Реферат: Оптимизация показателей

                                                                                                                                                      М

                                                                                     4

                (2)                                                                                                    

                                                                                                                                                                   6

                                                                                                                                                                

-9

                                                                                                         (3)

(1)

-4

                                                                                           10

                   

                                                                                                                         В                          М

                                                                                     4

                                             

                                    ( I )

-38/3

                (2)                                                                                                    

                                                                                                                                                                   6

                                                                                                                                                                

-9

                                                                                                         (3)

(1)

-4

В точці В, що є оптимальною за даних умов, перетикаються (I) відсічення та (1) обмеження. Знайдемо координати т.В

-3х1 + 9х2 = 38          х1=26/9

                                                                         т.В (26/9; 140/27)

10х1+ 6х2 = 60           х2=140/27                      F ( B) = 45 5/9

-1/12х3 – 17/18х6 = -8/9 – второе отсечение.

-1/12х3*(60 – 10х1- 6х2) – 17/18*(38 + 3х1 – 9х2) = -8/9

-2х1 + 9х2 = 40 – уравнение 2-го отсечения.

Х7= 40 + 2х1 - 92

                                                                                           10

                   

                                                                                                                         В                          М

                                                                                                                              С

                                                                                     4

                                             

                                   

-38/3

( II )                            (I)

                (2)                                                                                                     

                                                                                                                                                                   6

                                                                                                                                                               

-9

                                                                                                                           2 16/17

-20 (II)                                                                                            (3)

(1)

-4

                                                                                           10

                   

                                                                                                                         В                          М

                                                                                                                              С

                                                                                                                         D

                                                                                     4

(III)                                     

                                   

( II )                            (I)

                (2)                                                                                                    

                                                                                                                                                                   6

                                                                                                                                                                

-9

                                                                                                                           2 16/17

-20 (II)                                                                                            (3)

(1)

-4

Уравнение третьего отсечения:

-3/34х3 – 16/17х7 = -16/17

х7 находится из 2 го ограничения

-3/34 * ( 60 – 10х1 – 6х2) – 16/17*(40 + 2х1 – 9х2) = -16/17

-х1 + 9х2 = 42 – ур. Третьего отсечения

В т. D пересекаются (1) и (III)

10х1 + 6х2 = 60

-х1 + 9х2 = 42

х1=3; х2=5. F(D)=45

т.D (3;5)

Вывод:

 экономико-матем. модел. испольузется в экономике для решения различного рода заданий, для оптимизации их. В данной к.р. использованы симплекс метод,….. отсечения Гомори, двойной симплекс метод. Геометрическая интерпретация показывает весь ход решения.

Список використаної літератури:

1.   Кузнецов Ю.Н. “Математическое програмирование:(учебное пособие для экономических специальностей ”

2.   Оптимізація єкономічних показників з врахуванням умови цілочисленності: “Методичні вказівки до виконання курсової роботи з дисципліни “Економіко математичне моделювання для студентів економічних спеціальностей”(Викладач Іванов Л.П. –Чернігів: ЧТІ,1998-20с)”