6 Mixed-Integer Linear Programming.pdf

8/11/2019 6 Mixed-Integer Linear Programming.pdf

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Floudas: Nonlinear and Mixed-Integer OptimizationChapter 5

Mixed-Integer Linear Programming

Cheng-Liang Chen

PSELABORATORY

Department of Chemical EngineeringNational TAIWAN University

Chen CL 1

Modeling with 0 1 VariablesApplication: select among process units i P U

Let yi=

1 if unit i is selected

0 if unit i is NOT selectedi P U

Selectone & only oneunit iPU

yi = 1

Selectat most oneunit iPU

yi 1

Selectat least oneunit iPU

yi 1

Chen CL 2

Modeling with 0 1 VariablesApplication: select unit i if unit j is selected

yi= 1 if unit i P Uis selected, 0 otherwise

If unit j is selected (yj = 1)

then unit i should be selected too (yi= 1)

Others: not defined

yj yi 0 (check it)

Derivation: yi= 1 Pi is true; yj = 1 Pj is true

Pj Pi Pj Pi (?)

(1 yj) + yi 1 (?)

yj yi 0

Chen CL 3

Propositional Logic Expressions

NEGATION denoted as

AND denoted as

OR denoted as

EXCLUSIVE OR denoted as

Modeling with 0 1 Variables:

yi =

1 if clause Pi is true

0 if clause Pi is false

P1 P2 P1 P2 P1 P2 P1 P2 P1 P1 P2 P1 P2

1 1 1 1 1 0 1 0

1 0 0 1 0 0 0 1

0 1 0 1 1 1 1 1

0 0 0 0 1 1 1 0

Note: P1 P2 is equivalent to P1 P2


2/18

Chen CL 4

Modeling with 0 1 Variables

Proposition Mathematical Representation

1. P1 P2 P3 y1+ y2+ y3 1

2. P1 P2 P3

y1 1

y2 1

y3 1

3. P1 P2 (P1 P2) 1 y1+ y2 1 (y1 y2 0)

4.

(P1 P2)

(P1 P2) (P2 P1)

(P1 P2) (P2 P1)

y1 y2 0

y2 y1 0or y1= y2

5. (P1 P2 P3)

(P1 P3) (P2 P3)

y1 y3 0

y2 y3 0

6. P1 P2 P3 (exactly one) y1+ y2+ y3= 1

Chen CL 5

Blending Products Including Discrete Batch Sizes

The Batch Plant

Unit 1 has a max capacity of8, 000lb/day, and unit 2 of10, 000lb/day

To make1.0 lb of product 1 requires 0.4 lb ofA and 0.6 lb ofBTo make1.0 lb of product 2 requires 0.3 lb ofB and 0.7 lb ofC

A maximum of6, 000 lb/day ofB is available (no limit on A and C)

Net revenue of product 1 is $0.16/lb, and of product2 is $0.20/lb

Each product must be made in batches of2, 000lb

Q: How much of products 1 and 2 should be produced per day ?

Chen CL 6


MILP Formulation

x1, x2 thousands of pounds per dayfor products1and2

max f= 160x1+ 200x2

s.t. 0.6x1+ 0.3x2 6

xi= 2yi

0 2000y1 8000

0 2000y2 10000

yi: integers (0, 1, 2, . . .)

Chen CL 7


Solution

(y1, y2) = (2, 5) f = 2.8 103


3/18

Chen CL 8

Design of A Chemical Complex

Proposed Chemical complex

Objective: to maximize profit

Should chemical C be produced, and if so how much ?

Which process to build (II and III are exclusive) ?

How to obtain chemical B ?

Chen CL 9

Design of A Chemical ComplexData

Cost fixed ($) Variable ($/ton raw material)

Process I 1000 250

Process II 1500 400

Process III 2000 550

Price A 500/ton

B 950/ton

C 1800/ton

Conversions Process I 90% of A to B

Process II 82% of B to C

Process III 95% of B to C

Constraints Maximum supply of A: 16 tons/hr

Maximum demand of C: 10 tons/hr

Chen CL 10


Notation

YI = 1 if process I is selected and 0 if not

YII = 1 if process II is selected and 0 if not

YIII = 1 if process III is selected and 0 if not

P A purchases of A (tons/hr)

P B purchases of B (tons/hr)

SC sales of C (tons/hr)

BI production rate of B in process I (tons/hr)

BII consumption rate of B in process II (tons/hr)

BIII consumption rate of B in process III (tons/hr)

CII production rate of C in process II (tons/hr)

CIII production rate of C in process III (tons/hr)

Chen CL 11

Design of A Chemical ComplexFormulation

MB for B, C: BI+ P B = BI I+ BIII

CI I+ CIII = SC

Balance around I,II,III: BI = 0.9P A

CI I = 0.82BI I

CIII = 0.95BIII

Process can operate if built P A 16Y I

BI I (10/0.82)Y II

BIII (10/0.95)Y I I I

Demand limitation: SC 10

II and III are exclusive: Y II+ Y I I I 1

Profit to be maximized: J =

investment and operating costs

(1000Y I+ 250PA + 1500Y II+400BII+ 2000YIII + 550BIII)

(500PA + 950PB) raw materials

+1800SC sales

Model Size: 9 constraints; 8 continuous variables, 3binary variables


4/18

Chen CL 12


GAMS Code

$TITLE Design of a Chemical Complex

$offsymxref offsymlist

*filename: COMPLEX.gms

option optcr=0, limrow=0, limcol=0;

BINARY VARIABLES

YI denotes selection of process I when equal to one

YII denotes selection of process II when equal to one

YIII denotes selection of process III when equal to one ;

POSITIVE VARIABLES

PA purchases of A (tons per hr)

PB purchases of B (tons per hr)

SC sales of C (tons per hr)

BI production rate of B in process I (tons per hr)

BII consumption rate of B in process II (tons per hr)

BIII consumption rate of B in process III (tons per hr)CII production rate of C in process II (tons per hr)

CIII production rate of C in process II (tons per hr) ;

VARIABLE PROFIT objective function ;

EQUATIONS

E1 select at most one of process II and III

E2 mass balance for B

Chen CL 13

E3 mass balance for C

E4 mass balance around process I

E5 mass balance around process II

E6 mass balance around process III

E7 no purchases of A unless process I is selected

E8 no production of BII unless process II is selected

E9 no production of BIII unless process III is selected

OBJ objective function definition ;

E1 .. YII + YIII =L= 1 ;

E2 .. BI + PB =E= BII + BIII ;E3 .. CII + CIII =E= SC ;

E4 .. BI =E= 0.9 * PA ;

E5 .. CII =E= 0.82 * BII ;

E6 .. CIII =E= 0.95 * BIII ;

E7 .. PA =L= 16 * YI ;

E8 .. BII =L= (10/0.82) * YII ;

E9 .. BIII =L= (10/0.95) * YIII ;

* constraint (10) for max demand of C is declared as an upper bound here

SC.UP = 10 ;

O BJ .. PR OF IT =E = - ( 1 000 * YI + 2 50 * PA+ 15 00 * YI I + 4 00 * BII

+ 2 000 * Y II I + 5 50 * B II I )

- 500 * PA - 950 * PB

+ 1800 * SC ;

MODEL COMPLEX /ALL/ ;

SOLVE COMPLEX USING MIP MAXIMIZING PROFIT ;

Chen CL 14


Results

J = 459.3 $/hr

P A = 13.55 tons/hr

select process I and II

Chen CL 15

Planning of A Chemical Complex

Problem Description


5/18

Chen CL 16

Decide:

Which process to build (2 and 3 are not excluded) ? How to obtain chemical 2 ? How much should be produced of product 3 ?

Objective: to maximize net present value over a long range horizon consisting

of3 time periods of2, 3 and 5 years length

Assume:

Prices, demands, and cost coefficients remain constant during each timeperiod

Capacity expansions of the processes are allowed at the beginning of eachtime period

No limit of expansion frequencies, a limit on amount of money that can beinvested at each time period for expansions (20, 30, and 40 m$ for periods

1, 2, 3) The size of expansion may not exceed 100 kton/yr

Chen CL 17


Data

Fixed investment coefficients (105$/ton)

Period1 Period 2 Period 3

Process1 85 95 112

Process2 73 82 102

Process3 110 125 148

Variable investment coefficients (102$/ton)


Process1 1.38 1.56 1.78

Process2 2.72 3.22 4.60

Process3 1.76 2.34 2.84

Operating Expenses coefficients (102$/ton of main product)


Process1 0.4 0.5 0.6

Process2 0.6 0.7 0.8

Process3 0.5 0.6 0.7

Chen CL 18

Prices (102$/ton)


Chemical 1 4.0 5.24 7.32

Chemical 2 9.6 11.52 13.52

Chemical 3 26.2 29.2 35.2

Upper bounds for Purchase/Sales (kton/yr)


Chemical 1 6 7.5 8.6

Chemical 2 20 25.5 30

Chemical 3 65 75 90

Mass balance coefficients (to main product)

Chemical 1 Chemical 2 Chemical3

Process 1 1.11 1

Process 2 1.22 1

Process 3 1.05 1

Miscellaneous data

Existing capacity

kton/yr

Salvage value

factor

Working capital

factor

Process 1 0.1 0.15

Process 2 50 0.1 0.15

Process 3 0.1 0.15

Interest rate = 10%, Tax rate = 45%, Depreciation method: straight line

Chen CL 19


Formulation

Index notation

i process i= 1, NP

j chemical j = 1, NC

market = 1, NM

t time period t= 1, NT


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Chen CL 20

Plant capacities

Qit: total available capacity of process i in period t, t = 1, NT

Qi0: existing capacity at time t = 0

QEit: capacity expansion of process i installed in period t

QELit, QEUit : lower/upper bounds for capacity expansions

yit= {0, 1}: occurrence of expansions

yitQELit QEit QE

Uityit

yit= 0, 1 i= 1, NP t= 1, NT

Qit= Qi,t1+ QEit i= 1, NP t= 1, NT

Chen CL 21

Amount of chemicals being consumed and produced

Amounts of chemicals being consumed and produced

Li: index set of streams corresponding to inputs/outputs of process i

Wijt 0 i= 1, NP, j Li t= 1, NT

Stream mi Li: main product produced by process i, capacity

limitation

Wimit Qit i= 1, N P t= 1, NT

Note: Wimit= 0 if shut-down

Material balances in each plant (use constant characteristics of processi and chemical j )

Wijt = ijWimit i= 1, NP, j Li t= 1, N T

Chen CL 22

Purchases and sales ofN Cnodes of chemicals

(aL,Ujt , dL,Ujt : loewr/upper bounds for availabilities and demands)

aLjt Pjt aUjt

dLjt Sjt dUjt

j = 1, N C, = 1, N M, t= 1, NT

Mass balances on chemicals nodeI(j): index set of plants which consume chemical j

O(j): index set of plants which produce chemicalj

NM=1

Pjt+

kO(j)

Wkt =NM=1

Sjt +

kI(j)

Wkt j = 1, NC , t= 1, NT

Chen CL 23

Net present value of the project

it, it variable and fixed terms for the investment costs

it unit operating cost

jt, jt prices of sales and purchases of chemicalj in market

NP V = NTt=1

NPi=1

(itQEit+ ityit) NTt=1

NPi=1

itWimit

+NTt=1

NCj=1

NM=1

(jtSjt jtPjt)


7/18

Chen CL 28 Chen CL 29


8/18

Chen CL 28

TABLE ALPHA(I,T) variable investement coefficient

* (1e2 $/ton)

1 2 3

1 1.38 1.56 1.78

2 2.72 3.22 4.6

3 1.76 2.34 2.84

TABLE BETA(I,T) fixed investement coefficient

* (1e5 $)

1 2 3

1 85 95 1122 73 82 102

3 110 125 148

TABLE DELTA(I,T) operating expenses per unit of main product

* (1e2 $ /ton)

1 2 3

1 0.4 0.5 0.6

2 0.6 0.7 0.8

3 0.5 0.6 0.7

TABLE LAM(J,K,T) cost for purchase of one unit of chemical

* (1e2 $/ton)

* values should be left to zero if chemical J is not available

* f ro m ma rk et K d ur in g pe ri od T

1 2 3

1.1 4. 5.24 7.32

2.1 9.6 11.52 13.52

3.1

Chen CL 29

T AB LE PL B( J, K, T) pu rc ha se l ow er b ou nd s

* (kton/yr)

* both lower and upper bounds should be left to zero

* if chemical J is unavailable from market K during period T

1 2 3

1.1

2.1

3.1

TABLE PUB(J,K,T) purcase upper bounds

* (kton/yr)* both upper and lower bound should be left to zero

* if chemical J is unavailable from market K during period T

1 2 3

1.1 6. 7.5 8.6

2.1 20 25.5 30.

