Имитационное моделирование сложных систем. Духанов А.В - 59 стр.

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с3 = с6 + 7.
0100111010
000010000003
100000000008
010000000002
101000000006
100000000005
100000000007
100001000001
010001000102
101000100004
000100011001
3826571241
c
c
t
c
c
c
c
c
c
t
cctcccccct
Очередь:
10
.
Десятый столбец (с3):
c2 = c5 c4 + c3,
t2 = c3 * c3.
0101111010
100010000003
100000000008
010000000002
101000000006
100000000005
100000000007
100001000001
010001000102
101000100004
000100011001
3826571241
c
c
t
c
c
c
c
c
c
t
cctcccccct
  с3 = с6 + 7.
      t1 c 4      c 2 c1 c7 c5 c6 t 2 c8 c3
  t1 0 0           1 1 0 0 0 1 0 0            0
  c4 0 0           0 0 1 0 0 0 1 0            1
  c2 0 1           0 0 0 1 0 0 0 1            0
  c1 0 0           0 0 0 1 0 0 0 0            1
  c7 0 0           0 0 0 0 0 0 0 0            1
  c5 0 0           0 0 0 0 0 0 0 0            1
  c6 0 0           0 0 0 0 0 0 1 0            1
  t2 0 0           0 0 0 0 0 0 0 1            0
  c8 0 0           0 0 0 0 0 0 0 0            1
  c3 0 0           0 0 0 0 1 0 0 0            0
      0 1          0 1 1 1 0 0 1 0
Очередь: 10 .
Десятый столбец (с3):
  c2 = c5 – c4 + c3,
  t2 = c3 * c3.
        t1 c 4 c 2 c1 c7 c5 c6 t 2 c8 c3
   t1   0 0 1 1 0 0 0 1 0 0                       0
   c4   0 0 0 0 1 0 0 0 1 0                       1
   c2   0 1 0 0 0 1 0 0 0 1                       0
   c1   0 0 0 0 0 1 0 0 0 0                       1
   c7   0 0 0 0 0 0 0 0 0 0                       1
   c5   0 0 0 0 0 0 0 0 0 0                       1
   c6   0 0 0 0 0 0 0 0 1 0                       1
   t2   0 0 0 0 0 0 0 0 0 1                       0
   c8   0 0 0 0 0 0 0 0 0 0                       1
   c3   0 0 0 0 0 0 1 0 0 0                       1
        0 1 0 1 1 1 1 0 1 0

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