1、运筹学胡运权版第三章运输问题课后习题答案资料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A1, A2,A3的生产量、各销售点B1,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?表销地产地B1B2B3B4产量A141241116A22103910A38511622销量814121448解:一、该运输问题的数学模型为: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
2、业务。销地产地B1B2B3B4产量A141241116A28210392810A38511622销量814121448销地产地B1B2B3B4产量A141241116A28210239810A38511622销量814101448销地产地B1B2B3B4产量A14121041011 16 6A28210239810A38511622销量814101448销地产地B1B2B3B4产量A14121041011 16 6A28210239810A3814511146 22 8销量814101448销地产地B1B2B3B4产量A14121041011 16 6A28210239810A38145118
3、146 22 0销量81410 14 648销地产地B1B2B3B4产量A141210461011 16 0A282102398 10 0A38145118146 22 0销量81410 14 048此时得到一个初始调运方案(初始可行解):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-1=3+4-1=6).总运费为(目标函数值)2. 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B
4、1B2B3B4产量行差额A1412411160A221039101A385116221销量814121448列差额2513销地产地B1B2B3B4产量行差额A1412411160A2210398101A38145111462212销量814121448列差额2513销地产地B1B2B3B4产量行差额A1412411160A2210390101A38145118146221销量814121448列差额2513销地产地B1B2B3B4产量行差额A1412411160A282103890 10 21A38145118146221销量814121448列差额2513销地产地B1B2B3B4产量行差额A
5、1412124121116 47A2821032890 10 06A38145118146221销量814121448列差额2513销地产地B1B2B3B4产量行差额A14121244121116 07A2821032890 10 06A38145118146221销量8141214 048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-1=3+4-1=6)。总运费为(目标函数值):三、解的
6、最优性检验 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:(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:x11,x12,x22,x24,x31,x33。销地产地B1B2B3B4产量A1X1141210461116A2821023910A38145118622销量81412144811 = C11 + C23 - (C13 + C21) = 4 + 3 ( 4 + 2 ) =1销地产地B1B2B3B4产量A14X121210461116A2821023910A38145118622销量81412144812 = C12 + C
7、34 - (C14 + C32) = 12 + 6 ( 11 + 5 ) =2销地产地B1B2B3B4产量A141210461116A282X221023910A38145118622销量81412144822= C22 + C13 + C34 - (C23 + C14 + C32) = 10 + 4 + 6 ( 3 + 11 + 5 ) = 20 19 =1销地产地B1B2B3B4产量A1X1141210461116A2821023X24910A38145118622销量81412144824 = C24 + C13 - (C14 + C23) = 9 + 4 ( 11 + 3 ) = -1
8、销地产地B1B2B3B4产量A141210461116A2821023910A3X318145118622销量81412144831= C31 + C14 + C23 - (C34 + C13 + C21) = 8 + 11 + 3 ( 6 + 4 + 2 ) = 22 12 = 10销地产地B1B2B3B4产量A141210461116A2821023910A38145X33118622销量81412144833 = C33 + C14 - (C13 + C34) = 11 + 11 ( 4 + 6 ) =12由于24 = C24 + C13 - (C14 + C23) = 9 + 4 (
9、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:x11,x12,x22,x23,x31,x33。(伏格尔法)销地产地B1B2B3B4产量A1X1141212441116A2821032910A38145118622销量81412144811 = C11 + C24 - (C14 + C21) = 4 + 9 ( 11 + 2 ) = 0销地产地B1B2B3B4产量A14X121212441116A2821032910A38145118622销量81412144812 = C12 + C34
10、- (C14 + C31) = 12 + 6 ( 11 + 5 ) = 2销地产地B1B2B3B4产量A141212441116A282X221032910A38145118622销量81412144822 = C22 + C34 - (C24 + C32) = 10 + 6 ( 9 + 5 ) = 16 14 = 2销地产地B1B2B3B4产量A141212441116A28210X2332910A38145118622销量81412144823 = C23 + C14 - (C13 + C24) = 3 + 11 ( 4 + 9 ) = 14-13=1销地产地B1B2B3B4产量A1412
11、12441116A2821032910A3X318145118622销量81412144831 = C31+ C24 - (C21 +C34) = 8 + 9 ( 2 + 6 ) = 17-8 = 9销地产地B1B2B3B4产量A141212441116A2821032910A38145X33118622销量81412144833 = C33 + C14 - (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
12、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。销地产地B1B2B3B4产量A141210461116A2821023910A38145118622销量814121448构造方程组:u1 + v3 = c13 = 4u1 + v4 = c14 = 11u2 + v1 = c21 = 2u2 + v3 = c23 = 3u3 + v2 = c32 = 5u3 + 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 =
13、 10,u2 = -1,v1 = 3,将其代入非基变量检验数:ij=Cij - (ui + vj),得:11=C11 - (u1 + v1) = 4 ( 0 + 3 ) = 112=C12 - (u1 + v2) = 12 ( 0 + 10 ) = 222=C22 - (u2 + v2) = 10 ( -1 + 10 ) = 124=C24 - (u2 + v4) = 9 ( -1 + 11 ) = -131=C31 - (u3 + v1) = 8 ( -5 + 3 ) = 1033=C33 - (u3 + v3) = 11 ( -5 + 4 ) = 12与闭回路法计算的结果相同。(2)然后对
14、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,x21,x24,x32,x34。销地产地B1B2B3B4产量A141212441116A2821032910A38145118622销量814121448构造方程组:u1 + v3 = c13 = 4u1 + v4 = c14 = 11u2 + v1 = c21 = 2u2 + v4 = c24 = 9u3 + v2 = c32 = 5u3 + 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 =
15、-2,v1 = 4,将其代入非基变量检验数:ij=Cij - (ui + vj),得:11=C11 - (u1 + v1) = 4 ( 0 + 4 ) = 012=C12 - (u1 + v2) = 12 ( 0 + 10 ) = 222=C22 - (u2 + v2) = 10 ( -2 + 10 ) = 223=C23 - (u2 + v3) = 3 ( -2 + 4 ) = -131=C31 - (u3 + v1) = 8 ( -5 + 4 ) = 933=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
16、用最小元素法求得的初始方案中,由于240,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A141210461116A2821023910A38145118622表2销地产地B1B2B3B4产量A141210+246-21116A282102-230+2910A38145118622表3销地产地B1B2B3B4产量A141212441116A2821032910A38145118622调整后的结果如表3所示,此结果正好与使用伏格尔法求得的结果相同,因此最优性检验过程同前,由于非基变量的
17、检验系数都大于等于零,因此该方案是最优方案,最优解为:x13=12,x14=4,x21=8,x24=2,x32=14,x34=8。将最优解代入到目标函数中,得总运费为(目标函数值):P66: 9.解:首先列出这一问题的产销平衡表,见表1。 表1 销地产地B1B2B3B4产量A1A2A3317119432101085749销量3656一、该运输问题的数学模型为: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. 最小元素法第1步,从表1中找出最小运价为1,表示应先将A2的产品供应B1。在表中A2和B1的交叉格处填上3,得表2。将表2中的B1列运价划去,得表3表2 销地产地B1B2B3B4产量A1A2A333 1711943210108574 19销量3656表3 销地产地B1B2B3B4产量A1A2A333 1711943210108574 19销量3656第2步,在表3未划去的元素中再找出最小运价为2,确定A2多余的1 t物资供应B3 。得表4。将表4的A2行运价划去,得表5
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