research article mutation ant colony algorithm of milk...

Hindawi Publishing CorporationDiscrete Dynamics in Nature and SocietyVolume 2013 Article ID 418436 6 pageshttpdxdoiorg1011552013418436

Research ArticleMutation Ant Colony Algorithm of Milk-RunVehicle Routing Problem with Fastest CompletionTime Based on Dynamic Optimization

Jianhua Ma and Guohua Sun

School of Management Science and Engineering Shandong University of Finance and Economics Jinan 250014 China

Correspondence should be addressed to Jianhua Ma jianhuama126com

Received 20 December 2012 Accepted 28 February 2013

Academic Editor Xiang Li

Copyright copy 2013 J Ma and G Sun This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The objective of vehicle routing problem is usually to minimize the total traveling distance or cost But in practice there are a lotof problems needed to minimize the fastest completion time The milk-run vehicle routing problem (MRVRP) is widely used inmilk-run distribution The mutation ACO is given to solve MRVRP with fastest completion time in this paper The milk-run VRPwith fastest completion time is introduced first and then the customer division method based on dynamic optimization and splitalgorithm is given to transform this problem into finding the optimal customer order At last the mutation ACO is given and thenumerical examples verify the effectiveness of the algorithm

1 Introduction

The vehicle routing problem (VRP) was firstly broughtforward by Dantzig and Ramser in 1960 [1] With thedevelopment of VRP there exist several variations and spe-cializations Mostly VRP aims to minimize the total traveldistance (or travel time) and total cost But for distributionof fast foods express delivery and emergency supplies theseobjectives are not suitable and the completion time is moreimportant

Though the vehicle routing problem with time windowstakes the service time into consideration [2] it cannot solvethe problem when all customers want to be served as earlyas possible Nikolakopoulou et al solved a vehicle routingproblem by balancing the vehicles time utilization [3] Butit maybe takes a long time for some vehicles to completethe distribution tasks when the vehicles time utilizations arebalanced The VRP with fastest completion time was studiedto minimize the fastest completion time [4ndash6]

Traditional vehicle routing problem assumes that the totaldistribution can be completed by one vehicle in a roundtrip In practice the number of vehicles is limited and thereare many customers to be served So a round trip by one

vehicle is impossible and more round trips and vehicles areneeded such as the tobacco distribution with thousands ofcustomers As JIT production small-scaled and multiple-batch distributions are more popular in which one vehicleis needed to collect goods on multiple round trips This kindof vehicle routing problem is called milk-run vehicle routingproblem (MRVRP)

For problem aiming to minimize total travel distance ortotal cost MRVRP can be transferred into the traditionalVPR by increasing the routes number and allowing a vehicleserving customers in different routes But for problem tominimize the fastest completion time the completion timein the first period affects that in the second period It ismultiperiod optimization problem so the algorithm of theMRVRP with fastest completion time will be studied basedon dynamic optimization

VRP is an NP-hard problem [7] heuristics and evolu-tionary algorithms are used to solve VRP In this papermutation ant colony algorithm is used to solve MRVRP withfastest completion time In the next section we will give thedescription of the MRVRP with fastest completion time InSection 3 optimal divisionmethods of customer orders basedon dynamic optimization are given In Section 4 mutation

2 Discrete Dynamics in Nature and Society

Figure 1 The milk-run vehicle routes of two vehicles

ant colony algorithm is given to solve the problem In the lastsection a numerical example is given

2 MRVRP with Fastest Completion Time

Suppose that there is one depot serving n customers whichhave 119898 vehicles with capacity 119908 The demand of customer 119894is 119902119894(119894 = 1 119899) and satisfies (119896 minus 1)119908 le sum

119899

119894=1119902119894lt 119896119908

which means that for every vehicle it needs 119896 times to finishthe distribution tasks on average

Because the distances between the depot and the cus-tomers and the distances between the customers are differentthe travel time between any two nodes is different Supposingthat the traveling time between node 119894 and node 119895 is 119905

119894119895 119894 =

0 1 119899 119895 = 0 1 119899 node 0 is depot and the timessatisfy the triangle inequality

The vehicle routes satisfy the following

(i) Each customer is served by a certain vehicle(ii) One vehicle can serve many customers When the

demands of the customers exceed the vehicle capacitythe vehicle returns to the depot to unload and goesback to serve the next customer

(iii) The demands of the customers served by a vehiclecannot exceed 119896 times vehicle capacity

(iv) All vehicles should depart from the depot and returnto the depot

When the demands of the customers exceed the vehiclecapacity the customers should be divided into differentgroupsThe customers in the first group are served first thenthe second group until all the customers are served which isshown in Figure 1

For the MRVRP aiming to minimize total travel distanceor total cost because the traveling distance of a vehicle is thesum of the distances of each round and the total distance isthe sum of all vehicle traveling distances the MRVRP canbe transferred as the VRP with km vehicles The customersare divided into km routes and each vehicle serves 119896 routesThe total traveling distance of 119898 vehicles equals the totaldistance of km routes Then if the total distance of km routesis minimized the total traveling distance of119898 vehicles is alsominimized

