4. Milk Collection

7/29/2019 4. Milk Collection

1/14

Vol. 40, No. 2, MarchApril 2010, pp. 130143issn 0092-2102 eissn 1526-551X 10 4002 0130

informs

doi10.1287/inte.1090.0475 2010 INFORMS

A Milk Collection Problem withIncompatibility Constraints

Massimiliano CaramiaDipartimento di Ingegneria dellImpresa, Universit di Roma Tor Vergata, 00133 Roma, Italy,

[email protected]

Francesca GuerrieroDipartimento di Elettronica, Informatica e Sistemistica, Universit della Calabria, 87036 Rende, Italy,

[email protected]

The milk collection problem is well known in rural areas of the world. This paper considers this real-lifeproblem for an Italian dairy company that collects raw milk from farmers. In our milk collection problem, weaddress the constraint that some farms are small and inaccessible by large vehicles; moreover, these farmers

produce different milk types, and the tank trucks used for transporting the milk have multiple compartments.This generates the additional constraint that at most one milk type can be assigned to a tank compartment.The goal of this paper is to show how operations research techniques helped the company to improve itsdaily performance. In particular, we present a solution approach based on two mathematical formulations andlocal search, all embedded within a multiple-restart mechanism. The first mathematical formulation minimizesthe number of vehicles to be routed in the network; the second minimizes the tour length. We also discussexperiments we conducted as part of our case study and compare our solution with the process that the companyused previously.

Key words : industries: agriculture, food, transportation shipping; networks; graphs; heuristics.History : This paper was refereed. Published online in Articles in Advance December 30, 2009.

I

n this paper we study a real-life problem faced

by ASSO.LA.C., an Italian dairy company that col-lects raw milk from farmers. The considered milk col-

lection problem can be interpreted as a specialized

instance of the vehicle routing problem (VRP).

In the VRP, a set of (homogeneous) vehicles located

at a depot must be routed to serve geographically dis-

tributed customers. Routes must be optimized based

on one or more objective functions, e.g., the mini-

mization of the fleet size and (or) the total routing

cost. Each customer has a known demand and ser-

vice duration, and each must be visited only once by

exactly one vehicle. Each route starts and ends at a

depot, and a vehicles capacity may not be exceeded.The VRP is NP-hard and is usually solved by using

(meta)heuristics (Toth and Vigo 2002; Cordeau et al.

2000, 2002).

A general VRP assumption is that customers can

always be reached by the vehicles; i.e., the size of

a truck and (or) the location of the customer is not

relevant. In practice, this might not be true; indeed, a

truck with a trailer might not be able to reach some

customer locations. In such situations, the trailer must

be uncoupled from the complete vehicle (i.e., the truckplus the trailer) in ad hoc parking areas before vis-

iting these customer locations; after visiting the loca-

tions the trailer must be coupled again to the pure

truck (i.e., the truck without the trailer), forming a

complete vehicle that continues its tour. This prob-

lem is known as the truck and trailer routing problem

(TTRP); the milk collection problem is a generaliza-

tion of the TTRP.

Assume that the set S of customers is divided

into

customers Str who are accessible with or without

a trailer, and customers Swt who are only accessible without a

trailer.

To address the needs of both customer types, we

must consider a possible uncoupling of the trailer

from the complete vehicle at a parking place to serve

customers in Swt . Therefore, following the definition

in Chao (2002), we denote the section of the route in

which a trailer is assigned as the main tour of the

130


2/14

Caramia and Guerriero: Milk Collection Problem with Incompatibility ConstraintsInterfaces 40(2), pp. 130143, 2010 INFORMS 131

complete vehicle route (CVR). In a CVR, uncoupling a

trailer at a parking place and starting a subtour from

the point of the uncoupling is possible. In a subtour,

which must start and end at the same parking place,the company can service customers in Str and Swt .

Therefore, a CVR is formed by a main tour and possi-

bly a certain number of subtours. Following the TTRP

model of Chao (2002), the depot and any customer in

Str could be selected as a trailer parking place.

For a complete vehicle, the sum of all demands col-

lected on the CVR may not exceed the vehicle total

capacity C+C (the truck and trailer capacity, respec-

tively), and a subtour cannot exceed the remaining

pure truck capacity, which is C units when the truck

leaves the depot. It is important to note that current

models in the literature assume that at the parkingplaces, the transported commodities can be trans-

ferred between the truck and the trailer. This is not

possible in our model because of the heterogeneous

milk types.

Unlike a complete vehicle, a pure truck is always

able to service both customer types. Its corresponding

route is called a pure truck route (PTR); it starts and

ends at the depot, does not include any subtours, and

is limited by the pure truck capacity C.

Figure 1 illustrates a tour in which a complete vehi-

cle starts from the depot with a complete vehicle and

visits customers A, B, and C with the trailer; then, ata parking area, the trailer is uncoupled and the truck

serves a subset of customers in areas that are inacces-

sible to complete vehicles, forming a subtour (dashed

arcs). When the pure truck reaches the parking area

again, the trailer is reattached and the complete vehi-

cle goes back to the depot to complete its tour.

Depot

Parking area

A B

C

Figure 1: A TTRP solution is shown.

In general, the TTRPs objective is to find routes

that minimize the total travel cost or time, respecting

the aforementioned constraints, with a possible limi-

tation on the tour length. Solving the TTRP can alsoinclude finding the optimal number of subtours and

the location of the parking places for complete vehicle

routes.

