1 material to cover relationship between different types of models incorrect to round real to...

1

Material to Cover

relationship between different types of models

incorrect to round real to integer variables

logical relationship: site selection

weak and strong formulation: uncapacitated facility location problem

set covering problems: airline crew scheduling

generalized piecewise linear approximation

2

max/min z = c1x1 + c2x2 + … + cnxn

s.t.

ai1x1 + ai2x2 + … + ainxn bi, i = 1,…, m

0 xj uj, j = 1,…, n

xj integer for some or all j =1,…, n

{

}

Linear Integer Programming - IP

3

mixed IP (MIP): some xj Z#1, some xj #2

pure IP: all xj Z

binary decision variable: xj = 1 or 0 (e.g., a variable for a yes-no decision)

binary IP (BIP): all xj being binary

Linear Integer Programming - IP

#1 Z: the set of integers; Z+: the set of positive integers

#2 : the set of real numbers ; +: the set of positive real numbers

4

Motivation of Studying IP

integer variables in some context e.g., machine, manpower

logical relationship incorrect to round continuous variables

5

Incorrect to Round Continuous Variables

optimal LP solution

X1

X2

optimalIP solution

iso-cost line

usually all right to round in real life

problems with large xi

6

Example to Motivate IP

site selection: three designs A, B, C on sites 1, 2, 3, 4

total amount for investment: $100 M how to invest?

Option A1 A2 A3 A4 B1 B2 B3 B4 C1 C2 C3 C4

Net Income ($M) 6 7 9 11 12 15 5 8 12 16 19 20

Investment ($M) 13 20 24 30 39 45 12 20 30 44 48 55

7

Example to Motivate IP

I = {A, B, C, D}, J = {1, 2, 3, 4}

yij = 1 iff design i used at site j, iI, jJ

max z = iI pij yij

s.t. iI jJ aij yij 100

yij {0, 1}, iI, jJ

optimal solution: yA1 = yA3 = yB3 = yB4 = yC1 = 1; z* = 40

8

Example to Motivate IP

boss says NO!

at most one design at a site

a building at site 2 (required)

at most two designs at the three sites

design A considered for sites 1, 2, and 3 only if being used at site 4

how to model?

9

Example to Motivate IP

at most one design at a site and a building at site 2 (required)

yA1 + yB1 + yC1 1, yA2 + yB2 + yC2 = 1,

yA3 + yB3 + yC3 1, yA4 + yB4 + yC4 1

design A considered for sites 1, 2, and 3 only if being used at site 4

yA1 + yA2 + yA3 3yA4

10

Example to Motivate IP

at most two designs at the three sites

wi = 1, if design i is used, = 0, o.w., i = A, B, C

wA + wB + wC 2

yi1+yi2+yi3+yi4 4wi, i = A, B, C

optimal solution: yA1 = yA4 = yB2 = yB3 = 1; others = 0; z* = 37

11

Logical Constraints for Variables

n situations how to model (i) at most k of them hold, (ii) at least k of them hold, and (iii) exactly k hold

yj binary variables for j = 1 to n; yj = 1 if j holds, and = 0 otherwise

mutually exclusive yj: y1 + y2 + … + yn 1

at most k of yj = 1: y1 + y2 + … + yn k

at least k of yj = 1: y1 + y2 + … + yn k

exactly k of yj = 1: y1 + y2 + … + yn = k

12

Logical Constraints for Expressions

either-or constraints

either f1(x1, …, xn) b1 or f2(x1, …, xn) b2 or both

IP formulation: let y be a binary variable

M: a large positive number, practically “”

two constraints: f1(x1, …, xn) b1+My and f2(x1, …, xn) b2+M(1-y)

only one of these two being picked by the optimization procedure

13

Logical Constraints for Expressions

m constraints, at least k out of m being true f1(x1, …, xn) b1, …, fm(x1, …, xn) bm

modeling procedure m binary variables yi, one for each constraint

f1(x1, …, xn) b1+M(1-y1), …, fm(x1, …, xn) bm+M(1-ym)

y1 + … + ym k

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An Example of Logical Constraints for Expressions

single processor for three jobs, of processing times 3 hr, 5 hr, and 7 hr, respectively

objective: minimizing the total completion time of the three jobs

how to formulate it as an integer program? note. The IP is for the illustration of formulation.

The problem has very simple solution.

