4 Linear Programming.pdf

8/11/2019 4 Linear Programming.pdf

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Linear Programming

Cheng-Liang ChenPSELABORATORY

Department of Chemical EngineeringNational TAIWAN University


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Chen CL 1

Refinery SchedulingThe Refinery Process


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Chen CL 2

Refinery SchedulingProblem Data

Product Yields bbl/bbl crude Product Maximum

Fuel Process Lube Value Demand

Products/Crudes 1 2 3 4 4 $/bbl kbbl/wk

Gasoline 0.6 0.5 0.3 0.4 0.4 45.00 170

Heating Oil 0.2 0.2 0.3 0.3 0.1 30.00 85

Jet Fuel 0.1 0.2 0.3 0.2 0.2 15.00 85

Lube Oil 0.0 0.0 0.0 0.0 0.2 60.00 20

Operating Losses 0.1 0.1 0.1 0.1 0.1

Crude Cost $/bbl 15. 15. 15. 25. 25.Operating Cost $/bbl 5.0 8.5 7.5 3.0 2.5

Crude Supply kbbl/wk 100 100 100 200


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Chen CL 3

Refinery SchedulingProblem Analysis

For 1 kbbl/wk of crude 1, one can produce 0.6 kbbl/wk of

gasoline,

0.2 kbbl/wk of heating oil,

0.1 kbbl/wk of jet fuel,

with production cost of20 k$ per kbbl/wk.Values of products produced:

45(0.6) + 30(0.2) + 15(0.1) = 34.5 k$

Profit for 1 kbbl/wk of crude 1: 34.5 (15 + 5) = 14.5 k$

Similarly: profits of8.0, 4.5, 2.0, 8.5 k$ for crude options 2, 3, 4, 5

The usage preference for crude options are: 1>5>2>3> 4


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Chen CL 4


The only constraints that we should satisfy are supply constraints

and maximum demand constraints

Use most profitable crude (crude 1) as much as we can: Using 100kbbl/wk of crude 1 produces

60 kbbl/wk of gasoline,

20 kbbl/wk of heating oil,

10 kbbl/wk of jet fuel,

0 kbbl/wk of lube oil.

(not exceed maximum demands)

Use nextmost profitable crude (crude 4 in lube process)

C C


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Chen CL 5


Use nextmost profitable crude (crude 4 in lube process):

Using 100kbbl/wk of crude 4 produces (accumulated value)

60 + 40 = 100kbbl/wk of gasoline,20 + 10 = 30 kbbl/wk of heating oil,

10 + 20 = 30 kbbl/wk of jet fuel,

0 + 20 = 20 (max demand) kbbl/wk of lube oil.

(not exceed maximum demands except lube oil)

Use3rd most profitable crude (crude 2) as much as we can

Ch CL 6


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Chen CL 6


Use3rd most profitable crude (crude 2) as much as we can:

Using 100kbbl/wk of crude 2 produces

60 + 40 + 50 = 150 kbbl/wk of gasoline,20 + 10 + 20 = 50kbbl/wk of heating oil,

10 + 20 + 20 = 50 kbbl/wk of jet fuel,

0 + 20 + 0 = 20(max demand) kbbl/wk of lube oil.

(not exceed maximum demands except lube oil)

Use4th most profitable crude (crude 3) as much as we can

Ch CL 7


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Chen CL 7


Use4th most profitable crude (crude 3) as much as we can:

Using 66.67kbbl/wk of crude 3 produces

60 + 40 + 50 + 20 = 170 (max demand) kbbl/wk of gasoline,20 + 10 + 20 + 20 = 70 kbbl/wk of heating oil,

10 + 20 + 20 + 20 = 70 kbbl/wk of jet fuel,

0 + 20 + 0 + 0 = 20(max demand) kbbl/wk of lube oil.

(reaching maximum demand for gasoline and lube oil)

Ch CL 8


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Chen CL 8


We cannot use crude 4 in fuel process, because it also produces

gasoline (gasoline has reached maximum demand)

Summary:

Using (100, 100, 66.67, 0, 100) kbbl/wk of crudes 1, 2, 3, 4,5

Ch CL 9


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Chen CL 9

Refinery SchedulingLinear Programming Formulation

Max J =

product sales 4

p=1 vpQp production cost 5

r=1 Crxrs.t. xr Sr r= 1, 2, 3

x4+ x5 S4

Qp = ap1x1+ ap2x2+ ap3x3+ ap4x4+ ap5x5Qp Dp p= 1, . . . , 4

Qp 0 p= 1, . . . , 4

xr 0 r= 1, . . . , 5

Chen CL 10


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Chen CL 10

Refinery SchedulingGAMS Formulation

set C crudesset P products

data Sc CSUPPLY(C)

Dp PDEMAND(P)

vp PVALUE(P)

Cc TCOST(C)

apc YIELDs(P,C)

TCOST(C) sum of CCOST(C) and OCOST(C)(costs of crude c and operating cost)

PROFIT objective functionPRODUCTION(P) Qp equations

X4 total crude 4 usage

X4.UP = CSUPPLY(4-fuel), upper bound

X.UP(C) supply constraints onxc

Q.UP = PDEMAND(P)

Chen CL 11


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Chen CL 11

Refinery SchedulingGAMS Code

$TITLE Refinery Scheduling$OFFUPPER

$OFFSYMXREF OFFSYMLIST

OPTION SOLPRINT = OFF;

* Define index sets

SETS C Crudes /1*3, 4-fuel, 4-lube/

P Products /gasoline, heat-oil, jet-fuel, lube-oil/

*Define and initialize the problem data

PARAMETER CSUPPLY(C) Crude oil availabilities in kbbl per wk /1 100.0, 2 100.0, 3 100.0, 4-fuel 200.0, 4-lube 200.0/

CCOST(C) Crude oil costs in $ per bbl /

1 15.0, 2 15.0, 3 15.0, 4-fuel 25.0, 4-lube 25.0/

PDEMAND(P) Maximum product demands in kbbl per wk /

gasoline 170.0, heat-oil 85.0, jet-fuel 85.0, lube-oil

PVALUE(P) Product Values in $ per bbl /

Chen CL 12


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Chen CL 12

gasoline 45.0, heat-oil 30.0, jet-fuel 15.0, lube-oil 6

OCOST(C) Crude operating costs in $ per bbl /

1 5.0, 2 8.5, 3 7.5, 4-fuel 3.0, 4-lube 2.50/

TCOST(C) Total costs: crude plus operating;

TCOST(C) = CCOST(C) + OCOST(C);TABLE YIELDS(P, C) Yields of products for crudes

1 2 3 4-fuel 4-lube

gasoline 0.6 0.5 0.3 0.4 0.4

heat-oil 0.2 0.2 0.3 0.3 0.1

jet-fuel 0.1 0.2 0.3 0.2 0.2

lube-oil 0.0 0.0 0.0 0.0 0.2;

* Define the optimization variables

VARIABLES X(C) Crude oils used in kbbl per week

Q(P) Amounts of products produced in kbbl

X4 Total amount of crude 4 used in kbbl

PROFIT Total profit from product sales in k$;

POSITIVE VARIABLES X, X4, Q;

* Define constraints and objective function

EQUATIONS OBJFUN Objective function to be maximized

CRUDE4 Total crude 4 usage

PRODUCTION(P) Amounts of products produced;

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Chen CL 13

OBJFUN.. PROFIT =E= SUM(P, Q(P)*PVALUE(P)) - SUM(C, TCOST(C)*X(C));

PRODUCTION(P).. Q(P) =E= SUM(C, YIELDS(P,C)*X(C));

CRUDE4.. X4 =E= X("4-fuel") + X("4-lube");

* Define upper and lower bounds

* Upper bounds on amounts of product produced from their maximum demaQ.UP(P) = PDEMAND(P);

* Upper bounds on crude oil usages from their supplies

X.UP(C) = CSUPPLY(C);

X4.UP = CSUPPLY("4-fuel");

* Define model and solveMODEL SCHEDULE /ALL/;

SOLVE SCHEDULE USING LP MAXIMIZING PROFIT;

DISPLAY X.L, Q.L, PROFIT.L;

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Chen CL 14

Refinery SchedulingGAMS Results

x = [100, 100, 66.667, 0, 100]T (kbbl/wk, crude demands)

Q = [170, 70, 70, 20]T (kbbl/wk, gas., heating oil, jet fuel, lube oil)

profit = 3400 k$/wk

Chen CL 15


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Chen CL 15

Refinery SchedulingLinear Programming Formulation: Rearranged for MATLAB

Max J =

product sales 4

p=1vpQp

production cost 5

r=1Crxr

=

product sales 4

p=1

vp

5r=1

aprxr

Qp

production cost 5

r=1

Crxr

=5

r=1

(vpapr Cr) xr =5

r=1

frxr

Max J = vtA CT x = fTx

Chen CL 16


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Chen CL 16

s.t. xr Sr r= 1, 2, 3; x4+ x5 S4

Qp Dp p= 1, . . . , 4

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 1

a11 a15...

...

