第一章 线性规划及单纯形法、第二章 线性规划的对偶理论等习题解答.pdf

第一章线性规划及单纯形法1.用Xj(1.2.5)分别代表5中饲料的采购数，线性规划模型:min z=0.2x1+QJx2+0.4%3+0.3x4+0.8x5st.3%+2x2+%+6x4+18/700%+Q.5x2+0.2%3+2X4+/300.5%+%2+0.2a:3+2x4+O.8X5 100W，2,3,4,5,6)NO2.解：设再R954x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 2=%+%+七+/+/+4st.x1-x6 60%+2 70%+%2 60 x3+%4 50 x4+x5 20毛+430%(i=1,2.3.456)2 03.解：设用i=l,2,3分别表示商品A,B,C,j=l,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：max z=1000(xn+x12+x13)+700(a:21+2+/3)+600(x31+x32+st.xn+xn+x13 600 x21+x22+x23 1000X31+*32+X33 8010 xn+5x21+7%3i 40010 x12+5x22+7忍2 V 540010 x13+5x23+7%3 15008xn+6x21+5x31 20008x12+6x22+5x32 30008x13+6x23+5x33 15008石+6.2i+5*3i 0 58x12+6x22+5x328%13+6.23+5%0 58%2+692+5&28xn+6.21+5玉1 0(z=1,2.3 j=1,2,3)5.(1)max z=xt+x2st.6xx+10 x2 705103Vx2 2 8解：如图：由图可得：%*=(10,61；z*=16即该问题具有唯一最优解%*=(1，6尸无可行解(4)max z=541+6x2st.2%-x22-2%+3x2 0如图:由图知，该问题具有无界解。6(1)max z=3%i-4+2x3-4+4+Ojc 5+Ojcst.2%i+x 2-2x 3+%4-%4X1+%2%3+2x 4 2x 4+x 5-2xi+3x2+%3-x 4+x 4 F 6Xp%2%3%4 X 4%5%6 2 0=2=14二 6(2)max 2=2x；+2x2-3x3+3x 3+0 x 4st.xx+x 2+x3-x3=42xx+x2-x3+x3+x 4=6Xp X2，X 3，X 3，X 4 012 3 6 3 08 1-402(300007.1)系数矩阵A：*=20种组合0、o=(P1 P2 P3 p4 P5 p6)-b12|4|=忸 8 H=833 61-4=-54 w 0；4可构成基。0 0求比的基本解,勺2 3 6 9)1 0 0 0、8 1-4 10=0 1 0 16/3(B,b)=(3 0 0 0 7(0 0 1-7/6?yi=(0,16/3,-7/6,0,0,0)T同理 y2=(0,10,0,-7,0,0)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)TTy6=(0,0,3/2,0,8,0)Ty7=(1,0,-1/2,0,0,3)Ty8=(0,0,0,3,5,0)Ty9=(5/4,0,0,-2,0,15/4)yio=(0,3,-7/6,0,0,0)Tyn=(0,0,-5/2,8,0,0)Tyi2=(0,0,-5/2,3,5,0)Tyi3=(4/3,0,0,0,2,3/4)Tyi4=(0,10,0,-7,0,0)Tyi5=(0,3,0,0,7/3,0)Tyi6=(0,0,3/2,0,8,0)T基可行解：(每个x值都大于0),(y3,6,y8，yi2，yi3，y”，yi6)P4 P5为奇异，只有16个基。P2最优解：(y3，V6,yi5，yi6)Zmax=3 P2 P3 P4，P2 P3 P5，P3 P4 P5，解：（2）该线性问题最多有盘=6个基本解。基本解Z基本可行解最优解1XiX2X3X42-411/200V32/5011/503-1/30011/6401/220VV50-1/20260011VV1 0 6B=2 1 3=-5w08.基的定义 3 1 4一 1 02 1X1 X2 X3所对应的列向量可以构成基63(B,b)Xi X2 X3列向量构成=非基变量对应的向量构成=11 Fl o4-011 4一 3 5 一4 12 00-13/50 37/51 3/5B由N由1 0 62 1 33 1 42 0 0B对应的基解:(-13/5,37/5,0,0,3/5)9.解：(1)由图知：%*=(1,3/2)。