But for the MRVRP with fastest completion time thecompletion time of the whole distribution task is the com-pletion time of the last vehicle Suppose that the completion

time of vehicle 119897 is119879119897 119897 = 1 2 119898 and the completion time

of the whole distribution task is 119879 then we have

119879 = max119897=12119898

119879119897 (1)

For each vehicle the distribution task needs to completeseveral routes and the completion time is determined by thetime when last customer served Because there is order ofeach route the completion time of vehicle 119894 is the sumof everyroute completion time In the last route the time from thecustomer to the depot is not considered

If this problem is transferred to the problem with kmroutes in which each vehicle is in charge of 119896 routes thecompletion time is the sumof the travel times of previous 119896minus1routes and the completion time of the last routes Supposingthat the 119897th vehicle is in charge of the distribution task in 119896

routes the traveling time of the 119895th route is 119886119895 119895 = 1 2 119896

and the traveling time from the last served customer to thedepot is 119905 the completion time of this vehicle is

119879119897=

119896

sum

119895=1

119886119895minus 119905 (2)

The difficulty of solving VRP lies in too many arrays ofcustomer service order It is hard to solve VRP by dynamicprogramming since there are too many states For a givenarray of customers the problem is transferred to how todivide the customers into groups and the customers in thesame group are served by one vehicle The feasible groupsrsquodivision scheme is much less and it is easy to be solved bydynamic programming In this paper we will fix the servingarray first and then give the optimal division scheme andcalculate the fastest completion time Taking this completiontime as the objective we will determine the final customerserving order by ACO

3 Customer Division Method Based onDynamic Programming

The problem of dividing customers into groups is how todivide a given serving array 119894

1 1198942 119894

119899into m groups to

ensure that the completion time of all the distribution tasks isminimized and the total demands of the same group do notexceed kw The customers belonging to the same group areserved by one vehicle on the array order

Usually there are a lot of division schemes and theircompletion times are different Nikolakopoulou gave the splitmethod to minimize the total traveling distances by transfer-ring the dividing problem into the shortest-path problem [3]We gave the improved split method to minimize the fastestcompletion time by transferring the division problem intothe longest-edge shortest problem [5] The improved splitmethod spends a long calculation time as it repeats to findthe shortest pathsThen the customer division method basedondynamic programming is givenwhich spends less time [6]

For the MRVRP there are two divisions the first is todivide the customers into 119898 groups and the second is todivide the customers in the same groups into 119896 routes The

Discrete Dynamics in Nature and Society 3

aim of the first division is tominimize the longest completiontime and is solved with dynamic programming The aim ofthe second division is tominimize the total traveling time andis solved with split method

31 The First Division of Customers The problem of dividingcustomers into119898 groups is to determine119898minus1 cut points andcan be looked as at 119898-period decision problem In a periodthe number of customers served by a vehicle is determinedSo the state variable is the number of customers who havenot assigned denoted as 119904 and the number of vehicles whichhave not assigned denoted as 119909 Then the state variable is(119904 119909) and the initial state is (119899 119898) The decision variable isthe number 119911 of customers served by the next vehicle

In the next period the numbers of unassigned customersand vehicles are 119904minus119911 and119909minus1The state variable is (119904minus119911 119909minus1)The state variable and decision variable are both confinedby the vehicle capacity For a given (119904 119909) the demands ofassigned customer cannot exceed 119896 times vehicle capacitySupposing that 119889

119894(119894 = 1 119899) is the total demand from the

first customer to the 119894th customer we have

119889119899minus119904

le (119898 minus 119909) 119896119908 (3)

When the decision variable in the present period is 119911 andthe unassigned customers need 119909 minus 1 vehicles to distributethe total demand of the 119911 customers cannot exceed 119896119908 thatis

119889119899minus 119889119899minus119904+119911

le (119909 minus 1) 119896119908

119889119899minus119904+119911

minus 119889119899minus119904

le 119896119908

(4)

Denote decision variable set as Ω then

Ω=119911isinΖ | 119889119899minus 119889119899minus119904+119911

le(119909 minus 1) 119896119908 119889119899minus119904+119911minus 119889119899minus119904

le119896119908

(5)

Supposing that 119891(119904 119909) is the fastest completion time ofserving the 119904 customers with 119909 vehicles and ℎ(119904 119911) is thecompletion time if the former 119911 customers are served by avehicle we have

119891 (119904 119909) = min119911isinΩ

max (ℎ (119904 119911) 119891 (119904 minus 119911 119909 minus 1)) (6)

When there is a vehicle left if the total demands of 119904customers do not exceed kw the 119904 customers are served bythis vehicle and the completion time is ℎ(119904 119904) otherwise thecompletion time is infinite that is

119891 (119904 1) =

ℎ (119904 119904) if 119902119899minus119904+1

+ sdot sdot sdot 119902119899le 119908

+infin if 119902119899minus119904+1

+ sdot sdot sdot 119902119899gt 119908

(7)

The state variable and decision variable are discrete It canbe solved by enumeration method In a period the vehiclenumber decreases one unit and the unassigned vehicle isdetermined by the periods Then the number of states isdetermined by the number of unassigned customers thenumber of states of each period does not exceed 119899 Since theupper bound of vehicle capacity is 119896119908 there is also an upper