Because farms at which milk is collected are often

small and inaccessible by complete vehicles, milk col-

lection is an important application of the TTRP. Dairy

companies that collect milk are interested mainly in

two cost components: the cost of transport and the

cost of paying the farmers for the milk. Hence, to

maintain the same marginal profit, any increase in

the price paid to the farmers must be balanced by a

reduction in the cost of collecting the milk. Therefore,optimizing transport cost can allow the company to

pay a higher milk price to farmers, thus attracting

higher volumes and possibly producing economies of

scale in collecting milk (Butler et al. 2005).

This paper presents our experience as opera-

tions researchers in a real milk collection problem.

ASSO.LA.C. collects milk from farmers (we use the

terms customer and farmer synonymously) who live

in different towns and moves it from each produc-

tion site to a central warehouse. It uses a fleet of tank

trucks to collect milk (Figure 2). Unlike milk collec-

tion problems studied in the literature, the company

must collect different milk types; when milk of a par-

ticular type has been transferred into a compartment,

that compartment cannot be used to hold milk of a

different type until it has been sanitized and cleaned

at the central warehouse at the end of the day. This

constraint renders the studied milk collection problem

a generalization of the TTRP.

Furthermore, in the problem we consider, pure

trucks and trailers have heterogeneous capacities;

thus, it can be interpreted as a generalization of the

heterogeneous VRP (HVRP), which we denote as het-

erogeneous milk collection with heterogeneous fleet

(HMCHF). Baldacci et al. (2008) provide a recent sur-

vey on the HVRP.

Milk collection consists of multiple tasks: (1) milk

selection based on milk quality, (2) a prerefrigera-

tion process of the selected milk, (3) transfer of milk

into a tank by means of pumps capable of loading

1.1 tons in five minutes, and (4) writing of a report to


3/14

Caramia and Guerriero: Milk Collection Problem with Incompatibility Constraints132 Interfaces 40(2), pp. 130143, 2010 INFORMS

Figure 2: A milk collection tank truck consists of different compartments, each characterized by its capacity.

record information about the milk type and the quan-

tity picked up.

The collected milk is transported to the central

warehouse for inspection. As a tank truck arrives at

the warehouse, a member of a quality control unit

draws samples of the milk to check the correctness of

the information in the report. If a milk sample fails

the quality inspection, the milk is rejected, and the

rejection is reported to the corresponding farmer. The

truck tank is weighed each time a compartment is

emptied to determine the exact quantity of each milk

type collected. This information is important to the

quality control unit, which must determine a price for

each milk type.

In our HMCHF problem, we consider the following

set of operational constraints:

each node in the network can be a loading point

or a parking area,

the tank truck cannot contain a load greater than

its capacity,

multiple tank trucks can pick up milk from a par-

ticular farmer,

the time required for a tour cannot exceed the

work shift,

only one milk type can be assigned to a

compartment.

We propose a multistart optimization approach

based on two mathematical formulations and a local

search that exploits such formulations. The two math-

ematical formulations minimize the size of the fleet

and the total route length, respectively. This is one of

the strengths of our approach, because state-of-the-art

milk collection approaches do not consider the num-

ber of vehicles used to service clients.

Milk Collection Problem ReviewIn this section, we review previous decision support

system (DSS) work on the milk collection problem.

For example, Sankaran and Ubgade (1994) proposed

and implemented CARS, a DSS that utilizes route

construction heuristics for an instance with about 70

milk collection centers in Etah, India. Butler et al.(2005) showed how a DSS integrated with a geo-

graphic information system allows a scheduler to

interact with route optimization algorithms. Claassen

and Hendriks (2007) studied the development of a

pilot DSS for collecting goats milk and proposed an

operations research (OR)-based approach to support

the milk collection.

For information on algorithmic contributions to

the milk collection problem, we refer the reader to

Butler et al. (1997), Basnet et al. (1999), and Hoff

and Lkketangen (2007). Butler et al. (1997) proposed

to solve the problem for a dairy farm in Dublin,Ireland by using an extension of the symmetric travel-

ling salesman problem. Basnet et al. (1999) presented

an exact algorithm for a milk-tanker scheduling and

sequencing problem. They modeled a joint problem

of scheduling and sequencing as a linear integer pro-

gram with some nonlinear constraints and solved it

as a real problem for the New Zealand dairy industry;

they also present a fast heuristic algorithm.


4/14


Hoff and Lkketangen (2007) modeled the milk col-

lection problem as a TTRP and proposed a solution

algorithm based on tabu search. We also modeled our

milk collection problem as a TTRP. However, becauseour problem involves tank trucks with multiple com-

partments and a heterogeneous fleet of vehicles, we

developed a generalized TTRP model.

To solve the TTRP, Chao (2002) proposed a method

based on mathematical programming and local search,

Scheuerer (2006) defined two construction heuristics

and a tabu search approach, and Tan et al. (2006)

developed an evolutionary algorithm to cope with the

truck and trailer routing with two objective functions,

i.e., the number of trucks and the routing cost.

In the literature (Chao 2002), mathematical formu-

lation-based approaches for the TTRP define a gen-eralized assignment problem (Fisher and Jaikumar

1981) that assigns vehicles to customers with the goal

of minimizing the total distance that trucks travel.

They next find an optimal solution to the linear relax-

ation of the generalized assignment formulation, and

by exploiting some theoretical arguments (Chao et al.

