15.093: Optimization Methods. Lecture 9: Large Scale Optimization

15.093:

Optimization

Methods

1

Outline

Slide 1

1. Theideaofcolumngeneration 2. Thecuttingstock problem 3. Stochastic programming

2

Column

Generation

Slide 2

Forx 2< nandnlargeconsider theLOP:

min c 0 x s:t: Ax =b x 

0

Restricted problem X min cixi i2I X s:t: Aixi=b i2I x 

0

2.1

Two

Key

Ideas

Slide 3

GeneratecolumnsAj onlyasneeded.

Calculateminici ecientlywithoutenumeratingallcolumns.

3

The

Cutting

Stock

Problem

Slide 4

Companyhasasupplyoflargerollsofpaperofwidth W.

bi rollsofwidthwi
i= 1
:::
m needtobe produced.

Example: w=70inches,canbecutin3rollsofwidthw1 =17and1roll

ofwidthw2 =15,waste: 70; (3 17+1 15)=4 Slide 5 Givenw

1
:::
wmandW therearemanycuttingpatterns: (3
1)and(2
2)

forexample

3 17+1 15  70

m

X

aiwi W

i�1

3.1

Problem

Slide 6

Given wi
bi
i = 1
:::
m (bi: numberof rolls of width wi demanded,

andW (widthof largerolls):

Findhow to cutthe largerollsinorder to minimizethenumberofrolls

used.

3.2

Concrete

Example

Slide 7

WhatisthesolutionforW =70
w

1=21
w 2=11
b1=40
b2=40? feasiblepatterns: (2
2),(3
0),(0
6)

Solution1: (2
2): 20patterns20rollsused

Solution2: (3
0): 12,(0
6): 9,(2
2): 2patterns: 23rollsused

Slide 8 W =70
w 1=20
w =11
b =12 =17 1
b2 2 1 0 2 0 ;
3 0 ;
0 1 ;
1 1 ;
0 3 2 2 1 2 0 2 2 1 ;
; ; ; ;
1 3 ;





; Feasiblepatterns:









0 6 0 5 ; ; ; 1 4 x

1
:::
x15 =#offeasiblepatternsofthetype 0 4



;


0 6 ; min x1 +  +x 15         1 2 0 12 s:t: x1 +x2 +  +x 15 = 0 0 6 17 x1
:::
x15  0 Slide 9         0 0 3 12

Example: 2 +1 +4 = 7rollsused

6 5 0 17         0 0 3 12 4 ₄ + ₁ +4 ₀ = ₁₇ 9rollsused Anyideas? 1 0 ;

respectively
:::

3.3

Formulation

Slide 10

Decisionvariables:xj=numberofrollscutbypatternjcharacterizedbyvector

Aj: n P min_j xj �1 0 1 b1 n P B C Aj xj = @ ... A j�1 bm xj 0 (integer) Slide 11

Hugenumberofvariables.

Canweapplycolumngeneration,thatisgenerate thepatternsAj onthe

y?

3.4

Algorithm

Slide 12

Idea: Generatefeasiblepatternsasneeded.

0 1 0 1 0 1 0 1 b wWc 0 0 0 1 B 0 C B b Wc C B 0 C B 0 C B C
B w 2 B C B C

1) Startwithinitialpatterns: @ A @ A @ b Wc A
@ 0 A C
0 0 _w3 0 0 0 b wWc 4 Slide 13 2) Solve: x minx1 +  +xm 1 A 1 +  +xmAm=b xi 0 Slide 14

3) Computereduced costs cj= 1; p

0

Aj forallpatternsj

curren

Ifcj  0 t setofpatternsoptimal

Ifcs<0) xs needs toenter basis

How are we going to compute reduced costs cj = 1; p 0

Aj for all j? (huge

3.4.1

Key

Idea

Slide 15 4) Solve _m X � z =max piai i�1 m X s:t: wiai W i�1 ai 0
integer

Thisistheintegerknapsack problem

Slide 16

Ifz

�

 1) 1; p 0

Aj>08j) current solutionoptimal

Ifz

� >1

)9 s: 1; p 0

As <0) Variable xs becomes basic,i.e.,anew

patternAs willenterthebasis.

Performmin-ratiotestandupdate thebasis.

3.5

Dynamic

Programming

Slide 17 F(u) =max p1 a1 +  +pmam s:t: w1 a1 +  +wmam u ai 0
integer

Foru wmin,F(u) =0.

Foru wmin F(u)= max fpi+F(u; wi)g i�1::: m Why?