321 119911 119911 + 1

Figure 2 Split graph

bound of the number of customers to be served denoted as119892 The maximum iteration time of the method is 119898119899119892 Inpractice since the state variable should satisfy the inequality(3) and the decision variable should satisfy the inequality (4)the actual iteration time is much less than119898119899119892

In each period the completion time of every vehicle isneeded to calculate and the second division of customers isneeded

32 The Second Division of Customers Suppose that thecustomers served by a given vehicle are 119894

1 1198942 119894

119911 Because

the total demands exceed the vehicle capacity 119908 the routeshould be cut into smaller groups to ensure that the customersin the same group can be served by the same vehicle Thisis also a division problem which is different from that inSection 31 in the scale The scale is smaller and the objectiveis not to minimize the maximal completion but to minimizethe total milk run completion time

To get the division schemes a directed graph is con-structed The vertex set has the depot and the customersserved by the same vehicle For nodes 119895

1and 1198952 supposing

that 1198941198951

is in front of 1198941198952

the total demand of all customersbetween 119894

1198951

and 1198941198952

(Including customer 1198941198951

but not 1198941198952

) iscalculated If the total demand is less than119908 an arc from119895

1to

1198952is drawn The weight is the traveling time from the depot

in turn reach all customers between 1198941198951

and 1198941198952

(includingcustomer 119894

1198951

but not 1198941198952

) and then returnThe weighted directgraph is called split graph which is shown in Figure 2

The shortest distance from 1 to 119911 + 1 is the shortest traveltime of all distribution tasks and the fastest completion timeis this time minus the return time from customer 119894

119911to depot

Suppose the capacity of a vehicle is 15 and the orderof customers to be served by the vehicle is 1-2-5-3-4-7 thetraveling times between the nodes are shown in Figure 3where the numbers in the brackets are the demands

The total demand of customers 1 2 and 5 is no morethan 15 then an edge is drawn between node 1 and 3 withthe weight 90 The other weights are got similarly Aftercalculation the directed weighted graph is got in Figure 4

The shortest path is from node 1 to node 3 plus from node3 to node 0The optimal distance is 180 Customers 1 2 and 5construct the first route Customers 3 4 and 7 construct thesecond route The completion time is 155 (= 180 minus 25)

The algorithm to solve the shortest path problem suchas Dijkstra algorithm is polynomial Its complexity is 119874(1198992)where 119899 is the number of nodes For this problem themaximum number of nodes is 119892 and the complexity is119874(1198922)Because for every iteration the shortest path problem shouldbe solved to calculate the completion time of a vehicle thecomplexity of the whole algorithm is 119874(1198981198991198923)


4030

30

30

30

25

25

25

25

25

20

6 (7)

8 (3)20

20

20

7 (4)

4 (5)

3 (6)

5 (2)15

2 (3)

1 (6)

Figure 3 The distribution of customers

4 7521 0360 60 40 50 60 50

85

90

65

95

70

95

75

90

75


4 Mutation ACO Algorithm

In this section we will give an improved ACO to solve theoptimal customers array

41 ACO Algorithm Themain idea of ant colony algorithmsis tomimic the pheromone trail used by real ants as amediumfor communication and feedback among ants Basically theACO algorithm is a population-based cooperative searchprocedure that is derived from the behavior of real antsACO algorithms make use of simple agents called ants thatiteratively construct solutions to combinatorial optimizationproblemsThe key problem to solve VRPwith ACO is how anindividual ant constructs a complete solution by starting witha null solution and iteratively adding solution componentsuntil a complete solution is constructed The key problem ofACO is to determine the pheromone matrixThe pheromonematrix is (119899 + 1) times 119899 when there are 119899 customers wherethe last row stands for the information from the depot tothe customer and the 119894th row stands for the informationfrom customer 119894 to other customers Initially since for agiven customer there is the same possibility following othercustomers the pheromone matrix starts with equal probablematrix When 119905 = 0 the pheromone matrix is

119861119894119895 (0) =

0

1

119899 minus 1

sdot sdot sdot

1

119899 minus 1

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

1

119899 minus 1

1

119899 minus 1

sdot sdot sdot 0

1

119899

1

119899

sdot sdot sdot

1

119899

(119899+1)times119899

(8)

where 119887119894119895 119894 = 1 2 119899 + 1 119895 = 1 2 119899 is the pheromone

from 119894 to jIn the 119905th iteration there are 119898 ants and 119898 customer

arrays can be got Supposing that the fastest completion timeof each array is 119871

119896(119905) 119896 = 1 2 119898 the increased value of

pheromone in the 119905th iteration isΔ119887119894119895(119905) = sum

119898

119896=1Δ119887119896

119894119895(119905) where

Δ119887119896

119894119895(119905)

=

119876

119871119896 (119905)

if 119895 follows 119894 closely where119876 is a constant

0 otherwise

(9)Local update is performed during the ant constructive proce-dure in the following way

