Using fortran to Solve linear equations using gauss elimination and back substitution

In summary: The program will then proceed to the part where Z is assigned with the values of B, and then it will end. In summary, when lud is equal to zero, the values of AMD and ipvt will not be initialized and the program will end after assigning Z with the values of B.
  • #1
mmokhtar
1
0

Homework Statement



What will be the value of the variable ipvt and AMD when the input value of lud to the following subroutine (solver) is zero ?

subroutine Solver (A,B,N,lud,Z)
integer lda,N,ipvt(1000),info,lud,IDAMAX
&j,k,kp1,l,nm1,kb

double precision A(1000,1000),B(1000),Z(1000),t,AMD(1000,1000)
common/ludcmp/ipvt,AMD

nm1=N-1
do 5 i=1,N
Z(i)=B(i)
5 continue

if (lud.eq.0) goto 99

do 6 i=1,N
do 6 j=1,N
AMD(i,j)=A(i,j)
6 continue

info = 0
IF (nm1 .LT. 1) GO TO 70
DO 60 k = 1, nm1
kp1 = k + 1
C
C FIND l = PIVOT INDEX
C
l = IDAMAX(N-k+1,AMD(k,k),1) + k - 1
ipvt(k) = l
C
C ZERO PIVOT IMPLIES THIS COLUMN ALREADY TRIANGULARIZED
C
IF (AMD(l,k) .EQ. 0.0D0) GO TO 40
C
C INTERCHANGE IF NECESSARY
C
IF (l .EQ.k) GO TO 10
t = AMD(l,k)
AMD(l,k) = AMD(k,k)
AMD(k,k) = t
10 CONTINUE
C
C COMPUTE MULTIPLIERS
C
t = -1.0D0/AMD(k,k)
CALL DSCAL(N-k,t,AMD(k+1,k),1)
C
C ROW ELIMINATION WITH COLUMN INDEXING
C
DO 30 j= kp1, N
t = AMD(l,j)
IF (l .EQ.k) GO TO 20
AMD(l,j) = AMD(k,j)
AMD(k,j) = t
20 CONTINUE
CALL DAXPY(N-k,t,AMD(k+1,k),1,AMD(k+1,j),1)
30 CONTINUE
GO TO 50
40 CONTINUE
info= k
50 CONTINUE
60 CONTINUE
70 CONTINUE
Ipvt(N) = N
IF (AMD(N,N) .EQ. 0.0D0) info = N

if (nm1 .lt. 1) go to 130
do 120 k = 1, nm1
l = ipvt(k)
t = Z(l)
if (l .eq. k) go to 110
Z(l) = Z(k)
Z(k) = t
110 continue
call DAXPY (N-k,t,AMD(k+1,k),1,Z(k+1),1)
120 continue
130 continue
c
c now solve u*x = y
c
do 140 kb = 1, N
k = N + 1 - kb
Z(k) = Z(k)/AMD(k,k)
t = -Z(k)
call DAXPY (k-1,t,AMD(1,k),1,Z(1),1)
140 continue
return
end

SUBROUTINE DAXPY(N,DA,DX,INCX,DY,INCY)
* .. Scalar Arguments ..
DOUBLE PRECISION DA
INTEGER INCX,INCY,N
* ..
* .. Array Arguments ..
DOUBLE PRECISION DX(*),DY(*)
* ..
*
* Purpose
* =======
*
* DAXPY constant times a vector plus a vector.
* uses unrolled loops for increments equal to one.
*
* Further Details
* ===============
*
* jack dongarra, linpack, 3/11/78.
* modified 12/3/93, array(1) declarations changed to array(*)
*
* =====================================================================
*
* .. Local Scalars ..
INTEGER I,IX,IY,M,MP1
* ..
* .. Intrinsic Functions ..
INTRINSIC MOD
* ..
IF (N.LE.0) RETURN
IF (DA.EQ.0.0d0) RETURN
IF (INCX.EQ.1 .AND. INCY.EQ.1) THEN
*
* code for both increments equal to 1
*
*
* clean-up loop
*
M = MOD(N,4)
IF (M.NE.0) THEN
DO I = 1,M
DY(I) = DY(I) + DA*DX(I)
END DO
END IF
IF (N.LT.4) RETURN
MP1 = M + 1
DO I = MP1,N,4
DY(I) = DY(I) + DA*DX(I)
DY(I+1) = DY(I+1) + DA*DX(I+1)
DY(I+2) = DY(I+2) + DA*DX(I+2)
DY(I+3) = DY(I+3) + DA*DX(I+3)
END DO
ELSE
*
* code for unequal increments or equal increments
* not equal to 1
*
IX = 1
IY = 1
IF (INCX.LT.0) IX = (-N+1)*INCX + 1
IF (INCY.LT.0) IY = (-N+1)*INCY + 1
DO I = 1,N
DY(IY) = DY(IY) + DA*DX(IX)
IX = IX + INCX
IY = IY + INCY
END DO
END IF
RETURN
END