3.1

T AB LE GA MM A( J, K, T) sa le p ri ce s

* (1e2 $/ton)

* values should be left to zero if chemical J is not

* desired by market K during period T

1 2 3

1.1

2.1

3.1 26.20 29.20 35.20

TABLE SLB(J,K,T) sales lower bounds

* (kton/yr)

* both lower and upper bounds should be left to zero

Chen CL 30

* if chemical J is not desired by market K during period T

1 2 3

1.1

2.1

3.1

TABLE SUB(J,K,T) sales upper bounds

* (kton/yr)

* both upper and lower bounds should be left to zero

* if chemical J is not desired by market K during period T

1 2 3

1.1

2.13.1 65 75 90

*

* parameter evaluations

*

PARAMETERS

EP epsilon

ALPHAR(I,T) fixed investement coefficient depreciated etc

BETAR(I,T) variable investement coefficient depreciated etc ;

EP = 0.0001;

SCALAR FX project lifetime;

SCALARS Z1,Z2,INTER1;

PARAMETERS X(T),C1(I,T),C2(T),T1(T),T2(I,T),T3(I,T),T4(I),SCOEFF(T);

X(1)=0 ; LOOP (T $ (ORD(T) NE CARD(T)) , X(T+1) = X(T)+LENP(T) ) ;

FX=SUM(T,LENP(T));

Chen CL 31

INTER1=1/(1+INTR);

LOOP(T,

Z1=X(T)+1 ;

C2(T)=SUM (TT $((ORD(TT) GE Z1) AND (ORD(TT) LE FX)),INTER1**ORD(TT))

);

T1(T)=-INTER1**X(T);

T2(I,T)=WCAPF(I)*(T1(T)+INTER1**FX);

T3(I,T)=(1-SVALF(I))*TAX*C2(T)/(FX-X(T));

T4(I)=SVALF(I)*INTER1**FX;

C1(I,T)=T1(T)+T2(I,T)+T3(I,T)+T4(I);

ALPHAR(I,T)=-ALPHA(I,T)*C1(I,T);

BETAR(I,T)=-BETA(I,T)*C1(I,T);

LOOP(T $ (ORD(T) NE CARD(T)) , Z1=X(T)+1 ; Z2=X(T+1) ; SCOEFF(T)=SUM(TT

$ ((ORD(TT) GE Z1) AND (ORD(TT) LE Z2)) , INTER1**ORD(TT))*(1-TAX)

);

LOOP(T $ (ORD(T) EQ CARD(T)), Z1=X(T)+1 ; Z2=FX ; SCOEFF(T)=SUM(TT $

((ORD(TT) GE Z1) AND (ORD(TT) LE Z2)), INTER1**ORD(TT) )*(1-TAX)

) ;

LAM(J,K,T)=SCOEFF(T)*LAM(J,K,T);

GAMMA(J,K,T)=SCOEFF(T)*GAMMA(J,K,T);

DELTA(I,T)=SCOEFF(T)*DELTA(I,T);

*

* model definition

VARIABLES Q(I,T) capacities

QE(I,T) capacity expansions

Y(I,T) integer decision variables



9/18

Chen CL 32

W(I,*,T) flow rates

S(J,K,T) sales

P(J,K,T) purchases

NPV net present value;

POSITIVE VARIABLES Q,QE,W,S,P;

BINARY VARIABLES Y;

EQUATIONS CELBD(I,T) capacity expansion lower bounds

CEUBD(I,T) capacity expansion upper bounds

CAPBL(I,T) capacity balances PRCP(I,T) production cannot exceed capacity

PBAL(I,J,T) balances around processes

* constraints will be given as lower and upper bounds

BAL(J,T) balances for chemicals

OBJ net present value definition

NEXP(I) limitation of number of expansions

INVBD(T) bound on investment at each time period ;

SET JM(I,*) ; ALIAS (J,JP) ; JM(I,J) = YES $ (ORD(J) EQ JMM(I)) ;

CELBD(I,T) $ QELB(I,T) .. Y(I,T)*QELB(I,T) =L= QE(I,T);

CEUBD(I,T) $ (QEUB(I,T) NE INF) .. QE(I,T) =L= QEUB(I,T)*Y(I,T);

