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1
Appliquer la métthode Branch and Bound au problème de lassignation des tâches avec la matrice suivante: 1234 a11 88 b1415105 c33415 d10 2010 Calcul de borne supérieure: a1 + b2 + c3 + d4 = 11 + 15 + 4 + 10 = 40 a4 + b3 + c2 + d1 = 8 + 10 + 3 + 10 = 31*
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A = 1 A = 2 A = 3 A = 4 11+3+4+5= 23 8+3+3+5= 19 8+3+3+4= 18 * Cette branche semble la plus prometteuse * Calcul de la borne
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A = 1 A = 2 A = 3 A = 4 23 19 B = 1 B = 2 B = 3 8+14+3+4= 29 8+15+3+4= 30 8+10+3+3= 24 *
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A = 1 A = 2 A = 3 A = 4 23 19 B = 1 B = 2 B = 3 29 30 C = 1 C = 2 8+10+3+10= 31
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A = 1 A = 2 A = 3 A = 4 23 19 B = 1 B = 2 B=3, C=2, D=1 29 30 31 *
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A = 1 A = 2 A = 3 A = 4 23 19 B = 1 B = 2 B=3, C=2, D=1 30 31 C = 2 C = 3 8+14+3+20= 45 8+14+4+10= 36
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A = 1 A = 2 A = 3 A = 4 23 19 B = 1 B = 2 B=3, C=2, D=1 30 31 *
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A = 1 A = 2 A = 3 A = 4 23 19 B = 2 B=3, C=2, D=1 31 C = 1 C = 3 8+15+3+20= 46 8+15+4+10= 37
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A = 1 A = 2 A = 3 A = 4 23 19 B = 2 B=3, C=2, D=1 31 *
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A = 1 A = 2 A = 3 A=4, B=3, C=2, D=1 23 19 31 *
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A = 1 A = 2 A = 3 A=4, B=3, C=2, D=1 23 31 B = 1 B = 2 B = 4 8+14+3+10= 35 8+15+3+10= 36 8+5+3+3= 19 *
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A = 1 A = 2 A = 3 A=4, B=3, C=2, D=1 23 31 B = 4 C = 1 C = 2 8+5+3+10= 26
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A = 1 A = 2 A=3, B=4, C=1, D=2 23 * 26
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A = 1 A = 2 A=3, B=4, C=1, D=2 23 26 B = 1 B = 3 B = 4 11+14+4+10= 39 11+10+3+10= 34 11+5+3+4= 23 *
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A = 1 A = 2 A=3, B=4, C=1, D=2 23 26 B = 4 C = 1 C = 3 11+5+3+20= 39 11+5+4+10= 30
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A = 1 A = 2 A=3, B=4, C=1, D=2 23 26 *
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A = 1 A=3, B=4, C=1, D=2 26 B = 2 B = 3 B = 4 11+15+4+10= 40 11+10+3+10= 34 11+5+3+4= 23 *
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A = 1 A=3, B=4, C=1, D=2 26 B = 4 C = 2 C = 3 11+5+3+20= 39 11+5+4+10= 30
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A = 1 A=3, B=4, C=1, D=2 26
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A=3, B=4, C=1, D=2 26 La solution optimale est donc:
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