...

a41 a45

A

x1

x2

x3

x4

x5

x

S1

S2

S3

S4

D1

D2

D3

D4

B

Chen CL 17


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Chen CL 17

Refinery SchedulingMATLAB Code

YIELDS = [ 0.6 0.5 0.3 0.4 0.4 ;

0.2 0.2 0.3 0.3 0.1 ;

0.1 0.2 0.3 0.2 0.2 ;

0.0 0.0 0.0 0.0 0.2];

CSUPPLY = [100, 100, 100, 200 ];

PDEMAND = [170, 85, 85, 20 ];

PVALUE = [ 45, 30, 15, 60 ];

CCOST = [ 15, 15, 15, 25, 25];OCOST = [ 5, 8.5, 7.5, 3, 2.5];

TCOST = CCOST + OCOST ;

f = [- PVALUE*YIELDS + TCOST];

A = [1 0 0 0 0 ;

0 1 0 0 0 ;

0 0 1 0 0 ;0 0 0 1 1 ; YIELDS ] ;

B = [CSUPPLY, PDEMAND] ;

lb = zeros(5,1); ub = CSUPPLY;

x0 = [0; 0; 0; 0; 0]; %x0=[50;50;50;50;50];

options = optimset(LargeScale,off)[x,NPROFIT] = linprog(f,A,B,[],[],lb,ub,x0,options);

x, Q = YIELDS*x, PROFIT = - NPROFIT

x =

100.0000100.0000

66.6667

0.0000

100.0000

Q =170.0000

70.0000

70.0000

20.0000

PROFIT =

3.4000e+003

Chen CL 18


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Chen CL 18

A Heat Exchange Network Design ProblemStream F Cp (kW/K) Tin (K) Tout (K)

H1 20 700 410

H2 40 600 310H3 70 460 310

H4 94 360 310

C1 50 350 650

C2 180 300 400

Utility Temp. (K) Cost ($/KW-yr) Max available (kW)

Fuel 750 120

HP steam 510 90 1000

LP steam 410 70 500

Cooling water 300 325 15

min. temperature approach = 10 KH1 is not to be allowed to exchange any heat with C1

Chen CL 19


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Chen CL 19

Transshipment LP ModelHeat Contents of Process Streams

Interval Thot Tcold Heat Contents

H1 H2 H3 H4 C1 C2

700 6900 800 - - - - -

660 6501 1200 - - - 3000 -

600 5902 1800 3600 - - 4500 -

510 5003 1000 2000 - - 2500 -

460 4504 1000 2000 3500 - 2500 -

410 4005 - 2000 3500 - 2500 9000

360 3506 - 2000 3500 4700 - 9000

310 300

Chen CL 20


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Chen CL 20

Heat Exchanger Network DesignTransshipment Model for Heat Flows

Chen CL 21


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Condensed Transshipment LP ModelNotations

Assume : any pair of heat/cold streams can exchange heat

R15 heat residuals

Qfuel, QHP, QLP, QW : heating and cooling loads

Hk = {i| hot stream i supplies heat to interval k}

Ck = {j| cold stream j demands heat from interval k}

Sk = {m| hot utility msupplies heat to interval k}

Wk = {n| cold utility n extracts heat from interval k}S = set of hot utilities

W = set of cold utilities

Note : Rk= 0 means pinch point

Chen CL 22


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Condensed Transshipment LP ModelHeat Flows in Interval k

Par.s QHik heat contents of hot streami in interval k

QCjk heat contents of hot streami in interval k

cm, cn unit costs of hot streamm and cold streamn

Var.s QSm, QWn heat loads of hot (cold) utility m(n)Rk heat residual exiting intervalk

Chen CL 23


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Condensed Transshipment LP ModelLP Formulation

minmS

cmQSm +

nW

cnQWn

s.t. Rk1+

iHkQHik+

mSkQSm=Rk+

jCkQCjk +

nWkQWn

QSm 0, QWn 0, k= 1, , K 1

Rk 0, R0=RK= 0

Chen CL 24


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Expanded Transshipment LP ModelHeat Exchange Between Hot and Cold Streams

Chen CL 25


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Expanded Transshipment LP ModelHeat Flows Between Hot/Cold Streams in Interval k

H

k= {i| hot stream i supplies heat to interval k or at a higher interval}S

k= {m| hot utility msupplies heat to interval k or at a higher interval

Qijk=exchange of heat of hot stream i and cold stream j at interval k

Qmjk=exchange of heat of hot utility m and cold stream j at interval k

Qink=exchange of heat of hot stream i and cold utility n at interval kRik=heat residual of hot stream i exiting interval k

Rmk=heat residual of hot utility m exiting interval k

Chen CL 26


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Expanded Transshipment LP ModelLP Formulation

min mS

cmQSm +

nWcnQ

Wn

s.t. Ri,k1+ QHik=Rik+

jCk

Qijk+

nWk

Qink k K, i H

k

Rm,k1+ QSm=Rmk+ jCk Qmjk k K, m S

kiH

k

Qijk +

mS

k

Qmjk=QCjk k K, j Ck

iHkQink=Q

Wn k K, n Wk

Rik, Rmk, Qijk, Qmjk, Qink, QSm, Q

Wn 0 Ri0=RiK= 0

Qijk= 0 fori= 1, j = 1, k K (forbidden match)

QLij K

k=1

Qijk QUij i, j (lower/upper bound

Chen CL 27


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Expanded Transshipment LP ModelGAMS Code

$offsymxref offsymlist

* filename: TRANS.gms

* LP transhipment model with restricted matches

* Reference: Papoulias, S.A. and I.E. Grossmann, "A Structural Optimization

* Approach in Process Synthesis. II: Heat Recovery Networks", Computers and

* Chemical Engineering, 7, 707 (1983)

* Authors: Nikolaos V. Sahinidis and I.E. Grossmann

option limrow=0, limcol=0, solprint=off;

*================= input data

SETS H hot streams /1*4/

C cold streams /1*2/

S hot utilities /1*3/

W cold utilities /1*1/K temperature intervals (only an upper bound needed) /1*50/;

PARAMETER FCPH(H) FCp of hot streams

/ 1 20

2 40

3 70

4 94 /;

PARAMETER FCPC(C) FCp of cold streams

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/ 1 50

2 180 /;

PARAMETER TINH(H) inlet T of hot streams

/ 1 700

2 600

3 460

4 360 /;PARAMETER TOUTH(H) outlet T of hot streams

/ 1 420

2 310

3 310

4 310 /;

PARAMETER TINC(C) inlet T of cold streams

/ 1 350

2 300 /;PARAMETER TOUTC(C) outlet T of cold streams

/ 1 650

2 400 /;

PARAMETER TINS(S) hot util temp

/ 1 750

2 510

3 410 /;

PARAMETER TINW(W) cold util temp

/ 1 300 /;

PARAMETER DT minimum T approach ;

DT=10;

PARAMETER CS(S) hot utility cost

/ 1 120

2 90

3 70 /;

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PARAMETER CW(W) cold utility cost

/ 1 15 /;

PARAMETER AS(S) availability of hot util

/ 1 INF

2 1000

3 500 /;

PARAMETER AW(W) availability of cold util/ 1 INF /;

*===== bounds for heat exchange (imposed and restricted matches)

PARAMETER QLHC(H,C) Lower bound between hot and cold stream ;

PARAMETER QUHC(H,C) Upper bound between hot and cold stream ;

PARAMETER QLHW(H,W) Lower bound between hot stream and cold utility ;

PARAMETER QUHW(H,W) Upper bound between hot stream and cold utility ;

PARAMETER QLCS(C,S) Lower bound between cold stream and hot utility ;

PARAMETER QUCS(C,S) Upper bound between cold stream and hot utility ;

QLHC(H,C) = 0 ;

QUHC(H,C) = INF ; QUHC(1,1) = 0 ;

QLHW(H,w) = 0 ;

QUHW(H,W) = INF ;

QLCS(C,S) = 0 ;

QUCS(C,S) = INF ;

* these could also be inserted in table format ...

* imposed matches

* table QLHC(H,C)

* 1 2 3 4

* 1

* 2

* 3

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* 4

* 5 ;

*================= parameter evaluations

*=== temperature intervals

PARAMETERS HK(H,K) nonzero (1) elements define set H(k) - existence

CK(C,K) nonzero (1) elements define set C(k) - existenceSK(S,K) nonzero (1) elements define set S(k) - existence

WK(W,K) nonzero (1) elements define set W(k) - existence

QH(H,K) heat content of hot streams at each interval

QC(C,K) heat content of cold streams at each interval

QHT(H) total heat content of hot stream

QCT(C) total heat content of cold stream

TH(K) interval temperature (hot)

TC(K) interval temperature (cold) ;

PARAMETERS IS(S),IW(W),IH(H),IC(C);

* when these become 1, the corresponding stream inlet T

* has been considered for interval evaluations

HK(H,K)=0; CK(C,K)=0; SK(S,K)=0; WK(W,K)=0; QH(H,K)=0;

QC(C,K)=0; TH(K)=0; TC(K)=0;

PARAMETER KM actual temperature intervals ;

PARAMETERS IEND,AUX;

KM=0;IEND=0;IS(S)=0;IW(W)=0;IH(H)=0;IC(C)=0;AUX=0;

PARAMETERS HMAX,CMIN;

HMAX=SMAX(H,TINH(H)); CMIN=SMIN(C,TINC(C));

IS(S) $(TINS(S) GE HMAX)=1;

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IW(W) $(TINW(W) LE CMIN)=1;

LOOP (K $ (IEND EQ 0),

AUX=0;

LOOP (S $((IS(S) EQ 0) AND (TINS(S) GT AUX)), AUX=TINS(S));

LOOP (H $((IH(H) EQ 0) AND (TINH(H) GT AUX)), AUX=TINH(H));

LOOP (W $((IW(W) EQ 0) AND (TINW(W) GT AUX)), AUX=TINW(W)+DT);LOOP (C $((IC(C) EQ 0) AND (TINC(C) GT AUX)), AUX=TINC(C)+DT);

TH(K)=AUX;

TC(K)=TH(K)-DT;

IS(S) $(TINS(S) EQ AUX)=1;

IW(W) $(TINW(W) EQ (AUX-DT))=1;

IH(H) $(TINH(H) EQ AUX)=1;

IC(C) $(TINC(C) EQ (AUX-DT))=1;

IF (((SUM(S,IS(S))+SUM(H,IH(H))+SUM(W,IW(W))+SUM(C,IC(C))) EQ(CARD(S)+CARD(H)+CARD(W)+CARD(C))), IEND=1;KM=ORD(K)-1

)

);

*=== heat contents of streams

LOOP (K $ (ORD(K) LE KM),

HK(H,K)=1 $((TINH(H) GE TH(K)) AND (TOUTH(H) LT TH(K)));

CK(C,K)=1 $((TINC(C) LE TC(K+1)) AND (TOUTC(C) GT TC(K+1)));