CAPBL(I,T) .. Q(I,T) =E= EXCAP(I) $ (ORD(T) EQ 1) + Q(I,T-1) + QE(I,T);

PRCP(I,T) .. SUM(J $ JM(I,J), W(I,J,T)) =L= Q(I,T);

* equations are given as lower and upper bounds on the variables

Chen CL 33

P.LO(J,K,T) $ PLB(J,K,T) = PLB(J,K,T);

P.UP(J,K,T) $ (PUB(J,K,T) AND (PUB(J,K,T) NE INF)) = PUB(J,K,T);

S.LO(J,K,T) $ SLB(J,K,T) = SLB(J,K,T);

S.UP(J,K,T) $ (SUB(J,K,T) AND (SUB(J,K,T) NE INF)) = SUB(J,K,T);

PBAL(I,J,T) $ MI(I,J)

.. W(I,J,T) =E= ABS(MI(I,J))*SUM(JP $ JM(I,JP), W(I,JP,T));

BAL(J,T) .. SUM( K, P(J,K,T) $ LAM(J,K,T) - S(J,K,T) $ GAMMA(J,K,T) )

=E= SUM ( I, SIGN(MI(I,J)) * W(I,J,T) );

OBJ .. NPV =E= - SUM ( (I,T), ALPHAR(I,T)*QE(I,T) + BETAR(I,T)*Y(I,T)

+ DELTA(I,T)*SUM(J $ JM(I,J), W(I,J,T) ) )

+ SUM ( (J,K,T), GAMMA(J,K,T)*S(J,K,T) - LAM(J,K,T)*P(J,K,T) );

NEXP(I) $ (LNEXP(I) LT NPER) .. SUM(T, Y(I,T)) =L= LNEXP(I);

INVBD(T) $(LINVEST(T) NE INF) .. SUM ( I, ALPHA(I,T)*QE(I,T) +

BETA(I,T)*Y(I,T) ) =L= LINVEST(T);

* obtain solution using branch and bound

*

MODEL MULT/ALL/;

SOLVE MULT MAXIMIZING NPV USING MIP;

DISPLAY NPV.L, Y.L, QE.L, Q.L, W.L, S.L, P.L ;

Chen CL 34


Result

Period1 Period 2 Period3

Process 1 Capacity 7.75 7.75 7.75

Production 5.41 6.76 7.75

Process 2 Capacity 50.0 50.0 50.0Production 20.82 0 0

Process 3 Capacity 0 35.95 35.95

Production 0 30.72 35.95

Chen CL 35

Multi-product Batch Plant SchedulingBatch Scheduling and Planning Problems

A principal feature of batch plants is the production of multiple products using

the same set of equipment

The total time required to produce a set of batches (makespan or cycle time)depends on the sequence of production to minimize the makespan



10/18

Multi-product Batch Plant Scheduling

A 4-products 3-units Plant

Units

Products u1 u2 u3

p1 (i=1) 3.5 4.3 8.7

p2 (i=2) 4.0 5.5 3.5

p3 (i=3) 3.5 7.5 6.0

p4 (i=4) 12. 3.5 8.0


MILP Formulation

N, M numbers of products and units

Cj,k the time forj th slot of product leavingk unitti,k processing time forith product (pi) in kth unit (t3,2= 7.5h)

j,k processing time forj th slot of product in kth unit

jth slot of product cannot leave unit uk until it is processed, and it must haveleft unit u(k-1)

Cj,k Cj,k1+ j,k j = 1, N , k= 2, M (1)

jth slot of product cannot leave unit uk, until (j 1)th slot of product has leftand the former has been processed (C0,k= 0)

Cj,k Cj1,k+ j,k j = 1, N , k= 1, M (2)

Chen CL 38

jth slot of product cannot leave unit uk, until the downstream unit u(k+1) isfree ((j 1)th slot of product has left)

Cj,k Cj1,k+1 j = 1, N , k= 1, M 1 (3)

Note: Eq.s (3), Cj,k1 Cj1,k Eq.s (1) and (3) imply Eq.s (2) for k = 2, M

Only one product in every slot j of the sequenceXi,j= 1 ifith product (product pi) is in slot j of the sequence(If product pi is in slot j , then Xi,j = 1and Xi,1 = . . . = Xi,j1= Xi,j+1 = . . . = Xi,N = 0)