1995 show details), they round it to produce a pos-

sibly infeasible solution that first undergoes heuristic

route construction, and then they apply an improve-

ment process.

In this paper, we propose an approach that first

solves the farmer route assignment problem (FRAP)

assigning farmers to vehicles (each vehicle defines

a route) without considering the sequence in which

farmers are visited. The FRAPs goal is to minimize

the number of trucks used, satisfying the capacity

and demand constraints. Given a FRAP solution, the

route definition problem (RDP), which determines the

routes, is solved.

Unlike state-of-the-art approaches, the FRAP is not

followed directly by a route construction heuristic; the

RDP accomplishes this task. However, the RDP can

fail to find feasible routes. Thereby, the proposed local

search (1) retrieves feasibility and (2) allows the defi-

nition of proper constraints to be added to the formu-

lations to diversify solutions and possibly find better

solutions.

Another novelty of our proposed approach lies in

its iterative mechanism that permits information to be

tunneled from (to) the FRAP and the RDP to (from) the

heuristic solutions. Furthermore, the FRAP minimizes

the number of vehicles; competing approaches, how-

ever, minimize distance travelled, which we address

as the objective of the RDP.

Milk Collection with IncompatibilityConstraints

Proposed Approach

In this section, we present the strategy we adopted to

solve the HMCHF problem heuristically.

Assume a graph G = S s0A with vertex set

S s0 and arc set A. The special vertex s0 represents

the depot, where m trucks and m trailers are parked.

The trucks and trailers are considered homogeneous

with capacity Cand C

, respectively. Each vertex si Scorresponds to a customer i with a demand di 0.

A cost cij is associated with each arc si sj which

represents the nonnegative travel time or distance

from vertex si to vertex sj.

Our approach relies on the solutions of the two

mathematical formulations associated with the FRAP

and the RDP, respectively. The latter are extensively

reported in Appendices A and B.

In the FRAP, one is concerned with assigning farm-

ers to vehicles (each vehicle defines a route) consider-

ing capacity, demand, and milk-type constraints, and

neglecting the tour-length constraint. In particular, letI= I1 I2 be the fleet of trucks, where I1 is the set of

complete vehicles and I2 is the set of pure trucks; the

FRAP constraints are as follows.

1. For farmer s and milk type j, the farmer demand

Qjs must be satisfied.

2. If a positive amount of milk type j is loaded in

compartment k of tank truck i, milk type j is assigned

to compartment k of tank truck i, and this quantity

must not be larger than the capacity cik of compart-

ment k of tank truck i.

3. At least one truck must serve each farmer.

4. Complete vehicles must visit at least one farmer

s Str (this prohibits the undesirable assignment of a

complete vehicle to a set of hard-to-access farmers).

5. If a certain quantity of a milk type is loaded from

farmer s in a tank truck i, then tank truck i serves

farmer s.

6. At most, one milk type per compartment is

allowed.


5/14


The FRAP objective function is to minimize the

number of tank trucks used. Given the FRAP solu-

tion, the RDP defines a route for each vehicle with

the objective of minimizing the cumulative tour dura-tion. Let R be the number of routes defined by the

FRAP, and let Si, with i = 1 R, be the collection

of subsets of farmers assigned to each vehicle i I.

Let Si be the set of customers served by truck and

trailer, and let Si be the set of customers served by

truck only.

The RDP formulation works on the R routes by

separately solving (for each vehicle) a relaxation of

the traveling salesman problem. More specifically, if

i is a tour associated with a pure truck, then Si = Si .

The RDP constraints ensure that (1) a tour must start

and finish at the warehouse, and (2) if a tank truck

arrives at a farmer site, it must leave this site (flow

conservation constraints).

If tour i is associated with a complete vehicle, then

the related RDP constraints are split up into two

subsets of constraints, one for Si and the other for Si .

The rationale for dividing the RDP constraints related

to i into two parts is that a complete vehicle, unlike

a pure truck, must leave the trailer when it serves

farmers belonging to Swt ; therefore, the correspond-

ing tour must be formed by the proper composition

of subtours with farmers belonging to Str and Swt ,respectively.

The RDP objective function is to minimize the total

preparation time and transportation time of all the

tank trucks. The preparation time includes the time

needed for milk analysis, report writing, and pump

connection.

It is important to point out that the RDP for-

mulation might have the following two drawbacks,

which the heuristic functionalities discussed next

address:

1. The minimization of the total tour time might

return a duration for some tours that exceeds the time

T allowed for a working shift.

2. Subtours might be present.

The proposed heuristic has two main blocks,

PHASE I and PHASE II, and is based on the res-

olution of the FRAP and the RDP, respectively. We

refer to a complete execution of both PHASE I and

PHASE II as an iteration of the heuristic.

PHASE I calculates the optimal solution of the FRAP

formulation by obtaining the subset of farmers to be

served by each tank truck, respecting demand and

capacity constraints. However, because this assign-ment might not be feasible with respect to the max-

imum duration of a tour, PHASE II of the algorithm

verifies this by solving the RDP formulation.

Let us examine how the heuristic works depending

on the solution found by the RDP, concentrating on a

generic vehicle i.

