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#include #include using namespace std; double func(double x) { return x x x + x x x 1; } double derivative(double x) { return x x + x 1; } void newton(double x0, double epsilon) { double x1; int iteration = 0; do {
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| bet | 3/4 | | Sana | 18.05.2024 | | Hajmi | 6,35 Mb. | | #242306 |
Bog'liq SADULLAYEVA ASAL|
Maqsad funksiyasi qiymati
Maqsad funktsiyasining qiymatini Cb ustunini elementlar bo'yicha P ustuniga ko'paytirib, mahsulotlarning natijalarini qo'shib hisoblaymiz.
Maks P = (Cb 1 * P 01 ) + (Cb 11 * P 2 + (Cb 21 * P 3 = (0 * 78) + (0 * 109,2) + (2000 * 8,2) = 16400;
Baholangan nazorat o'zgaruvchilari
Biz har bir boshqariladigan o'zgaruvchi uchun hisob-kitoblarni o'zgaruvchi ustunidagi qiymatni elementlar bo'yicha, Cb ustunidagi qiymatga ko'paytirish, mahsulotlarning natijalarini umumlashtirish va ularning yig'indisidan maqsad funktsiyasi koeffitsientini ayirish orqali hisoblaymiz. bu o'zgaruvchi.
Maks x 1 = ((Cb 1 * x 1,1 ) + (Cb 2 * x 2,1 ) + (Cb 3 * x 3,1 ) ) - k x 1 = ((0 * 12.33) + (0 * 8,93) + (2000 * 0,27) ) - 1800 = -1266,67;
Maks x 2 = ((Cb 1 * x 1,2 ) + (Cb 2 * x 2,2 ) + (Cb 3 * x 3,2 ) ) - k x 2 = ((0 * 0) + (0 * 0) + (2000 * 1) ) - 2000 = 0;
Maks x 3 = ((Cb 1 * x 1,3 ) + (Cb 2 * x 2,3 ) + (Cb 3 * x 3,3 ) ) - k x 3 = ((0 * -1,67) + (0) * 9,33) + (2000 * 0,67) ) - 1500 = -166,67;
Maks x 4 = ((Cb 1 * x 1,4 ) + (Cb 2 * x 2,4 ) + (Cb 3 * x 3,4 ) ) - k x 4 = ((0 * 1) + (0 * 0) + (2000 * 0) ) - 0 = 0;
Maks x 5 = ((Cb 1 * x 1,5 ) + (Cb 2 * x 2,5 ) + (Cb 3 * x 3,5 ) ) - k x 5 = ((0 * 0) + (0 * 1) + (2000 * 0) ) - 0 = 0;
Maks x 6 = ((Cb 1 * x 1,6 ) + (Cb 2 * x 2,6 ) + (Cb 3 * x 3,6 ) ) - k x 6 = ((0 * -0,67) + (0) * -0,27) + (2000 * 0,07) ) - 0 = 133,33;
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Bosh sahifa
Aloqalar
Bosh sahifa
#include #include using namespace std; double func(double x) { return x x x + x x x 1; } double derivative(double x) { return x x + x 1; } void newton(double x0, double epsilon) { double x1; int iteration = 0; do {
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