119887119894119895 (119905 + 1) = (1 minus 120588) 119887

119894119895 (119905) + Δ119887119894119895 (

119905) (10)where 120588 is the evaporation coefficient

The heuristic information is in an (119899 + 1) times 119899matrix withthe last row 120578

119899+1119895= 119860119905

119895and the other rows 120578

119894119895= 119860119905

119894119895

where 119905119894119895is the traveling time between two nodes 119905

119895is the

average traveling time from customer 119895 to the depot and119860 isa constant depending on the situations

In a given service order the first served customer is deter-mined by probability The next served customer is selectedfrom the allowing set allowed

119894 which is the customers set

that can be served after 119894 The customer 119895 is chosen with theprobability 119901

119894119895 where

119901119894119895 (119905) =

(119887119894119895 (119905))

120572

(120578119894119895)

120573

sum119904isinallowed

119894

(119887119894119904 (119905))120572(120578119894119904)120573 119895 isin allowed

119894

0 119895 notin allowed119894

(11)

120572 is the importance of pheromone information and 120573 is theimportance of heuristics information After calculation of theprobability matrix the node is selected by the probability119901119894119895and the node is deleted from the allowing set Then a

customer service order is got with the pheromone

42 Mutation Operator ACO algorithm possibly runs intoprematurity just as other evolutionary algorithms The mainreason is the concentrations of pheromone which makesthe same solutions be got In reality the pheromone maybe changed by rain and other factors and this change mayhelp the ants find a new route A mutation operator will beintroduced into the pheromone of ACO to escape from localoptima and strengthen its global search ability

The initial pheromone matrixrsquos factors are the sameAs time goes on some elements become large but otherelements become small in the same row The pheromone isconcentrated which obstructs the ability to findmore optimalsolutionsThe concentration of a row is defined as the ratio ofthemaximum factor and the sumof the factors in the rowThe119894th row concentration 120583

119894(119905) is

120583119894 (119905) =

max119895=12119899

120591119894119895

sum119899

119895=1120591119894119895

119894 = 1 2 119899 + 1 (12)


The concentrations of different row are not the sameWhether tomutate a row should depend on the concentrationof the row When the concentrations of a row exceed thethreshold value the factors in the row should be mutated

For the row mutation the maximum factor decreasesrandomly and the decreased value is assigned to otherpositions Suppose 119895

0th factor is the maximum factor in the

119894th row and given a random number 119903 (0 le 119903 le 1) and arandom vector 119877

1times119899 A new row is got with the 119895

0th factor

multiplied by 119903 and the decreased value is assigned to otherpositions that is

1205911015840

119894= 1205911015840

119894+

1205911198941198950(1 minus 119903)

sum119899

119895=1119877119895

119877 (13)

Row mutation is a local mutation and only affects thechoice of routes partly When the matrix has high con-centration local mutation cannot ensure escaping the localoptima and matrix mutation is necessary The minimumconcentration of the row is the matrix concentration whichis ](119905)

] (119905) = min119894=12119899+1

120583119894 (119905) (14)

Whether to mutate a matrix should also depend on theconcentration of the matrix When the concentrations ofa matrix exceed the threshold value the matrix should bemutated For the matrix mutation every element in thematrix decreases randomly and the decreased value is againrandomly assigned A random number 119903 (0 le 119903 le 1) and arandom vector 119877

119899+1times119899are got New values are got with every

factor multiplied by 119903 Then the decreased value is assignedrandomlyrsquo that is

1205911015840= 119903 lowast 120591

1205911015840= 1205911015840+

sum119899+1

119894=1sum119899

119895=1(120591119894119895minus 1205911015840

119894119895)

sum119899+1

119894=1sum119899

119895=1119877119894119895

119877

(15)

Then a new pheromone matrix is got and the mutationcan ensure that the total pheromones do not change

The mutation ACO algorithm is as follows

Step 1 Initiation Determine the parameters119898 119879 120572 120573 120576 120601and 119901 Input the initial pheromone 120591

119894119895(0) and heuristics

matrix 120578119894119895and get the initial ants Give 119898 customers arrays

randomly as

119860 (0) = 1198601 (0) 1198602 (

0) 119860119898 (0) (16)

Step 2 For a given array assign the customers to the vehicleand get the corresponding completion time by dynamicprogramming and split algorithm Record the present bestroute 119887119890119904119905119905119903119894119901 and the fastest completion time 119891119894119899119894119904ℎ119905119894119898119890

Step 3 The local pheromone update is performed by all theants after each construction step on the formulas (9) and (10)

Step 4 Row Mutation Inspect whether the pheromoneconcentration of the row in the pheromone matrix is more

250249

23962

248754

2414

246665249873

252224

247577245295

247577

23962

25126

2444524623

24616524349244669

235084

244669

235084

230

235

240

245

250

255

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Figure 5 20 times calculation results

than 120576 if yes a random number 119903 is got If 119903 le 119901 a new rowis got by (13)

Step 5 Matrix Mutation Inspect whether the pheromonematrix concentration ismore than120601 if yes a randomnumber119903 is got If 119903 le 119901 a new pheromone matrix is got by theformula (15)

