Автоматизированный логико-вероятностный расчет надежности технических систем. Владов Ю.Р. - 27 стр.

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27
F
Л
=
1
[
2
\/ 2 ⋅
5
⋅
6
] \/ 1 ⋅
3
⋅
4
(
2
\/ 2 ⋅
5
⋅
6
) = (
2
\/ 2 ⋅
5
⋅
6
) (
1
\/ 1 ⋅
3
⋅
4
). (3.11)
Арифметизируем логическую функцию работоспособности:
F
ар
= (
2
+ 2 ⋅
5
⋅
6
) (
1
+ 1 ⋅
3
⋅
4
). (3.12)
Заменяя события их вероятностями, имеем:
Р = (
2
P+ Р
2
⋅
5
P ⋅
6
P) (
1
P + Р
1
⋅
3
P ⋅
4
P ). (3.13)
3.2.2.4 Сетевая структура с пятью каналами (SET.5)
Логическая функция работоспособности структуры запишется:
F
Л
= 1 ⋅ 2 \/ 4 ⋅ 3 \/ 5. (3.14)
Арифметезируя Fл, получаем:
F
ар
= 1 ⋅ 2 + 4 ⋅ 3 + 5 - 1 ⋅ 2 ⋅ 3 ⋅ 4 - 1 ⋅ 2 ⋅ 5 - 3 ⋅ 4 ⋅ 5 + 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 (3.15)
Заменяя события их вероятностями, получим
Р = (Р
1
⋅ Р
2
+ Р
4
⋅ Р
3
⋅ Р
5
) - (Р
1
⋅ Р
2
⋅ Р
3
⋅ Р
4
+ Р
1
⋅ Р
2
⋅ Р
5
+ Р
3
⋅ Р
4
⋅ Р
5
) +
+ Р
1
⋅ Р
2
⋅ Р
3
⋅ Р
4
⋅ Р
5
. (3.16)
3.2.2.5 Сетевая структура 6 канальная (SET.6)
Логическая функция работоспособности данной структуры запишется:
F
Л
= 1 ⋅ 2 \/ 5 \/ 4 ⋅ 3 \/ 1 ⋅ 6 ⋅ 3 \/ 4 ⋅ 6 ⋅ 2. (3.17)
Упростим Fл:
F
Л
=1 ⋅ 2 \/ 5 \/⋅3 (4 \/ 1 ⋅ 6) \/ 4 ⋅ 6 ⋅ 2 = 2 (1 \/ 4 ⋅ 6) \/ 5 \/ 3 (4 \/ 1 ⋅ 6) (3.18)
Арифметизируем Fл:
F
ap
= 2 (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) + 5 + 3 (4 + 1 ⋅ 6 - 4 ⋅ 1 ⋅ 6) – 2 (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) ⋅ 5 -
- 5 ⋅ 3 (4 + 1 ⋅ 6 – 4 ⋅ 1 ⋅ 6) – 3 (4 + 1 ⋅ 6 – 4 ⋅ 1 ⋅ 6) ⋅ 2 ⋅ (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) +
+ 2 (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) ⋅ 5 ⋅ 3 (4 + 1 ⋅ 6 – 4 ⋅ 1 ⋅ 6). (3.19) 
    FЛ = 1 [ 2 \/ 2 ⋅ 5 ⋅ 6 ] \/ 1 ⋅ 3 ⋅ 4 ( 2 \/ 2 ⋅ 5 ⋅ 6 ) = ( 2 \/ 2 ⋅ 5 ⋅ 6 ) ( 1 \/ 1 ⋅ 3 ⋅ 4 ).   (3.11)

     Арифметизируем логическую функцию работоспособности:

                                 Fар = ( 2 + 2 ⋅ 5 ⋅ 6 ) ( 1 + 1 ⋅ 3 ⋅ 4 ).                              (3.12)

     Заменяя события их вероятностями, имеем:

                             Р = ( P2 + Р2 ⋅ P5 ⋅ P6 ) ( P1 + Р1 ⋅ P3 ⋅ P4 ).                            (3.13)


     3.2.2.4 Сетевая структура с пятью каналами (SET.5)

     Логическая функция работоспособности структуры запишется:

                                         FЛ = 1 ⋅ 2 \/ 4 ⋅ 3 \/ 5.                                       (3.14)

     Арифметезируя Fл, получаем:

         Fар = 1 ⋅ 2 + 4 ⋅ 3 + 5 - 1 ⋅ 2 ⋅ 3 ⋅ 4 - 1 ⋅ 2 ⋅ 5 - 3 ⋅ 4 ⋅ 5 + 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5             (3.15)

     Заменяя события их вероятностями, получим

          Р = (Р1 ⋅ Р2 + Р4 ⋅ Р3 ⋅ Р5) - (Р1 ⋅ Р2 ⋅ Р3 ⋅ Р4 + Р1 ⋅ Р2 ⋅ Р5 + Р3 ⋅ Р4 ⋅ Р5) +
                + Р1 ⋅ Р2 ⋅ Р3 ⋅ Р4 ⋅ Р5.                                                                (3.16)


     3.2.2.5 Сетевая структура 6 канальная (SET.6)

     Логическая функция работоспособности данной структуры запишется:

                          FЛ = 1 ⋅ 2 \/ 5 \/ 4 ⋅ 3 \/ 1 ⋅ 6 ⋅ 3 \/ 4 ⋅ 6 ⋅ 2.                            (3.17)

     Упростим Fл:

          FЛ =1 ⋅ 2 \/ 5 \/⋅3 (4 \/ 1 ⋅ 6) \/ 4 ⋅ 6 ⋅ 2 = 2 (1 \/ 4 ⋅ 6) \/ 5 \/ 3 (4 \/ 1 ⋅ 6)          (3.18)


     Арифметизируем Fл:

Fap = 2 (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) + 5 + 3 (4 + 1 ⋅ 6 - 4 ⋅ 1 ⋅ 6) – 2 (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) ⋅ 5 -
     - 5 ⋅ 3 (4 + 1 ⋅ 6 – 4 ⋅ 1 ⋅ 6) – 3 (4 + 1 ⋅ 6 – 4 ⋅ 1 ⋅ 6) ⋅ 2 ⋅ (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) +
     + 2 (1 + 4 ⋅ 6 – 1 ⋅ 4 ⋅ 6) ⋅ 5 ⋅ 3 (4 + 1 ⋅ 6 – 4 ⋅ 1 ⋅ 6).                              (3.19)


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