SK(S,K)=1 $(((TINS(S) GE TH(K)) AND (TINS(S) LT TH(K-1))) OR

((TINS(S) GE TH(K)) AND (ORD(K) EQ 1)));

WK(W,K)=1 $(((TINW(W) LE TH(K)) AND (TINW(W) GT TH(K+1))) OR

((TINW(W) LE TH(K)) AND (ORD(K) EQ KM)))

);

LOOP (K,

LOOP (H $(HK(H,K) EQ 1),

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QH(H,K)=FCPH(H)*(TH(K)-TH(K+1));

IF ((TOUTH(H) GT TH(K+1)), QH(H,K)=FCPH(H)*(TH(K)-TOUTH(H)))

);

LOOP (C $(CK(C,K) EQ 1),

QC(C,K)=FCPC(C)*(TC(K)-TC(K+1));

IF ((TOUTC(C) LT TH(K)), QC(C,K)=FCPC(C)*(TOUTC(C)-TC(K+1)))

));

QHT(H) = FCPH(H) * ( TINH(H) - TOUTH(H) ) ;

QCT(C) = FCPC(C) * ( TOUTC(C) - TINC(C) ) ;

* ===================== model

POSITIVE VARIABLES

RH(H,K) heat residual of hot stream H exiting interval K

RS(S,K) heat residual of hot utility S exiting interval KQHC(H,C,K) heat exchange between hot H and cold C in interval K

QHW(H,W,K) heat exchange between hot H and cold util W in interval K

QSC(S,C,K) heat exchange between hot util S and cold C in interval K

QS(S) heating load for hot utility S

QW(W) cooling load for cold utility S ;

VARIABLE UTILITY total utility-cost ;

EQUATIONS

OBJ1 objective function for minimum utility consumption

HBAL(H,K) hot stream HEAT BALANCES

CBAL(C,K) cold stream HEAT BALANCES

SBAL(S,K) hot utility HEAT BALANCES

WBAL(W,K) cold utility HEAT BALANCES

hcIMPOSE(H,C) imposed match between hot and cold stream

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hcRESTRICT(H,C) restriction between hot and cold stream

hwIMPOSE(H,W) imposed match between hot stream and cold utility

hwRESTRICT(H,W) restriction between hot stream and cold utility

csIMPOSE(C,S) imposed match between cold stream and hot utility

csRESTRICT(C,S) restriction between cold stream and hot utility ;

OBJ1 .. UTILITY =E= SUM ( S, CS(S)*QS(S) ) + SUM ( W, CW(W)*QW(W) ) ;

HBAL(H,K) $ ( (TINH(H) GE TH(K)) AND (ORD(K) LE KM) ) ..

RH(H,K) - ( RH(H,K-1) $ (HK(H,K-1) EQ 1) )

+ SUM (C $ (CK(C,K) EQ 1), QHC(H,C,K) )

+ SUM (W $ (WK(W,K) EQ 1), QHW(H,W,K) ) =E= QH(H,K);

RH.FX(H,K) $ (ORD(K) EQ KM)= 0;

SBAL(S,K) $ ( (TINS(S) GE TH(K)) AND (ORD(K) LE KM) ) ..

RS(S,K) - ( RS(S,K-1) $ (TINS(S) GE TH(K-1)) )

+ SUM (C $ (CK(C,K) EQ 1), QSC(S,C,K) )

- ( QS(S) $ ( (TINS(S) EQ TH(K)) OR (ORD(K) EQ 1) ) ) =E= 0 ;

RS.FX(S,K) $ (ORD(K) EQ KM)= 0;

CBAL(C,K) $ ( CK(C,K) EQ 1 ) ..

SUM ( H $ (TINH(H) GE TH(K)), QHC(H,C,K) ) +

SUM ( S $ (TINS(S) GE TH(K)), QSC(S,C,K) ) =E= QC(C,K) ;

WBAL(W,K) $ ( WK(W,K) EQ 1 ) ..

SUM ( H $ (TINH(H) GE TH(K)), QHW(H,W,K) ) =E= QW(W) ;

QS.UP(S) $(AS(S) LT INF)=AS(S);

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QW.UP(W) $(AW(W) LT INF)=AW(W);

hcIMPOSE(H,C) $ (QLHC(H,C) GT 0) ..

QLHC(H,C) =L=

SUM ( K $ ( (ORD(K) LE KM) AND (CK(C,K) EQ 1) ), QHC(H,C,K) ) ;

hcRESTRICT(H,C) $ (QUHC(H,C) LT INF) ..SUM ( K $ ( (ORD(K) LE KM) AND (CK(C,K) EQ 1) ), QHC(H,C,K) )

=L= QUHC(H,C) ;

hwIMPOSE(H,W) $ (QLHW(H,W) GT 0) ..

QLHW(H,W) =L=

SUM ( K $ ( (ORD(K) LE KM) AND (WK(W,K) EQ 1) ) , QHW(H,W,K) ) ;

hwRESTRICT(H,W) $ (QUHW(H,W) LT INF) ..SUM ( K $ ( (ORD(K) LE KM) AND (WK(W,K) EQ 1) ) , QHW(H,W,K) )

=L= QUHW(H,W) ;

csIMPOSE(C,S) $ (QLCS(C,S) GT 0) ..

QLCS(C,S) =L=

SUM ( K $ ( (ORD(K) LE KM) AND (CK(C,K) EQ 1) ) , QSC(S,C,K) ) ;

csRESTRICT(C,S) $ (QUCS(C,S) LT INF) ..

SUM ( K $ ( (ORD(K) LE KM) AND (CK(C,K) EQ 1) ) , QSC(S,C,K) )

=L= QUCS(C,S) ;

MODEL TRANS / ALL /;

SOLVE TRANS MINIMIZING UTILITY USING LP;

DISPLAY UTILITY.L, RH.L, RS.L, QHC.L, QHW.L, QSC.L, QS.L, QW.L ;

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Expanded Transshipment LP ModelResults (H1-C1 is Forbidden)

fuel 3900 kW

high pressure steam 500 kW

cold utility 3800 kW

min utility cost 570, 000 $/year

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Linear Programming: Simple Examples

Case 1:

unique

solution

Min f= 4x1 x2+ 50

s.t. x1 x2 2

x1+ 2x2 8

x1, x2 0

Chen CL 37

S


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Case 2:

non-unique

solution

Min f= x1+ x2+ 30

s.t. x1 x2 2

x1+ 2x2 8

x1, x2 0

Chen CL 38

Li P i Si l E l


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Case 3:

unbounded

solution

Min f= 4x1 x2+ 50

s.t. x1 x2 2

x1, x2 0

Chen CL 39

Li P i Si l E l


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Case 4:

no feasible

solution

Min f= 4x1 x2+ 50

s.t. x1 x2 5

x1+ 2x2 8

x1, x2 0

Chen CL 40

MATLAB Pl f C 1


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MATLAB Plot for Case 1% LP: Case 1 - Single solution

[x1,x2] = meshgrid(0:0.1:6, 0:0.1:6);

f = -4*x1-2*x2+50;g1 = x1-x2-2; g2 = x1+2*x2-8;

g3 = -x1; g4 = -x2;

cla reset;

set(gca,XTick,[0,2,4,6],...YTick,[0,2,4,6],...

XTickLabel,{0,2,4,6},...

YTickLabel,{0,2,4,6},...

FontSize,16,LineWidth,3)

xLabel(\bf x_1,FontSize,16,Color,[0,0,0])

yLabel(\bf x_2,FontSize,16,Color,[0,0,0])

title(\bf LP Case 1: unique solution,...

FontSize,16,Color,[0,0,0])

Min f= 4x1 x2+ 50

s.t. x1 x2 2x1+ 2x2 8

x1, x2 0

Chen CL 41


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hold on

cv = [0 0];

fv = [25, 30, 35, 40, 45];

const1 = contour(x1,x2,g1,cv,r,LineWidth,3);const2 = contour(x1,x2,g2,cv,b,LineWidth,3);

const3 = contour(x1,x2,g3,cv,k,LineWidth,3);

const4 = contour(x1,x2,g4,cv,k,LineWidth,3);

fs = contour(x1,x2,f,fv,k-);

clabel(fs)

text(3.1, 1,\bf\leftarrow g_1:x_1- x_2=2,FontSize,12,

text(1.3,3.5,\bf\leftarrow g_2:x_1+2x_2=8,FontSize,12,

text( 1,0.2,\bf g_3,FontSize,12)

text(0.1, 2,\bf g_4,FontSize,12)plot( 4, 2,g,Marker,o,MarkerSize,6)

text( 1, 2,\it\bf FR, FontSize,14)

text(3.75,2.75,\it\bf Infeasible Region,FontSize,14)

hold off

Min f= 4x1 x2+ 50

s.t. x1 x2 2

x1+ 2x2 8x1, x2 0


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Chen CL 43

MATLAB S l ti f C 2


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MATLAB Solution for Case 2% Linear programming:

% Multiple solutions (Case 2)

%f = [-1 1];

A = [ 1 -1 ;

1 2];

B = [ 2; 8];

lb = zeros(2,1);

x0 = [ 0; 0];

%x0 = [ 0; 4];

options = optimset(LargeScale,off)

[x,fval,exitflag,output,lambda] =...linprog(f,A,B,[],[],lb,[],x0,options)

x0, x, fval = fval + 30

Min f= x1+ x2+ 30

s.t. x1 x2 2

x1+ 2x2 8x1, x2 0

x0 =

0

0

x =

2

0

fval =

28

x0 =

0

4

x =

3

1

fval =

28

Chen CL 44

A St d d Li P i P bl


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A Standard Linear Programming Problem

Min: f(x) =c1x1+ + cnxn =n

j=1

cjxj (f=cTx)

s.t. a11x1+ + a1nxn=b1 =

nj=1

a1jxj

(Ax=b)

am1x1+ + amnxn=bm =

nj=1

amjxj

xj 0, bj 0; (x 0 b 0)

