A stochastic multi-item lot-sizing problem with bounded number of - - PowerPoint PPT Presentation

a stochastic multi item lot sizing problem with bounded
SMART_READER_LITE
LIVE PREVIEW

A stochastic multi-item lot-sizing problem with bounded number of - - PowerPoint PPT Presentation

www.argon-consult.com A stochastic multi-item lot-sizing problem with bounded number of setups Sminaire des doctorants February 8th, 2017 Etienne de Saint Germain joint work with Frdric Meunier and Vincent Leclre Outline Business


slide-1
SLIDE 1

www.argon-consult.com Séminaire des doctorants

A stochastic multi-item lot-sizing problem with bounded number of setups

February 8th, 2017 Etienne de Saint Germain joint work with Frédéric Meunier and Vincent Leclère

slide-2
SLIDE 2

Outline

  • Business problem and model
  • Deterministic model
  • Stochastic model
  • Numerical experiments
slide-3
SLIDE 3

Business problem

  • Context: production function of the Supply Chain for one assembly line
  • Objective: reduction of holding costs
  • Main constraints: industrial flexibility and “high service level”
  • Typical horizon: 10 to 15 weeks
  • Typical time step: 1 week
  • Input data:

◮ a set of references r ∈ R ◮ capacity of the line ◮ demand for each reference and each week

2 / 20

slide-4
SLIDE 4

Classical problem: Capacitated Lot-Sizing Problem (CLSP) min

T

  • t=1
  • r∈R

(hr

tsr t + cr t xr t )

s.t. sr

t = sr t−1 + qr t − dr t

∀t, ∀r

  • r∈R

qr

t ≤ 1

∀t qr

t ≤ xr t

∀t, ∀r xr

t ∈ {0, 1}

∀t, ∀r qr

t , sr t ≥ 0

∀t, ∀r (CLSP)

  • with (r ∈ R for references, t ∈ [T] for weeks):

input data variables hr

t

holding cost sr

t

inventory level cr

t

setup cost qr

t

produced quantity dr

t

demand xr

t

setup variable

3 / 20

slide-5
SLIDE 5

Disadvantage of CLSP formulation (according to Argon Consulting)

  • Hard to compare holding costs and setup costs
  • Number of setups is a “technical” constraint

◮ Given by operational level ◮ Represent scheduling constraints (which are neglected at tactical level)

4 / 20

slide-6
SLIDE 6

Multi-item lot-sizing problem with bounded number of setups min

T

  • t=1
  • r∈R

(hr

tsr t +✟✟

✟ ❍❍ ❍

cr

t xr t )

s.t. sr

t = sr t−1 + qr t − dr t

∀t, ∀r

  • r∈R

qr

t ≤ 1

∀t qr

t ≤ xr t

∀t, ∀r

  • r∈R

xr

t ≤ N

∀t xr

t ∈ {0, 1}

∀t, ∀r qr

t , sr t ≥ 0

∀t, ∀r (P)

  • Bounded number of setups per week (N)
  • Easier for industrials to quantify holding costs and N

5 / 20

slide-7
SLIDE 7

Outline

  • Business problem and model
  • Deterministic model
  • Stochastic model
  • Numerical experiments
slide-8
SLIDE 8

The deterministic model is hard

  • (P) is NP-hard (reducing 3-PARTITION)

◮ Decide if there is a solution when N = 1 is polynomial ◮ Cases N = 1 and N = 2 still open

  • Continuous relaxation of (P) does NOT depend on N
  • 2 natural extended formulations:

◮ 1 binary variable xp,t where p ∈

R

N

  • per possible plan for a week

◮ 1 binary variable yq,r where q ∈ 2[T] per possible plan for a reference

references weeks yq,r xp,t

◮ Same continuous relaxations than compact formulation

6 / 20

slide-9
SLIDE 9

Outline

  • Business problem and model
  • Deterministic model
  • Stochastic model
  • Numerical experiments
slide-10
SLIDE 10

Need for backlog

  • Mathematical reason:

◮ In general, there is no feasible solution ◮ Simple example:

  • bounded capacity C
  • demand = Gaussian noise around forecast

infeasible region demand probability of realization forecast capacity

  • Industrial reason:

◮ Negative inventories are commercial constraints

= ⇒ “soft” constraints

◮ Firms can deliver late

7 / 20

slide-11
SLIDE 11

Stochastic model min E T

  • t=1
  • r∈R

(hr

sr

t + γbr t)

  • s.t.

sr

t = ˜

sr

t − br t

∀t, ∀r sr

t = sr t−1 + qr t − d r t

∀t, ∀r

  • r∈R

qr

t ≤ 1

∀t qr

t ≤ xr t

∀t, ∀r

  • r∈R

xr

t ≤ N

∀t xr

t ∈ {0, 1}

∀t, ∀r qr

t, ˜

sr

t, br t ≥ 0

∀t, ∀r σ (qr

t) , σ (xr t) ⊂ σ

  • (d r

0, . . . , d r t)r∈R

  • ∀t, ∀r

(S) with: γ backlog penalization, ˜ sr

t

inventory level, br

t

backlog quantity

8 / 20

slide-12
SLIDE 12

Stochastic model: size difficulty

  • Extensive formulation leads to a huge number of variables

t = 0 t = 1 t = 2 t = 3 1 1 1 1 2 2 2 2 1 1 2 2 1 2 ◮ example: for each references, 2 independent possibilities for demand

= ⇒ number of variables multiplied by

  • 2T|R|

◮ for T = 10, |R| = 10, numbers of variables ≈ 1030

  • Need for heuristics to solve (S)

◮ lot-size and cover-size ◮ open-loop feedback approach ◮ repeated two-stage stochastic programming approach

9 / 20

slide-13
SLIDE 13

Aside: computing lot-size and cover-size

  • Simplified model:

◮ Constant demand for each reference over time ◮ Aggregated demand

Inventory level of reference r:

¯ sr = 1

2 drT 2 r

time inventory level Tr −dr 2Tr cover-size lot-size

Corresponding program to solve:

min

  • r∈R

1 2hrdrTr s.t.

