1. Muammo: Transport masalasida shimoliy g‘arbiy burchak usulidan foydalanib boshlang‘ich tayanch reja matrisini topish kerak. 2. Shimoliy g‘arbiy burchak usuli: Bu usulda matrisning yuqori chap burchagidan boshlab, imkon qadar ko‘p yuklarni ajratib boramiz va har bir qator va ustun uchun talab va taklifni kamaytiramiz. 3. Berilgan talablar va takliflar: - Qatorlar (A_1, A_2, A_3) talablar: 4, 6, 5 - Ustunlar (B_1, B_2, B_3, B_4) takliflar: 3, 2, 2, 2 4. Shimoliy g‘arbiy burchak usulini qo‘llash: - A_1, B_1: minimal(4,3) = 3 - A_1, B_2: qolgan talab 4-3=1, taklif 2 minimal(1,2) = 1 - A_1, B_3: qolgan talab 0, shuning uchun 0 - A_2, B_2: talab 6, taklif 2-1=1 minimal(6,1) = 1 - A_2, B_3: talab 6-1=5, taklif 2 minimal(5,2) = 2 - A_2, B_4: talab 6-1-2=3, taklif 2-2=0 minimal(3,0) = 0 - A_3, B_3: talab 5, taklif 2-2=0 minimal(5,0) = 0 - A_3, B_4: talab 5, taklif 2 minimal(5,2) = 2 5. Natijada tayanch reja matritsi quyidagicha bo‘ladi: $$ \begin{bmatrix} 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} $$ 6. Berilgan variantlar orasida bu matrisga eng yaqin variant: $$ A = \begin{bmatrix} 3 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{bmatrix} $$ Bu variant shimoliy g‘arbiy burchak usuliga mos keladi. Javob: 3-chi variant.