1. Vehicle i is a pure truck. If the solution is a unique

tour, then the algorithm terminates; otherwise, the

algorithm tries to define a feasible solution by execut-

ing an edge-insertion routine. Its goal is to join each

subtour to the warehouse, if necessary, by inserting

the cheapest edges and deleting the heaviest ones.More formally, this routine, for each subtour i that

does not include the warehouse (node s0), inserts the

cheapest edges s0 s and s s, where s is a farmer in

subtour i, s is a neighbor of s and s is a neighbor of

s0. Finally, it deletes edges s0 s and ss. Figure 3

illustrates a solution of the RDP model.

2. Vehicle i is a complete vehicle. For each tour, the

RDP formulation works with farmers from Str and

Swt . A feasible tour for the TTRP requires a main tour

of the CVR and a number of subtours, each starting

from a farmer of the main tour. Therefore, if the solu-

tion to the RDP related to farmers in Str is a uniquetour, this is the main tour of the CVR. Otherwise, to

obtain such a main tour, the algorithm executes the

insertion routine. When the main tour of the CVR

has been defined, the least-cost edge-insertion routine

is invoked again to link each subtour to a farmer in

the main tour, thus obtaining a feasible solution. Such

farmers represent the parking areas of the complete

WarehouseWarehouse

Figure 3: The schematic illustrates the insertion routine. The left section

shows an RDP model solution with two subtours; one includes the ware-

house, the other does not. The right section shows a solution obtained by

applying the insertion routine.


6/14


vehicle tour. It is trivial that, if Si = for tour i, the

main tour of the CVR is a feasible solution for i.

In the proposed methodology, we do not permit the

possibility of visiting s Str on the subtours that visithard-to-access farmers in Swt .

If the solution obtained is such that time limit T is

obeyed for each tour, then the algorithm stops; oth-

erwise, PHASE I is run again with a limitation on

the number of farmers to be serviced by tank trucks.

Formally, if Il is the subset of tours violating the max-

imum shift duration allowed at the end of iteration l,

at iteration l + 1 the FRAP formulation is solved with

the following additional constraints:

sSzis fil 1 i Il (1)

where fil is the number of farmers assigned to tank

truck i by PHASE I at iteration l, and zis is a binary

variable that equals 1 if tank truck i services farmer s

and is zero otherwise.

The restart mechanism diversifies the solutions. To

provide counterbalancing, the algorithm tries to build

a new solution at l + 1 by not completely varying the

number of farmers assigned to each tour by the FRAP

at iteration l. It accomplishes this as follows: if Il is

the set of nonempty feasible tours at iteration l, the

FRAP formulation at iteration l + 1 is solved with the

additional constraints:sS

zis fil i Il (2)

To avoid cycling while trying to regain feasibility

i.e., to prevent PHASE I from returning the same solu-

tion on two alternating iterations, e.g., l 1 and l + 1

a tabu list in the algorithm is implemented. The assign-

ment of farmers to routes at iterations l 1 and l is

compared, and those farmers on route i at l 1 but not

on route i at iteration l, are marked as tabu to avoid

having a farmer return to route i at an immediate iter-

ation. Therefore, a tabu farmer-tour matrix must bemaintained; its generic entry si represents the num-

ber of iterations for which farmer s cannot be assigned

to tour i in the optimal solution of the FRAP in PHASE

I. Note that if at iteration l an entry si of the tabu

matrix is greater than zero, then variable zis is forced

to be zero in the solution of the FRAP at iteration l + 1.

For example, let us assume that tour i has farm-

ers 12510 assigned at iteration 1 and farmers

1234 assigned at iteration 2. Then farmers 5 and

10 are marked as tabu; i.e., they are not allowed to

return to tour i for number of iterations. In our

case study, we notice that must be very small, e.g., 23, compared with tenures of tabu search algo-

rithms for other combinatorial optimization problems.

It is worth noting that by imposing constraints (1)

and (2), the solution of PHASE I could become infea-

sible with respect to the capacity constraints. In this

case, the right sides of constraints (1) are increased

by one, the right sides of constraints (2) are decreased

by one, and PHASE I is solved again to retrieve fea-

sibility. This is done iteratively until a feasible solu-

tion is found. We note that the tabu list prevents

cycling among solutions during the search for a fea-sible solution.

The algorithm is stopped if the last 10 iterations

provide no solution improvement. Table 1 shows the

steps of the proposed algorithm.

Ensure: A set of tours respecting demand, capacity, tour duration, and

milk-type constraints.

Initialization

1: l = 1;

2: Il = ;

3: Il = ;

Main loop4: while not stopping rule do

5: PHASE I: Solve the FRAP mathematical formulation,

adding constraints (1)(2) if l > 1;

6: while PHASE I is infeasible do

7: Increase the right sides of constraints (1) by 1, and

decrease the right sides of constraints (2) by 1;

8: Solve the FRAP mathematical formulation,

adding constraints (1)(2);

9: end while

10: PHASE II: Solve the RDP formulation by using for each tour

the farmers assigned by PHASE I;

11: Possibly apply the edge-insertion routine

to retrieve a feasible tour;

12: if the stopping criterion is met then

13: STOP;14: else

15: Define sets Il and Il;

16: Define constraints (1) and (2);

17: Update farmer-tour entries of the tabu matrix;

18: Set variables zis = 0 if farmer s is tabu for tour i.

19: end if

20: end while

Table 1: This table lists the algorithm steps.


7/14


Experimental Analysis andComparison with Prior ProcessIn this section, we compare the solution we obtained

using our algorithm with the companys prior opera-

tional process. We implemented our algorithm using

the C language and the mathematical models using

the AMPL language; we also used the solver CPLEX

7.0 (http://www.ilog.com). The C code and the AMPL

implementation were run on a PC Pentium IV with

2.8 GHz and 512 MB of RAM.

Case Study Data

ASSO.LA.C. collects milk from 158 farmers in four

towns in Calabria, a region located in southern Italy.

