Продолжение табл. 3
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Вариант 5 |
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Вариант 15 |
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C1 |
C2 |
C3 |
C4 |
C5 |
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C1 |
C2 |
C3 |
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C4 |
C5 |
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Q1 |
0 |
1 |
0 |
1 |
0 |
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Q1 |
0 |
1 |
0 |
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1 |
0 |
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Q2 |
1 |
0 |
1 |
0 |
1 |
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Q2 |
1 |
0 |
1 |
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0 |
1 |
Q3 |
1 |
0 |
0 |
1 |
0 |
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Q3 |
1 |
0 |
1 |
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0 |
0 |
Q4 |
0 |
1 |
0 |
1 |
0 |
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Q4 |
0 |
1 |
0 |
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1 |
0 |
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Вариант 6 |
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Вариант 16 |
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C1 |
C2 |
C3 |
C4 |
C5 |
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C1 |
C2 |
C3 |
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C4 |
C5 |
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Q1 |
0 |
1 |
0 |
0 |
1 |
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Q1 |
0 |
1 |
0 |
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0 |
1 |
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Q2 |
1 |
0 |
1 |
1 |
0 |
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Q2 |
1 |
0 |
1 |
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1 |
0 |
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Q3 |
0 |
1 |
0 |
0 |
1 |
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Q3 |
1 |
0 |
0 |
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0 |
1 |
Q4 |
0 |
1 |
0 |
0 |
1 |
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Q4 |
0 |
1 |
0 |
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0 |
1 |
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Вариант 7 |
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Вариант 17 |
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C1 |
C2 |
C3 |
C4 |
C5 |
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C1 |
C2 |
C3 |
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C4 |
C5 |
Q1 |
0 |
1 |
0 |
1 |
0 |
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Q1 |
1 |
1 |
0 |
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1 |
0 |
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Q2 |
0 |
0 |
1 |
0 |
1 |
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Q2 |
0 |
0 |
1 |
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0 |
1 |
Q3 |
1 |
0 |
0 |
1 |
1 |
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Q3 |
1 |
0 |
0 |
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0 |
1 |
Q4 |
0 |
1 |
0 |
1 |
0 |
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Q4 |
0 |
1 |
0 |
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1 |
0 |
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Вариант 8 |
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Вариант 18 |
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C1 |
C2 |
C3 |
C4 |
C5 |
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C1 |
C2 |
C3 |
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C4 |
C5 |
Q1 |
0 |
1 |
0 |
0 |
1 |
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Q1 |
0 |
1 |
0 |
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1 |
0 |
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Q2 |
1 |
0 |
1 |
0 |
0 |
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Q2 |
0 |
0 |
1 |
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0 |
1 |
Q3 |
1 |
0 |
0 |
1 |
1 |
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Q3 |
1 |
0 |
0 |
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1 |
0 |
Q4 |
0 |
1 |
0 |
0 |
1 |
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Q4 |
0 |
1 |
0 |
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1 |
1 |
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Вариант 9 |
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Вариант 19 |
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C1 |
C2 |
C3 |
C4 |
C5 |
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C1 |
C2 |
C3 |
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C4 |
C5 |
Q1 |
1 |
0 |
0 |
1 |
0 |
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Q1 |
0 |
1 |
0 |
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1 |
0 |
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Q2 |
1 |
0 |
1 |
0 |
1 |
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Q2 |
1 |
0 |
1 |
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0 |
1 |
Q3 |
1 |
0 |
0 |
1 |
0 |
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Q3 |
1 |
0 |
0 |
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1 |
0 |
Q4 |
0 |
1 |
0 |
1 |
0 |
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Q4 |
0 |
1 |
0 |
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1 |
0 |
11
Окончание табл. 3
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Вариант 10 |
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Вариант 20 |
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C1 |
C2 |
C3 |
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C4 |
C5 |
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C1 |
C2 |
C3 |
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C4 |
C5 |
Q1 |
1 |
1 |
0 |
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1 |
0 |
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Q1 |
0 |
1 |
0 |
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1 |
0 |
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Q2 |
1 |
0 |
0 |
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0 |
1 |
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Q2 |
1 |
0 |
1 |
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0 |
1 |
Q3 |
1 |
0 |
1 |
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0 |
0 |
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Q3 |
1 |
0 |
0 |
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0 |
1 |
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Q4 |
0 |
1 |
0 |
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1 |
0 |
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Q4 |
0 |
1 |
0 |
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1 |
0 |
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Задача №4
РЕШЕНИЕ ЗАДАЧ ЛИНЕЙНОГО ПРОГРАММИРОВАНИЯ СИМПЛЕКС-МЕТОДОМ
Задание. Найти максимум целевой функции при заданных ограничениях.
