Tutorial: System-Oriented Inventory Models for Spare Parts · System-Oriented Inventory Models for...

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Tutorial: System-Oriented Inventory Models for Spare Parts Geert-Jan van Houtum YEQT VIII, 3-5 November, 2014 Eindhoven

Transcript of Tutorial: System-Oriented Inventory Models for Spare Parts · System-Oriented Inventory Models for...

Page 1: Tutorial: System-Oriented Inventory Models for Spare Parts · System-Oriented Inventory Models for Spare Parts Geert-Jan van Houtum ... for recoverable item control, Operations Research

Tutorial:System-Oriented Inventory Models for Spare Parts

Geert-Jan van Houtum

YEQT VIII, 3-5 November, 2014 Eindhoven

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Seminal paper:• Sherbrooke, C.C., METRIC: A multi-echelon technique

for recoverable item control, Operations Research 16, pp. 122-141, 1968

Two important books:• Sherbrooke, C.C., Optimal Inventory Modeling of

Systems: Multi-Echelon Techniques, Wiley, 1992. (2nd edition, Kluwer, 2004.)

• Muckstadt, J.A., Analysis and Algorithms for Service Parts Supply Chains, Springer, 2005.

Literature

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Based on:• Basten, R.J.I., and Van Houtum, G.J., “System-

oriented inventory models for spare parts”, Surveys in OR & MS 19, pp. 34-55, 2014

In progress:• Van Houtum, G.J., and Kranenburg, A.A., Spare Parts Inventory

Control under System Availability Constraints, International Series in OR & MS, Springer.

This tutorial

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1. Introduction2. Real-life networks3. Single-location model with backordering4. Single-location model with emergency shipments5. METRIC model6. Multi-location model with lateral and emergencyshipments7. Extensions8. Applications in practice9. Challenges for further research

CONTENTS

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1. Introduction

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11/7/2014 PAGE 5

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Total Cost of Ownership (TCO)

Total Cost of Ownership (TCO):The total costs during the whole life cycle(perspective: user of the system)

Needs andrequirements Design Production Exploitation Disposal

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TCO for an example system

0%

5%

10%

15%

20%

25%

30%

35%

40%

45%

Acquisitioncosts

Maintenancecosts

Downtimecosts

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Size of maintenance industry

• Worldwide revenues after-sales services: 1500 billion US Dollars per year (AberdeenGroup [2003])

• Sales of spare parts and services in US: 8% of GNP (AberdeenGroup [2003])

• Manufacturers in US, Europe and Asia generate 26% of their revenues via services (Deloitte [2006])

• At an airline, 10% of all costs is constituted by maintenance costs (Lam [1995])

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PAGE 911/7/2014

Long term trends

• Maintenance is outsourced to third party or OEM (pooling resources, pooling data, remote monitoring)

• More extreme: One sells function plus availability• Feedback to design (better systems, higher sustainability)

• Maintenance of complex systems becomes too complicatedfor users themselves

• Users require higher availabilities (less downtime)• Users look at TCO

/ School of Industrial Engineering

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Research topics

• Spare parts management• Condition based maintenance• Inventory models for spare parts and service tools• Scheduling of service engineers• Design of spare parts networks• Forecasting of failures• The effect of remote monitoring and diagnostics on total costs• Service contracts and customer differentiation• The effect of design decisions for new systems on their Total Cost of

Ownership• New business models for collaboration between users• Game-theoretic models on the relationship between OEM-s, third parties

and users• …

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Spare parts inventory models: • for critical components• of advanced capital goods• with service level constraints for system-

oriented service measures such as system availability or aggregate fill rate

Application in practice: Tactical planning level !

Focus of this tutorial

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2. Real-life networks

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Network ASML

60 warehouses5000 SKU’s

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Network ASML (cont.)