Chen CL 45


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2y1+ y2 4 2y1 y2 4

2y1 y24 2y1 y2 s1= 4, s1: surplus variable

2y1 y24 2y1 y2+ s2= 4, s2: slack variable

2y1 y2= 4, y2:unrestricted in sign 2y1 (y

+2 y

2) = 4, y

+2, y

2 0

max z min f= z


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Chen CL 47


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Max: z= 2y1+ 5(y+2 y

2)

s.t. 3y1+ 2(y+2 y2) + s1= 12

2y1+ 3(y+2 y

2) s2= 6

y1, y+2, y

2, s1, s2 0

Min: f= 2x1 5x2+ 5x3

s.t. 3x1+ 2x2 2x3+ x4= 12

2x1+ 3x2 3x3 x5= 6

xi 0

Chen CL 48

Basic Co ce ts Related to Li ea


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Basic Concepts Related to LinearProgramming Problems

All functions are linear

minimize convex (cost) function on a convex set, or maximize concave (profit) function on a convex set,

global optimum if optimum solution exists

The solution always lies on the boundary of the feasible

region if it exists

(some constraints are active at the optimum)


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Chen CL 50


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Max: z= 400x1+ 600x2

s.t. x1+ x2+ x3= 16

128

x1+ 114x2+ x4= 11

14x1+

124

x2+ x5= 1

xi 0

x3 = 16 x1 x2

x4 = 1 128

x1 114

x2

x5 = 1 114

x1 124

x2

z = 400x1+ 600x2

(xi 0) Infinite number of solutionswith different z values


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Chen CL 52

MATLAB Code


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MATLAB Code% ProfMax_V2.m (modified from ProfMax.m)

a1 = 0; a2 = 30;

b1 = 0; b2 = 30;x1 = [a1 : 0.1 : a2];

y1 = 16-x1;

y2 = 14-14/28*x1;

y3 = 24-24/14*x1;

y4 = 0*x1;

figure(name,axis property)hold on

plot(x1,y1,b,linewidth,3)

plot(x1,y2,r,linewidth,3)

plot(x1,y3,Color,[0,.5,0],linewidth,3)

set(gca,xlim,[a1 a2],ylim,[b1 b2],linewidth,3,fontsize,14)

xlabel(\bf x_1,fontsize,16)

ylabel(\bf x_2,fontsize,16)

title(\bf Profit maximization problem,fontsize,16)

text(1,15.3,\bf\leftarrow x_1+x_2=16, fontsize,12,Color,[0,0,1]

text(19,5.3,\bf 1/28x_1+1/14x_2=1, fontsize,12,Color,[1,0,0]

text(21,14-14/28*21+0.6,\bf\downarrow,fontsize,12,Color,[1,0,0]

Chen CL 53

text(7 24-24/14*7+0 3 \bf\leftarrow 1/14x 1+1/24x 2=1


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text(7,24 24/14*7+0.3, \bf\leftarrow 1/14x_1+1/24x_2 1 ,...

fontsize,12,Color,[0,.5,0]

text( 3, 5,\bf FR, fontsize,14)

text(15,20,\bf Infeasible Region,fontsize,14)

text( 0, 0.6,\bf 1, fontsize,12,Color,[1, 0, 0])

text( 0, 16,\bf 2, fontsize,12,Color,[0,.5,.5])

text( 0, 13,\bf 3, fontsize,12,Color,[1, 0, 0])

text( 0, 24,\bf 4, fontsize,12,Color,[0,.5,.5])

text( 16, 0.6,\bf 5, fontsize,12,Color,[0,.5,.5])

text( 28, 0.6,\bf 6, fontsize,12,Color,[0,.5,.5])

text(12.5,0.6,\bf 7, fontsize,12,Color,[1, 0, 0])text(3.5,11.1,\bf 8, fontsize,12,Color,[1, 0, 0])

text( 10, 4,\bf 9, fontsize,12,Color,[1, 0, 0])

text( 8, 10,\bf 10,fontsize,12,Color,[0,.5,.5])

Chen CL 54

LP Terminology


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LP Terminology Vertex (Extreme) Point:

Feasible Solution: Solution of constrained equationssatisfying nonnegativity conditions

Basic Solution: A solution of constrained equations obtained

by setting n m variables to zero.Variables set to zero are called nonbasic

Remaining ones are called basic

Basic Feasible Solution:A basic solution satisfying nonnegativity conditions on variables

Degenerate Basic Solution:

A solution with a zero-valued basic variable

Chen CL 55

Degenerate Basic Feasible Solution:


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Degenerate Basic Feasible Solution:

Optimum Solution:

A feasible solution with optimum cost function value

Optimum Basic Solution:

A basic feasible solution with optimum cost function value

Convex Polyhedron: If the feasible region for an LP problem

is bounded, it is called Convex Polyhedron

Basis: Columns of coefficient matrix Aof constrained equationscorresponding to basic variables are said to form a basis for the

mdimensional vector space. Any other mdimensional vector

can be expressed as a linear combination of the basis vectors.

Chen CL 56

Determination of Basic Solutions


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Determination of Basic Solutions

Max: z= 4x1

+ 5x2

s.t. x1+ x2 4

x1+ x2 6

x1, x2 0

Max: z= 4x1

+ 5x2

s.t. x1+ x2+ x3= 4

x1+ x2+ x4= 6

x1, x2, x3, x4 0

x1 x2 x3 x4 z

1 0 0 4 6 0 A

2 0 4 0 2 20 B3 0 6 2 0 E (x)

4 4 0 0 10 F (x)

5 6 0 10 0 24 D

6 1 5 0 0 29 C

Chen CL 57

Optimum Solution for LP Problems


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Optimum Solution for LP Problems

Theorem:Extreme points and basic feasible solutions

The collection of feasible solutions for an LP problem

constitutes a convex set whose extreme points

correspond to basic feasible solutions.

Chen CL 58

Theorem: basic theorem of linear programming


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Theorem: basic theorem of linear programming

Let the m n coefficient matrix A of constrainedequations have full row rank, rank(A) =m. Then

If there is a feasible solution, there is a basic feasible

solution (vertex), and

If there is an optimum feasible solution, there is an

optimum basic feasible solution.(one of the vertices of the convex polyhedron representing feasible soln)

Implications: optimum soln must be one of the basic

feasible solns (vertices) searching among finite basic feasible solns

(at most n!m!(nm)! vertices)


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Chen CL 60

Canonical Form


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Canonical Form

An m

n system of simultaneous equations withrank(A) = m is said to be in canonical form if eachequation has a variable (with unit coefficient) that doesNOT appear in any other equation

x1 +a1,m+1xm+1+ a1,m+2xm+2+ + a1,nxn =b1

x2 +a2,m+1xm+1+ a2,m+2xm+2+ + a2,nxn =b2... ... ...

xm

+am,m+1

xm+1

+ am,m+2

xm+2

+

+ am,n

xn

=bm

or I(m)x(m)+ Qx(nm) =b

Note: if x(nm)= 0 then x(m) =b

Chen CL 61

Tableau


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Tableau

x1 +a1,m+1xm+1+ a1,m+2xm+2+ + a1,nxn =b1

x2 +a2,m+1xm+1+ a2,m+2xm+2+ + a2,nxn =b2... ... ...

xm +am,m+1xm+1+ am,m+2xm+2+ + am,nxn =bm

Basic x1 x2 xm xm+1 xn RHS

x1 1 0 0 a1,m+1 a1,m+2 a1,n b1

x2 0 1 0 a2,m+1 a2,m+2 a2,n b2... ... ... ...

xm 0 0 1 am,m+1 am,m+2 am,n bm

Chen CL 62

Example: canonical form and basic feasible solution


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Example: canonical form and basic feasible solutionMin: f= 2x1 x2

s.t. 4x1+ 3x2+ x3= 12

2x1+ x2+ x4= 4

x1+ 2x2+ x5= 4, xi 0

Basic x1 x2 x3 x4 x5 b basic solution

x3 4 3 1 0 0 12 x1=x2= 0x4 2 1 0 1 0 4 x3= 12, x4= 4, x5= 4

x5 1 2 0 0 1 4 f= 0

One possible basic feasible solutionx1=x2= 0, x3= 12, x4= 4, x5= 4; f= 0 Q: optimum solution ?

We need to try another basic feasible solution!

Chen CL 63

Pivot Step: Interchange of Basic and


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Pivot Step: Interchange of Basic andNonbasic Variables

Pivot Step: starting from a basic feasible solution,

find another one to decrease cost by interchanging a

current basic variable with a non-basic variable

Pivot Row: the row used for elimination process

Pivot Column: the column on which the elimination

is performed

Chen CL 64

Pivot Operation


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Pivot OperationInterchange of Basic and Nonbasic Variables

Min: f= 2x1 x2

s.t. 4x1+ 3x2+ x3= 12

2x1+ x2+ x4= 4

x1+ 2x2+ x5= 4, xi 0

Basic x1 x2 x3 x4 x5 b basic solutionx3 4 3 1 0 0 12 x1=x2= 0

x4 2 1 0 1 0 4 x3= 12, x4= 4, x5= 4

x5 1 2 0 0 1 4 f= 0

x3 0 1 1 2 0 4 x2=x4= 0

x1 1 1

2 0 1

2 0 2 x3= 4, x1= 2, x5= 2

x5 0 3

2 0 12 1 2 f= 4

Chen CL 65

Basic Steps of The Simplex Method


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Basic Steps of The Simplex Method

Three basic questions: Is the current basic feasible solution optimal ?

If not,

How to choose a current nonbasic variable that should

become basic ?

Which variable from the current basic set should

become nonbasic ?