№1.
№3.
№5.
№7.
Z(x) x1 |
x2 |
x3 |
x4 2x5 max, |
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x 2x |
2x |
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6, |
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1 |
2 |
3 |
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x1 |
2x2 |
x3 |
x4 24, |
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2x x |
4x |
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x 30. |
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1 2 |
3 |
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5 |
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xi 0, i
Z(x) x1 x2 x3 x5 max,
x1 2x2 x3 x5 11,
2x1 1x2 x3 8,x1 x2 x3 x4 20. xi 0, i
Z(x) x1 2x2 x3 x4 x5 max,
2x1 x2 x3 3,x1 1x2 x3 x5 23,
x1 x2 x3 x4 18. xi 0, i
Z(x) x1 x2 2x3 x4 x5 max,
x1 x2 x3 10,
2x1 2x2 x3 x5 24,x1 x2 x3 x4 32. xi 0, i
Z(x) x1 2x2 x3 x4 max,
x1 x2 2x3 x4 10,
№2. x1 2x2 x3 14,2x1 x2 4x3 x5 12.
xi 0, i
Z(x) 2x1 x2 x3 x4 x5 max,
x1 2x2 x3 x5 9,
№4. x1 1x2 x3 2,x1 x2 x3 x4 16.
xi 0, i
Z(x) x1 2x3 x4 x5 max,
2x1 x2 x3 5, №6. x1 1x2 x3 x5 17,
x1 x2 x3 x4 26. xi 0, i
Z(x) x1 2x2 2x3 x4 x5 max,
x1 x2 x3 11,
№8. 2x1 2x2 x3 x5 31,x1 x2 x3 x4 40.
xi 0, i
12
Z(x) x1 2x2 2x3 x4 x5 max,
x1 x2 x3 5,
№9. 2x1 1x2 x3 x5 24,2x1 x2 x3 x4 38. xi 0, i
Z(x) x1 x2 x3 x4 x5 max,
x1 x2 x3 16,
№11. 2x1 x2 x3 x5 42,x1 x2 x3 x4 36. xi 0, i
Z(x) x1 x2 2x3 x4 x5 max,
x1 x2 x3 7,
№13. 2x1 x2 x3 x5 23,x1 x2 x3 x4 37.
xi 0, i
Z(x) 2x1 x2 2x3 x4 x5 max,
x1 x2 x3 6,
№15. x1 x2 x3 x5 18,x1 x2 x3 x4 34.
xi 0, i
Z(x) 2x1 x2 x3 x4 x5 max,
2x1 x2 x3 4,
№17. x1 x2 x3 x5 26,x1 x2 x3 x4 32.
xi 0, i
Z(x) 2x1 x2 x3 x4 x5 max,
2x1 x2 x3 4,
№19. x1 x2 x3 x5 26,x1 x2 x3 x4 32. xi 0, i
Z(x) x1 x2 x3 x4 x5 max,
x1 x2 x3 7,
№10. 2x1 1x2 x3 x5 34,x1 x2 x3 x4 21.
xi 0, i
Z(x) x1 x2 x3 x4 x5 max,
x1 x2 x3 9,
№12. 2x1 x2 x3 x5 29,x1 x2 x3 x4 39. xi 0, i
Z(x) 2x1 x2 2x3 x4 x5 max,
x1 x2 x3 6,
№14. x2 x3 x5 19,
x1 x2 x3 x4 33. xi 0, i
Z(x) 2x1 x2 2x3 x4 x5 max,
2x1 x2 x3 3, №16. x1 x2 x3 x5 23,
x1 x2 x3 x4 31. xi 0, i
Z(x) x2 x3 x4 x5 max,
2x1 x2 x3 1,
№18. x1 x2 x3 x5 14,x1 x2 x3 x4 27. xi 0, i
Z(x) x1 x2 x3 x4 x5 max,
x1 x2 x3 5, №20. 2x1 x2 x3 x5 17,
x1 x2 x3 x4 31. xi 0, i
13
Задача №5
ПОИСК МИНИМАЛЬНОГО ТЕХНОЛОГИЧЕСКОГО МАРШРУТА
Задание. На одном станке можно обрабатывать 5 видов деталей, затрачивая при этом одинаковое время на обработку одной детали каждого вида. В таблицах вариантов задана длительность переналадки станка для обработки j– й детали после i–й детали, i 1,5, j 1,5. Требуется найти последовательность обработки деталей, имеющую минимальную суммарную длительность переналадки.
Вариант 1
Вариант 2
Вариант 3
Вариант 4
14
Вариант 5
Вариант 6
Вариант 7
Вариант 8
Вариант 9
15