Z*=35/2;单纯形法：化为标准形如下:maxz=10 xx+5x2st.3xx+4x2+&=95%i+2x2+%=83a=1,2,3,4)NOc10500bCbXbXiX2X3Xr0X3341090Xr52018检验数1050000X3014/51-3/521/510X112/50-1/58/5检验数010-2-165X2015/14-3/143/210Xi10-1/73/70检验数00-5/14-25/14-35/2所以：x*=(l,3/2)7；z*=35/2;其中：(0,0,9,8)T 返 A(0,0)(8/5,0,21/5,OP 对应 8(8/5,0)(1,3/2,0,OP 对应 C(l,3/2)A点最大Z=8max z=2x1-x2-Ox3+0 x4st.3%1+5%2+%3=156玉+2x2+x4=24化为标准形：%G=L 2,3,4)0c2-100bCBXBXix2x3x40 x33510150 x4620124检验数2-1000 x3041-1/23Xi11/301/64检验数0-10-1/3-810.解0 点(0,0,15,24)A 点(4,0,3,0)Zmax=81)要使A(0,0)成为最优解则需C0 且 d=0 或 00 且 d5/2 且 Cd0；3)要使C(1,3/2)成为最优解则-5/2 -C/d 0；即 5/2 C/d 3/4 且 Cd0；4)要使D(0,9/4)成为最优解则C0 或 C=0,d0n.化为标准型：max 2=2%9+工st.3%i+/+&+%=6。X1-X2-2x3+15=10%+%2-+%6=20毛。=123,4,5,6)2 0c2-11000bCbXbXiX2X3X4x5X60X4311100600X51-12010100 x611-100120检验数2-1100000X404-51-30302Xi1-12010100X602-30-1110检验数01-30-20-200 x40011-1-2102Xi101/201/21/215-1X201-3/20-1/21/25检验数00-3/20-3/2-1/2-25x*=(15,5,0)。Z*=25;(2)max 2=2%i+3x2+5x3+Ox4+0 x5+0 x6+0 x7st.2%i+2%+3%+%=12%+2x2+2x3+x5=84x1+6x3+4=164x2+3x3+%7=12.a=1,2,3.7)10c2350000bCB XbXix2x3x4x5x6x70 x42231000120 x5122010080 x64060010160 x7043000112检验数235000000 x402010-1/2040 x5-1/32001-1/303/85 X32/301001/603/80 x7-24000-1/214检验数-4/33000-5/60-40/30 x410010-1/4-1/220 x52/30001-1/12-1/22/35 X32/301001/608/33 X2-1/21000-1/81/41检验数1/60000-11/24-3/4-49/30 x40001-2/3-1/81/412 Xi10002/3-1/8-3/415 X30010-11/41/223 X201003/4-3/16-1/83/2检验数0000-1/4-7/16-5/8-33/2/=(1,3/2,1,1,0,0)。Z*=33/2;(3)标准型：max 2=3%+5x23st.X1+&=42x2+/=123%x+2x2+/=18%(，=L2,3,4,540C35000bCbXbXiX2x3x4x50X31010040X402010120X53200118检验数350000X3101004X20101/206X5300-116检验数300-5/20-30X30011/3-1/32X20101/206Xi100-1/31/32检验数000-3/2-1-36x*=(2,6)。Z*=36;(4)标准型max z=x1+x2-x3-x4+x4-x5+x5-x6+x 6-x7-Mx8-Mx9+Oxlost.x1+x4-x4+x6-x6+x7=93%+一 4忍+2x 6-2x 6+x8=2%+2x3-%+X5+2/-2x 6+x9=64%+3x3+x10=12Xp 12%3%4%4%5%5%6%6 Xr%8%X10-c-11-1-11-11-11-M-M-M0bCbXbXiX2x3X4X4“X5X5“X6X6“X7x8X9X10-MX71001-1001-110009-Mx831-400002-201002-MX91200-112-2001060X10043000000000112检验数5M-1M+l-1-2MM-l1-MM+l1+M5M-11-5M000017M-MX70-1/34/31-1001/3-1/31-1/30028/3-1Xi11/3-4/300002/3-2/301/3002/3-Mx90-1/310/300-114/3-4/30-1/31016/30X10043000000000112检验数04/3-2/3M14/5M-7/3M-l1-MM-lM+l5/3M-1/3-5/3M+1/30-5/3M+1/300-41/3M-2/3-MX70-1/501-12/5-2/5-1/51/51-1/5-2/5031/5-1Xi11/5000-2/52/56/5-6/501/52/5014/5-1X30-1/10100-3/103/102/5-2/50-1/103/1008/50X10043/100009/10-9/10-6/56/503/10-9/10136/5检验数0-1/5M+11/100M-l1-M2/5M-17/10-2/5M+17/103/5-1/5M1/5M-3/50-6/5M-7/5M0-31M/5-22/5(5)max z=6石+lx2+10a:3+8x4st.5%i+6x2-4x3-4x4 205%i-3x2+2x3+8/254%-2x2+x3+3x4 0解：标准化：max 2=6%i+2x2+10 x3+8x4st.5再+6x2-4x3-4x4+/=205%i-3x2+2x3+8/+4=254%i-2x2+&+3x4+/=10 x.