X1,j+ X2,j+ . . . + XN,j = 1 j = 1, N (4)

Every product should occupy only one slot in the sequence

Xi,1+ Xi,2+ . . . + Xi,N= 1 i= 1, N (5)

For any given set ofXi,j (production sequence)ifith product is in slotj (Xi,j= 1), then the processing time ofjth-slot product

Chen CL 39

in unitk isti,k

j,k= X1,jt1,k+ X2,jt2,k+ . . . + XN,jtN,k j = 1, N , k= 1, M

MILP formulation:

min CN,M

s.t. Eq.s 3,4,5

Cj,k Cj,k1+Ni=1

Xi,jti,k j = 1, N k = 2, M

Cj,k Ci1,k+Ni=1

Xi,jti,k j = 1, N

Cj,k 0 and Xi,j binary



11/18


GAMS Code

$TITLE Multiproduct Batch Plant Scheduling

$OFFUPPER

$OFFSYMXREF OFFSYMLIST

OPTION SOLPRINT = OFF;

* Define product and unit index sets

SETS PI Product batches to be produced /p1*p4/

UK Four batch processing units in the plant /u1*u3/

J Slots for products in the sequence /1*4/;

ALIAS (I, J);

* Define and initialize problem data

TABLE T(PI,UK) Processing times of products on unit UK in h

u1 u2 u3

p1 3.5 4.3 8.7

p2 4.0 5.5 3.5

p3 3.5 7.5 6.0p4 12.0 3.5 8.0

PARAMETER TMIN(UK) Minimum of the processing times of products on UK;

TMIN(UK) = SMIN(PI, T(PI,UK));

PARAMETER TP(PI,UK) Processing times of products above TMIN on UK;

TP(PI,UK) = T(PI,UK) - TMIN(UK);

SCALAR N Number of products to be produced

M Number of units in the plant;

N = CARD(PI);

M = CARD(UK);

* Define optimization variables

VARIABLES X(PI,J) Product PI is in sequence slot J

C(I,UK) Completion time of the product in sequence slot I on unit UK

MSPAN Makespan or total time to produce all products;

POSITIVE VARIABLES C;

BINARY VARIABLES X;

* Define constraints and objective functionEQUATIONS OBJFUN Minimize makespan

ONEPRODUCT(J) Only one product should be in each slot

ONESLOT(PI) Only one slot should be assigned to each product

CEQ1(I,UK) Completion time recurrence Eqs. 8

CEQ2(I,UK) Completion time recurrence Eqs. 7

CEQ3(I,UK) Completion Time recurrence Eqs. 3;

OBJFUN.. MSPAN =E= SUM((I,UK) $(ORD(I) EQ N AND ORD(UK) EQ M), C(I,UK));

ONEPRODUCT(J).. SUM(PI, X(PI,J)) =E= 1;

ONESLOT(PI).. SUM(J, X(PI,J)) =E= 1;

CEQ1(I,"u1").. C(I,"u1") =G= C(I-1,"u1") $(ORD(I) GT 1) +

TMIN("u1") + SUM(PI, TP(PI,"u1")*X(PI,I));

CEQ2(I,UK) $(ORD(UK) GT 1)..

C(I,UK) =G= C(I,UK-1) + TMIN(UK) + SUM(PI, TP(PI,UK)*X(PI,I));

CEQ3(I,UK) $(ORD(I) GT 1 AND ORD(UK) LT M).. C(I,UK) =G= C(I-1,UK+1);

* Define model and solve

MODEL SCHEDULE /ALL/;

SOLVE SCHEDULE USING MIP MINIMIZING MSPAN;

DISPLAY X.L, C.L, MSPAN.L;

Chen CL 42

Multi-product Batch Plant SchedulingThe Optimal Schedule

X1,1= X2,4= X3,2= X4,3= 1, others = 0 product p1 is in slot j = 1, product p2 is in slot j = 4, to produce products in sequence: p1 - p3 - p4 - p2

Unit1 is processing during [0, 3.5] h. When p1 leaving u1 at t = 3.5 h, u1 starts processing p3 ([3.5, 7]) U1 holds p3 during [7, 7.8] because u2 is still processing p1

The finished batches of p1, p3, p2, and p4 are completed at times16.5 h, 23.3 h, 31.3 h, and 34.8 h (minimum makespan is 34.8 h)

