Solving fixed charge transportation problem with truck load constraint using metaheuristics

Abstract

Fixed charge transportation (FCT) problems addressed in literature assumed shipment between a source and a destination is fulfilled in a single lot. However, in reality the lot size may exceed the capacity of the carrier and hence the shipment needs to be executed by conducting more than one trip. This gives an increased fixed charge which is proportional to the number of trips performed. This paper proposes a special case of the FCT problem were the truck load constraint is considered and is referred as the fixed charge transportation problem with truck load constraints (FCT-TLC) problem. The objective considered in this problem is to minimize the total cost of transportation without violating the supply and demand constraints. The general FCT problem is classified as NP-hard and to solve this proposed problem with additional constraints, two metaheuristic algorithms are used. A Genetic Algorithm (GA) and a Simulated Annealing Algorithm (SAA) are proposed to solve the FCT-TLC problem and the performance of the algorithms is tested on twenty randomly generated problem instances. Detailed comparative study on the computational results obtained using GA and SAA are presented. Both metaheuristics show good results for solving the proposed problem. However, SAA outperformed GA for many problems with different truck load capacities. To test the performance of the proposed algorithms, comparison with approximate and lower bound solutions for the problem with a relaxed truck capacity constraint is also presented.

Appendices

Appendix A: Numerical illustration of GA for the FCT-TLC problem

This section illustrates the proposed GA for the FCT-TLC problem with an example problem given below:

The following data are given as input:

• ${p }={ 3, q}={ 3\,and\,W}={ 200}$

• ${C}_{{ij}} { , F}_{{ij}} {, S}_{i} {\,and\,D}_{j} ( \forall _{i} {, i }={ 1\,to\,p\,and\,}\forall _{j} {, j }= 1\,to\,q),$ are as given in Table 11.

Table 11 Cost matrix data

Table 12 shows the sample output at various stages (initial population module, Evaluation module, selection module, crossover module and new population generation module) in the generation of the new population from its previous one. The transportation plan for the first chromosome of the initial population is given in Table 13.

Table 12 New population generation from initial

Table 13 Transportation plan ${X}_{{ij}}$ for first chromosome of the initial population

Total cost of transportation for the first chromosome of the initial population is $Z~=~35{,}550$. The Optimal chromosome and transportation plan obtained after 50 iterations are given in Tables 14 and 15. Figure 8 illustrated the allocation procedure for the optimal chromosome.

Table 14 Optimal chromosome ${X}_{{k} _{{opt}}}$

Fig. 8

Allocation procedure for the optimal chromosome

Table 15 Optimal transportation plan ${{X}_ij _opt}$ of GA

The total transportation cost for the optimal transportation plan ${X}_{{{ij}} _{opt}}$ shown in Table 15 is ${Z}_{{opt}} ~=~28{,}350$.

Appendix B: Numerical illustration of SAA for the FCT-TLC problem

Detailed illustration on how the optimal transportation plan is achieved for FCT-TLC problem by SAA technique with an example problem given below:

Step 1 The input data of the illustrative problem are given below.

• ${p }={3,q}={3\, and \,W}={ 200}$

• ${C}_{{ij}} { , F}_{{ij}} {, S}_{i} {\,and\,D}_{j} {(}\forall _{i} {,i}={ 1\,to\,p\,and\,}\forall _{j} {, j }={ 1\,to\,q) }$ are as given in Table 11.

Step 2 SAA parameters are initialized.

Initial Temperature $T=475, \textit{ACCEPT}=0, \textit{TOTAL} = 0, \textit{FR}\_\textit{CNT}=0, \upalpha = 0.9$

Step 3 Generation of initial temporary seed and setting its solution

The initial seed ${X}_{k}$ is given in Table 16. The transportation plan ${X}_{{ij}}$ for the initial seed and its evaluation parameter Z are given in Table 17.

Set ${X}_{{k-g}} ={ X}_{k} {,X}_{{ij-g}}={ X}_{{ij}} {\,and\, Z}_{g}={ Z}$.