Step 6 According to the new pheromone matrix 119861119894119895(119905) and

the heuristics information matrix 120578119894119895 Give 119898 customers

arrays randomly as

119860 (119905 + 1) = 1198601 (119905 + 1) 1198602 (

119905 + 1) 119860119898 (119905 + 1) (17)

Step 7 Let 119905 = 119905+1 ACO procedure stops if 119905 = 119879 and outputthe finishtime and besttrip otherwise return to Step 2

5 Numerical Examples

In this section we will consider the emergency suppliesdistribution after the earthquake in Wenchuan After theearthquake the roads on earth are not fluentThe emergencysupplies are mainly transported by helicopter As the numberof helicopters is limited comparED to the broad place thehelicopter 119904 is needed for distribution many times Theproblem is a typical MRVRP with fastest completion time

Suppose there are 3 helicopters in charge of the distribu-tion of emergency supplies to 20 settlements The locationof material distribution center is (30 40) the demands andlocation of the settlements are given in Table 1

The capacity of a helicopter is 12tThe total demand is 92tEvery plane is needed to fly 3 times on average and distributeat most 5 settlements every time

The algorithm is realized by Scilab The parameters are asfollows the ants number is 10 and 120572 = 12 120573 = 05 119875 =

015 120576 = 08 120601 = 075 The iteration time is 400 Run theprogram 20 times for the same problem and get the fastestcompletion which times are as shown in Figure 5

Average fastest completion time is 2449978 their maxi-mum gap is 7291 and the gap is 31014 of the best schemeIt shows that the algorithm has good convergence

Best distribution route is as shown in Figure 6Its fastest completion time is 23508428 and total travel

time is 86566435Anew scheme tominimize total travel timeis as shown in Figure 7

Its shortest total travel time is 84589676 the fastestcompletion time of this customer array is 273379 and it is38295 minutes more than the first scheme And the total


Table 1 Locations and demands of settlement

Settlement 1 2 3 4 5 6 7 8 9 10Location (74 29) (64 26) (67 80) (88 15) (21 65) (72 42) (92 80) (46 38) (76 86) (30 46)

Demand 3 7 5 6 2 9 7 4 4 35Settlement 11 12 13 14 15 16 17 18 19 20Location (63 48) (23 11) (36 72) (29 54) (66 16) (10 10) (15 50) (10 10) (20 70) (70 12)

Demand 4 3 8 6 5 15 5 25 3 35

12

3

4

5

6

7

8

9

10 11

12

13

14

1516

17

18

19

21

20

Figure 6 Optimal scheme ofMRVRPwith fastest completion time

12

3

4

5

6

7

8

9

10 11

21

12

13

14

1516

17

18

19

20

Figure 7 Optimal scheme

flight time is reduced by 1976759 minutes In emergencymanagement effectively shortening the completion timeis very necessary and therefore the first scheme is morereasonable

6 Conclusions

In this paper the MRVRP with fastest completion time isproposed which has many applications in fast foods dis-tribution express delivery and emergency supplies Solvingthe problem is more difficult than the general VRP The keyproblem to solve MRVRP with fastest completion time is togive the division method for customer array The customerdivision method based on dynamic programming and splitmethod is given in this paper which can transfer MRVRPwith fastest completion time into the problem of finding theoptimal customer service order Then the problem is solvedwith mutation ACO

References

[1] G B Dantzig and J H Ramser ldquoThe truck dispatchingproblemrdquoManagement Science vol 6 pp 80ndash91 1960

[2] Q L Ding X P Hu and Y X Li ldquoHybrid ant colonysystem for vehicle routing problem with time windowsrdquo SystemEngineeringTheory and Practice vol 27 no 10 pp 98ndash104 2007

[3] G Nikolakopoulou S Kortesis A Synefaki and R KalfakakouldquoSolving a vehicle routing problem by balancing the vehiclestime utilizationrdquo European Journal of Operational Research vol152 no 2 pp 520ndash527 2004

[4] X Zhang J Ma W Liu and F Jin ldquoAnt colony algorithmfor vehicle routing problem with shortest completion timerdquo inProceedings of the 13th International Conference on IndustrialEngineering and Engineering Management vol 1ndash5 pp 12928ndash2933 2006

[5] J Ma and J Yuan ldquoAnt colony algorithm for multiple-depotvehicle routing problem with shortest finish timerdquo Communi-cations in Computer and Information Science vol 113 no 1 pp114ndash123 2010

[6] J Ma Y Fang and J Yuan ldquoMutation ant colony algorithm formultiple-depot multiple-types vehicle routing problems withshortest finish timerdquo Systems Engineering-TheoryampPractice vol31 no 8 pp 1508ndash1516 2011

[7] J K Lenstra and A H G Rinnooy Kan ldquoComplexity of vehiclerouting and scheduling problemsrdquo Networks vol 11 no 2 pp221ndash227 1981

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Figure 1 The milk-run vehicle routes of two vehicles

ant colony algorithm is given to solve the problem In the lastsection a numerical example is given

