1、1Study of the optimization of distribution routing problem based on the economy-matrix methodZhang Yu（School of Management, Xian Polytechnic University, Xian710048,China ）Abstract：On the study of the optimization of distribution routing problem, according to the status quo in China, this paper prese
2、nts the economy-matrix method to solve the problem and explains its solution steps in details and finally gives an example. The results demonstrates that the optimal or nearly optimal solutions to the distribution routing problem can be easily obtained by using economy-matrix method.Keywords：Distrib
3、ution routing; Economy-matrix method; Optimization; Furthest insert method1 、IntroductionWith the development of market economy and the specialization level of logistics technology, logistics and distribution industry has been developing rapidly. Logistics and distribution refers to in accordance wi
4、th users order, at the distribution center, picking and delivering goods1. In modern logistics system, logistics and distribution tends to develop toward low-volume, multi-species, and multi-batch time, which offers a personalized service .The distribution problem of modern logistics system involves
5、 many factors and the relationship is very complex. Among them, whether the choice of delivery lines is reasonable has a significant impact on accelerating the delivery speed, improving the quality of logistics services and reducing distribution costs. However, the operation status of most self- log
6、istics distribution center in our country is that: small delivery volume, too many end-customers, high shipping cost, lack of reasonable delivery planning; shortage of optimization software and logistics talent and so on. Therefore, finding a a simple and practical optimization method suitable for t
7、he current situation of logistics in our country has important significance.2、 Problem AnalysisThe transport of goods shipped from a central warehouse to the sub- library , and then the sub- library of vehicles shipped to retail outlets ( demand points) . General transport problems in operations res
8、earch from the central warehouse to the sub - library delivery arrangements can be resolved. This article focuses on the optimization algorithm from the sub- library delivery arrangements to retail outlets , through the development of rational delivery lines , fast and economical delivery of goods o
9、f sub - library users hands .In general, the delivery of goods is first from center warehouse to sub- warehouse, and then deliver to the 2customer and the delivery lines from sub- warehouse to customer can be tackled by optimization method. This paper mostly focuses on the application of optimizatio
10、n method from sub- warehouse to the customer.On the optimization problem of delivery route, experts at home and abroad propose a lot of theoretical methods, such as genetic method, tabu search method, simulated annealing method and so on2. These methods can solve complex distribution route optimizat
11、ion problem, but they have some drawbacks: the complexity of model, difficult solving steps, large amount of computation, which limit their promotion and application of logistics enterprises in China 3. In this paper, the economy-matrix method is used, which has the advantage of requiring less known
12、 conditions, easy operation and good optimization results. Even in the case of time limit or other restrictions, it still can be used to determine which vehicles to deliver goods to specific customer, as well as the optimal delivery route of each lorry, which is very suitable for the existing small
13、and medium sized logistics enterprises in our country 4. when the economy-matrix method is used, the conditions we need to know are as follows :the location of distribution center and the number of vehicles owned ;the location of each customer and their order . The optimization goals is to issue del
14、ivery mission statement to all vehicles involved (i.e. to determine the volume of goods and delivery routes, assign retail outlets for each vehicle), which enables the total cost of delivery to be minimum .3、Saving matrix analysisFirstly, the paper introduces the assumptions of this method. As the t
15、ransportation is a network structure, so it can be assumed that the distribution center and users to be a mathematical point on the net, the transport route between the distribution center and users can be assumes as a line. So, the shipping line optimization problem can be converted to find the bes
16、t path in the map spotted with points and lines. Set up the distribution center to be O point and its location coordinates is （x O，y O）; customers respective to be A , B , . N, and their position coordinates is respectively （x A，y A） 、（x B，y B）（ xN，y N）. The main solution steps of economy-matrix met
17、hod include:(1) Confirm the distance matrixConfirm the distance matrix is to confirm the distance between any two locations. The distance between any two points in the coordinate system are shown in equation 3-1.3（3-1）22),( MKMKyxDist In the equation, K、M can be taken any point between O and A N, so
