dev2.docx - Problème 3 Lot 1 Lot 2 Lot 3 Lot 4 Lot 5 Lot 6...

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Problème 3 Lot 1 Lot 2 Lot 3 Lot 4 Lot 5 Lot 6 Lot 7 Lot 8 Lot 9 Lot 10 Machine A 70 55 47 82 33 75 - 35 63 45 Machine B 80 45 34 67 40 65 40 30 40 50 Machine C - 50 38 73 - 50 50 25 32 60 Tableau 3.1 : temps de traitement (en minutes) Solutio n Obtenue à la question Méthode utilisée Temps utilisé sur la machine 1 Temps utilisé sur la machine 2 Temps utilisé sur la machine 3 #1 3.2b Méthode de pondération #2 3.3b Optimisation par objectif – minimisation de la somme pondérée des déviations #3 3.4 Optimisation par objectif – minimisation de la somme pondérée des déviations 3.1) Variables : X ij = 1 si le lot L i est traité sur la machine M j ; 0 sinon (Avec i = 1,…,10 et j = A, B, C) Contraires : Chaque lot doit être traité par exactement 1 machine : Xia + xib + xic = 1, pour i = 1,…,10 Claque machine a un temps d’utilisation disponible maximum : X1j + x2j +x3j + x4j +x5j +x6j +x7j +x8j +x9j +x10j ≤ 210 (minutes), pour j = A, B, C Binarité : Xij binaire, i = 1,…,10 et j = A, B, C Objectifs: Min Za = 70x1a + 55x2a + 47x3a + 82x4a + 33x5a + 75x6a + 35x8a + 63x9a + 45x10a
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  • Fall '18
  • X1, optimisation, Xia, Min Za, Min Zb, Min Zc

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