15

Definitions of Parameters and Variables

si: the processing start time of job i

ci: the completion time of job i

pi: the processing time of job i (i.e., p1 = 3, p2 = 5, p3 = 7)

C: total completion time

16

How About This?

if job 1 before job 2, and before job 3, C = 26

if job 1 before job 3, and before job 2, C = 28

if job 2 before job 1, and before job 3, C = 28

if job 2 before job 3, and before job 1, C = 32

if job 3 before job 1, and before job 2, C = 32

if job 3 before job 2, and before job 1, C = 34

17

How About This?

s1 s2 s3, C = 26 if (y12=1 & y23=1), C = 26

s1 s3 s2, C = 28 if (y13=1 & y32=1), C = 26

s2 s1 s3, C = 28 if (y21=1 & y13=1), C = 28

s2 s3 s1, C = 32 if (y23=1 & y31=1), C = 32

s3 s1 s2, C = 32 if (y31=1 & y12=1), C = 32

s3 s2 s1, C = 34 if (y32=1 & y21=1), C = 34

then setting conditions on yij …, which obviously not working

18

The Formulation

min c1 + c2 + c3,

s.t. c1-s1 = 3; c2-s2 = 5; c3-s3 = 7;

one of c1 s2 and c2 s1 holds;

one of c1 s3 and c3 s1 holds;

one of c2 s3 and c3 s2 holds;

ci 0, si 0, i = 1, 2, 3.

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The Formulation

min c1 + c2 + c3,

s.t. c1-s1 = 3; c2-s2 = 5; c3-s3 = 7;

c1 s2+My12; c2 s1+My21; y12+y21 = 1;

c1 s3+My13; c3 s1+My31; y13+y31 = 1;

c2 s3+My23; c3 s2+My32; y23+y32 = 1;

ci 0, si 0, i = 1, 2, 3;

yij {0, 1}, i j, and i, j = 1, 2, 3

yij = 1 if job j is before job i.

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Equivalence Between BIP and General PIP

BIP PIP BIP PIP a PIP of bounded integer variables BIP max 5x1 + 2x2

s.t. 2x1 + x2 15

x1 0, x2 Z+

conversion 0 x2

15

x2 = y1 + 2y2 + 4y3 + 8y4, yi binary

21

Fixed-Charge Problem

costs for having a facility at site j, j = 1 to n set up cost kj

variable cost cj per unit of capacity

capacity of the whole system C minimum cost site selection for the capacity co

nstraint

22

Fixed-Charge Problemn

j=1min fj(xj) where fj(xj ) = {

kj + cjxj, if xj > 0

0, if xj = 0

kj = set-up cost, cj = per unit cost

IP formulation:

minn

j =1 ( cjxj + kjyj )

s.t.xj Myj, j = 1, …, n;

j xj C;

yj {0,1}, j = 1, …, n;

xj 0, j = 1, …, n;

23

A More Realistic Fixed-Charge Problem

telecommunication network

source nodes S = {1, 3, 7}; destination nodes D = {2, 4, 5, 8}; transshipment node T = {6}

solid links: existing; dotted links: planned; total A = {1, 2, …, 17}

each link: (capacity, cost)

planned links A’ = {1, 2, …, 5}; fixed costs f1 = 8; f2 = 6; f3 = 9, f4 = 7, f5 = 7

min cost construction to satisfy the demands & flows

24

A More Realistic Fixed-Charge Problem

decision variables

xk: the amount of flow in link k

yk: the binary variable of constructing link k A'

parameters

bi: the demand of node i (source = -demand)

uk: the upper bound of flow of link k

fk: the fixed cost coefficient

ck: the variable cost coefficient

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A More Realistic Fixed-Charge Problem

min z = kA’fkyk + kA ckxk

s.t.

total in-flow – total out-flow = bi, conservation of flow node i

xk ukyk, capacity constraint (proposed) arc k A'

xk uk, capacity constraint (existing) arc k A

yk = 0 or 1, binary variable (proposed) arc k A'

xk 0, arc k A A'

26

Facility Location Problem

distributing goods to n customers possibly through m warehouses

warehouse i fixed cost fi

variable cost vi per unit capacity

maximum capacity ui

shipment cost cij per unit

from warehouse i to customer j

C1

.

.

.Cn

W1

.

.

.

