内容发布更新时间 : 2024/12/23 21:17:18星期一 下面是文章的全部内容请认真阅读。
运筹学(第3版) 习题答案 9
minZ?2x1?5x2 (6)
?x1?2x2?6??x1?x2?2?x,x?0?12
【解】无可行解。
运筹学(第3版) 习题答案 10
1.7 将下列线性规划化为标准形式 minZ?x1?6x2?x3?x1?x2?3x3?15 (1) ?
?5x1?7x2?4x3?32??10x1?3x2?6x3??5??x1?0,x2?0,x3无限制'''【解】（1）令x3?x3?x3,x4,x5,x6为松驰变量 ，则标准形式为 '''maxZ??x1?6x2?x3?x3'''?x1?x2?3x3?3x3?x4?15?'''?5x1?7x2?4x3?4x3?x5?32 ?'''??10x1?3x2?6x3?6x3?x6?5'''??x1,x2,x3,x3,x4,x5,x6?0minZ?9x1?3x2?5x3?|6x1?7x2?4x3|?20? (2) ?x1?5
??x1?8x2??8??x1?0,x2?0,x3?0【解】（2）将绝对值化为两个不等式，则标准形式为
运筹学(第3版) 习题答案 11
maxZ???9x1?3x2?5x3?6x1?7x2?4x3?x4?20??6x?7x?4x?x?201235? ?x?x?5?16??x?8x?82?1??x1,x2,x3,x4,x5,x6?0maxZ?2x1?3x2?1?x1?5 (3)???x1?x2??1?x?0,x?02?1【解】方法1：
maxZ?2x1?3x2?x1?x3?1?x?x?5 ?14??x1?x2?1??x1,x2,x3,x4?0??x1?1,有x1＝x1??1,x1??5?1?4 方法2：令x1??1)?3x2maxZ?2(x1??4?x1???1)?x2??1??(x1?x,x?0?12则标准型为
??3x2maxZ?2?2x1??x3?4?x1???x2?0??x1?x?,x,x?0?123
maxZ?min(3x1?4x2,x1?x2?x3)?x1?2x2?x3?30?(4) ?4x1?x2?2x3?15
??9x1?x2?6x3??5?x1无约束,x2、x3?0?【解】令y?3x1?4x2,y?x1?x2?x3,x1?x1??x1??，线性规划模型变为
运筹学(第3版) 习题答案 12
maxZ?y??x1??)?4x2?y?3(x1?y?x??x???x?x1123????x1???2x2?x3?30 ?x1???x1??)?x2?2x3?15?4(x1?9(x1??x1??)?x2?6x3??5??,x1??,x2、x3?0??x1标准型为
maxZ?y??3x1???4x2?x4?0?y?3x1?y?x??x???x?x?x?011235????x1???2x2?x3?x6?30 ?x1???4x1???x2?2x3?x7?15?4x1??9x1??9x1???x2?6x3?x8?5??,x1??,x2,x3,x4,x5,x6,x7,x8?0??x1
1.8 设线性规划
maxZ?5x1?2x2?2x1?2x2?x3?40 ?4x?2x?x?60?124?x?0,j?1,,4?j?21??20?取基B1??分别指出B1和B2对应的基变量和非基变量，求出基本?、B2＝??21?，40????解，并说明B1、B2是不是可行基．
【解】B1：x1、x3为基变量，x2、x4为非基变量,基本解为X=（15，0，10，0）T，B1是可行基。B2：x2、x4是基变量，x1、x3为非基变量，基本解X=（0，20，0，100）T，B2是可行基。
1.9分别用图解法和单纯形法求解下列线性规划，指出单纯形法迭代的每一步的基可行解对应于图形上的那一个极点．
maxZ?x1?3x2 (1)???2x1?x2?2
?2x1?3x2?12?x,x?0?12【解】图解法