节约里程法练习答案

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1、,单击此处编辑母版标题样式,,单击此处编辑母版文本样式,,第二级,,第三级,,第四级,,第五级,,,,*,单击此处编辑母版标题样式,,单击此处编辑母版文本样式,,第二级,,第三级,,第四级,,第五级,,,,*,节约里程法练习答案,练习,1,,有一配送中心（,Q,）要向,10,个用户配送，配送距离（公里）和需用量（吨）如下图所示。,,假设：采用最大载重量,2,吨、,4,吨、,8,吨三种汽车，并限定车辆一次运行距离,50,公里。,,用,节约里程法,选择最佳配送路线和车辆的调度。,配送网络图,Q,9,e,,1.4,8,6,10,7,6,f,,1.5,8,g,,0.6,3,2,9,h,,0.8,4,i

2、,,0.5,10,j,,0.6,7,a,,0.7,10,b,,1.5,9,c,,0.8,7,d,,0.4,8,5,5,4,4,11,8,第一步：选择初始方案：从,Q,点向各点分别派车送货。,P0,,,10,,,,9,,,,4,,,,7,,,,9,,,,5,,,,8,,,,14,,,,10,,,,5,,,,8,,,,18,,,,14,,,,9,,,,6,,,,8,,,,18,,,,17,,,,15,,,,13,,,,7,,,,3,,,,13,,,,12,,,,10,,,,11,,,,10,,,,6,,,,4,,,,14,,,,13,,,,11,,,,12,,,,12,,,,8,,,,2,,,,1

3、0,,,,11,,,,15,,,,17,,,,18,,,,18,,,,17,,,,11,,,,9,,,P1,,P2,,P3,,P4,,P5,,P6,,P7,,P8,,P9,,P10,,,7,,,,4,,,,8,,,,13,,,,15,,,,15,,,,15,,,,10,,,,11,,,,8,,,P1,,P2,,P3,,P4,,P5,,P6,,P7,,P8,,P9,,P10,,,,,,,,,0.7,,,,1.5,,,,0.8,,,,0.4,,,,1.4,,,,1.5,,,,0.6,,,,0.8,,,,0.5,,,,0.6,,,需要量,,最短距离矩阵,,第二步：作出最短距离矩阵，从配送网络图中列出

4、配送中心至用户相互间的最短距离矩阵 。,第三步：从最短矩阵中，计算用户相互间的节约里程。,15,,8,,11,,4,,7,,10,,0,,3,,6,,10,,0,,0,,0,,3,,9,,0,,0,,0,,0,,1,,5,,0,,0,,0,,0,,0,,4,,5,,9,,4,,0,,0,,0,,1,,2,,5,,P1,,P2,,P3,,P4,,P5,,P6,,P7,,P8,,P9,,P1,,P2,,P3,,P4,,P5,,P6,,P7,,P8,,P9,,P10,,13,,8,,1,,0,,0,,0,,0,,0,,9,,节约里程计算过程,,第四步：将节约里程按大小顺序排列分类。,Ⅰ,,a—b,,

5、15,,Ⅱ,,a—j,,13,,Ⅲ,,b—c,,11,,Ⅸ,,f—g,,5,,Ⅸ,,g—h,,5,,Ⅸ,,h—i,,5,,Ⅹ,,a—d,,4,,Ⅹ,,b—i,,4,,Ⅹ,,f—h,,4,,Ⅳ,,c—d,,10,,Ⅳ,,d—e,,10,,Ⅴ,,a—i,,9,,Ⅴ,,e—f,,9,,Ⅴ,,i—j,,9,,Ⅵ,,a—c,,8,,Ⅵ,,b—j,,8,,Ⅶ,,b—d,,7,,Ⅷ,,c—e,,6,,Ⅻ,,g—i,,2,,Ⅺ,,b—e,,3,,Ⅺ,,d—f,,3,,XIII,,c—j,,1,,XIII,,e—g,,1,,XIII,,f—i,,1,,节约里程顺序表,分类,,用户连接线,,节约里程,,分类,,用

6、户连接线,,节约里程,,第五步：按节约里程大小顺序，组成配送线路。,d,c,a,e,Q,f,g,h,b,j,i,8,8,8,7,3,4,5,4,4,7,10,初始配送路线图,,按上述方法，逐次选代，优化配送线路 。,d,c,a,Q,f,g,h,b,j,i,e,5,5,4,6,7,8,3,7,2,9,10,4,最佳配送线路,,练习,2,,设一配送中心向,13,个客户配送商品，配送中心及客户间的最短距离如表,1,所示，如果配送的车辆载重为,200,吨，那么利用节约法求解的配送路线的步骤如下：,,,第一步，计算配送中心到库户间的最短距离，画出距离表。因为本例已给出，所以可以直接进行第二步。,,,表,

