = MOD(N2,4)
IF (J .GE. 2) THEN
DO 20 I = 1, N1
Y(I) = ( (Y(I))
$ + X(J-1)*M(I,J-1)) + X(J)*M(I,J)
20 CONTINUE
ENDIF
C
C CLEANUP ODD GROUP OF FOUR VECTORS
C
J = MOD(N2,8)
IF (J .GE. 4) THEN
DO 30 I = 1, N1
Y(I) = ((( (Y(I))
$ + X(J-3)*M(I,J-3)) + X(J-2)*M(I,J-2))
$ + X(J-1)*M(I,J-1)) + X(J) *M(I,J)
30 CONTINUE
ENDIF
C
C CLEANUP ODD GROUP OF EIGHT VECTORS
C
J = MOD(N2,16)
IF (J .GE. 8) THEN
DO 40 I = 1, N1
Y(I) = ((((((( (Y(I))
$ + X(J-7)*M(I,J-7)) + X(J-6)*M(I,J-6))
$ + X(J-5)*M(I,J-5)) + X(J-4)*M(I,J-4))
$ + X(J-3)*M(I,J-3)) + X(J-2)*M(I,J-2))
$ + X(J-1)*M(I,J-1)) + X(J) *M(I,J)
40 CONTINUE
ENDIF
C
C MAIN LOOP - GROUPS OF SIXTEEN VECTORS
C
JMIN = J+16
DO 60 J = JMIN, N2, 16
DO 50 I = 1, N1
Y(I) = ((((((((((((((( (Y(I))
$ + X(J-15)*M(I,J-15)) + X(J-14)*M(I,J-14))
$ + X(J-13)*M(I,J-13)) + X(J-12)*M(I,J-12))
$ + X(J-11)*M(I,J-11)) + X(J-10)*M(I,J-10))
$ + X(J- 9)*M(I,J- 9)) + X(J- 8)*M(I,J- 8))
$ + X(J- 7)*M(I,J- 7)) + X(J- 6)*M(I,J- 6))
$ + X(J- 5)*M(I,J- 5)) + X(J- 4)*M(I,J- 4))
$ + X(J- 3)*M(I,J- 3)) + X(J- 2)*M(I,J- 2))
$ + X(J- 1)*M(I,J- 1)) + X(J) *M(I,J)
50 CONTINUE
60 CONTINUE
RETURN
END