Table 16 Initial seed ${X}_{k}$

Table 17 Transportation plan ${X}_{{ij}}$ for the initial seed and its evaluation parameter Z

Step 4 Generation of neighbourhood seed and evaluation

Generate a neighbourhood seed ${X}_{k}^{1}$ to a current seed ${X}_{k}$ using unary random mutation, Inverse mutation and pair wise interchange mutation. Table 18 provides the neighbourhood seeds ${X}_{k}^{1}$ and its evaluation parameter ${Z}^{{1}}$. Table 19 provides the best neighbourhood seed. Table 20 provides the transportation plan ${X}_{{ij}}^{1}$ for the best neighbourhood seed and its evaluation parameter ${Z}^{{1}}$.

Table 18 Neighbourhood seeds ${X}_{k}^{1}$ and its evaluation parameter ${Z}^{{1}}$

Table 19 Best neighbourhood seed ${X}_{k}^{1}$

Table 20 Transportation plan ${X}_{{ij}}^{1}$ for the best neighbourhood seed and its evaluation parameter ${Z}^{{1}}$

Table 21 Optimal seed ${X}_{{k-g}}$

Table 22 Optimal transportation plan ${X}_{{ij-g}}$ of SAA

Fig. 9

Allocation procedure for the optimal seed

Step 5 Calculation of uphill acceptance parameter Delta

The new seed is selected by calculating the value of the delta. Delta is the cost difference between the neighbourhood seed and the initial seed.

$$\begin{aligned} Delta= & {} {Z}^{1}- Z;\\ Delta= & {} 32{,}850 - 35{,}550 = -\,2700. \end{aligned}$$

Since $Delta < 0$ proceed Step 6.

Step 6 :

Downhill move

Assign ${X}_{k}={ X}{_{k}^{1}} {,X}_{{ij}}={ X}_{{ij}}^{1} {\,and\,Z}={Z}^{{1}}$ and ${\textit{ACCEPT}}={ \textit{ACCEPT} }+{ 1}$

Step 7 :

Since ${Z}={Z}_{g}$ goto Step 11.

Step 11 :

Set ${\textit{TOTAL}}={ 1}$,

Step 12 :

Check for termination of iteration at the set temperature

If (${\textit{TOTAL}}>{r})$ or (${\textit{ACCEPT}}>{r/2})$, then proceed to Step 13 else go back to Step 4 until it satisfies the condition in Step 12.

If the condition in Step 12 is not satisfied, execute Step 4. The optimal seed ${X}_{{k-g}}$ and optimal transportation plan ${X}_{{ij-g}}$ after decreasing the temperature form 475 to 20 for the example problem are given in Tables 21 and 22. Figure 9 shows the allocation procedure for optimal seed.

The total transportation cost for the optimal transportation plan ${X}_{{ij-g}}$ shown in Table 22 is ${Z}_{g}$ = 28,250.

Appendix C: Numerical illustration for lower bound and approximate solutions

This section illustrates the method use to calculate the lower bound and approximate solutions for the FCT-TLC problem under relaxed truck load constraint with an example problem given below. The following data are given as input:

• ${p }={ 3, q}={ 3 }$

• ${C}_{{ij}} { , F}_{{ij}} {, S}_{i} {\, and\, D}_{j} ( \forall _{i} {, i }={ 1\, \hbox {to}\, p\, and\, }\forall _{j} {, j }={ 1 to q) },$ are as given in Table 11.

Table 23 provides the equivalent variable cost matrix ${CF}_{{ij}}$ for plant to customer respectively.

Table 23 Equivalent variable cost ${CF}_{{ij}}$ matrix data

The optimal transportation plan $X_{{{ij}}^{{III}}_{opt}}$ of Problem III with the equivalent variable cost matrix is given in Table 24.

Table 24 Optimal transportation plan ${X}_{{{ij}}^{{III}} _{opt}}$ of problem III

Table 25 Optimal transportation plan ${X}_{{{ij}}^{{II}} _{opt}}$ of GA

Table 26 Optimal transportation plan ${X}_{{{ij}}^{{II}} _{opt}}$ of SAA

Substituting the results obtained in Eqs. (12) and (11), the lower bound and approximate solutions are found as follows:${Z}^{{{\prime }{\prime }}}$ = 14,578 (Eq. 12), ${Z}^{{{\prime }}}$ = 16,300 (Eq. 11).

The near optimal transportation plan obtained from the proposed GA and SAA for the above problem data is given in Tables 25 and 26.

The total transportation cost for the transportation plan shown in Table 25 is Z = 16,150 and Table 26 is Z = 16,150.