(j=1,2,3,4,5,6,7)0由表可得，?。且已(。因此问题的解无界。(6)化为:标准形：Z=-Zmax 2=-xx1-x2-x3C62108000bCbXbXiX2X3X4X5X6X70 x556-4-4100200X63-328010250X74-21300110检验数6210800000X521-208114600X6-510200-2510X34-21300110检验数-34220-2200-10-1000 x5110012120702X2-510201-2510X3-601702-320检验数7600-660-2234-210st.%-4%4-2%6=5%2+214+3x5+4=3%+2x4-15+64=5%(，=1,2,3,4,5,6空0(I)C-x-1-1000bCBXBXix2x3x4x5x6-XXi100-40-25-1x20102-313-1x30012-565检验数0004-4x-87-2x5x+81.1.XN7/2 时，检验数WO，最优解：(5,3,5,0,0)2.2.1WXV7/2 时，4-4X0由（I）得:C-x-1-1000bCB XbXix2x3x4x5x6-X Xi101/3-10/3-5/3020/3-1 x201-1/65/3-13/6013/60 x6001/61/3-5/615/6检验数00X/3-7/6-10X/3+5/3-5X/3-13/6020X/3+13/6%*=(20/3,13/6,0,0,0,5/6)。Z*=20X/3+13/6;3.X-2X+70检验数0,列系数W0,所以解无界。C-x-1-1000bCBXBXix2x3x4X5x6-XXi1200-6011-1x201/201-3/21/23/2-1x30-110-252检验数02x-200-6X-2511X+24.-3/2X4-4X0判断检验数的符号:C-x-1-1000bCB XbXix2x3x4x5x6-X Xi100-10/3-5/3020/3-1 x20105/3-13/6013/6-1 x60011/3-5/615/6检验数X/3-7/6-10X/3+5/3-5X/3-13/620X/3+13/6-10 x/3+5/3=0 卜=1/2-5x/3-13/6=0=-13/10-10 x/3+5/3=-5x/3-13/6 1x=2.3.%*=(20/3,13/6,0,0,0,5/6)4 Z*=20X/3+13/6;(1/2x2)-1.3WXl/2 时表(A)对-6X 讨论，令-6X=0 n X=01 0WXl/2时，检验数0c-x-1-1000bCBXBXix2x3x4X5x6-XXi1200-600110 x403/5-1/101-13/100013/100 x60-1/51/50-2/5112/5检验数02X-1-10-6X011X%*=(11,0,0/3/10,0,2/5)。Z*=11X;(oWxWl/2)2-1.3WX0又X5列的系数0,所以解无界3)-1.5X0，又X5的列的系数0,所以解无界(7)max 2=%+6x2+4x3st.X+2%+2%3 W134xx-x2+x3 1,x2 2,x3 3解：化为标准形:max 2=%+6x2+4x3-Mxw-Mxn-Mxnst.一%+lx2+2%+x44%1-12+%3x1+2x2+x3X%30(z=l,2,3.H)+4一马+再0+/+%n=13=20=17=1=2+/+芯2=3c164000000-M-M-MbCbXbXiX2X3X4X5X6X7X8X9Xi0XuX120X4-122100000000130X54-41010000000200X612100110000017MXi01000000001001MXii0100000-100102MXi200100000-10013检验数1+M6+M4+M000-MM-M0000X4-1021000200-2090X54010100-40040260X61010010200-2013MXi0100000-10010016X20100000-100102MXi200100000-10013检验数1+M04+M000-M6-M0-6-M00X4-1001000220-2-230X54000100-4104-1250X61000010210-2-110MXi0100000-10010016X20100000-1001024X