Optimizing a multi-stage production/inventory system by DC programming based approaches

Abstract

This paper deals with optimizing the cost of set up, transportation and inventory of a multi-stage production system in presence of bottleneck. The considered optimization model is a mixed integer nonlinear program. We propose two methods based on DC (Difference of Convex) programming and DCA (DC Algorithm)—an innovative approach in nonconvex programming framework. The mixed integer nonlinear problem is first reformulated as a DC program and then DCA is developed to solve the resulting problem. In order to globally solve the problem, we combine DCA with a Branch and Bound algorithm (BB-DCA). A convex minorant of the objective function is introduced. DCA is used to compute upper bounds while lower bounds are calculated from a convex relaxation problem. The numerical results compared with those of COUENNE (http://www.coin-or.org/download/binary/Couenne/), a solver for mixed integer nonconvex programming, show the rapidity and the ϵ-globality of DCA in almost cases, as well as the efficiency of the combined DCA-Branch and Bound algorithm. We also propose a simple heuristic algorithm which is proved by experimental results to be better than an existing heuristic in the literature for this problem.

Appendix

In this section we present the heuristic algorithm proposed by Hisao to solve problem (P _u ). The reader could refer [9] and [10] for more detail. The following notions are used

• $Q^{s}_{i,m_{i}}$ is the critical order value of Q,

• m _i (max) is the maximal value of m _i ,

• RG _k =[r _k−1,r _k ] is the kth critical interval of Q,

• Nc is the critical value number, $Nc\leq\sum_{i=1}^{N} m_{i}(\max)$,

• NumRg is the number of critical intervals,

• List is the set of index k such that the interval RG _k can be used to find the optimal solution.

Algorithm

Step 1 For i=1,2,…,N

Set m _i =1

(a) Compute the critical value

$$Q_{i,m_{i}}^{s}=\sqrt{\frac{d_{i}}{a_{i}( \frac{1}{R_{i}^{m_{i}}-1}-\frac{1}{R_{i}^{m_{i}+1}-1})}} $$

(b) If $Q_{i,m_{i}}^{s}<P_{r}$ then set m _i =m _i +1 and return (a) else set m _i (max)=m _i .

Step 2 Merge all the critical values and order them to establish a critical vector S=(s ₁,s ₂,…,s _Nc ) where s ₁≤s ₂≤⋯≤s _Nc ,s _Nc−1<P _r ,s _Nc >P _r .

Construct vector B=(b _j ) such that b _j =i if s _j is the critical value being given when computing m _i (max)

Step 3 Set m ¹=(1,1,…,1),M′=0,k=1,r ₁=s ₁,RG ₁=(0,r ₁].

Step 4 For j=2,3,…,Nc

Set M′′=(0,..,0,1,0,..,0), the only non zero element of M′′ is in the (b _j−1)th position.

If s _j−1<s _j then set r _k+1=s _j ,RG _k+1=[r _k ,r _k+1],m ^k+1=m ^k+M′+M ^′′′′,k=k+1,M′=0 else set M′=M′+M′′

Step 5 Set NumRg=k, List=∅

Compute

$$ Q^k=Q\bigl(m^k\bigr)=\sqrt{\frac{\alpha_1}{\alpha_2}} \quad \text{where } \alpha_1=\sum _{i=1}^N c_i+d_im^k_i,\ \alpha_2= \sum _{i=1}^N \biggl(b_i+\frac{a_i}{R_i^{m^k_i}-1}\biggr). $$

If Q ^k∈RG _k then List=List∪{k}

Step 6 If List=∅ then STOP without solution else Find f(Q _opt ,m _opt )=min{f(Q ^k,m ^k),k∈List}, STOP with the optimal solution (Q _opt ,m _opt ) and the optimal value f(Q _opt ,m _opt ).