transport masalasining optimal yechimini potensiallar usuli

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10-mavzu. amaliy mashg`ulit 1-masala. quyidagi transport masalasining optimal yechimini potensiallar usulidan foydalanib toping. tа’minоtchilаr istе’mоlchilаr tа’minоtchilаrdagi mahsulot zаhirasi b1 b2 b3 b4 a1 3 5 7 11 100 a2 1 4 6 2 130 a3 5 8 12 7 170 istе’mоlchilаrning tаlаbi 150 120 80 50 yechish. masalaning berilganlaridan foydalanib hisoblash jadvalini tuzamiz va boshlang‘ich bazis rejani "minimal xarajatlar" usulidan foydalanib topamiz. 1-jadval bj ai 150 120 80 50 100 3 20 5 80- θ 7 11 2 θ -7 130 1 130 4 6 2 -1 1 0 170 5 8 40+ θ 12 80- θ 7 50 1 θ=80 topilgan boshlang‘ich reja ushbu rejaga mos kelgan umumiy transport xarajati topilgan boshlang‘ich bazis rejani optimallikka tekshiramiz. buning uchun ta'minotchilarga mos ravishda iste'molchilarga mos ravishda potensiallarni mos qo‘yamiz hamda band kataklar uchun potensial tenglamalar tuzamiz: hosil bo‘lgan sistemaning aniq bir yechimini topish uchun deb qabul qilamiz va qolgan potensiallarning son …

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0 jadvaldan foydalanib band katakchalarga mos keluvchi potensial tenglamalar tuzib, potensiallarning son qiymatini topamiz: endi bo‘sh katakchalar uchun optimallik baholarini tuzamiz: bundan ko‘rinadiki, katakchadagi optimallik bahosi demak, reja optimal reja emas. katakchaga ni kiritib, bazis rejani optimal rejaga yaqinlashtirish mumkin. katakchaga θ ni kiritib, uni band katakchaga aytantiramiz va to‘rtburchakli yopiq kontur tuzamiz. θ ning son qiymati 20 ga teng bo‘ladi. (5) formulalar yordamida yangi bazis rejani aniqlaymiz. 3-jadval bj ai 150 120 80 50 100 3 5 20 7 80 11 -1 -7 130 1 130-θ 4 6 2 0 0 1 θ 170 5 20+ θ 8 100 12 7 50- θ -2 θ=50 yangi bazis rejani optimallikka tekshiramiz. buning uchun potensiallarning son qiymatini va bo‘sh kaktaklardagi optimallik baholarini jadvalning o‘zida hisoblaymiz. jadvaldan ko‘riladiki, demak, bazis reja optimal reja bo‘lmaydi. katakchaga θ sonni kiritib, yopiq kontur tuzamiz. θ ning son qiymatini topamiz. (5) formuladan foydalanib yangi bazis yechimni …

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0 230 160 tаlаb hаjmi 130 220 90 70 6. tа’minоtchilаr istе’mоlchilаr zаhirа hаjmi b1 b2 b3 b4 а1 а2 а3 5 3 1 4 2 6 3 5 3 4 5 2 160 140 60 tаlаb hаjmi 80 100 80 100 7. tа’minоtchilаr istе’mоlchilаr zаhirа hаjmi b1 b2 b3 b4 а1 а2 а3 4 6 3 2 3 2 3 5 6 1 6 3 70 140 80 tаlаb hаjmi 80 50 50 110 8. tа’minоtchilаr istе’mоlchilаr zаhirа hаjmi b1 b2 b3 b4 а1 а2 а3 6 5 3 7 1 2 3 4 6 2 3 2 180 90 170 tаlаb hаjmi 95 85 100 160 9. tа’minоtchilаr istе’mоlchilаr zаhirа hаjmi b1 b2 b3 b4 а1 а2 а3 8 4 1 3 1 9 5 6 4 2 7 3 180 140 200 tаlаb hаjmi 100 60 280 80 10. tа’minоtchilаr istе’mоlchilаr zаhirа hаjmi b1 b2 b3 b4 …

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vodlari qurilish ob`ektlаri zavodlarning ishlab chiqarish quvvati 18 13 11 15 500 12 21 16 14 850 10 16 14 15 600 betonga bo‘lgan tаlаb hаjmi 400 550 700 300 umumiy transport xarajatlarini minimallashtiruvchi tashish rejasini aniqlang. 23. 3 ta omborxonada guruch saqlanadi. ulardan birinchisida 135 tonna, ikkinchi va uchinchisida mos ravishda 165 va 160 tonnadan guruch zaxirasi mavjud. bu guruchlar 4 ta do‘konga yuboriladi. birinchi do‘konning guruchga bo‘lgan talabi 110 tonna, ikkinchisiniki 120 t., uchinchi va to‘rtinchi do‘konlarning talabi mos ravishda 110 tonna va 120 tonnani tashkil qiladi. 1 tonna mahsulotni tashish uchun sarf qilinadigan transport xarajatlari matritsasi quyidagi ko‘rinishga ega. qaysi omborxonani qaysi do‘koniga biriktirilganda sarf qilinadigan umumiy transport xarajatlari minimal bo‘ladi? 24. uchta fermer xo‘jaligidan 4 ta paxta tozalash zavodlariga paxta yuboriladi. fermer xo‘jaliklardagi paxta zaxirasi, paxta tozalash zavodlarining talabi va bir tonna paxtani tashish uchun sarf qilinadigan transport xarajatlari quyidagi jadvalda aks ettirilgan. fermer xo‘jaliklari paxta …

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10-mavzu. amaliy mashg`ulit 1-masala. quyidagi transport masalasining optimal yechimini potensiallar usulidan foydalanib toping. tа’minоtchilаr istе’mоlchilаr tа’minоtchilаrdagi mahsulot zаhirasi b1 b2 b3 b4 a1 3 5 7 11 100 a2 1 4 6 2 130 a3 5 8 12 7 170 istе’mоlchilаrning tаlаbi 150 120 80 50 yechish. masalaning berilganlaridan foydalanib hisoblash jadvalini tuzamiz va boshlang‘ich bazis rejani "minimal xarajatlar" usulidan foydalanib topamiz. 1-jadval bj ai 150 120 80 50 100 3 20 5 80- θ 7 11 2 θ -7 130 1 130 4 6 2 -1 1 0 170 5 8 40+ θ 12 80- θ 7 50 1 θ=80 topilgan boshlang‘ich reja ushbu rejaga mos kelgan umumiy transport xarajati topilgan boshlang‘ich bazis rejani optimallikka tekshiramiz. buning uchun ta'minotchilarga …

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