最新运筹学胡运权版第三章运输问题课后习题答案.docx

1、最新运筹学胡运权版第三章运输问题课后习题答案P66: 8某部门有3个生产同类产品的工厂（产地），生产的产品由4个销售点出售，各工 厂Ai, A，A3的生产量、各销售点 Bi，B2, B3, B4的销售量（假定单位为 t）以及各工厂 到销售点的单位运价（元/t）示于下表中，问如何调运才能使总运费最小？解：一、该运输问题的数学模型为:3 4min z 八 、 q 乂耳=4x11 12x12 4x13 11x14 2x21i 4 j 410x22 3x23 9x24 8x31 5x32 1 1x33 6X34%+X12+X13+ X14=16X21+X22+ X23+ X24=10X31+X32+X

2、33+ X34=22X11+X21+X31=8X12+X22+ X32=14X13+ X23+ X33=12X14+ X24+ X34=14XijX0,i=1.2,3;j =1,2,3,4X11XI2X13X14X21X22X23X24X31X32X33X34(1111111111111111111111111I11111I111丿7旳2可以证明约束矩阵的秩为r (A)=:6.从而基变量的个数为6.二、给出运输问题的初始可行解（初始调运方案）1.最小元素法思想：优先满足运价（或运距）最小的供销业务。销地产地Bi11-B2B3B4产量Ai41241116A28|21039108 2厂11A3；8

3、511622111销量814121448销地产地=1BiB4产量Ai4；121011116 610LA2-J1-0-19-_ 1_nA28109-1082 A381458622 01 14销量8 141014 648产地销地B1Bl1 1B3 !111B41产量A141241116 0110 116 1 11102_13-J9T28Tl U3 981 U XJI I21I11A3814115118|622 01丄11Ii14销量814 -1014 ； 048T1此时得到一个初始调运方案（初始可行解） ：X13 =10, X14 =6, X21 =8, X23 = 2, X32 = 14, X3

4、4 = 8,其余（非基）变量全等于零。此解满足所有约束条件， 且基变量（非零变量）的个数为6 （等于 m+n-仁3+4-仁6）.总运费为（目标函数值）3 4z -二 CjXji j=1=10 4 6 11 8 2 2 3 14 5 8 6 = 2462.伏格尔（Vogel）法伏格尔法的基本思想：运输表中各行各列的最小运价与次小运价之差值 （罚数）应尽可能地小。或者说：优先供应罚数最大行（或列）中最小运费的方格，以避免将运量分配到该行（或 该列）次小运距的方格中。销地产地B1B2B3B4产量行差额A1412411160A221039101A385116221销量814121448列差额2513销

5、地产地B1B2B3B4产量行差额A1412411160A221039101A38141511622 8141 21销量814121448列差额2513产地销地B11B2；1B3B4产量行差额A14124111601 11A221039101111A38J51160T4 _111822141销量814 1121448列差额25 113销地产地Bi丨|IIIB2B3丨iiiiiiB4产量行差额Ai4121241116 41271111 II |1A22I 10910 068281 II11 11IA31 8145116“ 0r111rr11*822141销量8 !11412 !L1448列差额2

6、!151 !13 Il I 销地产地B1B2B3B4产量行差额A1 1 4121241116 0127A421091Q 0.6828A3516CC 0L14r r为22141销量8141214 (48列差额2513此时得到一个初始调运方案 （初始可行解）：X13 = 12, X14 = 4, X21 = 8, X24 = 2, X32 = 14, X34 = 8其余（非基）变量全等于零。此解满足所有约束条件， 且基变量（非零变量）的个数为6 （等于 m+n-仁3+4-仁6）。总运费为（目标函数值）：3 4Z =、CjXj =12 4 4 11 8 2 2 9 14 5 8 6 = 244 三、

7、解的最优性检验1闭回路法（以下的闭回路都是顺时针方向）看非基变量的检验数是否满足：Gj _ 0.（1）首先对用最小元素法所确定的初始基本可行解进行检验。参见前面的计算结果，可知非基变量分别为： Xii, X12，X22, X24，X31, X33。产地销地B1B2B3B4产量A1X1141210461116厂A218 12101II239101A38145118622销量8141214485i = C11 + C23 - (C13 + C21) = 4 + 3 4 + 2 ) =1销地产地B1B2B3B4产量A14X12 厂.12一10 -46111116A2821 r111023111191

8、0A3814 -_ C L 1 斗 8622销量814121448(712 = Cl2 + C34 - (Cl4 + C32) = 12 + 6 - 11 + 5 ) =2产地销地B1B2B3B4产量A14121046 11116|! | jA282X22.一 10 -231 |9101111A38f14 15 =一一111_|8622销量814121448C22= C22 + Cl3 + C34 - (C 23 + C14 + C32)= 10 + 4 + 6 -( 3 + 11 + 5 ) = 20 T9 =1销地产地BiB2B3B4产量AiX11412106-1116T111A28210

9、21 3X24! 910l_A38145118622销量814121448C24 = C24 + C13 - (C14 + C23) = 9 + 4 11 + 3 ) = -1销地产地B1B2B3B4产量A141210厂46-| 11161li11A282102ri-3II1 910T1111A3X318 !145118I 6221jj销量814121448c31= C31 + C14 + C23 - (C34 + C13 + C21) = 8 + 11 + 3 ( 6 + 4 + 2 ) = 22 -12 = 10销地产地B1B2B3B4产量A141210厂.4 一6-1 11161L1A2

