The Fixed Charge Transportation Problem: A Strong Formulation Based On Lagrangian Decomposition and Column Generation

Transcription

1 The Fixed Charge Transportation Problem: A Strong Formulation Based On Lagrangian Decomposition and Column Generation Yixin Zhao, Torbjörn Larsson and Department of Mathematics, Linköping University, Sweden Column generation 2016

2 What is this talk about? Strong lower bounding for the fixed charge transportation problem by Lagrangian decomposition: supply and demand side copies of the shipping variables Dual cutting plane method (column generation)

3 What is this talk about? Strong lower bounding for the fixed charge transportation problem by Lagrangian decomposition: supply and demand side copies of the shipping variables Dual cutting plane method (column generation) Why? Lagrangian decomposition can give strong formulations Strong formulations of interest in column generation Combination not utilised in many papers Preliminary work to study the strength of the formulation: theoretically and empirically

4 Outline Introduction and background The fixed charge transportation problem Concluding comments

5 Lagrangian decomposition / Lagrangian relaxation Consider the problem (P) min{cx s.t. Ax b, Cx d, x X } = min{cx s.t. Ay b, Cx d, x = y, x X, y Y }, where Y is such that X Y

6 Lagrangian decomposition / Lagrangian relaxation Consider the problem (P) min{cx s.t. Ax b, Cx d, x X } = min{cx s.t. Ay b, Cx d, x = y, x X, y Y }, where Y is such that X Y Lagrangian decomposition (LD) min{cx + u(x y) s.t. Ay b, Cx d, x X, y Y } = min{(c + u)x s.t. Cx d, x X } + min{ uy s.t. Ay b, y Y }

7 Lagrangian decomposition / Lagrangian relaxation Consider the problem (P) min{cx s.t. Ax b, Cx d, x X } = min{cx s.t. Ay b, Cx d, x = y, x X, y Y }, where Y is such that X Y Lagrangian decomposition (LD) min{cx + u(x y) s.t. Ay b, Cx d, x X, y Y } = min{(c + u)x s.t. Cx d, x X } + min{ uy s.t. Ay b, y Y } Lagrangian relaxation w.r.t. one of the constraint groups, for example Ax b (LR) min{cx + v(ax b) s.t. Cx d, x X } = min{(c + va)x s.t. Cx d, x X } + vb, where v 0

8 Strength of bounds [Guignard and Kim(1987)]: The Lagrangian decomposition bound is as least as strong as the strongest of the Lagrangian relaxation bound when Ax b is relaxed the Lagrangian relaxation bound when Cx d is relaxed and there is a chance that it is stronger!

9 Related work Very few papers on Lagrangian decomposition and column generation: [Pimentel et al.(2010)]: The multi-item capacitated lot sizing problem Branch-and-price implementations for two types of Lagrangian relaxation and for Lagrangian decomposition Lagrangian decomposition: No gain in bound compared to Lagrangian relaxation when capacity is relaxed [Letocart et al.(2012)]: The 0-1 bi-dimensional knapsack problem and the generalised assignment problem Illustrates the concept No full comparison of bounds, conclusions not possible

10 Related work Very few papers on Lagrangian decomposition and column generation: [Pimentel et al.(2010)]: The multi-item capacitated lot sizing problem Branch-and-price implementations for two types of Lagrangian relaxation and for Lagrangian decomposition Lagrangian decomposition: No gain in bound compared to Lagrangian relaxation when capacity is relaxed [Letocart et al.(2012)]: The 0-1 bi-dimensional knapsack problem and the generalised assignment problem Illustrates the concept No full comparison of bounds, conclusions not possible Our work this far: Find an application where we gain in strength compared to the strongest obtainable from Lagrangian relaxation and investigate further...