300100000-10013检验数1+00000-M6410-6-4-MMM0X4000100-1221-2-240X500001004-41-44-1210X6000001121-1-2-191X1100000-10010016X20100000-1001024X300100000-10013检验数0000001640X80001/200-1/2111/2-1-120X5000210205-20-5290X6000-10120-1-20151X1100000-10010016X20101/200-1/2011/21/2-144X300100000-10013检验数000-30040-20X80001/401/4013/40-1-3/413/40X500031-100600-6240X7000-1/201/210-1/2-101/25/21X1100-1/201/200-1/2001/27/26X20101/401/4003/400-3/421/44X300100000-10013检验数000-10-2000-47即：X*=(7/2,21/4,3)。Z*=47;为最优解，但该问题具有无穷多最优解(8)max 2=再-X2+入3-3%3+%6 3%7st.3x3+/+%6=6%2+2%-X4=10-x1+x6=0 x3+x6+x7=6%。=1,2,7)2 0解：化为标准形：max z=x1-x2+x3-3x3+x5-x6-3x7-M%8-Mx9-M%10-Mxn st.3x3+/+4+/-6x2+2x3-x4+/=10-X+4+%io=0 x3+x6+x7+%ii=6x7(z=1,2,.11)0/=(0,-2,6,0,12,0,0)/；Z*=4;12.(1)解：标准形：1c1-11-31-1-3-M-M-M-MbCb XbXiX2X3X4x5X6X7X8X9XioXu-M X8003011010006-M X9012-1000010010-M Xio-100001000100-M Xu001001100016检验数1-MM-l6M+1-M-31+M-1+3M3+M00001 X30011/31/301/30002-M X9010-1-2/3-2/30-2/31006-M Xio-100001000100-M Xu0000-1/32/31-1/30014检验数1-MM-l0-M-32/3-MM-l/3M-30001 X300101/31/301/30002-1 X2010-1-2/3-2/3001006-M Xio-1000010-2/30100-M Xu0000-1/32/31-1/30014检验数1-M00-45/3M-2M-3001 X300101/20-1/21/200-1/20-1 X2010-1-101-110110-M Xio-10001/20-3/21/201-1/2-61 X60000-1/213/2-1/2003/26检验数1-M00-41/2M-1001 X3111000100-110-1 X2-200-100-2012-2-21 X5-200010-3102-3-12-1 x6-10000100100检验数-100-400-3max 2=2%i-%+2x3-Mx1-Mxs-Mx9 st.%+/+%+%7=6二2+为=0 x7(z=1,2,3.9)0由表可知此题解无界。c2-12000-M-M-MbCB XbXix2x3x4x5x6x7x8x9-M X7111-1001006-M X8-1010-100102-M X902-100-10010检验数2-1+3M2+M-M-M-M000-M X7103/2-101/210-1/26-M X8-1010-100102-1 X201-1/200-1/2001/20检验数205/2M+3/2-M-M1/2M-1/200-M X75/200-13/21/21-3/2-1/232 X3-1010-100102-1 x2-1100-1/2-1/201/21/21检验数5/2M+7/200-M3/2M1/2M02 Xi100-2/53/51/52/5-3/5-1/56/52 X3001-2/5-2/51/52/52/5-1/516/5-1 X2010-1/5-1/5-2/51/51/52/58/5检验数0007/5-3/5-6/52得一辅助问题：max w=-Mxy-Mxs-Mx9st.X1+%2+w 一%+X7=6=22x2-x3-x6+/=0(z=l,2,3.9)0C000000-1-1-1bCB XbXix2x3x4x5x6x7x8x9-1 X7111-1001006-1 x8-1010-100102-1 X902-100-10010检验数031-1-1-1000-1 x7103/2-101/210-1/26-1 x8-1010-1001020 x201-1/200-1/2001/20检验数005/2-1-11/200-3/2-1x75/200-13/21/21-3/2-1/230 x3-1010-1001020 