implicit real*8 (a-h,o-z)
parameter (m=5,n=5) i
integer it(0:n),il(0:m),ib(0:n),ir(0:m)
character*2 lt(0:n),ll(0:m),lr(0:m),lb(0:n)
dimension a(0:m+1,0:n+1),asaf(m,n),a1(m,m),a1i(m,m)
dimension a1ia2(m,n),a2(m,n),csaf(n),bsaf(m)
dimension ass(m,n),bi(m),tmp(n),ci(m+n)
common/bprint/ipt,nfile,ndig,npunch,jpt,mfile
common/temp/iterl
open (unit=10,file='lp.inp',status='old')
open (unit=20,file='lp.sto',status='unknown') o
open (unit=8,file='lp.out',status='unknown')
call dflt
read (10,990) ifr
990 format(i1) w
write (nfile,998) ifr
998 format(' ifr = ',i3)
do i=1,m a(i,0)=10.
end do
do j=1,n
a(0,j)=1.
end do
c Read in previously generated data in hex format
do i=1,m
read (20,4030) (a(i,j),j=1,n)
end do
do i=1,m
do j=1,n
ass(i,j)=a(i,j)
end do
end do
do i=1,m
write (nfile,4003) (a(i,j),j=1,n)
4030 format(5z16
end do
4003 format(1x,5f10.3)
c kpiv value will determine which pivot choice algorithm is used
do kpiv=1,2
write (nfile,9000) kpiv
9000 format(' kpiv = ',i1)
a(0,0)=0.
do i=1,m
do j=1,n
a(i,j)=ass(i,j)
end do
end do
do i=1,m
a(i,0)=10.
end do
do j=1,n
a(0,j)=1.
end do
call linproa(a,it,il,ir,ib,lt,ll,lr,lb,m,n,iter,ier,kpiv,
1asaf,a1,a1i,a2,csaf,bsaf,ci,bi,tmp,a1ia2,ifr)
write (*,1002) iter
write (nfile,1002) iter
1002 format(' Iteration count = ',i4)
end do
stop
end
****************************************************************
FILE LP.INP FOLLOWS
****************************************************************
0
****************************************************************
FILE LP.STO FOLLOWS
****************************************************************
BFDB87F487766000BFDE0919424C2A00BFCD92A48FD99900BFD6281AAAEBE500BFEECF44A6531A40
BFC411711741F200BFEAA6B367B1D400BFE26CF688891C80BFBC8DDDE1B29600BFE6255204E03940
BFD2F35A8C86F500BFDEBE3B58D9E080BFEFD1C1DA12DEC0BFEBED968A6406C0BFD2A8B63EA14100
BFD403B885512FC0BFD9B37235292E80BFEAC7505FC402C0BFC3B3F563B79400BFCBD9ED17EB1C00
BFDF80540D681040BFD771182C851C80BFE7AD4ABC413E80BFEC3B280A4F8140BFEE8454FC506480
****************************************************************
OUTPUT FILE FOLLOWS
****************************************************************
ifr = 0
-0.430 -0.469 -0.231 -0.346 -0.963
-0.157 -0.833 -0.576 -0.112 -0.692
-0.296 -0.480 -0.994 -0.873 -0.292
-0.313 -0.402 -0.837 -0.154 -0.218
-0.492 -0.366 -0.740 -0.882 -0.954
kpiv = 1
LP BEGINS
Pivot choice method = steepest ascent
Optimum reached
Objective function = 0.22546566E+02
Primal solution
x- 2 0.93598824E+01
x- 3 0.32740965E+00
u- 3 0.13706462E+01
u- 4 0.19458517E+01
x- 1 0.12859274E+02
Dual solution
y- 5 0.88070292E+00
y- 1 0.12842416E+01
y- 2 0.89712165E-01
v- 4 0.23158285E+00
v- 5 0.11384409E+01
Iteration count = 3
kpiv = 2
LP BEGINS
Pivot choice method = largest improvement at each iteration
Optimum reached
Objective function = 0.22546566E+02
Primal solution
x- 2 0.93598824E+01
x- 3 0.32740965E+00
u- 3 0.13706462E+01
u- 4 0.19458517E+01
x- 1 0.12859274E+02
Dual solution
y- 5 0.88070292E+00
y- 1 0.12842416E+01
y- 2 0.89712165E-01
v- 4 0.23158285E+00
v- 5 0.11384409E+01
Iteration count = 3