6.5 Freight Traffic Assignment Problems
6.5.3 Linear multicommodity minimum-cost flow problems
The linear multicommodity minimum-cost flow (LMMCF) problem can be formu- lated as the following LP model.
Minimize
k∈K
(i,j )∈A
ckijxkij subject to
{j∈V:(i,j )∈A}
xijk −
{j∈V:(j,i)∈A}
xj ik =
oik, ifi∈O(k),
−dik, ifi∈D(k), 0, ifi∈T (k),
i∈V , k∈K, xijk ⩽ukij, (i, j )∈A, k∈K,
k∈K
xijk ⩽uij, (i, j )∈A, (6.16) xijk ⩾0, (i, j )∈A, k∈K.
The LMMCF problem can be solved efficiently through a tailored Lagrangian pro- cedure. Letλij(⩾0) be the multipliers attached to constraints (6.16). The Lagrangian relaxation of the LMMCF problem is as follows.
Minimize
k∈K
(i,j )∈A
cijkxijk +
(i,j )∈A
λij
k∈K
xijk −uij
(6.17)
218 LONG-HAUL FREIGHT TRANSPORTATION subject to
{j∈V:(i,j )∈A}
xijk −
{j∈V:(j,i)∈A}
xj ik =
oik, ifi∈O(k),
−dik, ifi∈D(k), 0, ifi∈T (k),
i∈V , k∈K, (6.18) xijk ⩽ukij, (i, j )∈A, k∈K, (6.19) xijk ⩾0, (i, j )∈A, k∈K. (6.20) The relaxation (6.17)–(6.20), referred in the sequel to as the r-LMMCF problem, is made up of|K|independent single-commodity minimum-cost flow problems, since the sum
(i,j )∈Aλijuij in the objective function (6.17) is constant for a given set of multipliersλij,(i, j ) ∈ A. Therefore, thekth LMCF subproblem,k ∈ K, is as follows.
Minimize
(i,j )∈A
(ckij+λij)xkij (6.21) subject to
{j∈V:(i,j )∈A}
xijk −
{j∈V:(j,i)∈A}
xj ik =
oki, ifi∈O(k),
−dik, ifi∈D(k), 0, ifi∈T (k),
i∈V , (6.22) xijk ⩽ukij, (i, j )∈A, (6.23)
xijk ⩾0, (i, j )∈A, (6.24)
can be solved through the network simplex algorithm. Let LBkLMCF(λ),k ∈ K, be the optimal objective function value of the kth subproblem (6.21)–(6.24) and let LBr-LMMCF(λ)be the lower bound provided by solving the r-LMMCF problem. For a given set of multipliers, LBr-LMMCF(λ)is given by
LBr-LMMCF(λ)=
k∈K
LBkLMCF(λ)−
(i,j )∈A
λijuij.
Of course, LBr-LMMCF(λ)varies as the multiplierλchanges. The Lagrangian relax- ation attaining the maximum lower bound value LBr-LMMCF(λ)asλvaries is called thedual Lagrangian relaxation. The following property follows from LP theory.
Property. The lower bound provided by the dual Lagrangian relaxation is equal to the optimal objective function value of the LMMCF model, i.e.
maxλ⩾0{LBr-LMMCF(λ)} =z∗LMMCF.
LONG-HAUL FREIGHT TRANSPORTATION 219 Moreover, the dual Lagrangian multipliersλ∗ij,(i, j ) ∈ A, are equal to the optimal dual variablesπij∗,(i, j )∈A, associated with the relaxed constraints (6.16).
In order to compute the dual Lagrangian multipliers, or at least a set of multipliers associated with a ‘good’ lower bound, the classicalsubgradient procedure, already illustrated in Chapter 3, can be used.
Step 0. (Initialization). LetHbe a pre-established maximum number of subgradient iterations. Set LB = −∞,h = 1 andλ(h)ij =0,(i, j ) ∈ A. Set UB equal to the cost of the best feasible solution if any is available, or set UB= ∞, otherwise.
Step 1. (Calculation of a new lower bound). Solve the r-LMMCF problem usingλ(h) as a vector of multipliers. If LBr-LMMCF(λ(h)) >LB, set LB=LBr-LMMCF(λ(h)).
Step 2. (Checking the stopping criterion). If solutionxijk,(h),(i, j )∈ A,k ∈K, of the r-LMMCF problem satisfies the relaxed constraints (6.16), and
λ(h)ij
k∈K
xijk,(h)−uij
=0, (i, j )∈A,
STOP, the solution found is optimal (z∗LMMCF=LB). Ifxijk,(h),(i, j )∈A,k∈K, satisfies the relaxed constraints (6.16), update UB if necessary. If UB=LB,STOP, the feasible solution attaining UB is proved to be optimal. Ifh= H,STOP, LB represents the best lower bound available forzLMMCF∗ .