18、 ,In this way, a distance matrix of order N +1 is constructed. The relationship of distance between any two points, that is Dist (K, M) = Dist (M, K) ,can be easy to see that the distance matrix is a symmetric matrix . Therefore, the distance matrix can be simplified to a upper triangular or lower t
19、riangular matrix .(2) Confirm the economy matrix The economy-matrix refers to the accumulated savings of the truck when it comes to the joint delivery. the Saving can be measured in accordance with distance, time or money and in this paper it is established in accordance with distance. The routes of
20、 transport tools can be confirmed by its different trips via location lines. the trip of O - K -O shows that it begins distribution center ,then deliveries goods to customer and returns to the distribution center in the end .The Savings of S (K , M) shows the accumulated distance when combines the t
21、rip of O -K -O and O -M- O, and its formula is shown in equation 3-2.（3-2）MKDistOistKDistM, K，M 的取值范围同式 3-1，节约矩阵表见表 3-1。Table 3-1 the saving matrixClient A Client B Client I Client NClient A S(A，A)Client B S(B，A) S(B，B) Client I S(I，A) S(I， B) S(I，I) Client N S(N，A) S(N，B) S(N，I) S(N，N)(3) to classi
22、fy clients to different transport routesThe goal to classify clients to different transport routes is to maximize the accumulated distance while completing the delivery mission. Firstly, classify each consumer to a separate transport line, then combine the existing line with a certain point. The mer
23、ge priority is based on the save size of the two lines. The greater saving distance, then the higher the priority. The merger feasibility rule is that the total transport volume is no more than the maximum load of the specific lorry. The merger process is end when the merger process is not feasible,
24、 then it goes the merger of another car until all customer orders are classified to appropriate vechiles.(4) schedule the initial delivery lines for each vehicle4Schedule delivery lines is to determine the order of each vehicle in order to minimize the transport costs. As for practical application,
25、customer service often has certain limitations and it is called time window. The time window is divided into hard time window and soft time window. Hard time window means service time is given by the customer, otherwise the client may refuse the service ; soft time window allows late service , but t
26、he customer must be compensated for their loss6.In this paper, it is about optimization problem with soft time window, where the introduction of the concept of penalty function, that is, when the time window of a customers demands are not met , the specific vehicles must get punishment7. K-time wind
27、ow to set up a customer for (ETK ,LTK), if the vehicle to reach the K point in the ETK before , the vehicle must wait for this , the occurrence of the loss of opportunity cost ; if the vehicle after LTK to the K point , the service has been delayed .so it is required to pay a fine . Accordingly, the
28、 penalty function is set in equation 3-3. （3-KKKLTtLTtEtttP,0,)(3） In the above formula, is the time point when the vehicle arrives; and are penalty coefficients that Kt can be used by per unit of time, transport distance or transportation costs and their value are determined in accordance with the
29、degree of importance of delivery time required by the customer: The more important customers or the more demanding time , the value and the greater ; is the punished transport KKtPdistance or cost 8. In this paper transport distance functions as penalty function. When assigning the delivery routes,
30、the punished distance is added to the original transport distance, making the original route longer to be re-optimized. So the optimization process must be constantly approaching to meet the customers time requirements, so as to make the total transport distance constantly shorter, and ultimately th
31、e shortest delivery line must be completely or the maximum extent to meet customers delivery time requirements. After the introduction of punishment distance function, you can use the following method to schedule the initial delivery lines of the vehicle. It is worth mentioning that penalty function
32、 value must be taken into account during the process of line length calculation or its added distance9.The farthest insertion method. Approach to the distribution center for the initial point of the transport routes , have not been included in the customer point of the transport routes to the same v
33、ehicle to plan for the collection , respectively, estimated that every point of the collection is inserted into the existing line to any location ( except the starting point ) the length 5of the line after the increase in value , choose the minimum , that is to seek the insertion of the minimum adde