The towns are Cosenza (80), Catanzaro (8.7), ViboValentia (6.6), and Crotone (4.7) (see Figure 4); the

number following each town represents the per-

centage of the 158 farmers who live in that town.

The companys main warehouse is in the city of

Castrovillari.

Figure 4: This map shows where ASSO.LA.C. collects milk; circles in-

dicate the four towns in which farmers live and the companys main

warehouse.

Truck Trailer No. of Capacity of

ID vehicle Capacity capacity capacity compartments compartments

CV1 150 75 75 4 37.5/37.5/37.5/37.5

CV2 150 75 75 4 37.5/37.5/37.5/37.5

CV3 310 140 170 5 80/60/50/60/60

CV4 310 130 180 3 130/130/50

CV5 150 75 75 4 37.5/37.5/37.5/37.5

Table 2: This complete vehicle characteristics table shows capacities

expressed in quintals.

The milk types are denoted as high-quality,

standard-quality, sanitary-prescription, and origin-

protected milk. The fleet comprises complete vehicles

and pure trucks. Table 2 shows complete vehi-

cle characteristics, and Table 3 shows pure truckcharacteristics.

Tables 47 show the case studys main demand

nodes; note that to decrease the problem size to

improve tractability, we aggregated farmers closer

than two kilometers in a single node with a demand

equal to the sum of the demands of the clus-

tered nodes. The Complete vehicle column indi-

cates whether a complete vehicle can access the farm,

thereby indicating hard-to-access farms.

Finally, we assume that the preparation time ts is

known a priori for each farmer s and that the work

shift for drivers is eight hours.

Case Study Computational Results

The computational results demonstrate that the pro-

posed algorithm can provide tours such that the tank

trucks start their tours from Castrovillari and come

back to Castrovillari within the work shift. We used

No. of Capacity of

ID vehicle Capacity compartments compartments

PT1 140 3 46.7/46.7/46.7

PT2 140 3 46.7/46.7/46.7PT3 130 3 43.5/43.5/43.5

PT4 105 3 37.37/31

PT5 60 3 20/20/20

PT6 60 3 20/20/20

PT7 40 3 13.35/13.35/13.35

PT8 110 3 37/37/36

Table 3: This pure vehicle characteristics table shows capacities

expressed in quintals.


8/14


Node City Complete vehicle High quality Standard quality Sanitary prescription Origin protected

1 Castrovillari (1) Yes 2094 8

2 Castrovillari (2) Yes 855

3 Francavilla Marittima Yes 28

4 Spezzano Albanese Yes 703

5 San Lorenzo (1) No 1261 2243

6 San Lorenzo (2) No 58 787

7 Terranova da Sibari (1) No 505 4

8 Terranova da Sibari (2) No 262 16

9 Cassano allo Ionio Yes 26 3348

10 Tarsia (1) Yes 1077

11 Tarsia (2) No 375 24

12 Altomonte No 435 29

13 San Sosti No 84

14 Malvito No 376

15 Roggiano Gravina No 1353

16 San Marco Argentano Yes 104

17 Cerzeto No 5833

18 Santa Sofia DEpiro No 2519 Bisignano No 20 12

20 Luzzi (1) No 263 1272

21 Luzzi (2) No 86

22 Lattarico No 82 662

23 San Vincenzo la Costa No 175

24 Montalto (1) No 122 16 1267 645

25 Montalto (2) No 81 89

26 Castiglione Cosentino No 72

27 Rende (1) Yes 1204 239

28 Rende (2) Yes 1587 128

29 Cosenza (1) Yes 1936

30 Cosenza (2) Yes 526 119

31 Carolei No 47 3

32 Celico (1) No 389 61

33 Celico (2) No 146534 Spezzano Sila No 11 38 24

35 San Giovanni in Fiore Yes 209

36 Corigliano Yes 7193

37 Rossano Yes 625

38 Paludi No 3928

39 Crosia No 2

40 Calopezzati No 646

Table 4: The demand data for Cosenza are shown.

three complete vehicles, CV1, CV2, and CV4, to serve

farmers in Crotone, Vibo Valentia, and Catanzaro

areas that provide easy access to farmers. CV1 andCV2 serve Crotone and Vibo Valentia. CV4 is the

only vehicle that can be used to serve farmers in

Catanzaro to ensure that the working-shift duration

is not exceeded.

The algorithm uses all eight pure trucks to collect

milk from farmers in Cosenza. Note that the average

filling ratio, i.e., the average ratio between the load of

a truck and its capacity, is 95 percent.

To complete the study, we computed the distance

that each vehicle travelled and the transportation

cost (Table 8). ASSO.LA.C. provided both direct andindirect transportation costs. We used 1,189 seconds

and 20 iterations to find the solution.

Table 9 shows details of the proposed algorithms

performance. To show the effectiveness of constraints

(1) and (2) in the proposed method, we report the

results we achieved using three implementations of

the algorithm. The first implementation (Both in

Table 9) considers constraints (1) and (2), the second


9/14


Complete High Standard Sanitary Origin

Node City vehicle quality quality prescription protected

1 Catanzaro Yes 465

2 Lamezia Terme Yes 7672 432

Table 5: The demand data for Catanzaro are shown.