1

1. Normal delivery: 2 hrs.

3

3. Emergency replenishment: 48 hrs.

2

2. Lateral transshipment: 14 hrs.

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Network IBM (for next day deliveries)

905

947

900

892

895

894

902

904

480

8 9

1 23 4 5

13 14

11 12

24 25 2621 22 23

31 33 3427 28 29 30

885 886 887 888

879880

878

869 872873

864865 867

868

859 860 861 862 863

54 55

51 52

48 49 50

308310

307

303 304

300301 302

297298 299

294 296

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397 398399

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539 540 541542 543

544

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522 523524

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507 508 509510 511

495 496500

501502

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489 490 491492

481 482

453454

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377 378 379 380

366 367 368 369 372373

358

349 350 354

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340 341 346

331 334 335 337

323 324325 326 327 328

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551

549

472474 475

468 469 470 471

628 629

619 620 621

612613

605

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574575 576

567569

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553 554555

645 646 647 648 649

640 644

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631632 633

634

779

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797 799

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789 790791

786 787

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464 465 466 467460

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728

711 712 713

695 696

679 680

828 829 830

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370 371

359 360 361 362 363 364

351 352 353

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875 876 877

870 871

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404

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583805

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292801

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278 279 280 281282 283

258259 260 261 262 263

264 265

237 238 239 240 241 242 243 244

216 221 222

195 196 197 198 199 200201

202 203

173 174 175 176 177 178179

153 154 155 156

217 219 220

245 514

548

605

783

413

899

Madrid

Lyon

Rome

Coventry

Prague

874

793

Gothenburg

394

562

883

550

10

531

473

scenario 1

time constraint

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Network Nedtrain

Lateral supply out of QRS <2 hoursRegular replenishment 2-5/weekUrgent orders < 24 hrs

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User networks vs. OEM networks

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Examples of users maintaining their ownsystem

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Examples of OEMs maintaining soldsystem

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3. Single-location model withbackordering

(cf. Chapter 2 of Sherbrooke [1992], Single-item version: Feeney and Sherbrooke [1966])

3.1 Model description3.2 Overview of assumptions3.3 Evaluation3.4 Optimization3.5 Alternative optimization techniques and

service measures

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3.1. Model description

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Model

Warehouse

Demands(PoissonStreams)

Installed base

ready-for-use parts

failed parts

Repair shop

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• We use the terminology that is common for repairable LRU’s (Line Replaceble Units). The model is also applicable for consumable LRU’s.

• LRU’s are denoted as SKU’s (Stock-KeepingUnits) in this presentation.

• We have an infinite time horizon.• We look at the buy of the initial stock of spare

parts for all SKU’s.• No emergency shipments.

Model (cont.)

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Input variables:• I : Set of SKU's, SKU's are numbered 1,..., |I|• mi : Demand rate for SKU i (mi )• M iI mi : Total demand rate (M )• ti : Mean repair leadtime for SKU i (ti )• ci

h : Price of a part of SKU i (cih )

• EBOobj : Target level for aggregate meannumber of backorders

Notation

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Decision variables:• Si : Basestock level for SKU i (Si )• S (S1 S2 S|I|) : Vector with all basestock

levels denotes a solution

Notation (cont.)

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Output variables:• Ci(Si) ci

h Si : Inventory holding costs for spareparts of SKU i

• C(S) iI Ci(Si) iI cih Si : Total inventory

holding costs• EBOi(Si) : Mean number of backorders of SKU i• EBO(S) iI EBOi(Si) : Aggregate mean

number of backorders

Notation (cont.)

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Problem (P):min C(S)subject to

EBO(S) EBOobj

Si for all iI

Relation with availability:Availability(S) 1 – (EBO(S) / N)

where N is the number of machines

Optimization problem

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3.2. Overview of assumptions

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“ 1. Demands for the different SKU’s occur according to independent Poisson processes ”

• A failure of a component does not lead to additional failures of other components

• Assumption of Poisson demand processes is justifiedwhen:

- lifetimes of components are exponential, or- lifetimes are generally distributed and number of

machines is sufficiently large (a merge of many renewalprocesses gives a process that is close to Poisson)

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“ 2. For each SKU, the demand rate is constant ”

• Justified in case fraction of machines that is down, is always sufficiently small:

- either because downtimes are short in general- or downtimes occur only rarely

“ 3. Repair leadtimes for different SKU’s are independent and repair leadtimes of the same SKU are i.i.d. ”

• See repair leadtimes as planned repair leadtimes

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“ 4. A one-for-one replenishment strategy is applied for all SKU’s ”

