正在加载图片...
数规判 Integer programming(IP) 整数规划问题的求解方法 B12 Max Z=X,+ x2 14X+9X2≤51 分支定界法图解整数规划 -6X1+3X≤1 (3/2,10/3) X ≥2 X,≤2 Z1=29/6 B1:解 X,X,≥0 (2,23/9) Z11=41/9 B121 Max Z=X+ x2 B2:解 14X1+9X2≤51 (1,713) -6X1+3X2≤1 Z21=17/3 B12:解 X (3314,2) X2≤2 Z12=61114 X1,X2≥0 B122 Max Z=X1+ X2 B121:解 14X1+9X2≤51 B122:解 6X1+3X2≤1 (2,2) Z121=4 X ≤2 Z122=4 X2≤2 X1,Ⅹ2≥0 99 整数规划 Integer Programming(IP) 整数规划问题的求解方法 分支定界法图解整数规划 (3/2 ,10/3) Z1 = 29/6 B2:解 (1,7/3 ) Z21 = 17/3 B12 Max Z = X1 + X2 14X1 + 9X2 ≤ 51 - 6X1 + 3X2 ≤ 1 X1 ≥ 2 X2 ≤ 2 X1 , X2 ≥ 0 B12:解 (33/14,2 ) Z12 = 61/14 B121 Max Z = X1 + X2 14X1 + 9X2 ≤ 51 - 6X1 + 3X2 ≤ 1 X1 ≥ 3 X2 ≤ 2 X1 , X2 ≥ 0 B122 Max Z = X1 + X2 14X1 + 9X2 ≤ 51 - 6X1 + 3X2 ≤ 1 X1 ≤ 2 X2 ≤ 2 X1 , X2 ≥ 0 B121:解 (3,1 ) Z121 = 4 B122:解 (2,2 ) Z122 = 4 B1:解 (2,23/9 ) Z11 = 41/9
<<向上翻页向下翻页>>
©2008-现在 cucdc.com 高等教育资讯网 版权所有