2 MRVRP with Fastest Completion Time

Suppose that there is one depot serving n customers whichhave 119898 vehicles with capacity 119908 The demand of customer 119894is 119902119894(119894 = 1 119899) and satisfies (119896 minus 1)119908 le sum

119899

119894=1119902119894lt 119896119908

which means that for every vehicle it needs 119896 times to finishthe distribution tasks on average

Because the distances between the depot and the cus-tomers and the distances between the customers are differentthe travel time between any two nodes is different Supposingthat the traveling time between node 119894 and node 119895 is 119905

119894119895 119894 =

0 1 119899 119895 = 0 1 119899 node 0 is depot and the timessatisfy the triangle inequality

The vehicle routes satisfy the following

(i) Each customer is served by a certain vehicle(ii) One vehicle can serve many customers When the

demands of the customers exceed the vehicle capacitythe vehicle returns to the depot to unload and goesback to serve the next customer

(iii) The demands of the customers served by a vehiclecannot exceed 119896 times vehicle capacity

(iv) All vehicles should depart from the depot and returnto the depot

When the demands of the customers exceed the vehiclecapacity the customers should be divided into differentgroupsThe customers in the first group are served first thenthe second group until all the customers are served which isshown in Figure 1

For the MRVRP aiming to minimize total travel distanceor total cost because the traveling distance of a vehicle is thesum of the distances of each round and the total distance isthe sum of all vehicle traveling distances the MRVRP canbe transferred as the VRP with km vehicles The customersare divided into km routes and each vehicle serves 119896 routesThe total traveling distance of 119898 vehicles equals the totaldistance of km routes Then if the total distance of km routesis minimized the total traveling distance of119898 vehicles is alsominimized

But for the MRVRP with fastest completion time thecompletion time of the whole distribution task is the com-pletion time of the last vehicle Suppose that the completion

time of vehicle 119897 is119879119897 119897 = 1 2 119898 and the completion time

of the whole distribution task is 119879 then we have

119879 = max119897=12119898

119879119897 (1)

For each vehicle the distribution task needs to completeseveral routes and the completion time is determined by thetime when last customer served Because there is order ofeach route the completion time of vehicle 119894 is the sumof everyroute completion time In the last route the time from thecustomer to the depot is not considered

If this problem is transferred to the problem with kmroutes in which each vehicle is in charge of 119896 routes thecompletion time is the sumof the travel times of previous 119896minus1routes and the completion time of the last routes Supposingthat the 119897th vehicle is in charge of the distribution task in 119896

routes the traveling time of the 119895th route is 119886119895 119895 = 1 2 119896

and the traveling time from the last served customer to thedepot is 119905 the completion time of this vehicle is

119879119897=

119896

sum

119895=1

119886119895minus 119905 (2)

The difficulty of solving VRP lies in too many arrays ofcustomer service order It is hard to solve VRP by dynamicprogramming since there are too many states For a givenarray of customers the problem is transferred to how todivide the customers into groups and the customers in thesame group are served by one vehicle The feasible groupsrsquodivision scheme is much less and it is easy to be solved bydynamic programming In this paper we will fix the servingarray first and then give the optimal division scheme andcalculate the fastest completion time Taking this completiontime as the objective we will determine the final customerserving order by ACO

3 Customer Division Method Based onDynamic Programming

The problem of dividing customers into groups is how todivide a given serving array 119894

1 1198942 119894

119899into m groups to

ensure that the completion time of all the distribution tasks isminimized and the total demands of the same group do notexceed kw The customers belonging to the same group areserved by one vehicle on the array order

Usually there are a lot of division schemes and theircompletion times are different Nikolakopoulou gave the splitmethod to minimize the total traveling distances by transfer-ring the dividing problem into the shortest-path problem [3]We gave the improved split method to minimize the fastestcompletion time by transferring the division problem intothe longest-edge shortest problem [5] The improved splitmethod spends a long calculation time as it repeats to findthe shortest pathsThen the customer division method basedondynamic programming is givenwhich spends less time [6]

For the MRVRP there are two divisions the first is todivide the customers into 119898 groups and the second is todivide the customers in the same groups into 119896 routes The







119889119899minus119904

le (119898 minus 119909) 119896119908 (3)


119889119899minus 119889119899minus119904+119911

le (119909 minus 1) 119896119908

119889119899minus119904+119911

minus 119889119899minus119904

le 119896119908

(4)




le119896119908

(5)


119891 (119904 119909) = min119911isinΩ

max (ℎ (119904 119911) 119891 (119904 minus 119911 119909 minus 1)) (6)


119891 (119904 1) =

ℎ (119904 119904) if 119902119899minus119904+1

+ sdot sdot sdot 119902119899le 119908

+infin if 119902119899minus119904+1

+ sdot sdot sdot 119902119899gt 119908

(7)


321 119911 119911 + 1





1 1198942 119894

119911 Because




that 1198941198951

is in front of 1198941198952


1198951

and 1198941198952


but not 1198941198952


1to



and 1198941198952


1198951

but not 1198941198952



119911to depot






4030

30

30

30

25

25

25

25

25

20

6 (7)

8 (3)20

20

20

7 (4)

4 (5)

3 (6)

5 (2)15

2 (3)