x2-1100-1/2-1/201/21/21检验数5/200-13/200Xi100-2/53/51/52/5-3/5-1/56/50X3001-2/5-2/51/52/52/5-1/516/50 x2010-1/5-1/5-2/51/51/52/58/5检验数000000-1-1-10其最优解为X；=6/5,芯=16/5,4=8/5,%；=M=、=父=石=0;W*=0;其原问题的初始基本解可行解了=(6/5,16/5,8/5,0,0,0/由表知此题属于解无界c2-12000bCBXBXix2x3x4x5x62Xi100-2/53/51/5-1x3001-2/5-2/51/52x2010-1/5-1/5-2/5检验数0-334/5(2)大M法，先化为标准形：Z=-Z max 2=-4%-+0%3+0 x4-Mx5-Mx6 st.%+2x4+%=4x1+x2+/=34%+3x2x3+x6=6七”1,2,3,4,5,6)2 0C-4-100-M-MbCBXBXix2x3x4x5x60 x41201004-Mx51100103-Mx643-10016检验数5M-44M-1-M0009M0 x405/41/4105/2-Mx501/41/4013/2-4Xi13/4-1/4003/2检验数0M/4+2M/4-1003M/2+6-1x2011/54/502-Mx5001/5-1/511-4Xi10-2/5-3/500检验数00M/5-7/5-M/5-8/50M+2-1x2010110 x3001-15-4Xi100-12原问题唯一最优解000-39二阶段法：引入人工变量x5x6 得原问题的一个辅助问题：max w=-x5-x6st.%+2x4+x4=4%+%+/=34%i-3x2-x3+4=6%=123,4,5,6)2 0C0000-1-1bCBXBXix2x3x4x5x60 x41201004-Mx51100103-MX643-10016检验数54-100090 x405/41/4105/2-Mx501/41/4013/2-4Xi13/4-1/4003/2检验数01/41/4003/2-1x2011/54/502-Mx5001/5-1/511-4Xi10-2/5-3/500检验数001/5-1/501-1x2010110 x3001-15-4Xi100-12检验数000-10其最优解为X；=2,x；=l,x*=5,x；=M=X；=0;w=0;其原问题的初始基本解可行解x=(2,1,5,0,0,0/c-4-100bCBXBXix2x3x4-1x2001-150 x301011-4Xi100-12检验数000-39(3)min z=2玉3x2+x3st.xx+4x2+2x3 83%+2x2 6x.(j=l,2,3)0标准形：min z=-2x1+3x2-x3-Mxy-Mxs st.玉+4x2+2x3-x4+x7=83%i+2x2-x5+x8=6x.(j=l,2,3.8)0c-23-1000-M-MbCBXBXix2x3x4x5x6x7x8-Mx7142-100108-Mx83200-10016检验数-2-4M6M+32M-1-M-M0003x21/411/2-1/4001/402-Mx85/20-11/2-10-1/212检验数5/2M-11/40-M-5/21/2M+3/4-M003x2013/5-3/101/1003/10-1/109/5-2Xi10-2/51/5-2/50-1/52/54/5检验数00-18/513/1003x25/2100-1/2001/230 x450-21-20-124检验数0-103/20-M-M由表知此题解无界；(6)两阶段法：max z=-x7-%8st.%+4x2+2x3-%4+x7=83%+2x2-%5+x8=6%7(j=1,2,3.8)0C000000-1-1bCBXBXix2x3x4x5x6x7x8-1x7142-100108-1x83200-10016检验数462-1-10000 x21/411/2-1/4001/402-1X85/20-11/2-10-1/212检验数5/2M-11/40-M-5/21/2M+3/4-M000 x2013/5-3/101/1003/10-1/109/50Xi10-2/51/5-2/50-1/52/54/5检验数000000-1-1其最优解为%；=4/5,x；=9/5,%；%；=x；=x；=%；=0;攻*=0;c-23-1000bCBXBXix2x3x4x5x63x2013/5-3/101/1009/5-2Xi10-2/51/5-2/504/5检验数00-18/513/1003x25/2100-1/2030 x450-21-204检验数0-103/20由表知此题为解无界;(4)maxz=10%+15%+12%st.5%+3+%9-5xx+6x2+15x3 05化为标准形：max z=10 xi+15 x2+12%-Mxst.5%i+3x2+%+%5%i+6x2+15忍+%2%i+x2+x3-x6+x1x.