7、1,配送中心到客户间的最短距离表,,DC,1,2,3,4,5,6,7,8,9,10,11,12,13,1,12,0,,,,,,,,,,,,,2,8,9,0,,,,,,,,,,,,3,17,8,10,0,,,,,,,,,,,4,15,9,8,4,0,,,,,,,,,,5,15,17,9,14,11,0,,,,,,,,,6,20,23,15,20,16,6,0,,,,,,,,7,17,22,13,20,16,5,4,0,,,,,,,8,8,17,9,19,16,11,14,10,0,,,,,,9,6,18,12,22,20,17,20,16,6,0,,,,,10,16,23,14,22,19,9,

8、8,4,8,14,0,,,,11,21,28,18,26,22,11,7,6,13,19,5,0,,,12,11,22,14,24,21,14,16,12,5,7,9,13,0,,13,15,27,20,30,28,22,23,20,12,9,16,20,8,0,需求量,,48,36,43,92,57,16,56,30,57,47,91,55,38,第二步，根据最短距离表，利用节约法计算出用户间的节约里程，并由大到小排列，编制节约里程顺序表，如表2所示。,,ΔL=（L,a,+L,b,）－L,ab,,,1—2：L,1,+L,2,-L,12,=12+8-9=11,,1—3：L,1,+L,3,-L,

9、13,=12+17-8=21,,1—4：L,1,+L,4,-L,14,=12+15-9=18,,1—5：L,1,+L,5,-L,15,=12+15-17=10,,1—6：L,1,+L,6,-L,16,=12+20-23=9,,1—7：L,1,+L,7,-L,17,=12+17-22=7,,1—8：L,1,+L,8,-L,18,=12+8-17=3,,1—9：L,1,+L,9,-L,19,=12+6-18=0,,1—10：L,1,+L,10,-L,1、10,=12+16-23=7,,1—11：L,1,+L,11,-L,1、12,=12+21-28=5,,1—12：L,1,+L,12,-L,1、1

10、2,=12+11-22=1,,1—13：L,1,+L,13,-L,1、13,=12+15-27=0,,2—3：L,2,+L,3,-L,23,=8+17-10=15,,2—4：L,2,+L,4,-L,24,=8+15-8=15,,2—5：L,2,+L,5,-L,25,=8+15-9=14,,2—6：L,2,+L,6,-L,26,=8+20-15=13,,2—7：L,2,+L,7,-L,27,=8+17-13=12,,2—8：L,2,+L,8,-L,28,=8+8-9=7,,2—9：L,2,+L,9,-L,29,=8+6-12=2,,2—10：L,2,+L,10,-L,2、10,=8+16-14=

11、10,,2—11：L,2,+L,11,-L,2、11,=8+21-18=11,,2—12：L,2,+L,12,-L,2、12,=8+11-14=5,,2—13：L,2,+L,13,-L,2、13,=8+15-20=3,,3—4：L,3,+L,4,-L,34,=17+15-4=28,,3—5：L,3,+L,5,-L,35,=17+15-14=18,,3—6：L,3,+L,6,-L,36,=17+20-20=17,,……,,表,2,节约里程表,序号,路程,节约里程,,（,L,a,+L,b,）－,L,ab,序号,路程,节约里程,,（,L,a,+L,b,）－,L,ab,序号,路程,节约里程,,（,L,

12、a,+L,b,）－,L,ab,1,6,—,11,34,11,5,—,10,22,21,11,—,13,16,2,6,—,7,33,12,1,—,3,21,22,8,—,10,16,3,7,—,1,1,32,13,11,—,12,19,23,7,—,12,16,4,10,—,11,32,14,4,—,5,19,24,4,—,7,16,5,7,—,10,29,15,4,—,6,19,25,8,—,11,16,6,5,—,6,29,16,1,—,4,18,26,2,—,3,15,7,3,—,4,28,17,3,—,5,18,27,2,—,4,15,8,6,—,10,28,18,12,—,13,18,

13、28,7,—,8,15,9,5,—,7,27,19,10,—,12,18,29,6,—,12,15,10,5,—,11,25,20,3,—,6,17,…,…,…,,第三步，根据节约里程顺序表和配送中心的约束条件，绘制配送路线。其具体步骤如下：首先选择最节约里程的路段（6,—,11），然后是（6,—,7），由于配送路线必须包含DC，且每条循环路线上的客户需求量之和要小于200吨，在接下的选择中满足条件的只有路段（11,—,8），此时载重总量为193吨，因为在余下选择中没有满足条件的客户，所以，第一回合的配送路线为（DC,—,7,—,6,—,11,—,8,—,DC）。,按此方法类推，其余的配送路线分别是（DC,—,1,—,3,—,4,—,DC）、（DC,—,5,—,10,—,12,—,13,—,DC）、（DC,—,2,—,9,—,DC）。,,总路程为：（17+4+7+13+8）+（12+8+4+15）+(15+9+9+8+15)+(8+12+6)=170,,原路程为：,,2,×,（12+8+17+15+15+20+17+8+6+16+21+11+15） =362,,总共节约里程为：362－170=192,,或（33+34+16）+（28+21）+（22+18+18）+2=192,,