2) Найдем оптимальное целочисленное решение данной задачи методом ветвей и границ:

Полученный оптимальный базисный план для задачи ЛП-1 имеет вид:

Базис	B	1	2	3	4	5	6	7
2	5 2/5	0	1	0	0	1/5	0	3/5
1	4 2/5	1	0	0	0	1/5	0	- 2/5
3	3	0	0	1	0	0	0	-1
6	4	0	0	0	0	0	1	1
4	3/5	0	0	0	1	- 1/5	0	- 3/5
f(x)	59 4/5	0	0	0	0	2 2/5	0	2 1/5

ЗЛП-2	х1<=4	4
	- 2/5	0	0	0	0	- 1/5	0	2/5
Базис	B	1	2	3	4	5	6	7	8
2	5 2/5	0	1	0	0	1/5	0	3/5	0
1	4 2/5	1	0	0	0	1/5	0	- 2/5	0
3	3	0	0	1	0	0	0	-1	0
6	4	0	0	0	0	0	1	1	0
4	3/5	0	0	0	1	- 1/5	0	- 3/5	0
8	- 2/5	0	0	0	0	- 1/5	0	2/5	1
f(x)	59 4/5	0	0	0	0	2 2/5	0	2 1/5	0
Базис	B	1	2	3	4	5	6	7	8
2	5	0	1	0	0	0	0	1	1
1	4	1	0	0	0	0	0	0	1
3	3	0	0	1	0	0	0	-1	0
6	4	0	0	0	0	0	1	1	0
4	1	0	0	0	1	0	0	-1	-1
5	2	0	0	0	0	1	0	-2	-5
f(x)	55	0	0	0	0	0	0	7	12

ЗЛП-3	х1>=5
	- 3/5	0	0	0	0	1/5	0	- 2/5
Базис	B	1	2	3	4	5	6	7	8
2	5 2/5	0	1	0	0	1/5	0	3/5	0
1	4 2/5	1	0	0	0	1/5	0	- 2/5	0
3	3	0	0	1	0	0	0	-1	0
6	4	0	0	0	0	0	1	1	0
4	3/5	0	0	0	1	- 1/5	0	- 3/5	0
8	- 3/5	0	0	0	0	1/5	0	- 2/5	1
f(x)	59 4/5	0	0	0	0	2 2/5	0	2 1/5	0
Базис	B	1	2	3	4	5	6	7	8
2	4 1/2	0	1	0	0	1/2	0	0	1 1/2
1	5	1	0	0	0	0	0	0	-1
3	4 1/2	0	0	1	0	- 1/2	0	0	-2 1/2
6	2 1/2	0	0	0	0	1/2	1	0	2 1/2
4	1 1/2	0	0	0	1	- 1/2	0	0	-1 1/2
7	1 1/2	0	0	0	0	- 1/2	0	1	-2 1/2
f(x)	56 1/2	0	0	0	0	3 1/2	0	0	5 1/2

ЗЛП-4	х2<=4
	- 1/2	0	0	0	0	- 1/2	0	0	-1 1/2
Базис	B	1	2	3	4	5	6	7	8	9
2	4 1/2	0	1	0	0	1/2	0	0	1 1/2	0
1	5	1	0	0	0	0	0	0	-1	0
3	4 1/2	0	0	1	0	- 1/2	0	0	-2 1/2	0
6	2 1/2	0	0	0	0	1/2	1	0	2 1/2	0
4	1 1/2	0	0	0	1	- 1/2	0	0	-1 1/2	0
7	1 1/2	0	0	0	0	- 1/2	0	1	-2 1/2	0
9	- 1/2	0	0	0	0	- 1/2	0	0	-1 1/2	1
f(x)	56 1/2	0	0	0	0	3 1/2	0	0	5 1/2	0
Базис	B	1	2	3	4	5	6	7	8	9
2	4	0	1	0	0	0	0	0	0	1
1	5 1/3	1	0	0	0	1/3	0	0	0	- 2/3
3	5 1/3	0	0	1	0	1/3	0	0	0	-1 2/3
6	1 2/3	0	0	0	0	- 1/3	1	0	0	1 2/3
4	2	0	0	0	1	0	0	0	0	-1
7	2 1/3	0	0	0	0	1/3	0	1	0	-1 2/3
8	0	0	0	0	0	0	0	0	0	0
f(x)	54 2/3	0	0	0	0	1 2/3	0	0	0	3 2/3

ЗЛП-5	х1<5
	- 1/3	0	0	0	0	- 1/3	0	0	0	2/3
Базис	B	1	2	3	4	5	6	7	8	9	10
2	4	0	1	0	0	0	0	0	0	1	0
1	5 1/3	1	0	0	0	1/3	0	0	0	- 2/3	0
3	5 1/3	0	0	1	0	1/3	0	0	0	-1 2/3	0
6	1 2/3	0	0	0	0	- 1/3	1	0	0	1 2/3	0
4	2	0	0	0	1	0	0	0	0	-1	0
7	2 1/3	0	0	0	0	1/3	0	1	0	-1 2/3	0
8	0	0	0	0	0	0	0	0	0	0	0
10	- 1/3	0	0	0	0	- 1/3	0	0	0	2/3	1
f(x)	54 2/3	0	0	0	0	1 2/3	0	0	0	3 2/3	0
Базис	B	1	2	3	4	5	6	7	8	9	10
2	4	0	1	0	0	0	0	0	0	1	0
1	5	1	0	0	0	0	0	0	0	0	1
3	5	0	0	1	0	0	0	0	0	-1	1
6	2	0	0	0	0	0	1	0	0	1	-1
4	2	0	0	0	1	0	0	0	0	-1	0
7	2	0	0	0	0	0	0	1	0	-1	1
8	0	0	0	0	0	0	0	0	0	0	0
5	1	0	0	0	0	1	0	0	0	-2	-3
f(x)	53	0	0	0	0	0	0	0	0	7	5

ЗЛП-6	х1>=6
	- 2/3	0	0	0	0	1/3	0	0	0	- 2/3
Базис	B	1	2	3	4	5	6	7	8	9	10
2	4	0	1	0	0	0	0	0	0	1	0
1	5 1/3	1	0	0	0	1/3	0	0	0	- 2/3	0
3	5 1/3	0	0	1	0	1/3	0	0	0	-1 2/3	0
6	1 2/3	0	0	0	0	- 1/3	1	0	0	1 2/3	0
4	2	0	0	0	1	0	0	0	0	-1	0
7	2 1/3	0	0	0	0	1/3	0	1	0	-1 2/3	0
8	0	0	0	0	0	0	0	0	0	0	0
10	- 2/3	0	0	0	0	1/3	0	0	0	- 2/3	1
f(x)	54 2/3	0	0	0	0	1 2/3	0	0	0	3 2/3	0
Базис	B	1	2	3	4	5	6	7	8	9	10
2	3	0	1	0	0	1/2	0	0	0	0	1 1/2
1	6	1	0	0	0	0	0	0	0	0	-1
3	7	0	0	1	0	- 1/2	0	0	0	0	-2 1/2
6	0	0	0	0	0	1/2	1	0	0	0	2 1/2
4	3	0	0	0	1	- 1/2	0	0	0	0	-1 1/2
7	4	0	0	0	0	- 1/2	0	1	0	0	-2 1/2
8	0	0	0	0	0	0	0	0	0	0	0
9	1	0	0	0	0	- 1/2	0	0	0	1	-1 1/2
f(x)	51	0	0	0	0	3 1/2	0	0	0	0	5 1/2

ЗЛП-7	х2>4
	- 1/2	0	0	0	0	1/2	0	0	1 1/2
Базис	B	1	2	3	4	5	6	7	8	9
2	4 1/2	0	1	0	0	1/2	0	0	1 1/2	0
1	5	1	0	0	0	0	0	0	-1	0
3	4 1/2	0	0	1	0	- 1/2	0	0	-2 1/2	0
6	2 1/2	0	0	0	0	1/2	1	0	2 1/2	0
4	1 1/2	0	0	0	1	- 1/2	0	0	-1 1/2	0
7	1 1/2	0	0	0	0	- 1/2	0	1	-2 1/2	0
9	- 1/2	0	0	0	0	1/2	0	0	1 1/2	1
f(x)	56 1/2	0	0	0	0	3 1/2	0	0	5 1/2	0

Таким образом, оптимальный вариант получен при решении ЗЛП-2:

Базис	B	1	2	3	4	5	6	7	8
2	5	0	1	0	0	0	0	1	1
1	4	1	0	0	0	0	0	0	1
3	3	0	0	1	0	0	0	-1	0
6	4	0	0	0	0	0	1	1	0
4	1	0	0	0	1	0	0	-1	-1
5	2	0	0	0	0	1	0	-2	-5
f(x)	55	0	0	0	0	0	0	7	12

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