Chen CL 66

Example: steps of the Simplex method


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p p p

Max: z= 2x1+ x2 = f (min f)

s.t. 4x1+ 3x2 122x1+ x2 4

x1+ 2x2 4 x1, x2 0

z = 4 along CD

Chen CL 67

In standard LP form:


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Min: f = 2x1 x2

s.t. 4x1 + 3x2 + x3 = 122x1 + x2 + x4 = 4

x1 + 2x2 + x5 = 4

xi 0

NBs: x1 = 0 x2 = 0 (A)

BVs: x3 = 12 x4 = 4 x5 = 4f = 0 z = 0

Note: f 2x1 x2 = 0

Chen CL 68

Basic f x1 x2 x3 x4 x5 b bi/ai


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Basic f x1 x2 x3 x4 x5 b bi/aic

x3 0 4 3 1 0 0 12 12

4 = 3

x4 0 2 1 0 1 0 4 4

2

= 2

x5 0 1 2 0 0 1 4 4

1= 4

f 1 2 1 0 0 0 0

Cost row has nonzero entries only in nonbasic columns (x1, x2)

If all nonzero entries are nonnegative optimum soln (?)Currently, TWO negative entriesNOT optimum

Select x1 to become basic (?) elimination in x1 columnTo keep cost fcn right: f 2x1 x2+ 0x3+ 0x4+ 0x5 = 0

If letting x1 = 0, x2 = 0 1 thenf = 0 1; Ifx1 = 0 1, x2 = 0, f = 0 2

Take ratios of RHS par.s with positive elements in x1 col.

select x4 to become nonbasic (smallest ratio)pivot element is a21= 2 Constraint: 2x1+ x2+ 0x3+ x4+ 0x5 = 4;Ifx1 = 0 1, x2 = 0, x4 = 4 2; Ifx1 = 0 2, x2 = 0, x4 = 4 0

Chen CL 69

Basic f x1 x2 x3 x4 x5 b bi/ai


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x3 0 0 1 1 2 0 4

x1 0 1 1

2

0 1

2

0 2

x5 0 0 3

2 0 12 1 2

f 1 0 0 0 1 0 4

NBs: x2 = 0 x4 = 0 (D)BVs: x3 = 4 x1 = 2 x5 = 2

f = 4 z = 4

All coefficients in last row are nonnegativeno further decrease of cost is possible optimum

Cost coefficient corresponding to nonbasic variable x2 is zero

multiple solutions

Chen CL 70

What happens if we do not select a row with the least ratio as


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the pivot row ?

If interchange nonbasic var x1 with basic varx5 not feasible !


x3 0 4 3 1 0 0 12 12

4 = 3

x4

0 2 1 0 1 0 4 42

= 2

x5 0 1 2 0 0 1 4 4

1= 4

f 1 2 1 0 0 0 0

x3 0 0 5 1 0 4 4 not feasible !

x4 0 0 3 0 1 2 4 not feasible !

x1 0 1 2 0 0 1 4

f 1 0 3 0 0 4 8

Chen CL 71

Basic Ideas and Procedures


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of the Simplex Method

The method needs an initial basic feasible solution (canonical

form) to initiate the search for the optimum solution

type slack variable, = type: surplus variable ?? ( artificial variable)

The cost function must be expressed in terms of only the NBs

last row in Simplex tableau to eliminate BVs during pivot step

Coefficients in last row corresponding to BVs are zero, andall those corresponding to NBs are nonzero.

If any of the nonzero entries in last row is negative not optimumOtherwise, an optimum solution has been reached

Chen CL 72

If optimum is not reachedinterchange a current BV with a NB


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A NB with most negative coefficient is selected to become

basic (pivot column)Take ratios of RHS values with positive elements in pivot column,

identify the row with smallest ratio as the pivot row and the BV

associated with it should become NB

If all entries in pivot column are negative unbounded problem

After identifying pivot column and row

pivot step to obtain a new basic feasible solution

Repeat preceding steps until an answer is obtained

Chen CL 73

Derivation of the Simplex Method


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pSelection of A Basic Variable to Become Nonbasic

Assumexc is a NB tapped to become a BVcth NB column replace some current basic column

xcis to become a basic xc 0 (after interchange, allzero elements in cth column except unit element at one location)

x1 +a1,m+1xm+1+ + a1cxc+ + a1,nxn =b1

x2 +a2,m+1xm+1+ + a2cxc+ + a2,nxn =b2

xm +am,m+1xm+1+ + amcxc+ + am,nxn =bm

Chen CL 74

Pivoting new canonical form (still keep equality constraints):


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x1 + a1,m+1xm+1+ + a1,nxn =b1 a1cxc 0

x2 + a2,m+1xm+1+ + a2,nxn =b2 a2cxc 0

xm + am,m+1xm+1+ + am,nxn =bm amcxc 0

xi = bi aicxc 0 if aic0 and xc= biaic

br

arc= min

aic >0i = 1 m

bi

aic

maximum allowable xc

to keep all xi 0

Select the row index r giving the least ratio of RHS parameter

with positive element inc column

If all aic are non-positive in the cth column, then xc can be

increased indefinitelyunbounded problem !

Chen CL 75

Derivation of the Simplex Method


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pSelection of A Nonbasic Variable to Become Basic

xi = bi n

j=m+1

aijxj; i= 1 m

f =m

i=1 cixi+n

j=m+1 cjxj=

mi=1

ci

bi n

j=m+1

aijxj

+ n

j=m+1

cjxj

=m

i=1

bici f0

+n

j=m+1

cj m

i=1

aijci cj

xj

= f0+

n

j=m+1cjxj

Chen CL 76

Expressing cost function in terms of present NB


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cj: reduced cost coefficients (in last row ofSimplex tableau)

Forcj 0 unique optimum

If at least one cj = 0 a possibility of alternate optima

If the NB associated with the negative reduced cost coefficient cjcannot be made basis (all aij in c

j column are negative),

then the feasible solution is unbounded

Chen CL 77

Theorem: improvement of basic feasible solution


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Given a nondegenerate basic feasible solution with corresponding

cost function f0, suppose cj


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y pMax: z= 4x1+ 5x2 (min f= 4x1 5x2)

s.t. x1+ x2+ x3= 4

x1+ x2+ x4= 6, xi 0

Basic f x1 x2 x3 x4 b bi/aicx3 0 1 1 1 0 4

41= 4

x4 0 1 1 0 1 6 6

1= 6

f 1 4 5 0 0 0x2 0 1 1 1 0 4x4 0 2 0 1 1 2

22= 1

f 1 9 0 5 0 20

x2

0 0 1 12

1

2 5

x1 0 1 0 12

12 1

f 1 0 0 1292 29

Chen CL 79

Ex: Solution by the Simplex Method


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y

Max: z= 400x1+ 600x2 = f

s.t. x1+ x2+ x3= 16, 1

28

x1+ 1

14

x2+ x4= 11

14x1+ 124x2+ x5= 1 xi 0

Basic f x1 x2 x3 x4 x5 b bi/aicx3 0 1 1 1 0 0 16 16x4 0 1/28 1/14 0 1 0 1 14

x5 0 1/24 1/24 0 0 1 1 24f 1 400 600 0 0 0 0x3 0 1/2 0 1 14 0 2 4x2 0 1/2 1 0 14 0 14 28x5 0 17/336 0 0 7/12 1 5/12 140/17

f 1 100 0 0 8400 0 8400x1 0 1 0 2 28 0 4 x2 0 0 1 1 28 0 12 x5 0 0 0 17/168 2 1 3/14

f 1 0 0 200 5600 0 8800


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Chen CL 81

x(1) NBs: x2 = x4 = 0


82/202

x( ) NBs: x2=x4= 0

BVs: x1= 4, x3= 4 f= 4

or x(2) NBs: x3=x4= 0

BVs: x1= 3, x2= 2 f= 4

x =x(1) + (1 )x(2)

f(x) = 4 (z(x) = 4)

Chen CL 82

Example: Unbounded Problem


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Max: z=x1 2x2 = f

s.t. 2x1 x2 0

2x1+ 3x2 6 x1, x2 0

Basic f x1 x2 x3 x4 b

x3 0 2 1 1 0 0

x4 0 2 3 0 1 6

f 1 1 2 0 0 0

Chen CL 83

Simplex Algorithm: Summary


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Step 1: Start with an initial basic feasible solution

Step 2: The cost function must be in terms of only the NBs

Step 3: If all reduced cost coefficients for NBs are nonnegative

optimum

Step 4: If all elements in pivot column are negative

unbounded

Step 5: Complete the pivot step

Step 6: Identify basic and nonbasic variables, and their values.

Identify cost function value and go to Step 3

Chen CL 84

Two-Phase Simplex Method


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An initial canonical form (basic feasible point) is needed to initialize

Simplex

To obtain standard LP problem as a canonical form:

type: + slack variable (canonical form)

= type: + artificial variable (no physical meaning) type: surplus + artificial variable

Artificial variables ( AVs) augment the convex polyhedron of the

original problem. Initial basic feasible solution corresponds to anextreme point (vertex) located in the new expanded space

Extreme pt in expanded space extreme pt in original spaceall AVs will be nonbasic (zero value)

further movements are only among extreme pts of original space

Chen CL 85

Artificial cost Function


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Min: w =xn+1+ + xn+m = AV sxn+i

s.t. xn+1 +a11x1+ + a1nxn =b1

xn+2 +a21x1+ + a2nxn =b2...

xn+m +am1x1+ + amnxn =bm

w =m

i=1

bi n

j=1aijxj

=

m

i=1bi

n

j=1m

i=1aij

cj

xj

= c1x1 + + cnxn +

mi=1

bi

Chen CL 86

Phase I:


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to minimize w, original cost function is treated as a constraint

and the elimination step also executed for it

w is a sum of AVs, its minimum value is zero( AVs become nonbasic at the end of Phase I)

ifw = 0iterations continued for minimizing f ifw = 0no feasible solution exists for original problem