(z=l,2,3.7)0=9=15二5C101512000-MbCBXBXix2x3x4x5x6x70 x4531100090 x5-5615010015-Mx721100-1152M+10M+1512+M00-M010Xi13/51/51/50009/50 x50916110024-Mx70-1/53/5-2/50-112/503/5M0010Xi139/80015/80-1/800015/1012x309/1611/161/16003/2-Mx70-43/800-35/80-3/80-111/200013.(1)巧。；小 0,0,a2 0,0,cr2 014.C2-11000bCBXBXix2x3x4x5x60 x4311100600 x51-12010100 x601-1001202-110000 x404-51-30302Xi1-12010100 x602-30-111001-30-200 x40011-1-2102Xi101/201/21/215-1x201-3/20-1/21/2500-3/20-3/2-1/2-2515.C2-11000bCBXBXix2X3x4X5X60 x22/3101/3008/30 x5-4/3052/31014/30 x65/304-5/30129/3-1/30-4-5/300-1x22/3101/3008/31x3-4/15012/151/5014/150 x641/1500-33/15-4/5184/1544/15003/15-1/5016.鼻事是基,c=0,d=l5-OxO+5xl=Z?0 x2+5xq=-100 0 xl/5+5xl=gn ex：.cx-x)cx即 c*(x*_f)2 0(2)由(2)(1)得：(c*-c)(x*-%)2 0即得证。20.解(1)max z=Kcx,二1*是max 2=ex的最优解，即2*=x c又:入0为某一常数，入z*=x*入C即%*使问题maxw二入ex的最优解即max 2=入ex的最优解为x*(2)max 2=(c+A)x2*=(c+入)x*=ex+Kx第二章线性规划的对偶理论1.解：设X%，七分别表示A、B、C各产品的数量，Z表示总产值则：max 2=200%1+300%2+250%3st.3%+4%+2%3 V 602%+2%3 40%+3x2+3x3 2004%+为+3%3002%+2%+3y3 250 0,y2 0,y3 0经济解释：yi,y2,y3分别表示给别人代工时所得收入，对厂方而言，w越大越好，但定价 不能太高，要对方容易接受，应考虑使总收入即对方的总支出尽可能少才比较合理，厂方不 会吃亏，对方也容易接受。2.(1)min w=10%+20 y2st.+y2 10%+为212%+%2%，为20(2)max w=2%+3%+5 y3st.2%+3%+%V23y1+y2+4y3 25%+7%+6%44%2 y2 1乃 十%21%-为 2 0%，2 2 07%,%0-2%VO这与约束条件 不符该对偶问题无可行解原问题无最优解。4.证明：该线性问题的对偶问题为:min w=2%+%+2%st.%+%+2%21%-%+%V 2-%+%+为=1o,%自由，%V。易知y=（0,1,0）7是对偶问题的一个可行解,由对偶问题的对偶定理可得：0CX“。：2,2 口maxz 是原问题的一个可行解，Y=（0,1,0是对偶问题的一个可行解由对偶问题的推论2-3可得它们都有最优解。即得证。7.7.1）对偶问题：min w=5%+12%st.%+2y2 2 2%+3%2yi+4y230,为无符号限制2）由题知原问题的最优解为*=（320）。由互补松弛定理得：在对偶问题中对应第一，二个约束为紧，第三个约束条件为松,即，工+2工;2*y=43 1乃二对偶规划问题的最优解丁；a#3）影子价格为yi=4：8.解：先写出其对偶问题。min w=20%+20 y2 st.%+2y2 13%+%22%+8%2 33%+2y2 对偶规划问题的最优解=1.2,y；=0.2,由互补松弛定理,又y；，不为。，2%3+3%4=203思+2z=20代入约束条件，知第1,2约束条件成立严格不等式,原规划最优解中相应变量%；=芯=0则在原问题规划中对应的约束条件为紧，得=4=4原对偶规划问题的最优解%*=（，4,4）9.(1)min w=3%+6y2+2y3+2y4st.%+3y2+为 2 82%+%为+%+%3%+必+%*为4，乃4，为4，y4Vo(2)X=(1,1,2,0)T；Z*=20把X*=(1,1,2,0代入约束条件可得:X5%6 Xq 0,Xg 1；%=M=%=乃=。；凡+3%+0-0=8i=2%=2%=1%=-110.解(1)由表知歹1(112 0 A1/6 1/30、bB-N=”2-1/2令6=(1/2 0 xu-1/6 1/3-1.A勺/21/20、则7I。(a0、bnB；B%=(1/21-1/6 1/3JC2+(-4,-2)1-1nN)5/2、5/2,q J(52 0、=J 3,、=(4,2)n-4C2=-2G=6。3=10B 85=、j1 2、(3-1 17o A、jI=、10l/3Jbjn b原线性问题为：max 2=6%i-2x2+10 x3 st.x2+2x3 53xr-x2+x3 20(3)x*=(5/2,05/2)。