2. Алгоритм решения м-метода.
Задача 2.1
Решение.
М-задача
|
|
|
2 |
-3 |
6 |
1 |
0 |
0 |
-M |
|
Сx |
Бx |
|
|
|
|
|
|
|
|
|
1 |
|
24 |
2 |
1 |
2 |
1 |
0 |
0 |
0 |
24/2=12 |
0 |
|
22 |
1 |
2 |
4 |
0 |
1 |
0 |
0 |
22/4=5.5 |
-M |
|
10 |
1 |
-1 |
2 |
0 |
0 |
-1 |
1 |
10/2=5 |
|
F= |
-10M |
-M |
M |
-2M |
0 |
0 |
M |
0 |
|
|
|
+24 |
|
+4 |
-4 |
|
|
|
|
|
|
|
|
2 |
-3 |
6 |
1 |
0 |
0 |
|
Сx |
Бx |
|
|
|
|
|
|
|
|
1 |
|
14 |
3 |
0 |
0 |
1 |
0 |
-1 |
- |
0 |
|
2 |
-1 |
4 |
0 |
0 |
1 |
2 |
1 |
6 |
|
5 |
½ |
-1/2 |
1 |
0 |
0 |
-1/2 |
- |
|
F= |
44 |
4 |
0 |
0 |
0 |
0 |
-4 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2 |
-3 |
6 |
1 |
0 |
0 |
Сx |
Бx |
|
|
|
|
|
|
|
1 |
|
15 |
5/2 |
2 |
0 |
1 |
½ |
0 |
0 |
|
1 |
-1/2 |
2 |
0 |
0 |
½ |
1 |
6 |
|
11/2 |
¼ |
1/8 |
1 |
0 |
¼ |
0 |
|
F= |
68 |
2 |
8 |
0 |
0 |
2 |
0 |
|
|
|
|
|
|
|
|
|