Chen CL 43

Mathematical Description for MILP

min cTx + dTy

s.t. Ax + By b

x 0, x X Rn

y {0, 1}q

x a vector ofn continuous variables

y a vector ofq 0 1 variables

c,d (n 1), (q 1)vectors of parameters

A,B matrices of appropriate dimensions

b a vector ofp inequalities



12/18

Mathematical Description: Remark 1

A linear objective function and linear constraints in x and y

A mixed set of variables (continuous x and 0 1 y)

d= 0,B= 0 linear programming LP problem

c= 0,A= 0 integer linear programming ILPproblem

Major difficulty: combinatorial nature of domain ofy variables

Any choice of0 or 1 for elements ofy results in a LPproblem on x variables

Enumerating fully all possible combinations of0 1 variables for elements ofy 2q possible combinations !

Outline of MILPAlgorithms

Branch and Bound Methods

A binary tree to represent 0 1 combinations to partition feasible region into sub-domains systematically

Valid upper and lower bounds are generated at different levels ofthe binary tree

Chen CL 46

Outline of MILPAlgorithms

Cutting Plane Methods

New constraints (cuts) are generated and added which reduce

the feasible region until a 0 1 optimal solution is obtained

Decomposition Methods

Mathematical structure of the models is exploited via variable

partitioning, duality, and relaxation methods

Logic-Based Methods

Disjunctive constraints or symbolic inference techniques are

utilized which can be expressed in terms of binary variables

Chen CL 47


Basic Notion 1: Separation

Separation:Let an MILP model be denoted as (P) and let itsset of feasible solutions be denoted as F S(P). A set

of subproblems (P1), . . . , (Pn) of (P) is defined as aseparationof(P) if the following conditions hold:

A feasible solution of any of the subproblems(P1), . . . , (Pn) is a

feasible solution of(P); and

Every feasible solution of(P)is a feasible solution ofexactly one

of the subproblems



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Remark 1:

The above conditions imply that the FSs of the

subproblemsF S(P1), . . . , F S (Pn)are a partition of theFSs of(P), that is F S(P)

original problem(P)is called a parent node problemsubproblems are called children node problems



Remark 2: how to separate (P) into (P1), . . . , (Pn) ?

Most frequently used way:Considering contradictory constraints on a single binary variable

min 3y1 2y2 3y3+ 2x1

s.t. y1+ y2+ y3+ x1 2

5y1+ 3y2+ 4y3+ 10x1 10x1 0 y1, y2, y3= 0, 1

=

min 3y1 2y2 3y3+ 2x1

s.t. y1+ y2+ y3+ x1 2

5y1+ 3y2+ 4y3+ 10x1 10

y1= 0

x1 0 0 y2, y3 1

min 3y1 2y2 3y3+ 2x1

s.t. y1+ y2+ y3+ x1 2

5y1+ 3y2+ 4y3+ 10x1 10

y1= 1

x1 0 0 y2, y3 1

Chen CL 50


Basic Notion 2: Relaxation

Relaxation:

An optimization problem, denoted as (RP), is defined

as a relaxation of problem (P) if the set of feasible

solutions of(P)is a subset of the set of feasible solutionsof(RP)

F S(P) F S(RP)

Chen CL 51


Basic Notion 2: Relaxation

Remark 1:The above definition implies the following relationshipsbetween problem (P) and the relaxed problem (RP)

If(RP)has no feasible solution, then(P)has no feasible solution; Let the optimal solution of(P) be zPand the optimal solution

of(RP) be zRP. Then (for minimization problem)

zRP zP

That is, the solution of relaxed problem (RP) provides a lower

boundon the solution of problem (P) Ifan optimal solution of(RP) is feasible for problem (P),

thenit is an optimal solution of(P)


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Chen CL 56

G l B h d B d F kChen CL 57

G l B h d B d F k


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General Branch and Bound Framework

Step 4: Relaxation

Select a relaxation (RCS) of the current candidate subproblem

(CS), solve it and denote its solution by zRCS

Step 5: Fathoming apply the three fathoming criteria FC1: If (RCS) is infeasible, then the current (CS) has no

feasible solution. Go to step 2

FC2: IfzRCS z, then current (CS) has no feasible solution

better than the incumbent. Go to step 2

FC3: If optimal solution of (RCS) is feasible for (CS) (eg,

integral), then it is an optimal solution of(CS).

