运筹学课后习题及答案课件.ppt

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1、第二章线性规划2.1(1)maxz=2x1+x4x1+3x2≤122x1+x2≤84x1-x2≤8x1≥0,x2≥0448s.t.0x1x24x1+3x2=122x1+x2=84x1-x2=8由图知,有唯一最优解,x*=(9/4,1)T,z*=11/22.1(2)maxz=3x1+2x-x1+2x2≤43x1+2x2≤14x1-x2≤3x1≥0,x2≥014/327s.t.0x1x2由图知,有无穷多最优解,x*=α(4,1)+(1-α)(5/2,13/4)=(5/2+3α/2,13/4-9α/4),

2、z*=14α∈[0,1]-x1+2x2=43x1+2x2=14x1-x2=32.1(3)maxz=2x1+3x2x1-x2≤2-3x1+2x2≤4x1≥0,x2≥0s.t.此线性规划问题无界解220x1x2-3x1+2x2=4x1-x2=22.1(4)maxz=x1+x2x1-x2≥03x1-x2≤-3x1≥0,x2≥0s.t.220x1x23x1-x2=-3x1-x2=0此线性规划问题无可行解10500CBXBx1x2x3x4b00x3x4354210019838/5105000010x3x10

3、114/52/510-3/51/521/58/53/24010-216510x2x101105/14-1/7-3/142/73/2100-5/14-25/1435/22.4（1）解：首先化标准形式：∵σj≤0∴X*=(1,3/2,0,0)T，z*=35/2maxz=10x1+5x23x1+4x2+x3=95x1+2x2+x4=8x1,x2,x3,x4≥0单纯形表为：100200000CBXBx1x2x3x4x5b000x3x4x51121061000100015002001200500-20010

4、0200000000200x3x4x22/311/3001100010-1/601/6300200200450200600100/3000-100/3400000100200x3x1x2010001100-2/31-1/3-1/601/6500/3200400/3000-100/3-100/3140000/3∵σj≤0∴X*=(200，400/3，500/3,0,0)T，z*=140000/32.4（2）单纯形表为:解:大M法:变为标准形式并添加人工变量，则原线性规划问题化为：2.5（1）maxz=

5、3x1+2x2–Mx6-Mx7x1+2x2+x3=7x1-x2-x4+x6=1x1+x2-x5+x7=2x1,x2,x3,x4,x5,x6,x7≥0s.t.32000-M-MCBXBx1x2x3x4x5x6x7b0x3121000077-Mx61-10-101011-Mx71100-101223+2M20-M-M000x3031100623x11-10-1001--Mx70201-1111/205+2M03+M-M00x3001-1/23/29/233x1100-1/2-1/23/2-2x20101

6、/2-1/21/2-0001/25/232000-M-MCBXBx1x2x3x4x5x6x7b0x5002/3-1/313-3x1101/3-2/303-2x2011/31/302600-5/34/30130x50110153x11210070x40311060-4-30021续表∵σj≤0∴X*=(7,0,0,6,5,0,0)T，z*=21两阶段法:第一阶段:minw=x6+x7x1+2x2+x3=7x1-x2-x4+x6=1x1+x2-x5+x7=2x1,x2,x3,x4,x5,x6,x7≥0s

7、.t.0000011CBXBx1x2x3x4x5x6x7b0x31210000771x61-10-1010111x71100-10122-20011000x303110-10620x11-10-10101-1x70201-1-1111/20-20-11100x3001-1/23/21/2-3/29/230x1100-1/2-1/21/21/23/2-0x20101/2-1/2-1/21/21/2-0000011因为基变量中不含人工变量,因此进行第二阶段求解:32000CBXBx1x2x3x4x5b0

8、x3001-1/23/29/233x1100-1/2-1/23/2-2x20101/2-1/21/2-0001/25/20x5002/3-1/313-3x1101/3-2/303-2x2011/31/302600-5/34/30130x50110153x11210070x40311060-4-30021∵σj≤0∴X*=(7,0,0,5,6)T，z*=21解:变为标准形式并添加人工变量，则原线性规划问题化为：2.5（4）maxz=-x1-3x2-4x3-3x4-Mx7