11 Fixed charge transportation problem (FCTP) For each arc (i, j), i I, j J: u ij = min(s i, d j ) = upper bound c ij = unit cost for shipping f ij = fixed cost for shipping

12 Fixed charge transportation problem (FCTP) For each arc (i, j), i I, j J: u ij = min(s i, d j ) = upper bound c ij = unit cost for shipping f ij = fixed cost for shipping Variables: x ij = amount shipped from source i to sink j, i I, j J Concave cost function: g ij (x ij ) = { fij + c ij x ij if x ij > 0 0 if x ij = 0 i I, j J

13 Fixed charge transportation problem (FCTP) min s.t. g ij (x ij ) i I j J x ij = s i j J x ij = d j i I i I j J x ij 0 i I, j J Polytope of feasible solutions, minimisation of concave objective Optimal solution at an extreme point (can be non-global local optima at extreme points) MIP-formulation: A binary variable to indicate if there is flow on an arc or not

14 MIP-formulation of FCTP min s.t. ( ) cij x ij + f ij y ij i I j J x ij = s i j J x ij = d j i I x ij u ij y ij x ij 0 i I j J i I, j J i I, j J y ij {0, 1} i I, j J

15 The reformulation: variable splitting Supply and demand side duplicates of the shipping variables: xij s and xij d Introduce a parameter ν: 0 ν 1 min s.t. ν i I j J g ij (x s ij) + (1 ν) j J xij s = xij d i I, j J xij s = s i i I j J i I x d ij = d j j J x s ij, x d ij 0 i I, j J, i I g ij (x d ij )

16 The reformulation: inner representation Let each column correspond to an extreme point of a set X s i = {x s ij, j J j J x s ij = s i, 0 x s ij u ij, j J}, i I, or of a set X d j = {x d ij, i I i I x d ij = d j, 0 x d ij u ij, i I }, j J The flow from one source / to one sink is a convex combination of extreme point flows, introduce: λ s ip = convexity weight for extreme point p P s i and λ d jp = convexity weight for extreme point p P d j of set X s i, i I of set X d j, j J

17 The reformulation: column oriented formulation min s.t. ν i I p P s i j J g ij p P s i x s ijpλ s ip = λ s ip = 1 p P d j λ s ipxijp s x d ijpλ d jp i I + (1 ν) j J i I i I, j J g ij p P d j λ d jpxijp d p P s i λ d jp = 1 p P d j λ s ip 0 λ d jp 0 j J p P s i, i I p P d j, j J

18 The reformulation: approximating the objective The objective is bound below by its linearisation: ν g ij λ s ipx s ijp + (1 ν) g ij λ d jpx d ijp i I j J p P i s j J i I p P j d ν g ij (xijp) s λ s ip + (1 ν) g ij (xijp) d λ d jp i I j J j J i I p P s i p P d j What is lost by the linearisation?

19 The reformulation: arc cost True cost

20 The reformulation: arc cost True cost In LP-relaxation of MIP-formulation

21 The reformulation: arc cost True cost In LP-relaxation of MIP-formulation After our reformulation (here for supply side, similarly for sink side): Extreme points: true cost Non-extreme points: x s ij = p P s i : λs ip >0 xs ijp λs ip if x s ij > 0 and there is p P s i : λ s ip > 0 and xs ip = 0, then f ij is decreased by = p P s i : λs ip >0, xs ip =0 f ij λ s ip otherwise: true cost

22 The reformulation: arc cost True cost In LP-relaxation of MIP-formulation After our reformulation (here for supply side, similarly for sink side): Extreme points: true cost Non-extreme points: x s ij = p P s i : λs ip >0 xs ijp λs ip if x s ij > 0 and there is p P s i : λ s ip > 0 and xs ip = 0, then f ij is decreased by = p P s i : λs ip >0, xs ip =0 f ij λ s ip otherwise: true cost Example: For xij s = 4 = we have 3 x s ij1 = 6, λs i1 = 2 3 ; xs ij2 = 0, λs i2 = 1 3 ; = 1 3 f ij

23 The reformulation: column generation subproblem For a single source i I (and similarly for the sinks): ( ) min νg ij (xij) s α ij xij s β i s.t. j J j J x s ij = s i xij s u ij xij s 0 j J j J Concave minimization problem with an optimal solution at an extreme point of X s i MIP-formulation: Binary variable to indicate if there is flow on an arc or not

24 Theoretical strength [Guignard and Kim(1987)]: The Lagrangian decomposition bound (= convexification over source and sink side ) as least as strong as the strongest of the Lagrangian relaxation bound when Ax b is relaxed (= convexification over source side ) the Lagrangian relaxation bound when Cx d is relaxed (= convexification over sink side ) and there is a chance that it is stronger!