  • r∈R

1 Tr = N Tr > 0 ∀r

  • Closed-form expressions of solutions

ν∗

r =

1 T ∗

r =

N √ hrdr

  • p∈R
  • hpdp

and Cost = 1 2N

  • r∈R
  • hrdr

2

  • Closed-form expressions of solutions for stochastic case

10 / 20

slide-14
SLIDE 14

Strategy: lot-size and cover-size

  • Heuristic parameters: safety stock for each reference

foreach r ∈ R do Compute cover-size Tr / lot size ℓr = drTr; for week from 1 to T do foreach r ∈ R do Observe inventory level of r; if current inventory level < safety stock then case lot-size: produce quantity ℓr; case cover-size: produce the cumulated expected demand for the Tr next weeks;

  • Example:

◮ For a reference r, Tr = 2 weeks ◮ Expected demand is:

week t 1 2 3 4 5 expected demand ft 2 3 5 4 1

◮ If current inventory level < safety stock at week 1, we must produce:

f1 + f2 = 2 + 3 = 5 units of reference r

11 / 20

slide-15
SLIDE 15

Strategy: open-loop feedback approach

  • At week t:

◮ Observe current inventory level ◮ Solve deterministic version of (S) where the random variable d r

t is

replaced by the deterministic expected demand

  • It is a Mixed Integer Program
  • Almost program (P) but with backlog

◮ Set production decisions for week t

12 / 20

slide-16
SLIDE 16

Strategy: repeated two-stage stochastic programming approach

  • At week t:

◮ Observe current inventory level ◮ Construct a fan of demand scenarios to approximate the tree of scenarios

in (S)

t = 0 t = 1 t = 2 t = 3

Complete tree of scenarios

1 2 3

Fan of scenarios

1 2 3 ◮ Solve (S) ◮ Set production decisions for week t

13 / 20

slide-17
SLIDE 17

There is a lot of possible forecasts

  • Static deterministic forecast

◮ Expectation, median...

  • Adaptative deterministic forecast

◮ Autoregressive process, time series...

  • Stochastic forecast

◮ Tree of scenario, fan of scenarios...

Every strategy works even if we do not know distribution laws. We just need a forecast function!

14 / 20

slide-18
SLIDE 18

Outline

  • Business problem and model
  • Deterministic model
  • Stochastic model
  • Numerical experiments
slide-19
SLIDE 19

Simulations

Simulator

  • Start at week t = 0
  • For each week t:

time

past current

◮ Observe inventory level ◮ Use strategy and forecast to

compute the production planning

◮ Set production for current week

  • Return KPI

Strategy Forecast Data Demand realization Key performance indicators:

  • Holding costs
  • Backlog costs
  • Cycle service

level

  • Fill rate service

level

  • . . .

15 / 20

slide-20
SLIDE 20

Data of the instances

  • We use historical data from industrial
  • Numerical values:

◮ Horizon T = 13 weeks ◮ |R| = 30 references ◮ Demands 0 ≤ dr

t ≤ 4000 units

◮ Weekly capacity C ≈ 13000 units ◮ Weekly number of setups N = 10 ◮ Holding costs 50 ≤ hr

t ≤ 80 per units

16 / 20

slide-21
SLIDE 21

Building the distribution of the demand

  • Autoregressive Process (AR1).

For each reference r, dt+1 = ft+1+αet + (1 − α) ǫt+1

  • et+1

where (at week t):

◮ dt is the demand ◮ ft is the forecast ◮ et is the forecast error ◮ ǫt ∼ N (0, σft) is a white

noise

  • 2 parameters:

◮ α ∈ [0, 1] proportion

error/noise

◮ σ is the volatility.

time demand forecast α = 50% σ = 80% 17 / 20

slide-22
SLIDE 22

Results: holding costs for several realizations of demand

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1 · 106 2 · 106 3 · 106 4 · 106 α = 50% σ = 80% Fill rate service shortage Holding costs

Cover-size Open-loop feed back Lot-size Two-stages stochastic

18 / 20

slide-23
SLIDE 23

Results: holding costs for several values of volatility

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1 · 106 1.5 · 106 2 · 106 2.5 · 106 3 · 106 0.2 0.5 0.8 1.5 2 0.2 0.5 0.8 1.5 2 0.2 0.5 0.8 1.5 2 0.2 0.5 0.8 1.5 2 Fill rate service shortage Holding costs

Cover-size Open-loop feed back Lot-size Two-stages stochastic

19 / 20

slide-24
SLIDE 24

Results: holding costs for several values of volatility

Thanks for your attention!

20 / 20