• Justified in case:- there are no fixed ordering costs at all, or- fixed ordering costs are small relative to the prices of the

SKU’s

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3.3. Evaluation

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Evaluation can be done per SKU

Extra notation per SKU i :• Xi(t) : number of parts in repair at time t• Ii(t,Si) : number of parts on hand at time t• Bi(t,Si) : number of backordered demands at time t• Xi , Ii(t,Si), Bi(t,Si): corresponding steady-state variables

Extra notation

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Petri net of repair and demand fulfilment process

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States that can occur for (Xi(t),Ii(t,Si),Bi(t,Si)):(0, Si, 0) : Nothing in repair, Si good parts on stock

(1, Si-1, 0): 1 part in repair, Si-1 good parts on stock...(Si-1, 1, 0): Si-1 parts in repair, 1 good part on stock

(Si, 0, 0) : Si parts in repair, no parts on stock anymore

(Si+1, 0,1): Si+1 in repair, no stock, 1 request in backlog...

Possible states

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• Ii(t,Si) = (Si Xi(t))+

• Bi(t,Si) = (Xi(t) Si)+

• Stock balance equation:Xi(t) Ii(t,Si) Bi(t,Si) = Si

And thus also:• Ii(Si) = (Si Xi)+

• Bi(Si) = (Xi Si)+

• Xi Ii(Si) Bi(Si) = Si

Equations

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Palm’s Theorem (cf. Palm [1938]):If at a certain unit the arrival process of jobs is Poisson with rate and if the leadtimes for the jobs are independent and identically distributed random variables corresponding to any distribution with meanEW, then the steady state probability distribution for the number of jobs present in that unit is a Poissondistribution with mean EW.

• Developed for an M|G| queue• Result is easily seen for deterministic leadtimes• Generalizes Little’s law for this particular system

Palm's Theorem

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Application of Palm's Theorem

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Last step

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3.4. Optimization

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Problem (P):min C(S)subject to

EBO(S) EBOobj

Si for all iI

Kind of knapsack problem; hard to solve

Alternative problem

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Problem (Q):min C(S) iI ci

h Si

min EBO(S) iI EBOi(Si)subject to

S (S1 S2 S|I|) | Si for all i

Multi-objective programming problemWe derive so-called efficient solutions

Alternative problem (cont.)

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Input variables for Problem (Q):• |I| 3• m1 15, m2 5, m3 1, M 21 demands/yr• t1 t2 t3 1/6 yrs• c1

h € 1,000, c2h € 3,000, c3

h € 20,000

Extra input variable for Problem (P):• EBOobj 0.1

Example 1

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Example 1 (cont.)

Enumeration exercise:• Consider plausible

values for the basestock levels

• Compute C(S)• Compute EBO(S)• Plot them in a C(S)• vs. EBO(S) figure

S_{1} S_{2} S_{3} C(S) (€) EBO(S)

0 0 0 - 3,500

1 0 0 1.000 2,582

2 0 0 2.000 1,869

3 0 0 3.000 1,413

4 0 0 4.000 1,171

5 0 0 5.000 1,062

6 0 0 6.000 1,020

7 0 0 7.000 1,006

8 0 0 8.000 1,002

9 0 0 9.000 1,000

0 1 0 3.000 2,935

1 1 0 4.000 2,017

2 1 0 5.000 1,304

3 1 0 6.000 0,848

4 1 0 7.000 0,605

5 1 0 8.000 0,497

6 1 0 9.000 0,455

7 1 0 10.000 0,440

8 1 0 11.000 0,436

9 1 0 12.000 0,435

0 2 0 6.000 2,731

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Example 1 (cont.)

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

0 10 20 30 40 50 60 70 80 90

EBO

(S)

C(S) (x € 1,000)

Enumeration

Efficient solutions

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For Problem (Q), we obtain:• Efficient solutions• The whole efficient frontier• Observation: An optimal solution for

Problem (P) is efficient for Problem (Q), and vice versa

Optimal solution for Problem (P):• S (6, 2, 1)• C(S) € 32,000• EBO(S) 0.098

Example 1 (cont.)