3 Crotone Yes 1545

4 Crucoli No 1511

5 Cutro Yes 476

6 Isola Capo Rizzuto Yes 3058

7 Roccabernarda No 3221

8 Scandale No 1239

Table 6: The demand data for Crotone are shown.



9 Vibo Valentia Yes 11531 2

10 Cir Marina Yes 151

Table 7: The demand data for Vibo Valentia are shown.

(None) does not implement constraints (1) or (2),

and the third (Only (1)) uses only constraints (1).

For each iteration l in Table 9, we list (1) the num-

ber of tours obtained from PHASE I, (2) the total

tour length from PHASE II, (3) the fulfilment of the

maximum shift duration for each tour (yes means

that all the tours respect the maximum shift duration

allowed and no means that at least one tour has

a length greater than T), and (4) the average tight-

ness of constraints (1) and (2), computed, respectively,

as

iIlslacki/Il and

iIl

surplusi/Il, where slackiis the slack variable of constraint i Il and surplusiis the surplus variable of constraint i Il. The algo-

rithm takes advantage of both constraints to produce

a better-quality solution.

Figure 5 shows the cost trend (y axis indicates the

cost of the solution in euros) over different values

(x axis). As we mentioned in the previous section,

the algorithm reaches its best solution when the value

of is very small; e.g., = 2 ( is the number of

iterations within which a given farmer cannot belong

to a given tour).

Distance travelled Transportation cost

ID vehicle (km/day) (euro/day)

CV1 275 103CV2 282 106

CV4 274 114

PT1 186 84

PT2 198 86

PT3 176 83

PT4 166 81

PT5 135 77

PT6 135 76

PT7 141 79

PT8 158 81

Total 2126 667

Table 8: The data in this table illustrate case study results for our opti-

mized solution.

To show the improvements that the proposed

method produced, Table 10 reports how ASSO.LA.C.

operated prior to implementing our optimization

method; it uses the same parameters as Table 8 to

provide a fair comparison.

Note that the prior solution used 12 vehicles; our

optimized solution used 11 vehicles (it did not use

complete vehicle CV3), thus implying that the FRAP

objective function makes a difference. In the prior

solution, the daily transportation cost was 1,100 euros

and the trucks travelled 2,484 km.Comparing values in Tables 8 and 10, we see that

the dairy company attained a reduction of about

14.4 percent in the total distance that the tank trucks

600

700

800

900

1 2 3 4 5 6

Figure 5: The graph illustrates solution cost varying values. The y axis

indicates the cost in euros of the solution over different values reported

in the x axis; is the number of iterations within which a given farmer

cannot belong to a given tour.


10/14


Constraints No. of tours Total tour length Fulfilment of the maximum Average tightness Average tightness

Iterations (1) and (2) in PHASE I in PHASE II (km) shift duration for each tour of constraints (1) of constraints (2)

1 Both 10 2,268 No

None 10 2,268 No

Only (1) 10 2,268 No

2 Both 11 2,156 No 1.00 0.00

None 11 2,212 No

Only (1) 11 2,178 No 0.25

3 Both 12 2,148 Yes 0.67 0.00

None 11 2,178 No

Only (1) 12 2,154 Yes 0.33

4 Both 13 2,132 Yes 0.00

None 11 2,184 No

Only (1) 12 2,144 Yes

5 Both 11 2,209 No 0.00

None 12 2,144 Yes

Only (1) 11 2,182 No

6 Both 11 2,198 No 0.33 0.00None 12 2,154 Yes

Only (1) 11 2,192 No 0.00

7 Both 11 2,172 No 0.00 0.00

None 11 2,182 No

Only (1) 12 2,158 Yes 0.33

8 Both 12 2,154 Yes 0.33 0.00

None 11 2,192 No

Only (1) 12 2,148 Yes

9 Both 13 2,132 Yes 0.00

None 11 2,209 No

Only (1) 11 2,160 No

10 Both 11 2,126 Yes 0.00

None 12 2,172 Yes

Only (1) 11 2,178 No 0.00

11 Both 12 2,138 Yes 0.00

None 12 2,174 Yes

Only (1) 11 2,202 No 0.00

12 Both 11 2,146 No 0.67

None 11 2,178 No

Only (1) 12 2,172 Yes 0.33

13 Both 12 2,178 Yes 0.25 0.00

None 11 2,186 No

Only (1) 12 2,162 Yes

14 Both 12 2,198 No 0.00

None 11 2,192 No

Only (1) 12 2,158 Yes

15 Both 12 2,176 No 0.25 0.00

None 11 2,184 No

Only (1)

16 Both 12 2,174 Yes 0.25 0.00

None

Only (1)

Table 9: The data in this table show behavior of different implementations of the proposed algorithm. Note that

the asterisk means that at least one infeasibility in PHASE I has been met; i.e., the algorithm entered into the

while cycle at Step 6 (Table 1).


11/14


Constraints No. of tours Total tour length Fulfilment of the maximum Average tightness Average tightness

Iterations (1) and (2) in PHASE I in PHASE II (km) shift duration for each tour of constraints (1) of constraints (2)

17 Both 11 2,248 No 0.70

None

Only (1)

18 Both 11 2,202 No 0.50 0.57

None

Only (1)

19 Both 11 2,189 No 0.50 0.57

None

Only (1)

20 Both 12 2,152 Yes 0.75 0.00

None

Only (1)

Table 9: Continued.

covered. Furthermore, the average filling ratio of the

tank trucks, which represented 85 percent in the com-

panys prior solution, was 95 percent for our opti-

mized solution.