34、d value of the original length of the existing line ; And then compare the value of the smallest increase of all the collection , select the maximum value of which , and the value corresponding to the customer sites into the existing line , get a new itinerary . Repeat this process continues until a
35、ll the customer point of the line have all been included so far. Distance from the current line farthest clients are included in this method is called the farthest insertion method.The nearest insertion method .It is just the opposite process of the farthest insertion method. The nearest neighbor me
36、thod . This method startes from the distribution center, every step it extends trip by absorbing the nearest customer until absorbing all the customers belong to this line.(5)The optimization of initial delivery lines1After the use of the nearest neighbor method, you can use the dichotomy and tricho
37、tomy to make improvement in order to obtain more shorter transport distance or transportation costs.It is worth mentioning that the initial delivery line is the best one when we use the farthest insertion method or the nearest insertion method, so there is no need to make improvement 10.In this pape
38、r , we choose the farthest insertion method.4 、InstanceA companys distribution center (DC) received orders from 12 customers one morning. The coordinates of distribution center and customers are shown in Table 4-1. There are four trucks, each trucks maximum load is 225 units (1 unit = 0.1 tons). Det
39、ermine each trucks shipping lines by using economy matrix method.Table 4-1 clients coordinates and order formX Y OrderDC -12 0 74C1 -5 6 55C2 -15 7 68C3 -12 9 109C4 -3 15 81C5 0 20 41C6 2 17 74C7 4 7 52C8 6 1 806C9 6 15 40C10 7 20 103C11 9 7 75C12 -12 0 74解：运用式 3-1 计算确定出距离矩阵见表 4-2。The solution: use
40、equation 3-1 to calculate distance matrix as shown in Table 4-2 .Table 4-2 distance matrixDC C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12DC 0C1 12 0C2 8 9 0C3 17 8 10 0C4 15 9 8 4 0C5 15 17 9 14 11 0C6 20 23 15 20 16 6 0C7 17 22 13 20 16 5 4 0C8 8 17 9 19 16 11 14 10 0C9 6 18 12 22 20 17 20 16 6 0C10 16 2
41、3 14 22 19 9 8 4 8 14 0C11 21 28 18 26 22 11 7 6 13 19 5 0C12 11 22 14 24 21 14 16 12 5 7 9 13 0use equation 3-2 to calculate saving matrix as shown in table 4-2.Table 4-3 saving matrixC1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12C1 0C2 11 0C3 21 15 0C4 18 15 28 0C5 10 14 18 19 0C6 9 13 17 19 29 0C7 7 12 1
42、4 16 27 33 0C8 3 7 6 12 12 14 15 0C9 0 2 1 1 4 6 7 8 0C10 5 10 11 12 21 28 29 16 8 0C11 5 11 12 14 25 34 32 16 8 33 0C12 1 5 4 5 12 15 16 14 10 18 19 0 Assign clients to different transport vehicles .As shown in table 4-3, the largest saving is from customer 6 and 11, the savings amount is 34 and th
43、e total 7volume is 144, less than 225.so 6 and 11 is combined. 6 or 11 related to a maximum savings of 6 and 7, the savings amount to 33 among the total volume of 218, less than 225, so 11, 7, 6 merger. The largest savings and 11, 7, 6 10 and 11, savings of 33, the total volume of the four 287 great
44、er than 225, so the merging is not feasible and temporarily stop the merger. The final routes is 6, 7 and 11.Exclude customers 11,6 and 7, the next largest savings come from 3 and 4, savings is 28 and the total volume is 177, less than 225 , so 3 and 4 is combined . Considering 3 and 4, the largest
45、savings from 3 and 1, the savings amount is 21, among the total volume of 251 greater than 225, the merger is not feasible. 3 and 4, the largest savings from the 4 and 5, the savings is 19, among the total volume of 258 greater than 225, the merger is not feasible. 3 and 4, the largest savings from
46、the 2 and 3, the savings amount to 15, among the total volume of 232 greater than 225, the merger is not feasible. 3 and 4, the largest savings from 4 and 10, the savings amount to 12, among the total volume is 246 which is greater than 225, so the merger is not feasible. Considering 3 and 4, the la
47、rgest savings from 4 and 8, the savings amount to 12 and the total volume is 229 that is greater than 225, the merger is not feasible . The final route is 3 and 4.Exclude customers 3and 4, the next largest savings from 5 and 10, save 21 and the total volume is 150 which is less than 225 , so 5 and 1
48、0 merge . From 10 and 12 with 5 and 10 of the largest savings, savings is18 and the total volume is 225. So the merger is feasible and the final third route is 5, 10 and 12.Exclude customers 5, 10 and 12., the next largest savings from 1 and 2, the savings amount to11 and the total volume is 129, le
49、ss than 225 , so 1 and 2 combined . As for remaining clients 8 and 9, adding customer 8, the total volume is 181 , the merger is feasible . Add customer 9, the total volume is 221, the merger is feasible , so the final fourth route is 1, 2, 8 and 9 .Schedule initial routes for every truck. In this paper, we choose the farthest insertion method and it is also the best routes.For the first group 6,7, 11, the first trip only includes distribution centers, a length of 0 .
