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1、47页1.1b1 2 3 44321X12x1+x2=23x1+4x2=120X2用图解法找不到满足所有约束条件的公共范围，所以该问题无可行解47页1.1d1 2 3 454321-1-2-6 -5 -4 -3 -2 -1X2X12x1-x2=2-2x1+3x2=2无界解)(1.2（b）约束方程的系数矩阵 A= 1 2 3 4 2 1 1 2 P1 P2 P3 P4基基解是否可行解目标函数值X1 X2 X3 X4P1 P2-4 11/2 0 0否P1 P32/5 0 11/5 0是43/5P1 P4-1/3 0 0 11/6否P2 P30 1/2 2 0是5P2 P40 -1/2 0 2否P3
2、 P40 0 1 1是5最优解A=(0 1/2 2 0)T和(0 0 1 1)T49页13题设Xij为第i月租j个月的面积minz=2800x11+2800x21+2800x31+2800x41+4500x12+4500x22+4500x32+6000x13 +6000x23+7300x14s.t. x11+x12+x13+x1415 x12+x13+x14+x21+x22+x2310 x13+x14+x22+x23+x31+x3220 x14+x23+x32+x4112 Xij0用excel求解为：用LINDO求解： LP OPTIMUM FOUND AT STEP 3 OBJECTIVE
3、FUNCTION VALUE 1) 118400.0 VARIABLE VALUE REDUCED COST Z 0.000000 1.000000 X11 3.000000 0.000000 X21 0.000000 2800.000000 X31 8.000000 0.000000 X41 0.000000 1100.000000 X12 0.000000 1700.000000 X22 0.000000 1700.000000 X32 0.000000 0.000000 X13 0.000000 400.000000 X23 0.000000 1500.000000 X14 12.000
4、000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 -2800.000000 3) 2.000000 0.000000 4) 0.000000 -2800.000000 5) 0.000000 -1700.000000 NO. ITERATIONS= 3答 若使所费租借费用最小，需第一个月租一个月租期300平方米，租四个月租期1200平方米，第三个月租一个月租期800平方米，50页14题设a1，a2，a3, a4, a5分别为在A1, A2, B1, B2, B3加工的产品数量，b1，b2，b3分别为在A1, A2, B1加工的产
5、品数量，c1为在A2，B2上加工的产品数量。则目标函数为maxz= (1.25-0.25)( a1+a2+a3)+( 2-0.35) b3+( 2.8-0.5)c1 -0.05 (a1+b1)- 0.03 (a2+b2+c1)- 0.06 (a3+b3)-0.11(a4+c1)-0.05a5=0. 95a1+0. 97a2+0. 94a3+1.5b3+2.1c1-0.05b1-0.11a4-0.05a5s.t. 5a1+10b16000 7a2+b2+12c110000 6a3+8a34000 4a4+11c17000 7a54000 a1+a2-a3-a4-a5=0 b1+b2-b3=0 a
6、1，a2，a3, a4, a5, b1，b2，b3, c10用lindo求解得：LP OPTIMUM FOUND AT STEP 6 OBJECTIVE FUNCTION VALUE 1) 16342.29 VARIABLE VALUE REDUCED COST A1 1200.000000 0.000000 A2 0.000000 9.640000 A3 285.714294 0.000000 B3 10000.000000 0.000000 C1 0.000000 15.900000 B1 0.000000 0.230000 A4 342.857147 0.000000 A5 571.42
7、8589 0.000000 B2 10000.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.168000 3) 0.000000 1.500000 4) 0.000000 0.075000 5) 5628.571289 0.000000 6) 0.000000 0.008571 7) 0.000000 0.110000 8) 0.000000 -1.500000 NO. ITERATIONS= 6计算lindo截屏2.1a:对偶问题为：maxz=2y1+3y2+5y3s.t. y1+2y2+y32 3y3+y2+4
8、y32 4y1+3y2+3y3=4y10, y 20,y3无约束因为原问题的对偶问题的对偶问题仍是原问题，因此本问题的对偶问题的对偶问题为：minz=2x1+2x2+4x3s.t. x1+3x2+4x322x1+x2+3x33x1+4x2+3x3=5x1,x20,x3无约束81页2.12a)设x1,x2,x3分别为A，B，C产品数量maxz=3x1+x2+4x3s.t.6x1+3x2+5x3453x1+4x2+5x330x1,x2,x30用lomdo求解为LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 27.00000 VARIAB
9、LE VALUE REDUCED COST X1 5.000000 0.000000 X2 0.000000 2.000000 X3 3.000000 0.000000 X1,X2,X3 0.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.200000 3) 0.000000 0.600000 4) 0.000000 0.000000 NO. ITERATIONS= 2最大生产计划为A生产5个单位，C生产3个单位b) LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 27.00000 VARIABLE VALUE REDUCED COST X1 5.000000 0.000000 X2 0.000000 2.000000 X3 3.000000 0.000000 X1,X2,X3 0.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.200000 3) 0.000000 0.600000 4) 0.000000 0.000000 NO. ITERATIONS= 2 RANGES IN WHICH THE BASIS IS UNCHANGED:
