sub scal.for

来自「非线性回归问题SQP解法」· FOR 代码 · 共 32 行

		SUBROUTINE SCAL(N,M,NAMAX,MAMAX,G,C,A,B,OSCL,SCL,ZERO)
		DOUBLE PRECISION G(NAMAX,N),A(NAMAX,M),C(N),B(M),SCL(MAMAX),
	*	                 OSCL,ZERO
		OSCL=0.0
		DO 1 J=1,N
			IF(DABS(C(J)).GT.OSCL) OSCL=DABS(C(J))
1		CONTINUE
		OSCL=OSCL/FLOAT(M)
		IF (OSCL.LT.10.0) OSCL=1.0
		DO 2 J=1,N
			C(J)=C(J)/OSCL
			DO 2 I=1,N
				G(I,J)=G(I,J)/OSCL
2		CONTINUE
		DO 5 I=1,M
			SCL(I)=0.0
			DO 3 J=1,N
				IF (DABS(A(J,I)).GT.SCL(I)) SCL(I)=DABS(A(J,I))
3		CONTINUE
		IF (SCL(I).LT.ZERO) THEN
			QPINDX=2
			RETURN
		ENDIF
		IF (SCL(I).LT.10.0) SCL(I)=1.0
		DO 4 J=1,N
			A(J,I)=A(J,I)/SCL(I)
4		CONTINUE
		B(I)=B(I)/SCL(I)
5		CONTINUE
		RETURN
		END