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Problem (Q):min C(S) iI ci

a Si

min EBO(S) iI EBOi(Si)subject to

S (S1 S2 S|I|) | Si for all i

=> Problem (Q) is separable (cf. Fox, 1966)=> Efficient solutions via greedy algorithm

Alternative problem (cont.)

Separable

Separable

Cartesian product

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Greedy algorithm

An obvious efficient solution:• S =(0 0 0)• C(S) 0 => All other solutions: Higher costs• EBO(S) iI EBOi(0) iI mi ti

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Greedy algorithm (cont.)

Starting point

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

0 10 20 30 40 50 60 70 80 90

EBO

(S)

C(S) (x € 1,000)

EnumerationEfficient solutions

Next:Look for thesteepestdecrease inEBO(S) againstthe lowestincrease inC(S)

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Generation of a next efficient solution:• Current solution: S• If Si would be increased with 1 unit:

• Decrease for EBO(S) = i EBO(S) = EBOi(Si) = … (see proof of Lemma 3.2)

• Increase in C(S) = i C(S) = cih

• i : = i EBO(S) / i C(S)

• Pick the SKU with the largest i (‘biggest bang for the buck’)

Greedy algorithm (cont.)

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Greedy algorithm (cont.)

Set of efficient solutions

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Lemma 3.3Algorithm 3.1 generates efficient solutions for Problem (Q).

Proof: Via Fox (1966). Alternative: Directly via the logic that always the

most negative slope is chosen for the curve formed by the generated solutions.

Greedy algorithm (cont.)

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Application of greedy algorithm:

Example 1 (cont.)

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Example 1 (cont.)

We obtain a subset of all efficient solutions !

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Via greedy algorithm:• Heuristic solution for Problem (P) with

EBOobj 0.1 :- S (7, 3, 1), C(S) € 36,000, EBO(S) 0.031- Optimality gap = (36000-33000)/33000 = 9.1%

• Optimal solution for Problem (P) with EBOobj 0.031 :

- S (7, 3, 1), C(S) € 36,000, EBO(S) 0.031

Example 1 (cont.)

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Real-life data, 99 SKU’s

Example 2

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=> Smooth line, good heuristic solutions for Problem (P)

Example 2 (cont.)

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• Generates efficient solutions for Problem (Q)• Does not generate all efficient solutions, but a

certain subset of solutions• This subset has a certain robustness: small

changes in the input parameters (e.g., demand rates!) lead to small changes in the subset of generated solutions

• Leads to a heuristic solution for Problem (P). Conjecture: This solution is robust. (Notice: This does not hold for the exact solution of Problem (P).)

Conclusions w.r.t. greedy algorithm

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• Heuristic solution is optimal for Problem (P) for some specific values of EBOobj

• Generally, the heuristic solution for Problem(P) will be good when one has many SKU’s

• Simple and easy to understand• Easy to implement in practice• Requires little computational effort

Conclusions w.r.t. greedy algorithm(cont.)

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3.5. Alternative optimizationtechniques and service measures

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Alternative techniques:• Langrange relaxation• Dantzig-Wolfe decomposition-> Both give the same efficient solutions as greedyalgorithm

Alternative service measures:• Aggregate mean waiting time• Aggregate fill rate• …-> Works as long as you have the required convexityproporties

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Spare parts planning in practice:• Executed every 3 months, say• Per planning moment:

• Generation of new forecasts for the demand rates of all SKU’s

• Application of the greedy heuristic• Implementation of new solution:

• Both increasing and decreasing the stock of a SKU gives some costs

Why is robustness of the heuristicsolutions important?

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Show/prove that the greedy heuristic leads torobust solutions

Similarly for Dantzig-Wolfe decomposition (maybe easier when we go to dimension 3 or higher)

How to use of the model in a rolling horizon setting such that basestock levels do notchange too much

Open problems

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4. Single-location model withemergency shipments

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• Assumption for basic model:If a demand cannot be immediately fulfilled from stock, then the demand is backordered

• In several practical situations:- Downtimes of machines are very expensive- In case of a stockout, a demand will be satisfied

in an alternative way, i.e., via a fast repair procedure or via an emergency shipment

- 'lost sales' instead of 'backordering'

Emergency Shipments

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• tiem : Average time for an emergency shipment

for SKU i• ci

em : Cost of an emergency shipment for SKU i, minus cost of a normal repair (ci

em )• ci

h : Inventory holding cost per time unit per part of SKU i (ci

h )• Wi(Si) : Mean waiting time for a demand for

SKU i• W(S) iI (mi/M) Wi(Si) : Aggregate mean

waiting time for an arbitrary demand for allSKU’s together

Changes in model assumptions

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• Wobj : Target level for W(S)• Problem formulation:

• Link with availability:

Changes in model assumptions (cont.)