25 Theoretical strength: type of strength? The constraints in the MIP-formulation of the column generation subproblem (source side, similarly for the sink side): j J x s ij = s i x s ij u ij y s ij, j J y s ij {0, 1}, j J 0 x s ij, j J

26 Theoretical strength: type of strength? The constraints in the MIP-formulation of the column generation subproblem (source side, similarly for the sink side): j J x s ij = s i x s ij u ij y s ij, j J y s ij {0, 1}, j J 0 x s ij, j J Implied inequality: + u ij yij s s i j J

27 Theoretical strength: type of strength? The constraints in the MIP-formulation of the column generation subproblem (source side, similarly for the sink side): j J x s ij = s i x s ij u ij y s ij, j J y s ij {0, 1}, j J 0 x s ij, j J Implied inequality: + u ij yij s s i = j J Knapsack constraints over the binary variables At least (exactly?) that type of strength j J x s ij = s i xij s u ij yij, s j J u ij yij s s i j J y s ij {0, 1}, j J 0 x s ij, j J

28 Empirical strength Comparison: Convexification over both sides = Lagrangian decomposition Convexifiation over one side, see bounds obtained by [Roberti et al.(2015)] (A new formulation based on extreme flow patterns from each source; derive valid inequalities; exact branch and price; column generation to compute lower bounds)

29 Empirical strength Table: Instance characteristics instance total r. v. c. r. f. c. type ID size supply inf sup inf sup Roberti-set3 Table5-(1-10) Roberti-set3 Table6-(1-10) Roberti-set3 Table7-(1-10) r.v.c. = range of variable costs; r.f.c. = range of fixed costs Compare LBDs by comparing gaps (relative deviations) calculated as (UBD-LBD)/LBD in percent The UBDs used are the best known, most of them are verified to be optimal

30 Empirical strength instance cost gap (%) ID ratio (%) FCTP-LP Roberti et al. our Table Table Table Table Table Table Table Table Table Table AVG instance cost gap (%) ID ratio (%) FCTP-LP Roberti et al. our Table Table Table Table Table Table Table Table Table Table AVG instance cost gap (%) ID ratio (%) FCTP-LP Roberti et al. our Table Table Table Table Table Table Table Table Table Table AVG Average improvement in gap thanks to convexification, from first side + from second side Table5: 47% + 46% Table6: 43% + 45% Table7: 42% + 37% Both convexifications contribute!

31 Other observations this far Property of the dual function: Constant along the direction e Stabilsation: No improvement the opposite! Because of the property of the dual function: Tried regularisation instead to favour solutions with a small l 1 -norm

32 Conclusions this far: Strong formulation for the fixed charge transportation problem Empirically we gain significantly in strength from the convexification over both sides Further studies: Properties of the dual function, cf. experiences from subgradient methods? Understand the effects of stabilization. Customized techniques? How strength depends on instance characteristics

33 Bibliography Guignard M, Kim S (1987) Lagrangean decomposition: a model yielding stronger lagrangean bounds. Mathematical Programming 39 (2): Letocart L, Nagih A, Touati-Moungla N (2012) Dantzig-Wolfe and Lagrangian decompositions in integer linear programming. International Journal of Mathematics in Operational Research 4(3): Pimentel CMO, Alvelos FP, de Carvalho JMV (2010) Comparing Dantzig-Wolfe decompositions and branch-and-price algorithms for the multi-item capacitated lot sizing problem. Optimization Methods and Software 25(2): Roberti R, Bartolini E, Mingozzi A (2015) The fixed charge transportation problem: an exact algorithm based on a new integer programming formulation. Management Science, forthcoming. Thanks for listening!