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Multi-Product Lot-Sizing and Scheduling on Unrelated Parallel Machines

Multi-Product Lot-Sizing and Scheduling on Unrelated Parallel Machines. Mikhail Y. Kovalyov (presenter), Belarusian State University Alexandre Dolgui, Ecole des Mines de S.Etienne Anton V. Eremeev, Institute of Mathematics, SB RAS, Omsk. Problem formulation. Computational complexity.

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Multi-Product Lot-Sizing and Scheduling on Unrelated Parallel Machines

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  1. Multi-Product Lot-Sizing and Scheduling on Unrelated Parallel Machines Mikhail Y. Kovalyov (presenter), Belarusian State University Alexandre Dolgui, Ecole des Mines de S.Etienne Anton V. Eremeev, Institute of Mathematics, SB RAS, Omsk • Problem formulation. • Computational complexity. • Motivation. • Related studies. • Triangle inequality case. • Given number of products. • Perspectives for future research.

  2. 1. Problem formulation. R|slij,β|γ,βє{cntn,dscr}, γє{Cmax,Lmax}– 4 versions seq. and machine dep. n products, m machines pli – per unit processing requirement for product i on machine l, Di≤Xi≤Bi, Xi – total production ofi, q0li≤xli, xli – production quantity of i on l, Ci≤di, Ci– completion time for i, Mi – set of eligible machines for i, Nl – set of eligible products for machine l, nl=|Nl|, nmax=max{nl}. demand - variables min lot size Machine m … … sl0i i slij j sljk k Machine l … … time Machine 1

  3. 2. Computational complexity. TSP Hamiltonian Path of Minimum Weight R1|slij,β|γβє{cntn,dscr}, γє{Cmax,Lmax} R1|slij,β|γ -NPO-complete (no constant factor poly. approximation) (TSP - Orponen and Mannila, 1987) ∆TSPR1|∆slij,β|γ R1|∆slij,β|γis 220/219-non-approximable (∆TSP - Papadimitrou & Vempala, 2006)

  4. 3. Motivation. • Medium-range productionscheduling applications in: • textile industry (Silva,Magalhaes 2006; Taner,Hodgson,King, Schultz 2007); • metal production in foundries (dos Santos-Meza,dos Santos, Arenales 2002; de Araujo,Arenales,Clark 2008); • multi-product chemical plants(Bitran,Gilbert 1990; Lin,Floudas,Modi,Juhasz 2002; Shaik,Floudas,Kallrath,Pitz2007) slij– cleaning operations; Non- ∆slij– some chemicals have cleaning effect; dscr – production of granules in bags or packets; cntn – continuous production of granules; Cmax – latest plant completion time minimization; Lmax – equitable customer treatment w.r.t. due date satisfaction (if product=customer order).

  5. 4. Related studies. Monma, Potts, OR 1989: identical machines, ∆ case, each product i consists of Di distinct items having their own processing times and due dates, preemptions allowed (which makes the problem continuous), Length=O(mn2+Σ Di). Results: D.P. algorithms for single machine case (no pmtn can be assumed), NP-hardness for two machine case. OurLength=O(mn2). Brucker, Kovalyov, Shafransky, Werner, AOR 1998: discrete case, Bi =Di ,sequence independent setup times. Results: NP-hardness proofs, polynomial special cases, D.P. algorithms, (1+ε)-approximation algorithms.

  6. 5. Triangle inequality case. slij+sljk≥slik , l,i,j,k A Property 1. There exists an optimal solution for R|Δslij,β|γ, βє{cntn,dscr}, γє{Cmax,Lmax}, in which each product is produced in at most one lot on each machine. A schedule is fully specified by: for each machine: a set of products to be manufactured, their sequence and the corresponding lot sizes.

  7. 5. Triangle inequality case. Allocation 0-1 matrix Y={yli}: yli=1, if product i is manufactured on machine l, yli=0 otherwise. Feasible Y : {l | yli=1} є Mi & Σyli≥1, i. #feasibleY=O(2mnmax). Associated with Y: P(Y,l) – set of product permutations consistent with Y for each machine l. A Matrix of lot sizes X={xli} X consistent with Y: q0liyli ≤xli, lєMi, Di ≤∑xli≤Bi, i Schedule =(Y, π1,…,πm,X) #(m+1)-tuples(Y, π1,…,πm) = O(2mnmax(nmax!)m). A