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• Total costs:

• Formula for W(S):

• Expression for fill rate i(Si) (steady-state behavior is equivalent to behavior in Erlang loss system):

Evaluation

Erlang loss probability

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• Switch to related multi-objective problem:

• Karush (1957): Erlang loss probability is strictlyconvex and decreasing as a function of the number of servers

Optimization

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• Hence:- i(Si) is strictly concave and increasing- Wi(Si) is strictly convex and decreasing- Ĉi(Si) is convex- Ĉi(Si) may now be decreasing for smaller values of Si

• Define: Si,min = argmin Ĉi(Si)• Exclude solutions S with Si Si,min for some i• Then the remaining problem can be solved with a

greedy algorithm

Optimization (cont.)

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Optimization (cont.)

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5. METRIC model(cf. Sherbrooke [1968])

5.1 Model description5.2 Evaluation5.3 Optimization

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5.1. Model description

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0

|Jloc|

1Repair of

spare parts

LocalWarehouse

Machinesat customers

Machinesat customers

LocalWarehouse

Model

CentralWarehouse

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Input variables:• Jloc : Set of Local Warehouses (LW’s), LW’s are

numbered 1,..., | Jloc |• 0 : Index for Central Warehouse (CW)• J : Set of all warehouses, J Jloc

• I : Set of critical SKU's, SKU's are numbered1,..., |I|

• mi,j : Demand rate for SKU i at LW j (mi,j )

Notation

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Input variables (cont.):• ti,j : Order and ship time from CW to LW j for

SKU i (these times are deterministic; ti,j )• ti,0 : Mean repair leadtime for SKU i

(repair leadtimes are i.i.d.; ti,0 ) : Holding cost rate per part of SKU i ( )

: Target level for aggregate meannumber of backorders at LW j

Notation (cont.)

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Decision variables:• Si,j : Basestock level for SKU i at warehouse j

(Si,j )• Sj (S1, j S|I|,j) : Basestock vector for SKU i• S : Matrix with all basestock levels

Notation (cont.)

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Output and other variables (cont.):• EBOi,j(Si,0 ,Si,j) : Mean number of backorders for

SKU i at LW j;• EBOj(S0 ,Sj) : Aggregate mean number of

backorders at LW j

Notation (cont.)

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Output and other variables (cont.):• C(S) iI jJ Si : Total average costs (excl.

holding costs for pipeline stock)

Notation (cont.)

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Problem (R):min C(S)subject to

EBOj(S0 ,Sj) , jJloc,

Si,j for all iI and jJ

Remark: Sherbrooke [1968] had a constraint for sum of EBOj(S0 ,Sj). Then a greedy procedure can be applied to generate efficient solutions(after convexification).

Optimization problem

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5.2. Evaluation

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Observation 1:Evaluation can be done per SKU i(the SKU’s are only ‘connected’ via the

aggregate mean waiting times)

Exact Evaluation(due to Graves [1985])

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Observation 2:For each SKU i I :• LW places replenishment orders at CW j

according to a Poisson process with rate mi,j

• The total demand process at CW is a Poissonprocess with rate:

mi,0 jJloc mi,j

Exact evaluation (cont.)

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0

|Jloc|

1Repair of

spare parts

LocalWarehouse

Machinesat customers

Machinesat customers

LocalWarehouse

Exact evaluation (cont.)

CentralWarehouse

Order in which LW’s placereplenishment orders doesnot depend on the basestock levels !