Wm

27

Facility Location Problem distributing goods to n customers possibly through m warehouse

s data

dj : demand for customer j ui : maximum capacity at warehouse i fi : fixed cost to build warehouse i vi : variable cost/unit of capacity of warehouse i cij : variable cost/unit of goods sent from warehouse i to custome

r j decision variables

yi: build a warehouse at site i? (1 = yes, 0 = no)

zi : capacity (supply) of warehouse i xij: shipment from warehouse i to customer j

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Facility Location Problem

m

i

n

jijij

m

iii

m

iii xczvyf

1 111min

miy

njmix

miz

i

ij

i

,...,1 },1,0{

,...,1 ,,...,1 ,0

..., 1, ,0

warehousea of constraintcapacity

ouseeach warehcapacity

demands all meeting

1

1

,...,1 ,

,...,1 ,

,...,1 ,

miyuz

mizx

njdx

iii

n

jiij

m

ijij

29

Strong versus Weak Formulation – An Illustration through the Facility Location Problem

uncapacitated version of the facility location problem intuitively optimal to have each customer satisfied by

one warehouse simplified the formulation re-definition

cij: shipment cost to customer j, possibly including the variable cost of operating warehouse i for demand dj

xij: proportion of demand j satisfied by warehouse i

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Weak Formulation

miy

njmix

minyx

njx

xcyf

i

ij

n

jiij

m

iij

m

i

n

jijij

m

iii

,...,1 },1,0{

,...,1 ,,...,1 },1,0{

,...,1 ,

,...,1 ,1

min

ouseeach wareh of constraintcapacity 1

demands all meeting1

1 11

31

Strong Formulation

miy

njmix

njmiyx

njx

xcyf

i

ij

iij

m

iij

m

i

n

jijij

m

iii

,...,1 },1,0{

,...,1 ,,...,1 },1,0{

,...,1 ,,...,1 ,

,...,1 ,1

min

ouseeach wareh of constraintcapacity

demands all meeting1

1 11

32

Comparison of the Strong and Weak Formulations

strong: more constraints xij yi, i = 1, …, m; j = 1, …, n

weak: less constraints i xij nyi, j = 1, …, n

which one is better? strong: more precise constraints and possibly s

horter computation time

33

Covering Problems and Partitioning Problems

S: a set of m items Sj: a subset of S that includes one or more of the it

ems, j = 1, …, n cj: the cost of selecting subset j

selecting the minimum cost collection of subsets Sj to include elements of S set covering: fine as long as including all items of S set partitioning: each element of S is included exactly once

34

Airline Crew Scheduling (Set Covering Problem)

service network

group legs into tours according to constraints

LAX

SEA

CHI

DEN

DFW

35

Airline Crew Scheduling (Set Covering Problem)

a tour: feasible assignment for a crew, starting & ending at DFW

a leg: a flight scheduled between two cities covering 11 legs by 3 crews on 12 possible tours minimizing the total tour cost

Tour 1 2 3 4 5 6 7 8 9 10 11 12

DFW LAX 1 1 1 1DFW DEN 1 1 1 1DFW SEA 1 1 1 1

LAX CHI 2 2 3 2 3LAX DFW 2 3 5 5 CHI DEN 3 3 4CHI SEA 3 3 3 3 4DEN DFW 2 4 4 5DEN CHI 2 2 2

SEA DFW 2 4 4 5

SEA LAX 2

Cost 2 3 4 6 7 5 7 8 9 9 8 9

24 42

36

Airline Crew Scheduling (Set Covering Problem)

optimal solution: “Dead heading” on first leg

Min 2x1+3x2+4x3 + … + 8x11 + 9x12

s.t. x1 +x4 +x7 + x10 (DFW LAX)x2 +x5 +x8 + x11 (DFW DEN)x3 +x6 +x9 + x12 (DFW SEA)x4 +x7 +x9 + x10+ x12 (LAX CHI)x1 +x6 +x10 +x11 (LAX DFW)

x6 +x9 +x10 + x11+ x12

1(SEA LAX)

x1 +x2 +… + x12 = 3 (assign 3 crews)

xj {0,1}, j = 1,…,12

1

1

1 1 1

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The Days-Off Scheduling Problem