3.6

Example

Slide 18 max 11x1 + 7x2 + 5x3 +x4 s:t: 6x1 + 4x2 + 3x3 +x4  25 xi 0
xi integer F(0) =0 F(1) =1 F(2) =1 +F(1)=2 Slide 19 F(3) =max(5+F(0)�
1 +F(2))=5 F(4) =max(7+F(0)�
5 +F(1)
1 +F(3))=7 F(5) =max(7+F(1)�
5 +F(2)
1 +F(4))=8 F (6) =max(11+F (0) � 7 + F (2) 5 +F (3)1 +F (5))=11 F (7) =max(11+F (1) � 7 + F (2) 5 +F (3)1 +F (4))=12 F (8) =max(11+F (2)7 +F (4) � 5 + F (5)1 +F (7))=14 F (9) =11 +F (3)=16 F (10) =11 +F (4)=18 ( ) =11 + ( 6) =16 11

) F (25)=11+F (19)=11+11+F (13)=11+11+11+F (7)=33+12=45 � x =(4 0 0 1)

4

Stochastic

Programming

4.1

Example

Steel(lbs) Moldingmachine(hrs) Assemblymachine (hrs)

Demandlimit(tools/day)

Contributiontoearnings ($/1000units)

4.1.1

Random

data

Wrenches 1.5 1.0 0.3 15,000 $130* Pliers 1.0 1.0 0.5 16,000 $100 Slide 20 Cap. 27,000 21,000 9,000* Slide 21 max 130W+100P s:t: W  15 P  16 1:5W +P  27 W +P  21 0:3W + 0:5P 9 W
P 0 Slide 22 8 1 > < 8000 withprobability 2 Assemblycapacityisrandom:

> 1 : 10
000 withprobability 2 8 1 > < 160 withprobability 2 Contributionfromwrenches:

> 1

: 90 withprobability

2

4.1.2

Decisions

Slide 23

Need todecide steelcapacityinthecurrent quarter. Cost58$/1000lbs.

Soonafter,uncertainty willberesolved.

Next quarter,companywilldecideproductionquantities.

4.1.3

Formulation

1 8,000 2 10,000 3 8,000 4 10,000 160 0.25 160 0.25 90 0.25 90 0.25

DecisionVariables: S: steelcapacity,

Pi
Wi:i= 1
:::
4productionplanunderstate i. Slide 25

max ;58S+ 0:25Z 1 + 0:25Z2 + 0:25Z3 + 0:25Z4 s:t: Ass:1 0:3W1 + 0:5P1  8 Mol:1 W1 +P1  21 Ste:1 ;S+ 1:5W 1 +P1  0 W:d:1 W1  15 P:d:1 P1  16 Obj:1 ;Z 1 +160W1 +100P1 = 0 Slide 26 Ass: 2 0:3W2 + 0:5P2  10 Mol:2 W2 +P2  21 Ste:2 ;S+ 1:5W 2 +P2  0 W:d:2 W2  15 P:d:2 P2  16 Obj:2 ;Z 2 +160W2 +100P2 = 0 Slide 27 Ass:3 0:3W3 + 0:5P3  8 Mol:3 W3 +P3  21 Ste:3 ;S+ 1:5W 3 +P3  0 W:d:3 W3  15 P:d:3 P3  16 Obj:3 ;Z 3 +90W3 +100P3 = 0 Slide 28 Ass:4 0:3W4 + 0:5P4  10 Mol:4 W4 +P4  21 Ste:4 ;S+ 1:5W 4 +P4  0 W:d:4 W4  15 P:d:4 P4  16 Obj:4 ;Z 4 +90W4 +100P4 = 0 S
Wi
Pi 0

4.1.4

Solution

Slide 29 Solution:S =27
250lb. Wi Pi 1 15,000 4,750 2 15,000 4,750 3 12,500 8,500 4 5,000 16,000

4.2

Two-stage

problems

Slide 30

Randomscenarios indexed byw = 1
:::
k. Scenario w has probability

w.

Firststage decisions: x: Ax =b
x 

0

.

Second stagedecisions: yw: w= 1
:::
k. Constraints: Bwx +Dwyw=dw,yw

0

.

4.2.1

Formulation

Slide 31 min c 0 x +  1 f 0 kyk B 1 y 1 +  + kf 0 Ax =b 1 x + D 1 y 1 = d 1 B 2 x + D 2 y 2 = d 2 Slide 32 . . . . . . . . . Bkx + Dkyk =dk x
y 1
y 2
:::
yk 

0

: Structure: x y 1 y 2 y 3 y 4 Objective