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Line of analysis – Variables that are determined:• Ii,0(Si,0) : On-hand stock of SKU i at CW (in steady

state)• Bi,0(Si,0) : Number of backorders for SKU i at CW (in

steady state)• Bi,0

(j)(Si,0) : Number of backorders for SKU i of LW j at CW (in steady state)

• Ii,j(Si,0,Si,j) : On-hand stock of SKU i at LW j (in steady state)

• Bi,j(Si,0,Si,j) : Number of backorders for SKU i at LW j (in steady state)

Same variables with E added j: mean value

Exact evaluation (cont.)

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Given: i I

Define:Yi,j : Poisson distributed random variable with

mean mi,j ti,j for each warehouse j

Exact Evaluation (cont.)

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Step 1:• Xi,0 : Number of parts in repair pipeline at CW• Ii,0(Si,0) (Si,0 Xi,0) and Bi,0(Si,0) (Xi,0 Si,0)

• Repair shop at CW is as a M|G| queue, andthus Palm's theorem may be applied

• Thus: Xi,0 Yi,0

Exact Evaluation (cont.)

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Step 2:• Due to FCFS allocation:

Each backordered demand at CW is a demand from LW j with probability mi,j /mi,0

• Hence:

Exact Evaluation (cont.)

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Step 3:• Xi,j(Si,0) : Number of parts in replenishment

pipeline at LW j• Ii,j(Si,0,Si,j) (Si,j Xi,j(Si,0)) and

Bi,j(Si,0,Si,j) (Xi,j(Si,0) Si,j)• Due to deterministic order and ship time:

Xi,j(Si,0) at time t = Bi,0(j)(Si,0) at time tti,j

+ Demand in interval [t ti,j, t)• Hence: Xi,j(Si,0) Bi,0

(j)(Si,0) Yi,j

Exact Evaluation (cont.)

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Procedure:• Compute first two moments of Ii,0(Si,0) and Bi,0(Si,0)• Compute first two moments of Bi,0

(j)(Si,0) for each LW j:simple formulas available

• Compute first two moments of Xi,j(Si,0) for each LW j• Fit a negative Binomial distribution on first two moments

of Xi,j(Si,0) for each LW j• Compute first moments of Ii,j(Si,0,Si,j) and Bi,j(Si,0,Si,j)

=> Accurate and efficient approximation!

Appr. eval. via two-moment fits(cf. Graves [1985])

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Procedure:• Compute first moments of Ii,0(Si,0) and Bi,0(Si,0)• Compute first moment of Bi,0

(j)(Si,0) for each LW j:mean of Bi,0

(j)(Si,0) = (mi,j /mi,0) ( mean of Bi,0(Si,0) )• Compute first moment of Xi,j(Si,0) for each LW j• Fit a Poisson distribution on first moment of Xi,j(Si,0) for

each LW j• Compute first moments of Ii,j(Si,0,Si,j) and Bi,j(Si,0,Si,j)

=> Not always accurate!

METRIC approach (cf. Sherbrooke [1985])

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5.3. Optimization

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Greedy heuristic(cf. Wong et al. [2007])

Main idea:• Start with zero stock for all SKU’s at all LW’s• Use a steepest descent method to decrease costs• Use a greedy logic to get to a feasible solution

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Greedy heuristic (cont.)

• Distance of a solution S to the set of feasible solutions:

11/7/2014/ School of Industrial Engineering

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Greedy heuristic (cont.)

• Decrease in distance to feasible solutions per unit of increase in costs:

11/7/2014/ School of Industrial Engineering

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Greedy heuristic (cont.)

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Leads to: Heuristic solution (under exact evaluations)

No theoretical results like for single-location model

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Dantzig-Wolfe decompostion leads to:- Lower bound- Heuristic solution

Performance of greedy heuristic:• Small optimality gap• # SKU’s => Optimality gap • Computation time is relatively low

Greedy heuristic (cont.)

Under exact evaluations

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6. Multi-location model with lateraland emergency shipments

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Key features:• Multiple local warehouses (LW’s)• Each LW supports one or more groups of machines• Aggregate mean waiting time target per group• Central warehouse: outside scope of model, is

assumed to have infinite stock• Multiple SKU’s• All SKU’s are critical• In case of stockout: application of lateral

transshipment or emergency shipment• Two types of LW’s: Main and regular LW’s

General setting

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General setting (cont.)