(5,7)-cycle: 5 working days + 2 consecutive days off

7 days-off patterns parameters

ri = number of employees required on day i

cj = weekly cost of pattern j per employee

decisions xj = # of employees using days-off pattern j

38

The Days-Off Scheduling Problem

min z = cjxj

s.t. ( xj ) – xi – xi-1 ri , i = 1,…7

xj 0 and integer, j = 1,…,7; x0 = x7

7

j=1

7

j=1

solve problem to get x*j

minimum cost workforce W = x*j

7

j=1

39

The Days-Off Scheduling Problem

possibly to be solved as two LP compact expression

Minimize z = cx s.t.

rx

0011111

1001111

1100111

1110011

1111001

1111100

0111110

x 0 and integer

40

The Cutting Stock Problem

raw material: rolls of 25 ft

requirements 5-foot: 40 rolls

8-foot: 35 rolls

12-foot: 30 rolls

15-foot: 25 rolls

17-foot: 20 rolls

objective: using minimum # of 25-foot rolls

41

General Piecewise Linear Approximations

f (x )j j

jju

x

(d , f )1j 1j

(d , f )2j 2j

(d , f )3j 3j

(d , f )4j 4j

(d , f )5j 5j

fj(xj), 0 xj uj

r = number of grid points

(dij, fij) be ith grid point, i = 1…, r

42

Linear Transformation for xj

xj = idij

gj(xj) = ifij

i = 1, i 0, i = 1,…, r

not sufficient to guarantee that the solution is on one of the line segments

r

i=1

r

i=1

r

i=1

43

Additional Constraints for Piecewise

Linear Approximation condition: at most two positive i, and positive i‘s adjacent

1 ≤ y1

i ≤ yi-1 + yi , i = 2,…, r–1

r ≤ yr-1

y1 + y2 + · · · + yr-1 = 1

yi = 0 or 1, i = 1,..., r–1 not necessary to define ’s if minimizing a convex function or

maximizing a concave function

44

Approximation in Minimizing

a Convex Objective Function

20.251640y

4.5420x1

CBAOpoints

45

Approximation in Minimizing

a Convex Objective Function

all right to omit (4) if approximating a convex

objective function in minimization or a concave objective

function in maximization

46

Approximation in Minimizing

a Convex Objective Function

all right to omit (4) if approximating a convex

objective function in minimization or a concave objective

function in maximization

47

Special Non-linear Objective Functions

machines: A to D products: P, Q, R potential sales: P 100, Q 100, R 100

prod

mh timeP Q R available

time

A 20 10 10 2400

B 12 28 16 2400

C 15 6 16 2400

D 10 15 0 2400

48

Special Non-linear Objective Functions

nonlinear unit profit from the products

prod

salesP Q R

0-30 60 40 20

31-60 45 60 70

61-100 35 65 20

How to formulate the problem?

49

Special Non-linear Objective Functions

max Z = f1(P) + f2(Q) + f3(R) s.t. 20P + 10Q + 10R 2400 (mh A) 12P + 28Q + 16R 2400 (mh B) 15P + 6Q + 16R 2400 (mh C) 10P + 15Q 2400 (mh D) P 100, Q 100, R 100 (marketi

ng) P 0, Q 0, R 0 How to model f1(P), f2(Q), f3(R)

50

Special Non-linear Objective Functions

Pi: # of sales of product P in the ith price range

Qi: # of sales of product Q in the ith price range

Ri: # of sales of product R in the ith price range

object function: max Z = 60P1+45P2+ 35P3+40Q1+60Q2

+ 65Q3+20R1+70R2+20R3

51

Special Non-linear Objective Functions

Z = 60P1+45P2+ 35P3+40Q1+60Q2+ 65Q3

+20R1+70R2+20R3+ for P

0 P1 30, 0 P2 30, 0 P3 40 concave prices, no additional constraints

for Q 0 Q1 30, 0 Q2 30, 0 Q3 40

use second price segment only if Q1 = 30

use third price segment only if Q2 = 30

for R 0 R1 30, 0 R2 30, 0 R3 40

use second and third price segments only if R1 = 30

52

Special Non-linear Objective Functions

for Q yQ2 = 1 if sales in segment 2 are made

= 0 otherwise

yQ3 = 1 if sales in segment 3 are made

= 0 otherwise

30yQ2 Q1 30, 30yQ3 Q2 30yQ2, 0 Q3 40yQ3

for R yR2 = 1 if sales in segments 2 or 3 are made

= 0 otherwise

30yR2 R1 30, 0 R2 30yR2, 0 R3 40yR2

53

Other Examples

Traveling salesman problems sequence dependent setup times

assembly line balancing