' VJEZBA 5 LINEARNI KINETI^KI MODELI
' MODEL 3
DECLARE SUB gausl (n, ns, a())
DIM a(3, 4)
CLS
PRINT "MODEL 3"
n = 3: ns = 1
k1 = .15: km1 = .05
k2 = .22: km2 = .18
k3 = .28: km3 = .19
D = .01
A0 = 2
a(1, 1) = -(D + k1 + k3): a(1, 2) = km1: a(1, 3) = km3: a(1, 4) = -D * A0
a(2, 1) = k1: a(2, 2) = -(D + km1 + k2): a(2, 3) = km2: a(2, 4) = 0
a(3, 1) = k3: a(3, 2) = k2: a(3, 3) = -(D + km3 + km2): a(3, 4) = 0
CALL gausl(n, ns, a())
PRINT "Koncentracije komponenata A, B i C izra~unate GAUSSOVOM"
PRINT "eliminacijom iz dinami~ke bilanse za MEHANIZAM 2."
PRINT
PRINT "a="; a(1, 4); "(g/l)"
PRINT "b="; a(2, 4); "(g/l)"
PRINT "c="; a(3, 4); "(g/l)"
selek3 = a(2, 4) / a(3, 4)
PRINT
PRINT "Brzina razredjenja ="; D; "(1/h)"
PRINT "SELEKTIVNOST 3 ="; selek3
END

SUB gausl (n, ns, a())
'Gaussova eliminacija za rjesavanje linearnog
'sistema jednadzbi  A(n,n)X(n,ns)=B(n,ns)
'rjesenja su stupci (n+1) do (n+ns) matrice A
'pocetni elementi matrice A se nakon pozivanja GAUSL gube
n1 = n + 1
nt = n + ns
IF i = 1 THEN GOTO 5650
FOR i = 2 TO n
ip = i - 1
i1 = ip
x = ABS(a(i1, i1))
FOR j = i TO n
IF ABS(a(j, i1)) < x THEN GOTO 5520
x = ABS(a(j, i1))
ip = j
5520 NEXT j
IF ip = i1 THEN GOTO 5590
FOR j = i1 TO nt
x = a(i1, j)
a(i1, j) = a(ip, j)
a(ip, j) = x
NEXT j
5590 FOR j = i TO n
x = a(j, i1) / a(i1, i1)
FOR k = i TO nt
a(j, k) = a(j, k) - x * a(i1, k)
NEXT k
NEXT j
NEXT i
5650 FOR l = 1 TO n
i = n1 - l
FOR k = n1 TO nt
a(i, k) = a(i, k) / a(i, i)
IF i = 1 THEN GOTO 5730
i1 = i - 1
FOR j = 1 TO i1
a(j, k) = a(j, k) - a(i, k) * a(j, i)
5730 NEXT j
NEXT k
NEXT l
END SUB