Main LW’s

Regular LW’s

Two types of LW’s: main and regular LW’s• Only main locals are suppliers for lateral transshipment• This distinction is made to facilitate implementation in practice• Pooling: full, partial, none

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Output variables: (basic output)Demand stream for SKU i at LW j :• i,j(Si) : Fraction satisfied by LW j itself• i,j,k(Si) : Fraction satisfied by main LW k via a

lateral transshipment• Ai,j(Si) : Total fraction satisfied by a lat. transshipm.,

Ai,j(Si) = kK\{j} i,j,k(Si)• i,j(Si) : Fraction satisfied by an emerg. shipment It holds that: i,j(Si) + Ai,j(Si) + i,j(Si) = 1

To be determined

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Done per given SKU iI

Via Markov process:• States xi (xi,1 xi,2 xi,|J|), where xi,j denotes

the on-hand stock at LW i• Steady-state distribution xiis determined

numerically• The i,j(Si), i,j,k(Si), i,j(Si) follow from the xi• Number of states: jJ (Si

Exact evaluation

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Given: J , K , k3 Si,1 , Si,2 , Si,3 0

EXAMPLE

0,1,0

0,0,0

1,1,0

1,0,0 2,0,0

2,1,0

t 1t

t1t 1t

1t1t

Mi,1Mi,3Mi,2Mi,1Mi,3Mi,2

Mi,1Mi,3 Mi,1Mi,3

Mi,2 Mi,1Mi,3

Mi,2 Mi,2

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We obtain:i,1(Si) 1,0,0 1,1,0 2,0,0 2,1,0,i,1,2(Si) 0,1,0, i,1(Si) 0,0,0, …

EXAMPLE (cont.)

0,1,0

0,0,0

1,1,0

1,0,0 2,0,0

2,1,0

t 1t

t1t 1t

1t1t

Mi,1Mi,3Mi,2Mi,1Mi,3Mi,2

Mi,1Mi,3 Mi,1Mi,3

Mi,2 Mi,1Mi,3

Mi,2 Mi,2

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Approximate evaluation: Principle(cf. Kranenburg and Van Houtum, 2009)

• Normal delivery by A• If A out of stock: lateral transshipment from B• Then B observes extra demand: overflow demand• Approximation: Overflow demand processes are

Poisson processes• Iterative procedure

A B

Lateral transshipmentNormal delivery

Principle: Decompose network into single-location models

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• Efficient and accurate

• One can apply the procedure to manystructures for the lateral transsipments!

• In use at ASML since 2005• 60 stockpoints, 5000 SKU’s, say• Optimization: Greedy heuristic

Approximate evaluation

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7. Extensions

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• Exact recursion like for METRIC model• Efficient and accurate appr. Eval. Based on

two-moment fits

Multi-echelon, multi-indenture systems

Seal6

Casing7

Bearing8

Pump3

Rotor9

Stator10

El.mo.4

Pump unit 11

Pump3

Rotor11

Stator12

El.mo5

Pump unit 22

Fire Extinguishing System

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Easy to incorporate for:• Single-location model with backordering• METRIC model with (s,Q) rule at central

warehouse and given Q’s

Hard to incorporate for models with emergencyshipments

Batching

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Standard way: Critical levels

Drawbacks:• More complicated optimization problem; no greedy

heuristic available• Critical levels are too strict for very low demand

rates• Critical levels may not be accepted in practice

Challenge: Find better ways for the differentiationwhen demand rates are low

Multiple demand classes

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8. Applications in practice

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Advanced software packages:• MCA Solutions: Founded by Morris Cohen &

Associates• Xelus• Network Neighborhood: Developed by IBM• …Own developments:• US Coast Guard: Development with Deshpande et al.

