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Bajarish: Avval simpleks jadvalini tuzib olamiz

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Algoritm va loyihalash

4.Quyidagi va 4.1 laboratoriya topshiriqlarida berilgan masalalarni simpleks usulda yechish algoritmini va dasturini keltiring. Misollar: 1.

Bajarish:
Avval simpleks jadvalini tuzib olamiz:

Erkli o’	x	y	a	b	c	Z
a	100	60	1	0	0	25000
b	40	75	0	1	0	7000
c	1440	1680	0	0	1	300000
F	-2500	-2775	0	0	0	0

1-qadam:
25000/60=416,67;

7000/75=93,3;
300000/1680=178,57

Erkli o’	x	y	a	b	c	Z
a	100	60	1	0	0	25000
b	40/75	1	0	1/75	0	7000
c	1440	1680	0	0	1	300000
F	-2500	-2775	0	0	0	0

Erkli o’	x	y	a	b	c	Z
a	68	0	1	-60/75	0	19400
b	8/15	1	0	1/75	0	7000/75
c	544	0	0	-22,4	0	143200
F	-1020	0	0	37	0	259000

2-qadam:
19400/68=289.3;

7000/40=175; +
143200/544=263.2

Erkli o’	x	y	a	b	c	Z
a	68	0	0	-60/75	0	19400
b	1	15/8	0	1/40	0	175
c	544	0	0	-22.4	1	143200
F	-1020	0	0	37	0	259000

Erkli o’	x	y	a	b	c	Z
A	0	-1275	0	0.9	0	7500
B	1	15/8	0	1/40	0	175
C	0	-120	0	-36	1	48000
F	0	1912.5	0	25.5	0	437500

4.Quyidagi va 4.1 laboratoriya topshiriqlarida berilgan masalalarni simpleks usulda yechish algoritmini va dasturini keltiring.
Misollar:
1. 100x+60y=<25000 0<=x<=2400;
40x+75y=<7000 0<=y<=2400
1440x+1680y=<300000
F=2500x+2775y ->max c1=15, c2=25. “F -Maqsad funksiyasi”
#include
using namespace std;
int main(){
int x1,x2,z;
int max=0;
for(int i=1; i<=100; i++)
for(int j=1; j<=100; j++)
if(100*i+60*j<=25000 && 40*i+75*j<=7000 && 1440*i+1680*j<=300000){
z=2500*i+2775*j;
if(z>max){
max=z;
x1=i;
x2=j;
}
}
cout<<"x1 = "<Download 0,62 Mb.
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