Phase II: same as before

Chen CL 87

Use AVs for Type Constraints


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Max: z=y1+ 2y2 = f

s.t. 3y1+ 2y2 12

2y1+ 3y2 6

y1 0, y2:free in sign

y2 = y+2 y

2 x2 x3;

x4 : slack;

x5 : surplus;

x6 : AV

Chen CL 88

Min: w=x6= 6 2x1 3x2+ 3x3+ x5


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s.t. 3x1+ 2x2 2x3+ x4= 12

2x1+ 3x2 3x3 x5+ x6= 6

f x1 2x2+ 2x3= 0, xi 0

Basic f w x1 x2 x3 x4 x5 x6 b bi/aicx4 0 0 3 2 2 1 0 0 12

122 = 6

x6 0 0 2 3 3 0 1 1 6 6

3= 2

f 1 0 1 2 2 0 0 0 0w 0 1 2 3 3 0 1 0 6

x4 0 0 5

3 0 0 1 2

3 23 8

82/3= 12

x2 0 0 2

3 1 1 0 13

13 2

f 1 0 13

0 0 0 23

2

3 4

w 0 1 0 0 0 0 0 1 0 Phase I ENDx5 0 0

52 0 0

32 1 1 12

x2 0 0 3

2 1 1 1

2 0 0 6 f 1 0 2 0 0 1 0 0 12 Phase II END

Chen CL 89

Use of AVs for Equality Constraints


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Max: z=x1+ 4x2 = f

s.t. x1+ 2x2 5

2x1+ x2= 4

x1 x2 0 (infeasible)

Chen CL 90

Min: f= x1 4x2


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s.t. x1 + 2x2 + x3 = 5

2x1 + x2 + x5 = 4

x1 x2 x4 + x6 = 3

xi 0

Basic f w x1 x2 x3 x4 x5 x6 b bi/aicx3 0 0 1 2 1 0 0 0 5

51= 5

x5 0 0 2 1 0 0 1 0 4 42= 2x6 0 0 1 1 0 1 0 1 3

31= 3

f 1 0 1 4 0 0 0 0 0w 0 1 3 0 0 1 1 0 7

x3 0 0 0 3

2 1 0 12 0 3

x1 0 0 1 1

2 0 0 1

2 0 2x6 0 0 0

32 0 1

12 1 1

f 1 0 0 72 0 0 1

2 0 2w 0 1 0 32 0 1

32 0 1 Phase I END

Chen CL 91

Use of AVs: Unbounded Problems


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Min: f= 3x1+ 2x2

s.t. x1 x2 0

x1+ x2 2 x1, x2 0


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Chen CL 94

Implication of Degenerate Basic Feasible Solution


95/202

Max: z= x1 +4x2

s.t. x1 +2x2 5

2x1 +x2 4

2x1 +x2 4

x1 x2 1

Min: f= x1 4x2

s.t. x1 +2x2 +x3 = 5

2x1 +x2 +x4 = 4

2x1 +x2 x5 +x7 = 4

x1 x2 x6 +x8 = 1

xi 0

Chen CL 95

Basic f w x1 x2 x3 x4 x5 x6 x7 x8 b bi/aic


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x3 0 0 1 2 1 0 0 0 0 0 5 5

1= 5

x4 0 0 2 1 0 1 0 0 0 0 4 4

2= 2

x7 0 0 2 1 0 0 1 0 1 0 4 4

2= 2

x8 0 0 1 1 0 0 0 1 0 1 1 1

1= 1

f 1 0 1 4 0 0 0 0 0 0 0

w 0 1 3 0 0 0 1 1 0 0 5x3 0 0 0 3 1 0 0 1 0 1 4

43

x4 0 0 0 3 0 1 0 2 0 2 2 2

3

x7 0 0 0 3 0 0 1 2 1 2 2 2

3x1 0 0 1 1 0 0 0 1 0 1 1

f 1 0 0 5 0 0 0 1 0 1 1

w 0 1 0 3 0 0 1 2 0 3 2


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Chen CL 97



98/202

x3, x4: slack; x5, x6: surplus; x7, x8: artificial

3rd tableau: basic variablex4 has zero value

degenerate basic feasible solution

x5

should become basic

x5

is pivot column

The element in pivot column (x5) and the degenerate variable row

(x4, the 2nd row) is positive (1)

the (2nd) row must always be the pivot row (min ratio is zero)

2nd row is pivot row (x4 will become a NB)new basic feasible solution will be degenerate (final tableau)

Chen CL 98



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The only way the new feasible solution can be nondegenerate is

when the element in pivot column and the degenerate variable rowis negative

The final solution:

NBs: x4=x6=x7=x8= 0

BVs: x1= 53, x2=

23, x3= 2, x5= 0

f= 133 z= 13

3

Chen CL 99

Alternate Simplex Method:Th Bi M M th d


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The Big-M MethodFor , = type constraints, augment fwith AVs having Big coeff.s

Max: z=y1+ 2y2

s.t. 3y1+ 2y2 12

2y1+ 3y2 6

y1 0, y2: unrestricted in sign Max: z=x1+ 2x2 2x3

s.t. 3x1+ 2x2 2x3+ x4= 12

3x1+ 3x2 3x3 x5+ x6= 6, xi 0

Min: f = x1 2x2+ 2x3+ M x6 (M= 10?)

= x1 2x2+ 2x3+ 10(6 2x1 3x2+ 3x3+ x5)

= 60 21x1 32x2+ 32x3+ 10x5

Chen CL 100

Basic f x1 x2 x3 x4 x5 x6 b bi/aic12


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x4 0 3 2 2 1 0 0 12 12

2 = 6

x6 0 2 3 3 0 1 1 6 6

3= 2

f 1 21 32 32 0 10 0 60x4 0

53 0 0 1

23

23 8

82/3= 12

x2 0 2

3 1 1 0 13

13 2

f 1 13 0 0 0 23

323 4

x5 0 5

2 0 0 3

2 1 1 12

x2 0 3

2 1 1 1

2 0 0 6

f 1 2 0 0 1 0 10 12

Chen CL 101

Simplex Method in Matrix Form


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Min: f = cTx

s.t. Ax b

x 0

Min: f = cTxs.t. Ax + xs = b

x 0, xs 0

Chen CL 102


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0 A I

1 cT 0

f

x

xs

=

b

0

Basic f x xs RHS

xs 0 A I b

f 1 cT 0 0

Chen CL 103


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b =

A I

xxs

=

B N

xBxN

= Bx

B+ N x

N=0 = BxB x

B = B1b

f = cTB

xB

+ cTN

xN

= cTB

xB

= cTB

B1b

Chen CL 1040 B

f

=

b


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1 cT

B

xB

0

fxB

= 0 B1 cT

B

1b0

= cTBB1 1

B1 0b

0

= cT

BB1b

B1b

Chen CL 105

f


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0 A I

1 cT

0

f

x

xs

=

b

0

cT

BB1 1

B1 0

1 cT cTBB1A cTBB10 B1A B1

fxxs

=cTBB1b

B1b

Basic f x xs RHS

xB

0 B1A B1 B1b

f 1 cT cTB

B1A cTB

B1 cB

B1b

Chen CL 106

Example


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Min: f= 6 4 0 0 0

x1

x2

x3

x4

x5

s.t.9 4 1 0 05 8 0 1 0

1 1 0 0 1

x1

x2x3

x4

x5

=360400

55


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Chen CL 108

0 1 8


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B1 =

0 1383

0 13 53

1 53 523

xB

= B1b =

0 13

83

0 13 53

1 5

3 52

3

360

400

55

=

403

1253

220

3

f = cT

BxB

=

6 4 0

403

1253

2203

= 740

3


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111/202

Chen CL 111

Procedures


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Initialization:

Initial canonical form

Improvement:

Compute cj dj, select xEwith most negative cj dj

Select xV

with least ratio min bki

aiE

, aiE

>0

Replace aV

ofxV

with aE

ofxE

find the new Bnew

Optimality Check:

compute new cj dj by using Bnew stop ifcj dj 0 j N

xB

=B1newb, f =c

BxB


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Chen CL 113


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xB

=

x3

x4

x5

xN =

x1

x2 c

B =

0

0

0

cN

=

6

4

B=

1 0 0

0 1 0

0 0 1

B1 =

1 0 0

0 1 0

0 0 1


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116/202


117/202

Chen CL 117


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xN

= x2x3

cT

N cT

BB

1

N = 4 0 6 0 019 0 0

5

9 1 019 0 1

4 1

8 01 0

=

4 0

83 23

=

43

23

43 < select x2 as xE

Chen CL 118


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xB = x1

x4x5

= B1b B1a

Ex

E

= 19 0 0

59 1 0

19 0 1

360

400

55

19 0 0

59 1 0

19 0 1

4

8

1

xE

= 40 49xE

200 52

9xE15 59xE

ifx

E = min

40

4/9= 90, 20052/9,

155/9 = 27

= 27

x5 = 0 (=xV, become a NB)

Chen CL 119


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xB

=

x1

x4

x2

xN =

x3

x5 c

B =

6

0

4

cN

=

0

0

B=

9 0 4

5 1 8

1 0 1

B1 =

15 0

45

35 1

325

15 0 9

5

Chen CL 120


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xN

=x3

x5

cTN

cTB

B1N =

0 0

6 0 4

1

5 0 4

535 1

325

15 0 9

5

1 00 00 1

= 0 0 25 125 = 25 125

Chen CL 121

B1b

15 0 45

3 1 32

360

400


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xB

= B 1b=

35 1

325

15 0 9

5

400

55

=

28

44

27

=

x1

x4

x2

(x3=x5= 0)

f = cTB

xB

=

6 0 42844

27

= 276

Chen CL 122

Quadratic Programming


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Min: f = cTx + 12xTQx

s.t. Ax b x 0 Min: f = cTx + 12x

TQx

s.t. Ax + s = b x 0

L = cTx + 12xTQx + T(Ax + s b) Tx

KKC: L = c + Qx + AT = 0

Q AT I 0A 0 0 I

x

s

=

cb

Chen CL 123

Add Artificial Variables, v, on these equality constraints,

Use two-phase method to solve the problem:


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Min: w=n

i=1

vi

s.t. Q AT I 0 I

A 0 0 I 0

x

s

v

=cb

Chen CL 124

QP: Example


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Min: f= 3x1 x1x2+ x21+ 2x

22

s.t. 3x1+ x2 5

x1, x2 0

Min: z=

3 0x1

x2

+ 12 x1 x2 2 1

1 4

x1x2

s.t. 3 1x1

x2 5

x1

x2

0

0


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127/202

Chen CL 127

Duality: Start from Lagrangian andThe Kuhn-Tucker Conditions (Min,)


