MUHAMMAD AL-XORAZMIY NOMIDAGI TOSHKENT AXBOROT TEXNOLOGIYALARI UNIVERSITETI
1-Amaliy ish
Mavzu: Masalalar yechish
Bajardi:
317-21 guruhi talabasi Turdiyev Elbek
Toshkent – 2023
1-topshiriq variantlari
Variant parametrlarini quyidagicha aniqlang: n1={60/3}+1=1; n2={60/5}+1=1; n3={60/7}+1=5, bu yerda N talabalarning potokdagi nomeri. {N/3} bu N sonini 3 ga bo’lgandagi qoldig’i.
Tenglamani hosil qildim:
Tenglamaning yechim joylashgan oralig’ini toping. Vatarlar usuli yordamida ildizlarni e=10-5 aniqlik bilan hisoblash algoritmini tuzing. Bunday aniqllika erishish uchun zarur qadamlar sonini aniqlang.
Quyida Vatarlar usuli yordamida hisoblash algoritmi tuzilgan, JAVA dasturlash tilidan foydalanilgan:
import java.util.Scanner;
public class Hosila {
public static float f(float x) {
return (float) (Math.pow(x, 3) + Math.pow(x, 2) - x - 5);
}
public static float f1(float x) {
return (float) (3*Math.pow(x, 2) + 2*x - 1);
}
public static float f2(float x) {
return 6*x + 2;
}
public static void main(String[] args) {
Scanner kiritish = new Scanner(System.in);
float a, b, eps = 0.00001f;
float[] x = new float[1000];
System.out.print("Birinchi oraliq : ");
a = kiritish.nextFloat();
System.out.print("Ikkinchi oraliq: ");
b = kiritish.nextFloat();
int sum = 0;
if (f1(a)*f2(a) > 0) {
x[0] = a;
} else {
x[0] = b;
b = a;
}
x[1] = x[0] - f(x[0]) / (f(b) - f(x[0]));
int i = 0;
while (Math.abs(x[i+1] - x[i]) > eps) {
i++;
x[i+1] = x[i] - f(x[i]) / (f(b) - f(x[i]));
sum++;
}
System.out.println("x: " + x[i]);
System.out.println("f(x): " + f(x[i]));
System.out.println(sum + " marta urunish");
kiritish.close();
}
}
Dastur Natijasi:
Ushbu masalani Nyuton usuli orqali yechish algoritmini tuzing. Ikkala usulda ham yechimlarni oling. Ko’rib chiqilgan usullar samaradorligini taqqoslang.
Quyida Nyuton usuli orqali yechish algoritmini tuzdim, JAVA dasturlash tilidan foydalanilgan:
import java.util.Scanner;
public class Nyuton {
public static double f(double x) {
return Math.pow(x, 3) + Math.pow(x, 2) - x - 5;
}
public static double f1(double x) {
return 3 * Math.pow(x, 2) + 2 * x - 1;
}
public static double f2(double x) {
return 6 * x + 2;
}
public static void main(String[] args) {
Scanner kiritish = new Scanner(System.in);
System.out.print("Birinchi oraliq : ");
double a = kiritish.nextDouble();
System.out.print("Ikkinchi oraliq : ");
double b = kiritish.nextDouble();
double eps = 0.00001;
double[] x = new double[100];
x[0] = 0;
int sum = 0;
if (f(a) * f2(a) > 0) {
x[0] = a;
} else {
x[0] = b;
}
x[1] = x[0] - f(x[0]) / f1(x[0]);
int i = 0;
while (Math.abs(x[i + 1] - x[i]) > eps) {
i++;
x[i + 1] = x[i] - f(x[i]) / f1(x[i]);
sum++;
}
System.out.println("x=" + x[i]);
System.out.println("f(x)=" + f(x[i]));
System.out.println(sum + " marta qadamlar soni");
}
}
)
Dastur Natijasi:
Bundan ko’rishimiz mumkinki Vatar usuli ishonchli lekin unda qadamlar soni biroz ko’proq, Nyuton usulida esa dasturimiz minimal qadamlar evaziga yaxshi natija oldi nyuton usuli ancha afzalroq ekan.
Chiziqli dasturlash masalasi hosil qilindi:
Quyida egizak masala hosil qilindi:
CHDM ni simpleks usulida hisoblang. Topilgan yechimlardan iqtisodiy tahlil o’tkazing. Tegishli matematik modelni ishlab chiqarishni takomillashtirish bo’yicha tavsiyalarni shakllantiring.
|
f
|
x1
|
x2
|
x3
|
x4
|
x5
|
x6
|
|
|
175
|
15
|
10
|
5
|
1
|
0
|
0
|
17 1/2
|
|
190
|
10
|
4
|
12
|
0
|
1
|
0
|
47 1/2
|
|
227
|
4
|
15
|
10
|
0
|
0
|
1
|
15 1/8
|
|
0
|
-1800
|
-2000
|
-1500
|
0
|
0
|
0
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
f
|
x1
|
x2
|
x3
|
x4
|
x5
|
x6
|
|
|
23 2/3
|
12 1/3
|
0
|
-1 2/3
|
1
|
0
|
- 2/3
|
1 8/9
|
|
129 4/9
|
8 8/9
|
0
|
9 1/3
|
0
|
1
|
- 1/4
|
14 1/2
|
|
15 1/8
|
1/4
|
1
|
2/3
|
0
|
0
|
1/9
|
56 3/4
|
|
30266 2/3
|
-1266 2/3
|
0
|
-166 2/3
|
0
|
0
|
133 1/3
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
f
|
x1
|
x2
|
x3
|
x4
|
x5
|
x6
|
|
|
1 8/9
|
1
|
0
|
- 1/7
|
1/9
|
0
|
-0
|
-14 1/5
|
|
112 1/3
|
0
|
0
|
10 5/9
|
- 5/7
|
1
|
2/9
|
10 2/3
|
|
14 5/8
|
0
|
1
|
5/7
|
-0
|
0
|
1/9
|
20 4/5
|
|
32697 2/7
|
0
|
0
|
-337 5/6
|
102 5/7
|
0
|
64 6/7
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
f
|
x1
|
x2
|
x3
|
x4
|
x5
|
x6
|
|
x1
|
3 3/8
|
1
|
0
|
0
|
1/9
|
0
|
-0
|
|
x3
|
10 2/3
|
0
|
0
|
1
|
- 1/9
|
1/9
|
0
|
|
x2
|
7 1/8
|
0
|
1
|
0
|
0
|
- 1/9
|
1/9
|
|
L
|
36297 3/7
|
0
|
0
|
0
|
79 1/2
|
32
|
71 4/5
|
|
|
|
|
|
|
y1
|
y2
|
y3
|
|
|