Step 3. (Updating the multipliers). Determine, for each(i, j ) ∈A, the subgradient of the relaxed constraint:
sij(h)=
k∈K
xijk,(h)−uij, (i, j )∈A.
Then set
λ(h+ij 1)=max(0, λ(h)ij +β(h)sij(h)), (i, j )∈A, whereβ(h)is a suitable coefficient
β(h)= α
h, (i, j )∈A, (6.25)
withαarbitrarily chosen in the interval(0,2]. Alternatively, if a feasible solution the problem is available, set
β(h)= α(UB−LBLMCF(λ(h)))
(i,j )∈A(sij(h))2 , (i, j )∈A,
with UB equal to the objective function value of the best feasible solution available.
Seth=h+1 and go back to Step 1.
The subgradient method converges toz∗LMMCFprovided that the variations of the multipliers are ‘small enough’ (this assumption is satisfied if, for example, Equation (6.25) is used). However, this assumption makes the algorithm very slow.
220 LONG-HAUL FREIGHT TRANSPORTATION Table 6.4 Demand (in tons)dkt, k=1,2, t=1,2, in the Exofruit problem.
t=1 t=2
k=1 18 000 18 000 k=2 12 000 14 000
Table 6.5 Maximum amounts available (in tons)okt, k=1,2, t=1,2, in the Exofruit problem.
t=1 t=2
k=1 26 000 20 000 k=2 14 000 13 000
Table 6.6 Purchase prices (in euros/ton)pkt, k=1,2, t=1,2, in the Exofruit problem.
t=1 t=2
k=1 500 700
k=2 600 400
If the solutionxijk,(h),(i, j ) ∈ A,k ∈ K, of the r-LMMCF problem satisfies the relaxed constraints (6.16), it is not necessarily the optimal solution for the LMMCF problem. Asufficient(butnot necessary) condition for it to be optimal is given by the complementary slackness conditions:
λ(h)ij
k∈K
xijk,(h)−uij
=0, (i, j )∈A. (6.26)
If relations (6.26) are not satisfied, then the feasible solutionxijk,(h),(i, j ) ∈ A, k∈K, is a simply acandidate optimalsolution.
Ifh = H, the solution attaining LB could be infeasible for the LMMCF, or, if feasible, may not satisfy the complementarity slackness conditions. In any case, if subproblems (6.21)–(6.24) are solved by means of the network simplex method, a basic (feasible or infeasible) solution for the LMMCF model is available. In fact, the basic variables of the|K|subproblems (6.21)–(6.24) make up a basis of the LMMCF problem. The basic solution obtained this way can be used as the starting solution for the primal or dual simplex method depending on whether the solution is feasible for the LMMCF problem or not. If initialized this way, the simplex method is particularly efficient since the initial basic solution provided by the subgradient algorithm is a good approximation of the optimal solution.
LONG-HAUL FREIGHT TRANSPORTATION 221
1
2
3
4
5
6
Figure 6.13 Graph representation of the Exofruit problem.
Exofruit Ltd imports to the EU countries several varieties of tropical fruits, mainly coming from Northern Africa, Mozambique and Central America. The company pur- chases the products directly from farmers and transports them by sea to its warehouses in Marseille (France). The goods are then stored in refrigerated cells or at room tem- perature. As purchase and selling prices varies during the year, Exofruit has to decide when and how much to buy in order to satisfy demand over the year. The problem can be modelled as an LMMCF problem. In what follows, a simplified version of the problem is examined. It is assumed that a single source exists, products are grouped into two homogeneous groups (macro-products), and the planning horizon is divided into two semesters. Let dkt andokt, k = 1,2, t = 1,2, be the demand and the maximum amount (in tons) of macro-productkavailable in semestert, respectively (see Tables 6.4 and 6.5). Purchase prices (in euros/tons)pkt, k =1,2, t =1,2, of macro-productkin semestert are reported in Table 6.6.
The transportation costv of 1 ton of a macro-product is equal to €100, while the stocking costwof 1 ton of a macro-product is€100 per semester. Finally, the maximum quantityqof goods that can be stored in a semester is 8000 tons.