These costs imply that our proposed approach

produces a yearly transportation cost savings of

156,000 euros and an additional cost reduction of

about 10,000 euros (because of using one less truck

and its constituent devices installed to monitor the

temperature of the milk during transport). Reduc-

ing transportation cost by 166,000 euros in our 158-

farmer case study is proportional to the cost savingsdescribed by Butler et al. (2005), in which an Irish

Distance travelled Transp. cost

ID vehicle (km/day) (euro/day)

CV1 260 100

CV2 334 118

CV3 260 112

CV4 245 100

PT1 192 86

PT2 210 89

PT3 180 84

PT4 163 80

PT5 155 82

PT6 153 80

PT7 155 82

PT8 177 87

Total 2484 1100

Table 10: The data in this table illustrate case study results for the prior

solution.

dairy company with about 800 farmers reduced itsannual costs by 635,000 euros.

SummaryThis paper studies a problem faced by an Italian dairy

company that collects milk from farmers. We mod-

eled the problem as a generalization of the TTRP and

show how OR techniques improved the companys

performance.

The problem has specific constraintsfarmers pro-

duce different types of milk, and each tank truckmust have different compartments to hold these dif-

ferent milk types; this imposes the limitation that

at most one milk type can be assigned to a tank-truck

compartment. We propose an optimization approach

based on the joint use of mathematical programming

and local search, present experiments we performed

as part of the case study, and compare the perfor-

mance of our proposed method with the process that

the company used prior to implementing our pro-

posal. The results show that our process results in cost

savings of approximately 166,000 euros per year.

Appendix A. The Farmer RouteAssignment Problem FormulationLet

i I be tank truck i in fleet I;

fleet I= I1 I2, where I1 is the set of complete

vehicles and I2 is the set of pure trucks;


12/14


j J be milk type j in the milk-type set J;

s, s S, be farmers s and s in the set S of

farmers;

set S of farmers be split into Str and Swt , with Stras the set of farmers reachable by complete vehicles

and Swt reachable by pure trucks only;

k Ki be compartment k in tank truck i;

Ki =Ki K

i , where K

i is the set of compartments

on the truck component of complete vehicle i and Kiis the set of compartments on the trailer component

of complete vehicle i;

Qjs , be the amount of milk type j associated with

farmer s; and

cik , be the capacity of compartment k in tank

truck i.

We define a (real) variable yikjs 0, which is asso-ciated with the quantity of milk type j loaded in

compartment k of tank truck i from farmer s. Vari-

able yikjs > 0 means that tank truck i had to load a

certain quantity of milk type j from farmer s. We

then define a binary variable zis to keep track of this

farmer; i.e.,

zis =

1 if tank truck i serves farmer s,0 otherwise

In addition, xikj and Ri denote two binary variables

such that

xikj =

1 if milk type j is assigned to

compartment k of tank truck i,

0 otherwise

Ri =

1 if tank truck i is used,0 otherwise.

Therefore, the overall formulation of the FRAP is

miniI

Ri

iI

kKi

yikjs =Qjs j J s S (A1)

sS

yikjs cikxikj i I j J k Ki (A2)

iI

zis 1 s S (A3)

sStr

zis Ri i I1 (A4)

yikjs ziscik i I k Ki j J s S (A5)

sSzis RiS i I (A6)

jJ

xikj 1 i I k Ki (A7)

yikjs = 0 i I1 k Ki j J s Swt (A8)

xikj 01 i I k Ki j J

yikjs + i I k Ki j J s S

Ri 01 i I

Constraints (A1) say that for farmer s and milk

type j, demand Qjs must be satisfied. Constraints (A2)

impose the restriction that if a positive amount ofmilk type j is loaded in compartment k of tank truck

i, then xikj = 1; i.e., milk type j is assigned to com-

partment k of tank truck i, and this quantity must

not be larger than the capacity cik of compartment k

of tank truck i. Constraints (A3) mean that at least

one truck must serve each farmer. Constraints (A4)

state that complete vehicles (if utilized) must visit

at least one farmer in Str. Constraints (A5) establish

that if a certain quantity of a milk type is loaded from

farmer s in a tank truck i (i.e., yikjs 0), then tank

truck i serves farmer s; i.e., zis = 1. Inequalities (A6)

are strictly related to the objective function. Indeed,when at least one customer is assigned to vehicle i,

then Ri must equal one and contributes to the value

of the objective. Constraints (A7) are the milk-type

incompatibility constraints. Recall that by constraints

(A2), if the quantity

sSyikjs of milk type j loaded

in compartment k of tank truck i is greater than zero,

the binary variable xikj is forced to assume a value

equal to one. The FRAP formulation is completed by

constraints (A8), which ascertain that milk cannot be

transferred between the truck and trailer because of

the existence of varying milk types.

Appendix B. The Route DefinitionProblem FormulationLet wiss be a binary variable, which is defined for each

tank truck i I and for each farmer s and s. We have

wiss =

1 if tank truck i travels from s to s

0 otherwise.