Z*=40;x4=x5=0%=%=3%=6 pi=4即%-%=-2 n%=2?%+%=10|%=4y*=(4,2)t w*=40；11.解(1)设占%，3分别表示甲、乙、丙各产品的数量，Z表示总产值则:max 2=3%+2x2+x3st.龙i+2x2+x3 4002玉+/+2x3 1 0,0此时，yi,y2分别表示出租A,B设备所得利润，*由(1)中的最优表得为=1/3,即如出租A设备可获得1000/3元，而1000/3350 所以不合算。12.解：(1)由影子价格的定义可得：8Z 3Z.7，力.1(2)由(1)可知yi只与匕的值有关.,当xi的系数由3变为x的系数1/3时，的值并不 发生变化；xi不可能在最优基中出现，J x也不可能在最优基中出现z=GXj=wj=i j=iV当目标函数由2=tgXj变为z=f2GXj时，q.增大了两倍 j=i j=i(3)e影子价格Yi也增大了两倍。;(4)不会。13.解:1)解:maxz先将问题化为标准式1。石5%2 4%+0%4+04 2%2+3%2+/二3-4冗2/+%5=10取初始正则基B=（P4 P5）=I王”1,2,3,4,5）2。则原问题已化为关于基B的典式,c23100BCBXBXix2x3x4x50 x4-3-2310-30 x5-40-201-10检验数-10-5-40000 x4-9-2013/2-18-4x22010-1/25检验数-2-500-220-10Xi12/90-1/9-1/62-4x20-4/912/9-1/61检验数0-41/90-2/9-7/324可得原问题的最优解为：x=(2,0,1,0,0)T,Z*=-24,Z=24,y*=(2/9,7/3),w*=24.(2)将问题化为：max z=一2石-3x2-4x3St.-%2%2 _%3+*4=-3-2%+%3%3+%5=4%”L2,3,4,5)2。C32100BCBXBXix2x3x4x50 x4-1-2-110-30X521-301-4检验数-2-3-4000 x40-5/21/21-1/2-13Xi1-1/23/20-1/22检验数0-4-10-12502x201-1/5-2/51/52/53Xi1014/10-1/5-4/1011/5检验数00-9/5-8/5-1/5-28/5X*=(ll/5,2/5,0)r；Z*=-28/5;Z*=28/5(3)max 2=-2xx-x2 st.-3为一+%-4%-3x3+4%+2x2=3二-6+/=3%(，二123,4,5,6)2 0C350000BCBXBXix2x3x4x5x60 x4-3-10100-30 x5-403010-60X61200013检验数-2-100000 x4z3-10100-30X34/3010-1/3020 x61200013检验数-2-10000-2Xi11/30-1/30010 x30-4/914/9-1/302/30X605/301/3012检验数0-1/30-2/3002(4).化为:max 2=-3xr-2x2-x3st.X+忍+%=6Xy-%+4 4X?+X3+/=3%(i=123,4,5,6)2 0由表知，此题无可行解。14.(1)由表知，C1C2为基变量的系数c-3-2-1000bCBXBXix2x3x4x5x60 x411110060 x5-10A010-40X60-11001-3检验数-3-2-10000 x40101102-1x31010-1040X6-1-10011-7检验数-2-200-100 x40101102-1x30A1001-3-3Xi1100-1-17检验数0000-3-20 x4001111-1-2x201-100-13-3Xi1010-104检验数0000-3-2 ACf=max+-3/2 1/2 J=3C2 g 6,+oo)1400,1600最优解将发生变化;AZ?2=+00g1300,+oo)f 100350 AZ?3=max-.-l=-50a入+-f 1001A,=50 4=300,400(3).4=500 e300,400AO=(O,O/5O)Ab=B-1A/7=1/2 0-2)-3 1 100 0 1-o-0150=300)1500150 7C38000bCBXBXiX2x3x4x53Xi101/202-2000 x400-311020008x201001500检验数00-3/20-20 x5-1/20-1/4011000 x450-1/21010008x21/21400400检验数-10-200-3200 x=(0,400)r；Z*=3200;15.1).Ci变化时，-5+C-0 14 70b0M：-4.2/?74.8Z?C23100bCBXBXix2x3x4x52Xi1111030 x5031117检验数0-3-1-20-6X*=(300,0,7)7；Z*=6;(3)把x*=(6,0,0,0,10尸代入约束条件-+3%2V-6+0 2不成立,该问题的最优解将发生变化,&-2x2+/=-2C2-11000BCBXB