The problem can be formulated as an LMMCF problem with two commodities (one for each macro-product) on the directed graph shown in Figure 6.13. In such a representation,
• vertices 1 and 5 represent the source and the destination of macro-product 1, respectively;
• vertices 2 and 6 represent the source and the destination of macro-product 2, respectively;
• vertices 3 and 4 represent the warehouse in the first and in the second semester, respectively;
222 LONG-HAUL FREIGHT TRANSPORTATION
• arc (1,3) has a cost per unit of flow equal toc113=p11+v=600 euros/ton and a capacity equal tou13 =o11=26 000 tons;
• arc (1,4) has a cost per unit of flow equal toc114=p12+v=800 euros/ton and a capacity equal tou14 =o12=20 000 tons;
• arc (2,3) has a cost per unit of flow equal toc223=p21+v=700 euros/ton and a capacity equal tou23 =o21=14 000 tons;
• arc (2,4) has a cost per unit of flow equal toc224=p22+v=500 euros/ton and a capacity equal tou24 =o22=13 000 tons;
• arc (3,4) represents the storage of goods for a semester and is therefore associ- ated with a cost per unit of flow equal toc134=c234=w=100 euros/ton and a capacity ofu34=q=8000 tons;
• arc (3,5) has a zero cost per unit of flow and a capacity equal tou35 =d11= 18 000 tons;
• arc (4,5) has a zero cost per unit of flow and a capacity equal tou45 =d12= 18 000 tons;
• arc (3,6) has a zero cost per unit of flow and a capacity equal tou36 =d21= 12 000 tons;
• arc (4,6) has a zero cost per unit of flow and a capacity equal tou46 =d22= 14 000 tons.
The problem is formulated as follows.
Minimize
600x131 +800x141 +100x341 +700x232 +500x224+100x342 subject to
x351 +x451 =36 000, x131 −x134−x351 =0, x141 +x134−x451 =0,
x362 +x462 =26 000, x232 −x236−x342 =0, x242 +x234−x462 =0,
LONG-HAUL FREIGHT TRANSPORTATION 223 x131 ⩽26 000,
x141 ⩽20 000, x351 ⩽18 000, x451 ⩽18 000, x232 ⩽14 000, x242 ⩽13 000, x362 ⩽12 000, x462 ⩽14 000, x134+x342 ⩽8000,
x131, x141 , x341, x351, x451, x232, x242, x234, x362, x462 ⩾0.
By relaxing in a Lagrangian fashion the constraintx134+x342 ⩽8000 with a multi- plierλ34, the problem decomposes into two single-commodity linear minimum-cost flow problems. By initializing the subgradient algorithm withλ(0)34 =0 and using the updating formula (6.25) withα=0.05, the procedure provides, after 20 iterations, a lower bound LB=€40 197 387 (λ(20)34 =99.887). At the end, the procedure con- verges toλ∗34 =100, which corresponds to an optimal objective valuezLMMCF∗ equal to€40 200 000. Subproblemk=1 has, forλ∗34, two optimal basic solutions
x131,∗=x141,∗=x351,∗=x451,∗=18 000, x1,34∗=0 and
x131,∗=26 000, x141,∗=10 000, x341,∗=8000, x351,∗=18 000, x451,∗=18 000, while subproblemk=2 has a single optimal solution equal to
x232,∗=x242,∗=13 000, x342,∗=1000, x362,∗=12 000, x462,∗=14 000.
By combining the two partial solutions, the following two solutions are obtained:
x131,∗=x141,∗=x351,∗=x451,∗=18 000, x341,∗=0,
x232,∗=x242,∗=13 000, x342,∗=1000, x362,∗=12 000, x462,∗=14 000 and
x131,∗=26 000, x1,∗14 =10 000, x341,∗=8000,
x351,∗=18 000, x1,∗45 =18 000, x232,∗=x242,∗=13 000, x342,∗=1000, x2,∗36 =12 000, x462,∗=14 000.
The former solution has an objective function value equal to€41 000 000 and is feasible, but does not satisfy the complementarity slackness conditions (6.26), while
224 LONG-HAUL FREIGHT TRANSPORTATION
B
A C A
C
B
End-of-line terminal End-of-line terminal
End-of-line terminal End-of-line terminal End-of-line terminal End-of-line terminal D
Break-bulk terminal
Figure 6.14 Two alternative service networks for a three end-of-line terminal transportation system.
the latter is infeasible. It can be easy verified that the optimal solution is a convex combination of the two previous solutions and corresponds to
x131,∗=25 000, x1,∗14 =11 000, x341,∗=7000, x351,∗=18 000, x451,∗=18 000, x232,∗=12 000, x2,∗24 =13 000, x342,∗=1000, x362,∗=12 000, x2,∗46 =14 000.