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The Kuhn Tucker Conditions (Min,)

Min: f(x)s.t. gj(x) 0 j = 1, , m

hk(x) = 0 k= 1, , ; x 0

L(x, ) = f(x)+

mj=1

jgj(x) +

k=1

m+khk(x)

KTC: gj(x) 0 j = 1, , m (x is feasible)

hk(x) = 0 k= 1, ,

jgj(x

) = 0 j = 1, , m

xf(x) +

mj=1

jxgj(x) +

k=1

m+kxhk(x) = 0

j 0; m+k unrestricted in sign

Chen CL 128

Saddle Point Theorem (Min,)

L( ) L( ) L( )


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L(x, ) L(x, ) L(x, )

Proof:

Min: f(x) (x 0)

s.t. gj(x) 0 j = 1, , m (skip h now)

L(x, ) = f(x)+m

j=1 jgj(x)NC on x:

Lxi

x,

=

fxi

x,

+j

j

gjxi

x,

= 0

x 0 Lxix, 0 xi Lxix, = 0

(ifxi >0

Lxi

x,

= 0)

(ifxi = 0

Lxi

x,

>0)

L(x, ) L(x, )

Chen CL 129

0 df(x, ) =i

f

xi

x,

dxi (min prob)


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i x ,

= i j j gj

xix, dxi=

j

ji

gjxi

x,

dxi

= j jdgj(x,

) 0

j 0 j

NC: j 0 Ljx,

=gj 0 j Ljx,

= 0

L(x, ) L(x, ) L(x, ) L(x, ) L(x, )

Chen CL 130

Saddle Point and Properties (Min,)


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xL(x, ) = 0 L(x

, ) = 0

xL(x, ) 0 L(x

, ) 0

Chen CL 131

The Max-Min Problem (Min,)


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L(x

, ) L(x

,

) L(x,

)(x, ) defines a saddle pt of the Lagrangian

If define the Lagrangian in terms of alone

L() = minx

L(x, ) = L(x, )

sinceL() is a maximum w.r.t. at the optimum

max

L() = max

minx

L(x, ) gives (x, )

similary min

x

L(x) = min

x

max

L(x, ) gives (x, )

Chen CL 132

The Primal and Dual Problems (Min,)


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Min:f(x)

s.t. gj(x) 0 j = 1, , m

hk(x) = 0 k= 1, ,

L(x, ) = f(x)+

mj=1

jgj(x) +

k=1

m+khk

Primal minx

max

L(x, ) = minx

L(x) x 0

Dual max

minx

L(x, ) = max

L() 0

Chen CL 133

Duality (Min,)


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Min: f(x)s.t. gj(x) 0 j = 1, , m

L(x, ) = f(x)+m

j=1jgj(x)

L(x, ) L(x, ) =f(x)

L(x, )

max

minx

L(x, )

L(x,)() x,

minx

max

L(x, ) L(x,

)f(x)

x,

() f(x) f(x)


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136/202

Chen CL 136

A Linear Programming Example (Min,)


137/202

Min: fp(x) = 4x1 x2+ 50

s.t. g1(x) = x1 x2 2

g2(x) = x1+ 2x2 8

g3(x) = x1+ x2 10g4(x) = 5x1+ x2 5

c = 41 A =

1 1

1 2

1 1

5 1

b =

2

8

10

5

Chen CL 137

Dual Problem:

Max: L() = 21 82 103 54+ 50

( )


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s.t. g1() = 1+ 2+ 3 54 4

g2() = 1+ 22+ 3+ 4 1i 0

=

73

53 0 0

TL() = 32

xL(x, ) =

41

+ 1 1 1 51 2 1 1

7/3

5/3

0

0

=

00

1,

2= 0 x

1 x

2 = 2 x

1+ 2x

2 = 8

x =

4

2

fp(x) = 3 2


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140/202


141/202

Chen CL 141

Duality in Separable Programming (I)


142/202

Min: f(x) =

n

i=1 aixi+1

2

bix2i (separable obj)

s.t. gj(x) =n

i=1

cjixi j = 1, , m

Li() = minxi aixi+12bix2i +m

j=1jcjixi

xiLi() = ai+ bixi+m

j=1

jcji = 0

xi = ai mj=1 jcjibi

(xi xi xui; xi = x

i ifbi= 0)

Max: Li() with quadratic obj and linear constraints

Chen CL 142

Duality in Separable Programming (II)


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Min: f(x) =

n

i=1 aixi+1

2

bix2i (separable obj)

s.t. gj(x) =n

i=1

cjixi+

1

2djix

2i

j = 1, , m

Li() = minxi aixi+12bix2i +m

j=1j cjixi+12djix2i

xiLi() = ai+ bixi+m

j=1

j(cji+ djixi) = 0

xi = ai mj=1 jcjibi+

mj=1 jdji

(xi xi xui; xi = x

i if denominator = 0)

Max: Li() with quadratic obj and quadratic constraints

Chen CL 143

Duality in Linear Programming(A Simpler Approach Again)


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Primal Min: fp=c1x1+ + cnxn=n

j=1

cjxj =cTx

s.t. a11x1+ + a1nxn b1

(Ax b)

am1x1+ + amnxn bm

Dual Max: zd= b1y1 bmym= m

i=1 biyi=bTy

s.t. a11y1+ + am1ym c1

(ATy c)

a1ny1+ + amnym cn

Chen CL 144

Example: Dual of an LP Problem

Min: fp= 5x1+ 2x2


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fp 1 + 2

s.t. 2x1+ x2 9

x1 2x2 2

3x1+ 2x2 3

x1, x2 0

Max: zd= 9y1 2y2 3y3

s.t. 2y1+ y2 3y3 5

y1 2y2+ 2y3 2

y1, y2, y3 0

Chen CL 145

Treatment of Equality ConstraintsMin: fp= x1 4x2

s t x1 + 2x2 5


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s.t. x1+ 2x2 5

2x1+ x2= 4

x1 x2 1, x1, x2 0

Min: fp= x1 4x2

s.t. x1+ 2x2 52x1+ x2 4

2x1 x2 4

x1+ x2 1, x1, x2 0

Max: zd= 5y1 4y2+ 4y3+ y4

s.t. y1+ 2y2 2y3 y4 1

2y1+ y2 y3+ y4 4, y1, y2, y3, y4 0


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Chen CL 147

Primal Solution from Dual Solution


148/202

Theorem: Relation between primal and dual

Letxandy be in the constraint set (feasible points) of primal anddual problems, respectively. Then the following relations hold:

zd(y) fp(x) If zd = fp, then x and y are solutions for primal and dual

problems If primal is unbounded, the corresponding dual is infeasible, and

vice versa

If primal is feasible and dual is infeasible, then primal is

unbounded and vice versa

Chen CL 148

a11y1+ + am1ym c1 x1 a1ny1+ + amnym cn xn


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x1(a11y1+ + am1ym) + +xn(a1ny1+ + amnym) c1x1 cnxn

y1

b1 (a11x1+ + a1nxn) + +ym (am1x1+ + amnxn)

bm

c1x1 cnxn

b1y1 bmym zd: dual profit fcn

c1x1+ + cnxn fp: primal cost fcn

zd fp x and y with fp=zd minimize fp while maximizing zd

maximum dual profit fcn is also the minimum primal cost fcn

Chen CL 149

Theorem: Primal and dual solutions

Let both primal and dual have feasible points. Then both have

solution in x and y respectively and zd(y) =fp(x)


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y (y) p( )

Theorem: Solution of primal from dualIf theith dual constraint is a strict inequality at optimum, then the

corresponding ith primal variable is NB, i.e., it vanishes. Also,

if the ith dual variable is basic, then the ith primal constraint is

satisfied at equality.

Ifmi=1

aijyi > cj then xj = 0

If yi > 0 then

nj=1

aijxj = bi

Chen CL 150

Primal and Dual Solutions: Example


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Minx: fp= 5x1+ 2x2 (fp = 18)

s.t. 2x1+ x2 9

x1 2x2 2

3x1+ 2x2 3, x1, x2 0

Basic fp x1 x2 x3 x4 x5 b bi/aic

x2 0 0 1 0.2 0.4 0 1x1 0 1 0 0.4 0.2 0 4

x5 0 0 0 0.8 1.4 1 13

fp 1 0 0 1.6 1.8 0 18

Chen CL 151


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Dual: Max: zd= 9y1 2y2 3y3

s.t. 2y1+ y2 3y3 5

y1 2y2+ 2y3 2

Max: zd= 9y1 2y2 3y3s.t. 2y1+ y2 3y3 y4+ y6= 5

y1+ 2y2 2y3+ y5= 2

y4: surplus, y5: slack, y6: artificial

Chen CL 152

Basic zd w y1 y2 y3 y4 y5 y6 b bi/aic

y6 0 0 2 1 3 1 0 1 5 5

2

y5 0 0 1 2 2 0 1 0 2


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y5

zd 1 0 9 2 3 0 0 0 0

w 0 1 2 1 3 1 0 0 5

y1 0 0 1 0.5 1.5 0.5 0 0.5 2.5 2.5

0.5

y5 0 0 0 2.5 3.5 0.5 1 0.5 4.5 4.5

2.5

zd 1 0 0 2.5 16.5 4.5 0 4.5 22.5

w 0 1 0 0 0 0 0 1 0 I ENDy1 0 0 1 0 0.8 0.4 0.2 0.4 1.6

y2 0 0 0 1 1.4 0.2 0.4 0.2 1.8

zd 1 0 0 0 13 4 1 4 18 II END

BVs: y1= 1.6, y2= 1.8

NBs: y3= 0, y4= 0, y5= 0

Max: zd = 18 (= fp )


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Chen CL 154

Use of Dual Tableau to Recover PrimalSolution


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Primal and Dual tables of previous example:

Basic fp x1 x2 x3 x4 x5 b bi/aic

x2 0 0 1 0.2 0.4 0 1

x1 0 1 0 0.4 0.2 0 4

x5 0 0 0 0.8 1.4 1 13fp 1 0 0 1.6 1.8 0 18

Basic zd w y1 y2 y3 y4 y5 y6 b bi/aic

y1 0 0 1 0 0.8 0.4 0.2 0.4 1.6

y2 0 0 0 1 1.4 0.2 0.4 0.2 1.8

zd 1 0 0 0 13 4 1 4 18 II END

Chen CL 155

Some notes:

Elements in row of dual tableau match elements in last column

of primal tableau


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of primal tableau

Reduced cost coefficients in final primal tableau match the dual

variables

To recover primal variables from final dual tableau, we use

reduced cost coefficients in columns corresponding to slack or

surplus variables

Reduced cost coefficient in column y4 is precisely x1 and that incolumny5 is precisely x2

Reduced profit coefficients corresponding to slack and surplus

variables in final dual tableau give values of primal variables

Chen CL 156

Theorem: recovery of primal solution from dual tableauLet dual of the standard primal (min cTx s.t. Ax b) be solvedby standard Simplex method. Then value of ith primal variable


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equals the reduced cost coefficient of the slack or surplus variable

associated with the ith dual constraint in the final dual tableau.In addition, if a dual variable is nonbasic, then its reduced profit

coefficient equals the value of slack or surplus variable for the

corresponding primal constraint.

