Attached is the original code for help (program issue)

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There are 9 components in logistics; why can the program only handle the first three of the mixed components? If you can help me with this, I’ll treat you to a meal in Beijing. QQ: 534989768. The procedure is as follows:
SUBROUTINE MODEL (NSIN, NINFI, SIN1, SIN2, SIN3, SIN4, SINFI, NSOUT, NINFO, SOUT1, SOUT2, SOUT3, SOUT4, SINFO, NSUBS, IDXSUB, ITYPE, NINT, INT, NREAL, REAL, IDS, NPO, NBOPST, NIWORK, IWORK, NWORK, WORK, NSIZE, SIZE, INTSIZ, LD)
C IMPLICIT NONE
C C Declare variables used in dimensioning
C INTEGER NSUBS, NINT, NPO, NIWORK, NWORK, NSIZE
C #include “ppexec_user.cmn”
EQUIVALENCE (RMISS, USER_RUMISS)
EQUIVALENCE (IMISS, USER_IUMISS)
C REAL*8 B(1), IB(1)
EQUIVALENCE (B(1), IB(1))
#include “dms_ncomp.cmn”
#include “shs_stwork.cmn”
#include “dms_global.cmn”
#include “dms_ppwork.cmn”
#include “dms_ipwork.cmn”
C C Declare arguments
C INTEGER IDXSUB(NSUBS), ITYPE(NSUBS), INT(NINT), IDS(2,13), NBOPST(6,NPO), IWORK(NIWORK), INTSIZ(NSIZE), NSIN, NINFI, NSOUT, NINFO, NREAL, LD, J
REAL*8 SIN1(1), SIN2(1), SIN3(1), SIN4(1), SOUT1(1), SOUT2(1), SOUT3(1), SOUT4(1), WORK(NWORK), SIZE(NSIZE), SINFI, SINFO
C C Declare local variables
C INTEGER IMISS, IC, IO2, DMS_KFORMC, DMS_KCCIDC, DMS_KNCIDC, DMS_KNCID
INTEGER IASH, IH2, IN2, IS, ICL2, IH2O, K, DMS_IFCMNC, ICOAL, I
INTEGER LI1, LI2, LI, LIDSCC
REAL*8 REAL(NREAL), RMISS, SPLIT(9)
CHARACTER*8 ASH, C
C C C Begin executable code
C SPLIT(1)=1
SPLIT(2)=1
SPLIT(3)=1
SPLIT(4)=0.5
SPLIT(5)=1
SPLIT(6)=0.5
SPLIT(7)=0.5
SPLIT(8)=0.5
SPLIT(9)=0.5
C Split the first inlet into the first and second outlets
C SOUT1(1) = SIN1(1) * REAL(1)
SOUT1(2) = SIN1(2) * REAL(2)
SOUT1(3) = SIN1(3) * REAL(3)
SOUT2(4) = SIN1(4) * REAL(4)
SOUT3(5) = SIN1(5) * REAL(5)
SOUT4(6) = SIN1(6) * REAL(6)
SOUT1(7) = SIN1(7) * REAL(7)
SOUT1(8) = SIN1(8) * REAL(8)
SOUT1(9) = SIN1(9) * REAL(9)
C SOUT2(I) = SIN1(I) – SOUT1(I)
C Copy all intensive variables
C DO 200 I = NCOMP_NCC+2, NCOMP_NCC+9
SOUT1(I) = SIN1(I)
200 CONTINUE
DO 300 I = NCOMP_NNCC+2, NCOMP_NNCC+9
SOUT1(I) = SIN1(I)
300 CONTINUE
RETURN
END