1 (6)


4 7521 0360 60 40 50 60 50

85

90

65

95

70

95

75

90

75





119861119894119895 (0) =

0

1

119899 minus 1

sdot sdot sdot

1

119899 minus 1


1

119899 minus 1

1

119899 minus 1

sdot sdot sdot 0

1

119899

1

119899

sdot sdot sdot

1

119899

(119899+1)times119899

(8)






119898

119896=1Δ119887119896

119894119895(119905) where

Δ119887119896

119894119895(119905)

=

119876

119871119896 (119905)


0 otherwise


119887119894119895 (119905 + 1) = (1 minus 120588) 119887

119894119895 (119905) + Δ119887119894119895 (



119899+1119895= 119860119905


119894119895= 119860119905

119894119895


119895is the





119894119895 where

119901119894119895 (119905) =

(119887119894119895 (119905))

120572

(120578119894119895)

120573


119894

(119887119894119904 (119905))120572(120578119894119904)120573 119895 isin allowed

119894


(11)





119894(119905) is

120583119894 (119905) =

max119895=12119899

120591119894119895

sum119899

119895=1120591119894119895

119894 = 1 2 119899 + 1 (12)







0th factor


1205911015840

119894= 1205911015840

119894+

1205911198941198950(1 minus 119903)

sum119899

119895=1119877119895

119877 (13)


] (119905) = min119894=12119899+1

120583119894 (119905) (14)




1205911015840= 119903 lowast 120591

1205911015840= 1205911015840+

sum119899+1

119894=1sum119899

119895=1(120591119894119895minus 1205911015840

119894119895)

sum119899+1

119894=1sum119899

119895=1119877119894119895

119877

(15)






randomly as

119860 (0) = 1198601 (0) 1198602 (

0) 119860119898 (0) (16)




250249

23962

248754

2414

246665249873

252224

247577245295

247577

23962

25126

2444524623

24616524349244669

235084

244669

235084

230

235

240

245

250

255

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20






arrays randomly as

119860 (119905 + 1) = 1198601 (119905 + 1) 1198602 (

119905 + 1) 119860119898 (119905 + 1) (17)
















Demand 4 3 8 6 5 15 5 25 3 35

12

3

4

5

6

7

8

9

10 11

12

13

14

1516

17

18

19

21

20


12

3

4

5

6

7

8

9

10 11

21

12

13

14

1516

17

18

19

20



6 Conclusions


References















Volume 2014




Journal of











Journal of


Function Spaces






Algebra















119889119899minus119904

le (119898 minus 119909) 119896119908 (3)


119889119899minus 119889119899minus119904+119911

le (119909 minus 1) 119896119908

119889119899minus119904+119911

minus 119889119899minus119904

le 119896119908

(4)




le119896119908

(5)


119891 (119904 119909) = min119911isinΩ

max (ℎ (119904 119911) 119891 (119904 minus 119911 119909 minus 1)) (6)


119891 (119904 1) =

ℎ (119904 119904) if 119902119899minus119904+1

+ sdot sdot sdot 119902119899le 119908

+infin if 119902119899minus119904+1

+ sdot sdot sdot 119902119899gt 119908

(7)


321 119911 119911 + 1





1 1198942 119894

119911 Because




that 1198941198951

is in front of 1198941198952


1198951

and 1198941198952


but not 1198941198952


1to



and 1198941198952


1198951

but not 1198941198952



119911to depot






4030

30

30

30

25

25

25

25

25

20

6 (7)

8 (3)20

20

20

7 (4)

4 (5)

3 (6)

5 (2)15

2 (3)

1 (6)


4 7521 0360 60 40 50 60 50

85

90

65

95

70

95

75

90

75





119861119894119895 (0) =

0

1

119899 minus 1

sdot sdot sdot

1

119899 minus 1


1

119899 minus 1

1

119899 minus 1

sdot sdot sdot 0

1

119899

1

119899

sdot sdot sdot

1

119899

(119899+1)times119899

(8)






119898

119896=1Δ119887119896

119894119895(119905) where

Δ119887119896

119894119895(119905)

=

119876

119871119896 (119905)


0 otherwise


119887119894119895 (119905 + 1) = (1 minus 120588) 119887

119894119895 (119905) + Δ119887119894119895 (



119899+1119895= 119860119905


119894119895= 119860119905

119894119895


119895is the





119894119895 where

119901119894119895 (119905) =

(119887119894119895 (119905))

120572

(120578119894119895)

120573


119894

(119887119894119904 (119905))120572(120578119894119904)120573 119895 isin allowed

119894


(11)





119894(119905) is

120583119894 (119905) =

max119895=12119899

120591119894119895

sum119899

119895=1120591119894119895

119894 = 1 2 119899 + 1 (12)







0th factor


1205911015840

119894= 1205911015840

119894+

1205911198941198950(1 minus 119903)

sum119899

119895=1119877119895

119877 (13)


] (119905) = min119894=12119899+1

120583119894 (119905) (14)