[2006] • ASML: Planning algorithm developed with TU/e• …

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9. Challenges

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The full problem

Supply spare parts

Central Stockpoint

LocalStockpoint

Customers with contracts

Reg. repl.:1-2 weeks

Emergency Shipm.: 1-2 days

Reg. repl.:1-2 weeks

LocalStockpoint

Lateral Shipments: A few hours

Customers with contracts

Direct sales

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Motivation: • One has visibility of actual stocks• Condition monitoring data• Service contracts are for 1/2/3 years

Needed:• Approximate dynamic programming, …• Interaction between tactical and operational planning

Dynamic decisions

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Example of condition monitoring data

MACHINE NUMBER

TIMESTAMPMACHINE TYPE

CUSTOMER ID

P955 P956 P957 P958 P959 P960 P961

M0005      07‐Jan‐09 T0007 C1058 ‐6.8961 ‐6.8860 ‐6.8910 ‐6.9177 ‐6.8542 ‐6.8860 ‐3.7979M0005      13‐Feb‐09 T0007 C1058 ‐7.3892 ‐7.3831 ‐7.3862 ‐7.4487 ‐7.3123 ‐7.3805 ‐4.3121M0005      16‐Feb‐09 T0007 C1058 ‐7.4847 ‐7.4738 ‐7.4792 ‐7.5021 ‐7.4451 ‐7.4736 ‐4.3567M0005      17‐Feb‐09 T0007 C1058 ‐7.5400 ‐7.5320 ‐7.5360 ‐7.5974 ‐7.4640 ‐7.5307 ‐4.3823M0005      19‐Feb‐09 T0007 C1058 ‐7.5305 ‐7.5201 ‐7.5253 ‐7.5479 ‐7.4915 ‐7.5197 ‐4.3793M0005      01‐Mar‐09 T0007 C1058 ‐7.6871 ‐7.6767 ‐7.6819 ‐7.7032 ‐7.6489 ‐7.6761 ‐4.5276M0005      07‐Mar‐09 T0007 C1058 ‐7.7783 ‐7.7686 ‐7.7734 ‐7.7954 ‐7.7414 ‐7.7684 ‐4.6072M0005      10‐Mar‐09 T0007 C1058 ‐7.7222 ‐7.7109 ‐7.7165 ‐7.7396 ‐7.6819 ‐7.7107 ‐4.6219M0005      06‐Apr‐09 T0007 C1058 ‐8.0890 ‐8.0804 ‐8.0847 ‐8.1049 ‐8.0527 ‐8.0788 ‐4.9698M0006      06‐Apr‐09 T0007 C1058 ‐7.9700 ‐7.9602 ‐7.9651 ‐7.9858 ‐7.9337 ‐7.9597 ‐4.9725M0006      22‐Apr‐09 T0007 C1058 ‐8.1674 ‐8.1635 ‐8.1654 ‐8.2214 ‐8.0994 ‐8.1604 ‐5.1675M0006      01‐Jun‐09 T0007 C1058 ‐8.5568 ‐8.5504 ‐8.5536 ‐8.5730 ‐8.5239 ‐8.5484 ‐5.6113M0006      09‐Jun‐09 T0007 C1058 ‐8.5599 ‐8.5553 ‐8.5576 ‐8.6090 ‐8.4949 ‐8.5519 ‐5.6860M0006      22‐Jul‐09 T0007 C1058 ‐8.7384 ‐8.7350 ‐8.7367 ‐8.7869 ‐8.6755 ‐8.7312 ‐6.2033M0006      24‐Jul‐09 T0007 C1058 ‐8.8330 ‐8.8257 ‐8.8293 ‐8.8461 ‐8.8004 ‐8.8232 ‐6.1697M0006      14‐Aug‐09 T0007 C1058 ‐8.9163 ‐8.9142 ‐8.9153 ‐8.9639 ‐8.8553 ‐8.9096 ‐6.3815M0006      14‐Aug‐09 T0007 C1058 ‐8.9392 ‐8.9371 ‐8.9381 ‐8.9871 ‐8.8776 ‐8.9323 ‐6.3531M0006      16‐Aug‐09 T0007 C1058 ‐8.8599 ‐8.8574 ‐8.8586 ‐8.9078 ‐8.7988 ‐8.8533 ‐6.4169

11/7/2014 PAGE 116

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/ School of Industrial Engineering PAGE 11711/7/2014

• Coupling with service tools planning

• Other performance measures: Long delays vs. Short delays

• Setting of leadtimes for procurement/repair

• Choice of transport modes

• …

Further challenges