If zRCS z, then record this solution as the new incumbent

(z =zRCS). Go to step 2

General Branch and Bound Framework

Step 6: Separation

Separate the current candidate subproblem (CS) and add its

children nodes to the list of candidate subproblems. Go to step

2

Remark 1:

In step 1, one can use prior knowledge on an upper

bound and initialize the incumbent to such an upper

bound

Remark 2: Node Selection

Depth-first search (Last-In-First-Out)

Breadth-first search

Chen CL 58

Branch and Bound Based on LinearProgramming Relaxation

Chen CL 59

MILP

min cTx + dTy

s.t. Ax + By b

x 0, x X Rn

y {0, 1}3

(RCS)01=

z(RCS)01

min cTx+ dTy

s.t. Ax+ By b

x 0, x X Rn

0 y1, y2, y3 1

(RCS)11

=z(RCS)11

min cTx + dTy

s.t. Ax+ By b

x 0, x X Rn

y1= 0; 0 y2, y3 1

(RCS)12

=z(RCS)12

min cTx+ dTy

s.t. Ax+ By b

x 0, x X Rn

y1= 1; 0 y2, y3 1

(RCS)01 z(RCS)01: (LP relaxation of the MILP )

a lower bound on optimal solution of the MILP

(optimal and terminate if all y sare 0 1)

(RCS)11 z(RCS)11: (LP relaxation of the (CS))an upper bound on the solution of(RCS)11

Chen CL 60

B h d B d B d LiChen CL 61

BB M th d B d LP R l ti


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Branch and Bound Based on Linear

Programming Relaxation: Illustration

min 2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1, y2, y3= 0, 1

relax of=root node

min 2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

0 y1, y2, y3 1

x1 = 0

(y1, y2, y3) = (0.6, 1, 1)

z = 6.8

BB Method Based on LPRelaxation:

Illustration The MILP problem:

minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1, y2, y3 {0, 1}

Relaxed problem:minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10x1 0

y1, y2, y3 [0, 1]

z = 6.8, x1= 0, y= [0.6, 1, 1]6.8< z = +=separate problem into subproblems 1,2

Chen CL 62

BB Method Based on LPRelaxation:Illustration

Consider breadth first search, at level 1:

Relaxed problem of subproblem 1:

minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1 = 0

y2, y3 [0, 1]

z= 5, x1= 0, y= [0, 1, 1]

optimal solution of (RP) is feasible for problem (P), and5< z = + z = 5

Chen CL 63


Consider breadth firstsearch, at level 1:


minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1 = 1

y2, y3 [0, 1]

z = 6.667, x1= 0, y= [1, 0.333, 1]

6.667< z

= 5 = separate subproblem 2 into subproblems 3,4

Chen CL 64

BB Method Based on LP RelaxationChen CL 65

BB Method Based on LP Relaxation


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Illustration At level 2:


minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1 = 1

y2 = 0

y3 [0, 1]

z= 6, x1= 0, y= [1, 0, 1]optimal solution of (RP) is feasible for problem (P), and6< z = 5 z = 6


Illustration At level 2:


minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1 = 1

y2 = 1

y3 [0, 1]

z = 6.5, x1= 0, y= [1, 1, 0.5]6.5< z = 6 =separate subproblem 4 into subproblems 5,6

Chen CL 66


At level 3:

Subproblem 5minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1 = 1

y2 = 1

y3 = 0

z= 5, x1= 0, y= [1, 1, 0]optimal solution of (RP) is feasible for problem (P), and

5> z = 6 Stop

Chen CL 67


At level 3:

Subproblem 6:minx,y

2x1 3y1 2y2 3y3

s.t. x1+ y1+ y2+ y3 2

10x1+ 5y1+ 3y2+ 4y3 10

x1 0

y1 = 1

y2 = 1

y3 = 1

Infeasible=Stop

Chen CL 68

Branch and Bound Based on LinearChen CL 69

Branch and Bound Based on Linear


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Branch and Bound Based on LinearProgramming Relaxation: Illustration

Binary Tree for Breadth First Search

Branch and Bound Based on LinearProgramming Relaxation: IllustrationBinary Tree for Depth First Search with Backtracking

Chen CL 70

Thank You for Your Attention

Questions Are Welcome

6 Mixed-Integer Linear Programming.pdf

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