10、82102! 3! 910厂111A38145X33V十-8622销量814121448C33 = C33 + C14 - (C13 + C34) = 11 + 11 4 + 6 ) =12由于024 = C24 + C13 - (C14 + C23)= 9 + 4 -11 + 3 ) = -1 0 ,所以当前方案不是最优方案。(2)然后对用伏格尔法所确定的初始基本可行解进行检验。参见前面的计算结果，可知非基变量分别为：Xu, X12，X22, X23, X31, X33。(伏格尔法)产地销地BiB2B3B4产量AiX1141212441116LA28 21032:910H- 1A381451

11、18622销量814121448(711 = C11 + C24 - (Cl4 + C21)= 4 + 9 - 11 + 2 ) = 0销地产地B1B2B3B4产量A14X12712124411161A28211111032! 91011A3814丄5 118622销量81412144852 = C12 + C34 - (C14 + C31)= 12 + 6 - 11 + 5 ) = 2销地产地B1B2B3B4产量A141212441116A282X2229101 iI |A3814；5118；6122L. 1销量814121448C22 = C22 + C34 - (C24 + C32)=

12、10 + 6 -( 9 + 5 ) = 16 T4 = 2产地销地B1B2B3B4产量A141212 4 亠4 _1111611!1A28210X23:32 _:9101A38145118622销量814121448C23 = C23 + C14 - (C13 + C24) = 3 + 11 _ ( 4 + 9 ) = 14-13=1销地产地B1B2B3B4产量A141212441116A8 |f1210r - 329101 iA3X31：8145-118622销量814121448o3i = C31+ C24 - (C21 +C34)= 8 + 9 ( 2 + 6 ) = 17-8 = 9销

13、地产地B1B2B3B4产量A1412124411I161 LA28210:329101 A38145X33!. 118622销量814121448C33 = C33 + Ci4 - (C13 + C34) = 11 + 11 _( 4 + 6 ) = 22-10 = 12由于所有非基变量的检验数都大于零，说明当前方案是最优方案，最优解为:X11 = 12, X14=4，X21=8， X24=2， X32 = 14， X34=8。2位势法(1)首先对用最小元素法所确定的初始基本可行解进行检验。参见前面的计算结果，可知 基变量分别为：X13, X14, X21, X23，X32, X34。产地销地

14、B1B2B3B4产量A141210461116A2821023910A38145118622销量814121448构造方程组：广 U1 + V3 = C13 = 4U1 + V4 = C14 = 11U2 + V1 = C21 = 2U2 + V3 = C23 = 3U3 + V2 = C32 = 5U3 + V4 = C34 = 6令自由变量ui = 0 ,将其代入方程组，得：U1 = 0 , V3 = 4 , V4 = 11 , U3 = -5 , V2 = 10, U2 = -1 , Vi = 3，将其代入非基变量检验数: qj=Cij - (Ui + Vj)，得：oi1=C11 - (

15、U1 + V1)= 4 -( 0 + 3 ) = 1Of2 = C12 -(U1 + V2)= 12 -( 0 + 10 ) = 2(22 = C22 -(U2 + V2)= 10 -( -1 + 10 ) = 1(24 = C24 -(U2 + V4)= 9 一( -1 + 11 ) = -1(31 =C31 - (U3 + V1)= 8 _( -5 + 3 ) = 10C33 = C33 -(U3 + V3)= 11 -( -5 + 4 ) = 12与闭回路法计算的结果相同。(2)然后对用伏格尔法所确定的初始基本可行解进行检验。参见前面的计算结果，可知基变量分别为：x13, x14, x2

16、1 , x24, x32, x34o产地7地B1B2B3B4产量A141212441116A2821032910A8145118622销量814121448构造方程组：厂 U1 + V3 = C13 = 4U1 + V4 = C14 = 11U2 + V1 = C21 = 2U2 + V4 = C24 = 9U3 + V2 = C32 = 5L U3 + V4 = C34 = 6令自由变量U1 = 0 ,将其代入方程组，得：U1 = 0 , V3 = 4 , V4 = 11 , U3 = -5 , V2 = 10, U2 = -2 , V1 = 4，将其代入非基变量检验数: (j =Cij -

17、 (Ui + Vj),得：(5f1=C11 - (U1 + V1) = 4 -( 0 + 4 ) = 0Ot2 = C12 - (U1 + V2)= 12 -( 0 + 10 ) = 2(2 = C22 -(U2 + V2)= 10 -( -2 + 10 ) = 2(3 = C23 -(U2 + V3)= 3 -( -2 + 4 ) = -1(1 =C31 -(U3 + V1)= 8 -( -5 + 4 ) = 9C83 = C33 - (U3 + V3)= 11 -( -5 + 4 ) = 12与闭回路法计算的结果相同。四、解的改进（用闭回路法调整）在使用最小元素法求得的初始方案中，由于 0240,说明当前方案不是最优，需要改进或调整。见表1中非基变量X24所在的闭回路，调整量为 = min2,6 = 2。调整过程见表 2：表1销地产地B1B2B3B4产量A141210厂