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1、第一章线性规划及单纯形法1 .用Xj (j=1.2)分别代表5中饲料的采购数，线性规划模型： min z 0.2x1 0.7x2 0.4x3 0.3x4 0.8x5st 3K 2x2 x3 6x4 +18x5700x1 0.5x2 0.2x3+2x4 x5300.5x1 x2 0.2x3+2x4 0.8x5 100xj(j 1,2,3,4,5,6) 02 .解：设x1x2x3x4x5x6x表示在第i个时期初开始工作的护士人数，Z表示所需的总 人数，则min z x x2 x3 x4 x5 x6st. x1 x6 60x1 x2 70x2 x3 60x3 x4 50x4 x5 20x5 x6 3
2、0 %(i 1,2.3.4.5.6) 03 .解：设用i=1, 2, 3分别表示商品A, B, C, j=1, 2, 3分别代表前，中， 后舱，Xij表示装于j舱的i种商品的数量，Z表示总运费收入则：maxz 1000(x1 x12 x13) 700(x21 x22 x23) 600(x31 x32 x33)st. x11 x12 x13600x21 x22 乂231000x31 x32 乂3380010x11 5x21 7x3140010x12 5x22 7x32 540010x13 5x23 7x33 15008x116x215x3120008x126x225x3230008x136x23
3、5x3315008x116x215x310.158x126x225x328x136x235x330.158x126x225x328x116x215x310.18x136x235x33xij0(i1,2.3.j1,2,3)5. (1)(3)无可行解(10,6)Tmax z x1 x2st. 6x1 10x2 120x1 x2 705 x1 103 x2 8解：如图：由图可得：*T*x (10,6); Z 16*max z 5x1 6x2st. 2x1 x2 22x1 3x22x1, x2 0如图:XI2X1-X2=2由图知，该问题具有无界解。6 (1)max z 3xi 4x22x3 5x4 5
4、x 4 0x 5 0xst -2 Xi2X3 X4 X4X1-2X1X2X32X42X 4 +X 5143x2X3X4x 4-xX1, X2,X3,X4,X4, X5,(2)maXz2X2x2 3x33x 3stXiX22X1X3X30X 44X2X1,X2,X3,X3,X3 X3+X4X4012(P1 p2P3 P4 P5 P 6)7.1)系数矩阵A:c；20种组合B112540; - B1可构成基。求B的基本解,12369100081410 =01016/33000001-7/6(B, b)= y1= (0,16/3,-7/6，0, 0,0) T同理y2= (0,10,0,-7, 0,0)
5、 Ty3= (0,3, 0,0, 7/2, 0) Ty4= (7/4 , -4,0,0, 0,21/4) Ty5= (0,0, -5/2,8, 0, 0) T解：（2）该线性问题最多有C26个基本解基本解Z基本可行解最优解1XiX2X3X42-411/200V32/50p1/5013-1/30011/6401/220VV50-1/2；02 二60011VVy6=(0,0,3/2,0, 8, 0) Ty8=(0,0,0, 3，5,0)Tyi0=：(0,3,-7/6, 0,0，0)yi2=：(0,0,-5/2，3,5,0) Tyi4=：(0,10,0,-7,0,0) Tyi6=：(0,0,3/2,
6、0,8,0) T基口行解：(4系个x值考他大十0)T最优解：(y3, y6,yi5, yi6)(y3,y7= (1, 0, -1/2, 0, 0, 3) T y9= (5/4, 0, 0, -2, 0, 15/4) T yii= (0, 0, -5/2, 8, 0, 0) Tyi3=yi5=(4/3, 0,(0, 3,0, 0, 2, 3/4)0, 0, 7/3, 0)Zmax=3p2 P3 P4, p2 P3 P5, p3 p4 p5,y6,卵，yi2, yi3,yi5, yi6)P2P4 P5为奇异,只有16个基。5 08.基的定义X1 X2 X由Xi X2 X3列向量构成所对应的列向量可
7、以构成基(B,b)=1061100-13/5213401037/531420013/5由非基变量对应的向量构成1 0 6.B对应的基解：(-13/5 , 37/5 , 0, 0, 3/5)9.解：(1)5X1-2X2=8T2由图知：x (1,3/2)T; Z 35/2;单纯形法：化为标准形如下:max z10x1 5x2st. 3xi4x2 x395x12x2 x48xi(i1,2,3,4)0C10500bCbXbXiX2X3Xr0X3341090Xr52018检验数1050000X3014/51-3/521/510 1Xi12/50-1/58/5检验数010-2-165X2015/14-3/
8、143/210 1Xi10-1/73/70检验数00-5/14-25/14-35/2*T*所以：x (1,3/2); Z 35/2;其中：(0,0,9,8) T 对应 A(0,0)(8/5,0,21/5,0) T 对应 B(8/5,0)(1,3/2,0,0) T 对应 C(1,3/2)9.2 )/(4A0)st.3x16x15x22x2x3x41524化为标准形：x/i1,2,3,4)0C2-100bCbXbXiX2X3X40X33510150点(0, 0,15, 24)0X4620124检验数2-1000X3041-1/23A点(4, 0,3, 0)Xi11/301/64Zmax=8检验数0
9、-10-1/3-810.解 1)要使A (0, 0)成为最优解则需.A点最大Z= 8max z 2x1 x2 0x3 0x4C 0且d 0;2)要使B (8/5, 0)成为最优解则C 0 且 d=0 或 C0 且 d0;3)要使C (1, 3/2)成为最优解则-5/2 -C/d -3/4 且 Cd0;即 5/2 C/d 3/4 且 Cd0;4)要使D (0, 9/4)成为最优解则11.(1枇为标准型：maxz 2x1 x2 x3st. 3x1 x2 x3 x4x1 x2 2x3x5x1x2x3C0 或 C=0,d06010x 20%(i 1,2,3,4,5,6) 0C2-11000bCbXbx
10、iX2X3I*，1X5 1X60X4311100600X51-12010100X6h1-1I 0 I0 I120检验数2-1100000X41。4-51-30302Xi11-12I 0 I1 I0100X602-30-1110检验数101-3口-2 10-200X4I 0011-1-2102Xi101/201/21/215-1X2【01-3/2【0 :-1/21/25检验数00-3/20-3/2-1/2-25*_ _ T_ *_x(15,5, 0)T;Z25;(2)max z 2x1 3x2 5x3 0x4 0x5 0x6 0x7 st. 2x, 2x2 3x3 x412x1 2x2 2x3x584% 6x34164x2 3x3x7 12xi(i 1,2,3.7) 0C2350000bCb XbXiX2X3X4X5X6X70 X42231000120 X5122010080 X64060010160 X7043000112检验数235000000 X402010-1/2040 X5-1/32001-1/303/85 X32/3
