ZADATAK 2.
Resenje:
Funkcija cilja: F(X) = 4 x1 + 12 x2 + x3 pa je sistem
ogranicenja: 2 x1 + x3 24 4 x1 + x2 + x3 36 => - 4 x1 - x2 - x3
- 36 x2 + x3 18 => - x2 - x3 - 18
Primarni matematicki model cini funkcija cilja: minF(X) = 4 x1 +
12 x2 + x3uz sistem ogranicenja : 2 x1 + x3 24 - 4 x1 - x2 - x3 -
36 - x2 - x3 - 18i uslove negativnosti:
x1 0, x2 0, x3 0
Dualni matematicki model ima 3 promenljive y1, y2 i y3 :max(Y) =
24 y1 - 36 y2 - 18 y3a broj ogranicenja duala je 3:2 y1 - 4 y2 4 -
y2 - y3 12 y1 - y2 - y3 1
Kako funkcija cilja ima m = 3 promenljive, a sistem ogranicenja
ima k = 3 nejednacine, dodaju se 3 izravnavajuce promenljive, pa je
matematicki model:
max(Y) max(y1, y2, y3, y4, y5, y6) = 24 y1 - 36 y2 - 18 y3 + 0
y4 + 0 y5 + 0 y6uz sistem ogranicenja: 2 y1 - 4 y2 + y4 = 4 - y2 -
y3 + y5 = 12 y1 - y2 - y3 + y6 = 1
i uslove negativnosti: y1 0, y2 0 , y3 0 , y4 0, y5 0 , y6
0.
0. INTERACIJA0.1. oznake elemenata
tabeleST(0)bjb0b1b2b3b4b5b6
BBYBYB0YB1YB2YB3YB4YB5YB6
b4y4y40a11a21a31a41a51a61
b5y5y50a12a22a32a42a52a62
b6y6y60a13a23a33a43a53a63
j - bj0 b01b12b23b34b45b56b6
ST(0)bjb0b1b2b3b4b5b6
BBYBYB0YB1YB2YB3YB4YB5YB6
b4y4y40a41a42a43a44a45a46
b5y5y50a51a52a53a54a55a56
b6y6y60a61a62a63a64a65a66
j - bj0 b01b12b23b34b45b56b6
0.2. popunjavanje elemenata tabeleST(0)bj024-36-18000
BBYBYB0YB1YB2YB3YB4YB5YB6
0y442-40100
0y5120-1-1010
0y611-1-1001
0j - bj0-243618000
0.3. odredjivanje vodece kolone I vodeceg
redaST(0)bj024-36-18000
BBYBYB0YB1YB2YB3YB4YB5YB6YB0/YB1
0y442-401002
0y5120-1-1010
0y611-1-10011
0j - bj0-243618000
1. INTERACIJA1.1. oznake elemenata
tabeleST(1)bj024-36-18b4b5b6
BBYBYB0YB1YB2YB3YB4YB5YB6
0y4y40a41a42a43a44a45a46
0y5y50a51a52a53a54a55a56
24y1Y10a11a12a13a14a15a16
j - bj0 b01b12b23b34b45b56b6
1.2. popunjavanje elemenata tabeleST(1)bj024-36-18000
BBYBYB0YB1YB2YB3YB4YB5YB6
0y420-2210-2
0y5120-1-1010
24y111-1-1001
0j - bj24012-60024
1.3. odredjivanje vodece kolone I vodeceg reda
ST(1)bj024-36-18000
BBYBYB0YB1YB2YB3YB4YB5YB6YB0/YB3
0y420-2210-21
0y5120-1-1010-12
24y111-1-1001-1
0j - bj24012-60024
2. INTERACIJA2.1. oznake elemenata
tabeleST(2)bj024-36-18b4b5b6
BBYBYB0YB1YB2YB3YB4YB5YB6
-18y3Y30a31a32a33a34a35a36
0y5y50a51a52a53a54a55a56
24y1Y10a11a12a13a14a15a16
j - bj0 b01b12b23b34b45b56b6
2.2. popunjavanje elemenata tabeleST(2)bj024-36-18000
BBYBYB0YB1YB2YB3YB4YB5YB6
-18y310-110,50-1
0y5130-200,51-1
24y121-200,500
0j - bj300603018
OPTIMALNO RESENJEMaksimalna vrednost funkicje cilja (Y)
je:max(Y) = (Y*) = 30za Y* = (y1*, y2*, y3*) = (2, 0, 1)
Minimalna vrednost funkcije cilja F(X) je:minF(X) = F(X*) =
30zaX* = (X1*, X2*, X3*) = (3, 0, 18)
PROVERAZa x1 = 3, x2 = 0 i x3 = 18 funkcije cilja je:F(x) = 4 x1
+ 12 x2 + x3 = 4 3 + 12 0 + 18 = 30
a nejednacina sistema ogranicenja su :
2 x1 + x3 = 2 3 + 18 = 24 244 x1 + x2 + x3 = 4 3 + 0 + 18 = 30
36 x2 + x3 = 0 + 18 = 18 18