SUBROUTINE DSCAL(N,DA,DX,INCX)
* .. Scalar Arguments ..
DOUBLE PRECISION DA
INTEGER INCX,N
* ..
* .. Array Arguments ..
DOUBLE PRECISION DX(*)
* ..
*
* Purpose
* =======
*
* DSCAL scales a vector by a constant.
* uses unrolled loops for increment equal to one.
*
* Further Details
* ===============
*
* jack dongarra, linpack, 3/11/78.
* modified 3/93 to return if incx .le. 0.
* modified 12/3/93, array(1) declarations changed to array(*)
*
* =====================================================================
*
* .. Local Scalars ..
INTEGER I,M,MP1,NINCX
* ..
* .. Intrinsic Functions ..
INTRINSIC MOD
* ..
IF (N.LE.0 .OR. INCX.LE.0) RETURN
IF (INCX.EQ.1) THEN
*
* code for increment equal to 1
*
*
* clean-up loop
*
M = MOD(N,5)
IF (M.NE.0) THEN
DO I = 1,M
DX(I) = DA*DX(I)
END DO
IF (N.LT.5) RETURN
END IF
MP1 = M + 1
DO I = MP1,N,5
DX(I) = DA*DX(I)
DX(I+1) = DA*DX(I+1)
DX(I+2) = DA*DX(I+2)
DX(I+3) = DA*DX(I+3)
DX(I+4) = DA*DX(I+4)
END DO
ELSE
*
* code for increment not equal to 1
*
NINCX = N*INCX
DO I = 1,NINCX,INCX
DX(I) = DA*DX(I)
END DO
END IF
RETURN
END

INTEGER FUNCTION IDAMAX(N,DX,INCX)
* .. Scalar Arguments ..
INTEGER INCX,N
* ..
* .. Array Arguments ..
DOUBLE PRECISION DX(*)
* ..
*
* Purpose
* =======
*
* IDAMAX finds the index of element having max. absolute value.
*
* Further Details
* ===============
*
* jack dongarra, linpack, 3/11/78.
* modified 3/93 to return if incx .le. 0.
* modified 12/3/93, array(1) declarations changed to array(*)
*
* =====================================================================
*
* .. Local Scalars ..
DOUBLE PRECISION DMAX
INTEGER I,IX
* ..
* .. Intrinsic Functions ..
INTRINSIC DABS
* ..
IDAMAX = 0
IF (N.LT.1 .OR. INCX.LE.0) RETURN
IDAMAX = 1
IF (N.EQ.1) RETURN
IF (INCX.EQ.1) THEN
*
* code for increment equal to 1
*
DMAX = DABS(DX(1))
DO I = 2,N
IF (DABS(DX(I)).GT.DMAX) THEN
IDAMAX = I
DMAX = DABS(DX(I))
END IF
END DO
ELSE
*
* code for increment not equal to 1
*
IX = 1
DMAX = DABS(DX(1))
IX = IX + INCX
DO I = 2,N
IF (DABS(DX(IX)).GT.DMAX) THEN
IDAMAX = I
DMAX = DABS(DX(IX))
END IF
IX = IX + INCX
END DO
END IF
RETURN
END




Homework Equations


Solving linear equations using
1- Gauss elimination
2- Backsubstitution


The Attempt at a Solution



What will be the value of the variable ipvt and AMD when the input value of lud to the subroutine (solver) is zero ?
 
Physics news on Phys.org
  • #2
When the input value of lud is zero, the program will skip the part where AMD is assigned with the values of A. Therefore, the value of AMD will be zero and ipvt will not be initialized.
 

FAQ: Using fortran to Solve linear equations using gauss elimination and back substitution

What is Fortran and how is it used to solve linear equations?

Fortran is a programming language commonly used in scientific computing. It is particularly well-suited for solving mathematical problems, such as linear equations. Fortran has built-in functions and data types that make it efficient for solving these types of problems.

What is gauss elimination and how does it work?

Gauss elimination is a method for solving systems of linear equations. It involves systematically eliminating variables by using row operations, such as adding or subtracting rows, to create a simpler system of equations. This process continues until the system is reduced to a triangular form, making it easier to solve for the remaining variables.

What is back substitution and when is it used?

Back substitution is a technique used to find the values of the variables in a system of linear equations after the system has been reduced to a triangular form. This method involves solving for the variables starting with the last equation and working backwards. It is used when the system of equations has been reduced to a triangular form, which is often the case after using gauss elimination.

What are the advantages of using Fortran to solve linear equations?

Fortran is a highly efficient language for solving mathematical problems, including linear equations. It has built-in functions and data types specifically designed for these types of calculations, making it faster and more accurate than other programming languages. Additionally, Fortran is widely used in scientific computing, so there is a large community of resources and support available for those using it.

Are there any limitations to using Fortran for solving linear equations?

While Fortran is a powerful language for solving mathematical problems, it does have some limitations. One potential limitation is that it may be more difficult to learn for those without programming experience. Additionally, Fortran may not be the best choice for solving very large systems of equations, as it may not be as efficient as other languages for these types of problems.

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