  8. 5. Triangle inequality case. Two-stage solution procedure: Enumeration of Y, and given Y, m-tuples of permutations (π1,…,πm), πlєP(Y,l), l=1,…,m. Given (m+1)-tuple (Y, π1,…,πm), solvelot-sizing subproblem - LP with O(mn) variables: Minimize Cmax (Lmax), subject to t(πl,l)+∑iєN_lplixli≤ Cmax, (…- di^l_k≤Lmax), l=1,…,m, Di ≤∑lєM_ixli≤Bi, i=1,…,n, q0liyli≤xli≤Diyli, l=1,…,m, i=1,…,n. Variables: Cmax(Lmax) and {xli}. Rational if β=cntn, integer if β=dscr. total setup time

  9. 5. Triangle inequality case. Statement 1. Problem R|Δslij,β|γ, βє{cntn,dscr}, γє{Cmax,Lmax}, can be solved in O(τβ2mnmax(nmax!)m) time, where τβ– solution time for LP (β=cntn) or ILP (β=dscr) with O(mn) variables and O(m+n) constraints. min lot size If minimum lot processing times q0li pli and setup times satisfy certain inequalities (similar to ∆), the two-stage procedure can be modified to be used for the case, in which ∆ for slij is violated.

  10. 5. Triangle inequality case. DP1 for R|Δslij,β|Cmax (modified from Held and Karp 1962, for ∆TSP): T(l,S,i) – minimum total setup time on machine l for processing products of set SєNlprovided that iєS is processed last. Initialization: T(l,S,i)=sl0ifor S={i}, iєNl, l=1,…,m. Recursion: T(l,S,i)=minjєS\{i}{T(l,S\{i},j)+slji} Minimum total setup times T*(l,Y) can be computed in O(m(nmax)2 2nmax) time. Statement 2. Problem R|Δslij,β|Cmax is solved by DP1 in O(τβ2mnmax+m(nmax)2 2nmax) time (reduced by a factor of (nmax!)m comparing with an enumeration of all permutations).

  11. 6. Given number of products. Statement 3. R|slij=s,β|γ, βє{cntn,dscr}, γє{Cmax,Lmax}, is NP-hard, even if n=2,q0li=0, pli=pl, Di=D, Bi=∞, di=d, and any two pli differ by at most a factor of 2. Proof:Bounded Partition: Given 2k+1 positive integer numbers e1,...,e2kand E, which satisfy ∑el=2E and E/(k+1)< el<E/(k-1), l=1,...,2k, is there a subset XєK:={1,...,2k} such that ∑lєX el=E? Calculate A=∏er. Instance of R|slij=s,β|γ, βє{cntn,dscr},γє{Cmax,Lmax}: n=2, m=2k, slij=A, q0li=0, pli=A/el, Di=E, Bi=∞, di=2A. Bounded Partition has solution Cmax≤2A (Lmax≤0). (slij=A each machine l can process at most el units of the same product within the remaining A available time units X=Machine-Set-For-Product-1, ∑iєX el≥E, ∑iєK\X el≥E).

  12. 6. Given number of products. DP2 for R|slij,dscr|Cmax: (assigns product lots to machines 1,…,m in this order) Cl(z1,…,zn,j,t) – minimum Cmax value for a partial schedule, in which zi units of product i, i=1,…,n, are processed on machines 1,…,l, product jєNlis processed last on machine l, and the last unit of this product completes at time t, t≤nmax(pmaxBmax+smax). Initialization:C0(z1,…,zn,j,t)=0 for (z1,…,zn,j,t)=(0,…,0), and C0(z1,…,zn,j,t)=∞ for (z1,…,zn,j,t)≠(0,…,0). Recursion:Cl(z1,…,zn,j,t)=min miniєN_{l-1}U{0},t {Cl-1(z1,…,zn,i,t)}, if (j,t)=(0,0), miniє(N_lU{0})\{j},δє{q^o_{lj},…,z_j}{max{t,Cl(z1,…, zj-δ,…,zn,i,t-(slij+δplj)}, if (j,t) ≠(0,0). no product is assigned

  13. 6. Given number of products. Statement 4. Problem R|slij,dscr|Cmax is solved by DP2 in O(m(nmax)3(Bmax+1)n+1(smax+pmaxBmax)) time, which is pseudopolynomial for a given n and linear in m. unusual for batch scheduling problems Reduced running time:

  14. 6. Given number of products. DP3 for R|slij,dscr|Lmax: (modification of DP2 for Cmax) Difference: t≤nmax(pmaxBmax+smax)+dmax. Statement 5. Problem R|slij,dscr|Lmax is solved by DP3 in O(m(nmax)3(Bmax+1)n+1(smax+pmaxBmax+dmax)) time.

  15. 7. Perspectives for future research.

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