# Strukturalanmagan masalalarni yechish prinsiplari

 Sana 13.01.2020 Hajmi 58.57 Kb.

Mavzu;Strukturalanmagan masalalarni yechish prinsiplari

Misol:Shahar rahbariyati jamoat transportlaridan zarar ko’ra boshladi. Ushbu muammoni yechish variantlarini baholash uchun 4 ta eksportan iborat komissiya tuzildi.

Quydagi variantlar keltirilgan.

1. Transport solig’ini oshirish.

2. Qatnovlar sonini yo’nalish bo’yicha kamaytirish.

3. Yo’l haqii oshirish.

4. Jamoat transportlarini(metro, avtobus , ford) hajmini kengaytirish.

m=4va n=4 holati uchun afzallik metodi yordamida maqsadlar og’irligini topamiz.

1. Boshlang’ich afzalliklar matritsasi:
 Ej/Zj Z1 Z2 Z3 Z4 E1 8/12 3/12 1 E2 4/12 7/12 6/12 E3 9/12 5/12 3/12 E4 11/12 6/12 9/12

 Ej/Zj Z1 Z2 Z3 Z4 E1 7/12 1 3/12 E2 5/12 5/12 7/12 E3 0 7/12 2/12 E4 9/12 5/12 10/12

 Ej/Zj Z1 Z2 Z3 Z4 E1 9/12 1 2/12 E2 3/12 6/12 8/12 E3 0 6/12 1/12 E4 10/12 4/12 11/12

 Ej/Zj Z1 Z2 Z3 Z4 E1 /12 1 3/12 E2 5/12 5/12 7/12 E3 0 7/12 2/12 E4 9/12 5/12 10/12

#include

#include

using namespace std;

int main(){

float Z1[4][4]={1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};

float Z2[4][4]={0,1/30,1/10,0,0,1/30,1/10,0,0,1/30,1/10,0,0,1/30,1/10,0};

for(int i=0;i<4;i++){

for(int j=0;j<4;j++){

cin>>Z1[i][j];

}

}

cout<<"Z1"<

for(int i=0;i<4;i++){

for(int j=0;j<4;j++){

cout<

}

cout<

}

for(int i=0;i<4;i++){

for(int j=0;j<4;j++){

cin>>Z2[i][j];

}

}

cout<<"Z2"<

for(int i=0;i<4;i++){

for(int j=0;j<4;j++){

cout<

}

cout<

}

float f[2][4] = {0,0,0,0,0,0,0,0};

for(int i=0;i<4;i++){

for(int j=0;j<4;j++){

f[0][i]+=Z1[i][j];

f[1][i]+= Z2[i][j];

}

}

float Q[2][4];

for(int i=0;i<4;i++){

for(int j=0;j<4;j++){

Q[0][j]=f[0][j]/30;

Q[1][j]= f[1][j]/30;

}

}

float W[4]={0,0,0,0};

for(int j=0;j<4;j++){

W[j] = Q[0][j]+Q[0][j];

cout<<"W["<

}

return 0 ;

}

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