1205911015840= 119903 lowast 120591

1205911015840= 1205911015840+

sum119899+1

119894=1sum119899

119895=1(120591119894119895minus 1205911015840

119894119895)

sum119899+1

119894=1sum119899

119895=1119877119894119895

119877

(15)






randomly as

119860 (0) = 1198601 (0) 1198602 (

0) 119860119898 (0) (16)




250249

23962

248754

2414

246665249873

252224

247577245295

247577

23962

25126

2444524623

24616524349244669

235084

244669

235084

230

235

240

245

250

255

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20






arrays randomly as

119860 (119905 + 1) = 1198601 (119905 + 1) 1198602 (

119905 + 1) 119860119898 (119905 + 1) (17)
















Demand 4 3 8 6 5 15 5 25 3 35

12

3

4

5

6

7

8

9

10 11

12

13

14

1516

17

18

19

21

20


12

3

4

5

6

7

8

9

10 11

21

12

13

14

1516

17

18

19

20



6 Conclusions


References















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










4030

30

30

30

25

25

25

25

25

20

6 (7)

8 (3)20

20

20

7 (4)

4 (5)

3 (6)

5 (2)15

2 (3)

1 (6)


4 7521 0360 60 40 50 60 50

85

90

65

95

70

95

75

90

75





119861119894119895 (0) =

0

1

119899 minus 1

sdot sdot sdot

1

119899 minus 1


1

119899 minus 1

1

119899 minus 1

sdot sdot sdot 0

1

119899

1

119899

sdot sdot sdot

1

119899

(119899+1)times119899

(8)






119898

119896=1Δ119887119896

119894119895(119905) where

Δ119887119896

119894119895(119905)

=

119876

119871119896 (119905)


0 otherwise


119887119894119895 (119905 + 1) = (1 minus 120588) 119887

119894119895 (119905) + Δ119887119894119895 (



119899+1119895= 119860119905


119894119895= 119860119905

119894119895


119895is the





119894119895 where

119901119894119895 (119905) =

(119887119894119895 (119905))

120572

(120578119894119895)

120573


119894

(119887119894119904 (119905))120572(120578119894119904)120573 119895 isin allowed

119894


(11)





119894(119905) is

120583119894 (119905) =

max119895=12119899

120591119894119895

sum119899

119895=1120591119894119895

119894 = 1 2 119899 + 1 (12)







0th factor


1205911015840

119894= 1205911015840

119894+

1205911198941198950(1 minus 119903)

sum119899

119895=1119877119895

119877 (13)


] (119905) = min119894=12119899+1

120583119894 (119905) (14)




1205911015840= 119903 lowast 120591

1205911015840= 1205911015840+

sum119899+1

119894=1sum119899

119895=1(120591119894119895minus 1205911015840

119894119895)

sum119899+1

119894=1sum119899

119895=1119877119894119895

119877

(15)






randomly as

119860 (0) = 1198601 (0) 1198602 (

0) 119860119898 (0) (16)




250249

23962

248754

2414

246665249873

252224

247577245295

247577

23962

25126

2444524623

24616524349244669

235084

244669

235084

230

235

240

245

250

255

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20






arrays randomly as

119860 (119905 + 1) = 1198601 (119905 + 1) 1198602 (

119905 + 1) 119860119898 (119905 + 1) (17)
















Demand 4 3 8 6 5 15 5 25 3 35

12

3

4

5

6

7

8

9

10 11

12

13

14

1516

17

18

19

21

20


12

3

4

5

6

7

8

9

10 11

21

12

13

14

1516

17

18

19

20



6 Conclusions


References















Volume 2014




Journal of











Journal of


Function Spaces






Algebra















0th factor


1205911015840

119894= 1205911015840

119894+

1205911198941198950(1 minus 119903)

sum119899

119895=1119877119895

119877 (13)


] (119905) = min119894=12119899+1

120583119894 (119905) (14)




1205911015840= 119903 lowast 120591

1205911015840= 1205911015840+

sum119899+1

119894=1sum119899

119895=1(120591119894119895minus 1205911015840

119894119895)

sum119899+1

119894=1sum119899

119895=1119877119894119895

119877

(15)






randomly as

119860 (0) = 1198601 (0) 1198602 (

0) 119860119898 (0) (16)




250249

23962

248754

2414

246665249873

252224

247577245295

247577

23962

25126

2444524623

24616524349244669

235084

244669

235084

230

235

240

245

250

255

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20






arrays randomly as

119860 (119905 + 1) = 1198601 (119905 + 1) 1198602 (

119905 + 1) 119860119898 (119905 + 1) (17)
















Demand 4 3 8 6 5 15 5 25 3 35

12

3

4

5

6

7

8

9

10 11

12

13

14

1516

17

18

19

21

20


12

3

4

5

6

7

8

9

10 11

21

12

13

14

1516

17

18

19

20



6 Conclusions


References















Volume 2014




Journal of











Journal of


Function Spaces






Algebra













Demand 4 3 8 6 5 15 5 25 3 35

12

3

4

5

6

7

8

9

10 11

12

13

14

1516

17

18

19

21

20


12

3

4

5

6

7

8

9

10 11

21

12

13

14

1516

17

18

19

20



6 Conclusions


References















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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