1、运筹学 习题一答案习题一1.2【解】设x1、x2、x3分别为产品A、B、C的产量,则数学模型为1.3 【解】 第一步:求下料方案,见下表。方案一二三四五六七八九十十一十二十三十四需要量B1:2.7m21110000000000300B2:2m01003221110000450A1:1.7m00100102103210400A2:1.3m01120010130234600余料0.600.30.700.30.70.610.10.900.40.8第二步:建立线性规划数学模型设xj(j=1,2,,14)为第j种方案使用原材料的根数,则(1)用料最少数学模型为用单纯形法求解得到两个基本最优解X(1)=(
2、 50 ,200 ,0 ,0,84 ,0,0 ,0 ,0 ,0 ,0 ,200 ,0 ,0 );Z=534X(2)=( 0 ,200 ,100 ,0,84 ,0,0 ,0 ,0 ,0 ,0 ,150 ,0 ,0 );Z=534(2)余料最少数学模型为用单纯形法求解得到两个基本最优解X(1)=( 0 ,300 ,0 ,0,50 ,0,0 ,0 ,0 ,0 ,0 ,200 ,0 ,0 );Z=0,用料550根X(2)=( 0 ,450 ,0 ,0,0 ,0,0 ,0 ,0 ,0 ,0 ,200 ,0 ,0 );Z=0,用料650根显然用料最少的方案最优。1.4 【解】设xj、yj(j1,2,6)分
3、别为16月份的生产量和销售量,则数学模型为(1)最优解X=(800,1000,1000,0,1000,1000)Y=(1000,1000,0,1000,1000,1000)Z=310000【另解】变量设置如表月份123456月产量X1X2X3X4X5X6月销量Y1Y2Y3Y4Y5Y6月初库存Z1Z2Z3Z4Z5Z6Zi+1=xi+zi-yi(1) maxz=350y1-300x1+340y2-330x2+350y3-320x3+420y4-360x4+410y5-360x5+340y6-300x6x1-y1-z2=-200Z2+x2-y2-z3=0Z3+x3-y3-z4=0Z4+x4-y4-z
4、5=0Z5+x5-y5-z6=0z6+x6-y61000y1-x1200y2-x2-z20y3-x3-z30y4-x4-z40y5-x5-z50y6-x6-z60X1800X2+Z21000X3+Z31000X4+Z41000X5+Z51000X6+Z60并且ai70,原问题无可行解。两阶段法第一阶段:数学模型为C(j)000001R. H. S.RatioBasisC(i)X1X2X3X4X5X6X3053100091.8X40-56010015MX612100-1152.5C(j)-Z(j)-2-10010514X1013/51/50009/5X4009110024X610-1/5-2/50-117/5C(j)-Z(j)01/52/5010因为X60,原问题无可行解。图解法如下: (4) 【解】大M法。X7是人工变量,数学模型为Cj425000MR.H.S.RatioCBXBX1X2X3X4X5X6X7