5.9. Типовой расчет
В данном наборе представлено 30 вариантов задач для студентов. Требования для студентов:
- задачи 1 и 2 необходимо решать геометрическим методом;
- задачи 3 и 4 – простым симплекс – методом;
- задачи 5 и 6 – методом искусственного базиса;
- в задачах 7 и 8 необходимо для заданных задач построить
двойственные им задачи;
- задачи 9 и 10 нужно решать методом обратной матрицы.
Вариант 1
1. f(X)= x1 + x2 max
0 x1 1
0 x2 2
0 x1 + x2 3
- 1 x1 – x2 0.
2. f(X) = x1 + x2 + x3 max
x1 + x2 + 2x3 = 12
x1 + x2 2
x1 + 2x2 3
xj 0, j = 1,2.
f(X) = 6x1 + 2x2 + x3 min
2x1 + x2 + x3 = 4
x1 – 2x2 + x4 =5
xj 0, j = 1,2,3,4.
f(X) = 2x1 – x2 + 3x3 – 2x4 + x5 max
-x1 + x2 + x3 = 1
x1 – x2 + x4 =1
x1 + x2 + x5 =2
xj 0, j = 1,2,3,4,5.
f(X) = -3x1 + x2 + 3x3 – 10x4 max
x1 + 2x2 – x3 + x4 = 0
2x1 – 2x2 + 3x3 + 3x4 = 9
x1 – x2 + 2x3 – x4 = 6
xj
0, j = 1,2,3,4.
f(X) = x2 - x3 min
-2x1 + x2 + x3 – 2x4 = 1
3x1 + x2 + x4 = 3
xj 0, j = 1,2,3,4.
7. f(X)= x1 - 2x2 - x3 max
x1 - 2x2 - x3 3
x1 - x2 1
x1 + 2x2 - 3x3 = 2
xj 0, j = 1, 2.
8. f(X)= 2x1 + 4x2 + 9x3 min
x1
- 3x3
-2
x1 + x2 + 2x3 - 3
xj 0, j = 1, 2, 3.
9. f(X) = 6x1 + 2x2 + x3 min
2x1 + x2 + x3 = 4
x1 – 2x2 + x4 =5
xj 0, j = 1,2,3,4.
f(X) = 2x1 – x2 + 3x3 – 2x4 + x5 max
-x1 + x2 + x3 = 1
x1 – x2 + x4 =1
x1 + x2 + x5 =2
xj 0, j = 1,2,3,4,5.
Вариант 2
f(X)= x1 +x2 + x3 max
x1 + 4x2 + x3 =12
3x1 + x2 6
3x1 + 2x2 2
xj 0, j = 1, 2, 3.
f(X)= x2 – x1 max
x1 + 3x2 12
3x1 – x2 6
3x1 + 4x2 0
xj 0, j = 1, 2.
f(X) = -x1 + x2 max
x1 – x2 + x3 = 1
x1 – 2x2 + x4 = 2
xj 0, j = 1,2,3,4.
f(X) = -x1 + 3x2 + 2x3 min
x1 + x2 + 2x3 -5
2x1 – 3x2 + x3 3
2x1 – 5x2 + 6x3 5
xj 0, j = 1,2,3.
5. f(X) = 2x1 + x2 - x3 min
x1 + 2x2 + x3 = 4
x1 + x2 2
x1 + x2 1
xj 0, j = 1,2,3.
6. f(X) = 3x1 + 7x2 + 6x3 + 5x4 max
4x1 + 3x2 + 2x3 – x4 =7
x1 – 2x2 -5x3 – 3x4 =-12
xj 0, j = 1,2,3,4.
7. f(X)= 2x1 + x2 – x3 min
2x1 – x2 + 3x3 5
x1 - 2x2 = 3
3x1 + x2 - x3 1
xj 0, j = 1,2,3.
8. f(X) = 6x1 + x2 + 4x3 - 5x4 max
3x1 + x2 - x3 + x4 = 4
5x1 + x2 + x3 - x4 = 4
xj 0, j = 1,2,3,4.
9. f(X) = -x1 + x2 max
x1 – x2 + x3 = 1
x1 – 2x2 + x4 = 2
xj 0, j = 1,2,3,4.
10. f(X) = -x1 + 3x2 + 2x3 min
x1 + x2 + 2x3 -5
2x1 – 3x2 + x3 3
2x1 – 5x2 + 6x3 5
xj 0, j = 1,2,3.
Вариант 3
f(X)= x1 – 4x2 min
x1 + 2x2 4
x1 – 2x2 -1
0 x1 3
x2 0.
f(X)= 4x1 + 6x2 max
2x1 + x2 16
x1 + x2 10
0 x1 7
0 x2 6
xj 0, j = 1, 2.
f(X) =2x3 – x4 min
x2 + x4 5
x1 + 2x2 + x3 8
2x3 + x5 10
xj 0, j = 1,2,3,4,5.
f(X) = x4 – x2 min
x1 + x2 – 2x3 = 3
x2 – 2x3 + x4 = 1
xj 0, j = 1,2,3,4.
f(X) = 2x1 + x2 - x3 max
-x1 + 2x2 + x3 = 3
x1 + x2 2
x1 + 3x2 5
xj 0, j = 1, 2,3.
f(X) = 2x1 + 3x2 min
x1 - x2 + x3 2
2x1 - x2 - x3 3
xj 0, j = 1, 2,3.
7. f(X)= 3x1 + x2 max
x1 + 2x2 2
x1 - 2x2 + x3 = 0
x2 + x3 = 3
xj 0, j = 1, 2.
8. f(X) = - x1 + 2x2 - x3 min
x1 + 4x2 + x3 = 5
x1 - 2x2 + x3 = -1
xj 0, j = 1, 2,3.
9. f(X) =2x3 – x4 min
x2 + x4 5
x1 + 2x2 + x3 8
2x3 + x5 10
xj 0, j = 1,2,3,4,5.
10. f(X) = x4 – x2 min
x1 + x2 – 2x3 = 3
x2 – 2x3 + x4 = 1
xj 0, j = 1,2,3,4.