Supply Chain Management Chapter 5: Application of ILP. Unified optimization methodology. Beun de Haas

Supply Chain Management

Chapter 5:

Unified Optimization Methodology for

Operational Planning Problems

What to do when ILP takes too much

computation time?

Application of ILP

• Timetable Dutch Railways (NS)

• Bus and driver scheduling at Connexxion, GVU, Qbuzz

• Production planning at Hoogovens

• Advanced blending at Shell

• Plant location at Interbrew

• Cutting plastics at Exxon Mobil

• Crew planning at American Airlines Emerging area: Healthcare

Links:

• Customers of AIMMS

• http://www.ompartners.com/

Unified optimization methodology

= Column generation

• Faster than straightforward ILP, often better than heuristic

• Possibility to find very good (but not necessary optimal) solution with quality measure

• Many successful applications in SCM

• More advanced

Beun de Haas

Day

4 hours Friday

8 hours Thursday

4 hours Wednesday

8 hours Tuesday

6 hours Monday

Wed, thu, fri 4

2 hours 3

….. …..

….. etc

Mon, tue, thu 6

3 hours 2

Days Revenue

Duration Job

Mon, wed, fri 5

2 hours 1

The Beunhaas problem

• Beun de Haas is an independent entrepreneur.

• Clients contact him for small jobs.

• For each job j is given:

– the reward (cj); – the time it takes (a_j);

– the days on which they can be done

• Planning period: days 1, . . . , T.

• Beun has Qt time on day t (t = 1, . . . , T). • Goal.Choose and plan the work to earn as much as possible.

Advanced ILP formulation

•

Formulation with day plans.

•

A day plan for day

t

is a set of jobs that Beun

can do on day

t

.

•

S

is the set of feasible day plans for day

t

•

The reward of day plan

j

is equal to

C

•

Use a binary variable:

– x_jt = 1 if day plan jfrom S_t is the chosen day plan for day t (and 0 otherwise).

Decomposition!!!!!

Job at most once

One plan per day

ILP with day plans

• Disadvantage: solving ILP may take a long time

• Solution: relax integrality constraints, LP-relaxation.

• Disadvantage: There are so many possible day plans

• Solution:Consider only interesting day plans

Job at most once

One plan per day

Column generation for LP

1. Start with a small set of day plans

2. Solve LP-relaxation.

3. Find out if there is a new dayplans that can improve the solution (= pricing)

4. No ⇒optimum found

5. Yes ⇒add plan to model and go to 2.

Pricing= Lagrangean subproblem

•

Finding out if there are day plans to improve

solution

•

Recall: variable can improve solution if and

only if reduced cost are positive

•

Pricing problem:

– Find day plan with maximal reduced cost

• If maximum > 0, add day plan

• Otherwise stop

Pricing (2)

•

So, find a `optimal’ dayplan for each day

t

– Yi = 1if job i is selected, 0 otherwise – Reduced cost

– Day plan has to fit

t n

1 i

i i n

1 i

i

iy y

c −

∑

∑

= =

∑

≤

n

1 i

t i

iy Q

Pricing problem for given

t

}

1

,

0

{

y

Q

y

a

subject to

y

)

c

(

max

i n

1 i

t i i

i i n

1 i

i

∈

≤

π

−

∑

∑

= =

Finding integral solution

•

Solve LP-relaxation = upper bound

•

Then:

– Rounding

– Solve ILP with resulting set of columns

– Branch-and-price

Decomposition

• Master IP:

– Variables: 0/1 (or integer) for selection of feasible sub plans

– In principle huge number of variables

• Solve Master LP:

– Only include restricted collection of variables,i.e. only interesting variables

• Solution procedure:

1. Start with feasible solution from Master LP

2. Solve LP and find shadow prices

3. Consider sub problem to find interesting variables (with negative reduced cost for minimization problem)

4. Go to 2. if interesting columns (with negative reduced cost for minimization problem) were found

Master LP

Master IP

Lagrangean Subproblem (pricing problem)

Shadow prices

Feasible Subplan

Feasible Plan

Depot

Local delivery problem

– Chemicals are transported by truck from depot to 13 customers

– What should be the routes of the truck such that the cost are minimal?

Local delivery problem

• Master problem: Is every customer visited?

– Decision variablexrindicates selection of route r

– cr: cost of router

– ajr=1 if customeriin route r

r

x

i

x

a

x

c

r n

i

r ir

r r

all

for

}

1

,

0

{

all

for

1

s.t.

min

1

∈

≥

∑

∑

=

Local delivery problem (2)

•

Subproblem: Routes for individual trucks.

Solving the sub problem:

Maximize

corresponds with minimize reduced cost route

route in customer

i

−

c

π

∑

∑

π

−

route in customer

i route

c

Gate assignment at Schiphol

1.

ILP: 80 flights

2.

column generation: 560 flights = one day at

Gate assignment: Wat is het

probleem?

• We hebben een verzameling vluchten:

– Aankomst- en vertrektijd

– Type vliegtuig

– Herkomst en bestemming

– Eventuele voorkeuren van de maatschappij

– Grondafhandelaar

• En we hebben een verzameling gates:

– Mogelijke regio's (Schengen/EU/Non-EU)

– Mogelijke vliegtuigtypes (grootte)

– Mogelijke grondafhandelaren

• Gezocht: een optimale planning

Twee fasen aanpak

•

Gateplannen: verzameling vluchten op een

gate

1.

Zoek voor elke groep van `gelijke’ gates een

even groot aantal gateplannen.

2.

Koppel de gateplannen aan fysieke gates.

Kostenfunctie

•

Robuuste oplossingen

•

Variantie: (tussentijden)

•

Of soortgelijke functie

•

Eventueel correcties voor:

– Vluchten zelfde maatschappij

– Vluchten zelfde grondafhandelaar

– Betrouwbaarheid maatschappij

Gate assignment at Schiphol fase 1

•

Master problem:

– Variables are plans for one gate

– Each flight is on exactly one gate

– Flight assigned to gate of correct type

– Other constraints …..

– Maximize robustness

•

Subproblem:

Kolom generatie fase 1

• Model

– xsgeeft selectie van plan s aan

– cs: kosten van plans

– ais=1 als vluchti in plan s

s x

i x a

x c

s s

s is

s s

∀ ∈

∀ =

∑

∑

} 1 , 0 {

n beperkinge overige

1 s.t.

min

Rekenresultaten

•

Een dagje Schiphol:

– Looptijd LP: 70-234 seconden

– Looptijd ILP: 5-333 seconden

Production scheduling

• 6 types of tires, planning for 8 weeks

• demands are given

• limited machine capacity

• cost: set up cost and inventory cost

Master problem: overall machine capacity

Production Schedule

Tire 1

Production Schedule

Tire 6

………..

Cutting stock

•

Iron bars of given length

•

Orders for small bars of different lengths

•

How to cut given orders from minimal

number of bars?

Decomposition:

•

Master LP: production of all ordered formats

•

Lagrangean subproblem: patterns for

Customer order

8 60 cm

22 48 cm

16 42 cm

13 35 cm

23 31 cm

24 27 cm

32 19 cm

Number

mi

Length

bi

Lengte staaf L: 1 meter

ILP with patterns

• Disadvantage: solving ILP may take a long time

• Solution: relax integrality constraints, LP-relaxation.

• Disadvantage: There are so many possible patterns

• Solution:Consider only interesting patterns:

Column generation.

Column generation for LP

1. Start with a small set of patterns

2. Solve LP-relaxation.

3. Find out if there is a new pattern that can improve the solution

4. No ⇒optimum found