13/14


Let Succs and Preds be the set of successors and

predecessors, respectively, of farmer s. Starting from

the definitions of Si and Si , we can define the RDP

constraints separately for each tour i. In particular, ifi is a tour associated with a pure truck, then the RDP

constraints are

sSuccs0Si

wis0s = 1 i I (B1)

sPreds0Si

wiss0 = 1 i I (B2)

sSuccsSi

wiss = 1 i I s Si (B3)

s

PredsSi

wiss = 1 i I s Si (B4)

Constraints (B1) and (B2) say that a tour must start

and finish at warehouse s0. Constraints (B3) and (B4)

are flow-conservation laws; i.e., if a tank truck arrives

at a farmer site, it must leave this site.

As we discussed in the Milk Collection with Incom-

patibility Constraints section, if tour i is associated with

a complete vehicle, then the RDP constraints split up

into two subsets, one related to Si obtained by replac-

ing Si with Si in constraints (B1)(B4) and the other

related to Si defined by constraints (B3)(B4) in which

Si is replaced by Si .The objective function of RDP takes the following

form:

miniI

sS

zists +sS

s S

dss wiss

v

This equation comprises two parts; the first is

the total preparation time, i.e., the time needed for

the milk analysis, writing the report, and connect-

ing pumps to tanks. For this first part, we must con-

sider which tank truck serves which farmer. If tank

truck i serves farmer s, then zis is equal to one. When

this happens, we consider the contribution given bythe preparation time ts of farmer s. Summing up the

total number of farmers, we have the total prepa-

ration time. Note that this contribution depends on

the number of tank trucks that services a certain

client.

The second part refers to the transportation time.

When wiss = 1, then tank truck i serves both farmers

s and s. Summing all the farmers, multiplying by the

distance dss between s and s, and dividing by the

average truck speed v, we obtain the total transporta-

tion time.

AcknowledgmentsWe thank the anonymous referees for their valuable sug-gestions and the ASSO.LA.C. Dairy Company for its kindcollaboration in this research project.

References

Baldacci, R., M. Battarra, D. Vigo. 2008. Routing a heterogeneousfleet of vehicles. B. Golden, S. Raghavan, E. Wasil, eds. TheVehicle Routing Problem: Latest Advances and New Challenges.Springer, Heidelberg, Germany, 328.

Basnet, C., L. R. Foulds, J. M. Wilson. 1999. An exact algorithm fora milk tanker scheduling and sequencing problem. Ann. Oper.Res. 86 559568.

Butler, M., P. Herlihy, P. B. Keenan. 2005. Integrating informationtechnology and operational research in the management ofmilk collection. J. Food Engrg. 70(3) 341349.

Butler, M., H. P. Williams, L.-A. Yarrow. 1997. The two-period trav-elling salesman problem applied to milk collection. Comput.Optim. Appl. 7(3) 291306.

Chao, I.-M. 2002. A tabu search method for the truck and trailerrouting problem. Comput. Oper. Res. 29(1) 3351.

Chao, I. M., B. L. Golden, E. A. Wasil. 1995. An improved heuristicfor the period vehicle routing. Networks 26(1) 2544.

Claassen, G. D. H., T. Hendriks. 2007. An application of specialordered sets to a periodic milk collection problem. Eur. J. Oper.Res. 180(2) 754769.

Cordeau, J. F., M. Gendreau, G. Laporte, J. Y. Potvin, F. Semet. 2000.Classical and modern heuristics for the vehicle routing prob-lem. Internat. Trans. Oper. Res. 7 285300.

Cordeau, J. F., M. Gendreau, G. Laporte, J. Y. Potvin, F. Semet. 2002.A guide to vehicle routing heuristics. J. Oper. Res. Soc. 53(5)512522.

Fisher, M. L., R. Jaikumar. 1981. A generalized assignment heuristicfor vehicle routing. Networks 11(2) 109124.

Hoff, A., A. Lkketangen. 2007. A tabu search approach formilk collection in western Norway using trucks and trailers.Proc. Sixth Triennial Sympos. Transportation Anal. (TRISTAN VI),Phuket Island, Thailand.

Sankaran, J. K, R. R. Ubgade. 1994. Routing tankers for dairy milkpickup. Interfaces 24(5) 5966.

Scheuerer, S. 2006. A tabu search heuristic for the truck and trailerrouting problem. Comput. Oper. Res. 33(4) 894909.

Tan, K. C., Y. H. Chew, L. H. Lee. 2006. A hybrid multi-objectiveevolutionary algorithm for solving truck and trailer vehiclerouting problems. Eur. J. Oper. Res. 172(3) 855885.

Toth, P., D. Vigo, eds. 2002. The Vehicle Routing Problem. SIAMMonographs on Discrete Mathematics and Applications,Philadelphia.

Camillo Nola, Chief Executive Officer, ASSO.LA.C.

Societ Cooperativa Agricola, Contrada Ciparsia,

87012 Castrovillari (CS), Italy, writes: I certify that


14/14


in 2005, the ASSOLAC dairy company, Castrovil-

lari (CS) Italy, in collaboration with the Faculty of

Engineering, University of Calabria, Italy, initiated a

project to minimize routing costs deriving from milkcollection while also improving the number of cus-

tomers to be serviced: it is well known that optimiz-

ing transport cost can allow the company to pay a

higher milk price to farmers, thus attracting higher vol-

umes. The method based on mathematical program-

ming and local search heuristic proposed has signif-

icantly improved the overall transportation cost andalso the number of vehicles employed in servicing the

current clients; therefore, this clearly enables the sec-

ond objective of enlarging the set of customers.

4. Milk Collection

Documents

Transcript of 4. Milk Collection