Chen CL 157

Use of Final Primal Tableau to RecoverDual Solution: Example


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Min: fp= x1 4x2s.t. x1+ 2x2 5

2x1+ x2= 4

x1 x2 1

x1, x2 0

Min:fp = x1 4x2

s.t. x1 + 2x2 + x3 = 5

2x1 + x2 + x4 = 4

2x1 + x2 x5 + x7 = 4

x1 x2 x6 + x8 = 1


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Chen CL 159

y1= 0, for 1st constraint (reduced cost coeff ofx3, slack var.)

y2= 53, for 2nd constraint (reduced cost coeff ofx4, slack var.)


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y3= 0, for 3rd constraint (reduced cost coeff ofx5, surplus var.)

y4= 73, for 4th constraint (reduced cost coeff ofx6, surplus var.)

dual variable for equality constraint 2x1+ x2= 4 is y2 y3= 53

y4 = 7

3 is the dual variable for the 4th constraint written as

x1+ x2 1 and NOT forx1 x2 1

Chen CL 160

Use of Final Primal Tableau to RecoverDual Solution: Example (2)


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Min: fp= x1 4x2

s.t. x1+ 2x2 5

2x1+ x2= 4

x1 x2 1x1, x2 0

Min: fp

= x1

4x2

s.t. x1 + 2x2 + x3 = 5

2x1 + x2 + x5 = 4

x1 x2 x4 + x6 = 1


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Chen CL 162

slack: x3 surplus: x4 artif.: x5, x6

BVs: x1= 53, x2=

23, x3= 2


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NBs: x4 = x5 = x6 = 0

Min: fp = 133 (same result)

y1= 0, for 1st constraint (reduced cost coeff ofx3, slack var.)

y2= 53, for 2nd constraint (reduced cost coeff ofx5, artif var.)

y3= 73, for 3rd constraint (reduced cost coeff ofx4, surplus var.)

Same resultwe do not have to replace an equality constraint bytwo inequalities in standard Simplex method

Reduced cost coefficient corresponding to the artificial variable

associated with the equality constraint gives the value of dual

variable for the constraint


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Chen CL 164

NC: cj+mi 1

yiaij+ vj = 0 j = 1 n (b)


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i=1

yi n

j=1

aijxj bi = 0 i= 1 m (c)

vjxj = 0 xj 0 j = 1 n (d)

yi 0 i= 1 m (e)

vj 0 j = 1 n (f)

(b) cj+mi=1

yiaij = vj

mi=1

yiaij cj (g)

yjs are feasible soln for dual constraints


166/202

Chen CL 166

Post-Optimality Analysis


167/202

Sensitivity Analysis:

the study of discrete parameter changes

Changes in the cost function coefficients, cj

Changes in the resource limits, bi

Changes in the constraint coefficients, aij

The effect of including additional constraints

The effect of including additional variables

Chen CL 167

Changes in Resource Limits


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Theorem: Lagrange Multiplier Values

In the final tableau, the Lagrange multiplierfor theith constraintequals the reduced cost coefficient in the slack or artificial

variable column associated with the ith constraint

For a minimization problem with type or equality constraintsei: RHS parameter, yi: Lagrange multiplier

z

ei= yi 0

Chen CL 168

Recovery of Lagrange Multipliers from Final Tableau


169/202

Max: z = 5x1 2x2 (= f)

s.t. 2x1+ x2 9

x1 2x2 2

3x1+ 2x2 3

x1, x2 0

BV: x1= 4, x2= 1, x5= 13

NB: x3= 0, x4= 0

z = 18

Chen CL 169

Basic z x1 x2 x3 x4 x5 b bi/aic

x3 0 2 1 1 0 0 9 9

2

x 0 1 2 0 1 0 2 6


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x4 0 1 2 0 1 0 2 3

x5 0 3 2 0 0 1 3 z 1 5 2 0 0 0 0

x3 0 0 5 1 2 0 5 5

5

x1

0 1

2 0 1 0 2

x5 0 0 4 0 3 1 9

z 1 0 8 0 5 0 10

x2 0 0 1 0.2 0.4 0 1

x1 0 1 0 0.4 0.2 0 4

x5 0 0 0 0.8 1.4 1 13

z 1 0 0 1.6 1.8 0 18

c

1 c

2 c

3 c

4 c

5


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Chen CL 171

Recovery of Lagrange Multipliers from Final Tableau


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Max: z =x1+ 4x2 (= f)

s.t. x1+ 2x2 5

2x1+ x2= 4

x1 x2 1

x1, x2 0

Max: z= x1 + 4x2

s.t. x1 + 2x2 + x3 = 52x1 + x2 + x5 = 4

x1 x2 x4 + x6 = 1

xi 0

Chen CL 172


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?? ??

BV: x1= 53, x2= 23, x3= 2

NB: x4= 0, x5= 0, x6= 0

z = 133

Chen CL 173

Basic z w x1 x2 x3 x4 x5 x6 b bi/aic

x3 0 0 1 2 1 0 0 0 5 5

1

x5 0 0 2 1 0 0 1 0 4 4

2

0 0 1 1 0 1 0 1 1 1


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x6 0 0 1 1 0 1 0 1 1 1

1

z 1 0 1 4 0 0 0 0 0w 0 1 3 0 0 1 0 0 5

x3 0 0 0 3 1 1 0 1 4 4

3

x5 0 0 0 3 0 2 1 2 2 2

3

x1 0 0 1 1 0 1 0 1 1 z 1 0 0 5 0 1 0 1 1

w 0 1 0 3 0 2 0 3 2

x3 0 0 0 0 1 1 1 1 2

x2 0 0 0 1 0 2

313

23

23

x1 0 0 1 0 0 1313

13

53

z 1 0 0 0 0 7353

73

133

w 0 1 0 0 0 0 1 1 0 I,II END

Chen CL 174

1st constraint ( 5) y1 = 0 (c3 in slack var col x3)

2nd constraint ( = 4) y2 = 5 (c5 in art var col x5)


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2nd constraint ( = 4) y2 = 3 ( c5 in art var colx5)

3rd constraint ( 1) y3 = 73 (c6 in art var colx6)or3rd constraint ( 1) y3 =

73

z

e1 = 0

z

e2 =

5

3

z

e3 =

7

3

Ife2: 5 4 z: 13

3 13

3 + (53) (1) = 6

Ife3: 1 2 z: 133 13

3 + (73) (1) = 2

Chen CL 175

Ranging Right-hand Side Parameters

When RHS of a constraint is changed


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constrained boundary moves parallel to itself

changing feasible region(iso-cost lines do not change)

If the changes are within a certain range, the sets of basic and

nonbasic variables do not change Limits on changes in resources

Let k be possible change in RHS bk of kth constraint. If ksatisfies the following inequalities, then no more iterations of the

Simplex method are required to obtain solution for the alteredproblem

maxi

bi

aij; aij >0

i mini

bi

aij; aij


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a type constraint,

or the artificial variable for an equality constraint, or typeconstraint

k: possible change in RHS ofkth constraint;

the slack or the artificial variable for kth constraint determines

index j of the column whose elements are used in the above

inequalities

Note: new RHS parameters bi due to a change ofk in bk

bi =bi+ ka

ij i= 1 m

Chen CL 177

Ranges for Resource Limits: type


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Max: z = 5x1

2x2

(=

f)

s.t. 2x1+ x2 9

x1 2x2 2

3x1+ 2x2 3, x1, x2 0

Max: z= 5x1 2x2

s.t. 2x1 + x2 + x3 = 9x1 2x2 + x4 = 2

3x1 + 2x2 + x5 = 3, xi 0

Chen CL 178


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Chen CL 179


x2 0 0 1 0.2 0.4 0 1

x1 0 1 0 0.4 0.2 0 4


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x1 0 1 0 0.4 0.2 0 4

x5 0 0 0 0.8 1.4 1 13z 1 0 0 1.6 1.8 0 18

For 1st constraint, x3 is the slack variable

j for1 is x3 (the third column)

No negative element in 3rd column

no upper limit on 1

Chen CL 180

max

1

0.2,

4

0.4,

13

0.8

1

5 1


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4 b1 max

4

0.2,

13

1.4

2 min

1

0.4

9.286 2 2.5

7.286 b2 4.5max

13

1

3

13 3

10 b1

Chen CL 181

New values for design variables if2x1+ x2 910 (1= 1)

x2 = b1 =b1+ 1a

13

1 + (1)(0 2) 1 2


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= 1 + (1)(0.2) = 1.2

x1 = b2 =b2+ 1a23

= 4 + (1)(0.4) = 4.4

x5 = b3 =b3+ 1a

33

= 13 + (1)(0.8) = 13.8 (F)

Note: other variables remain nonbasic (= 0)


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Chen CL 183

New values for design variables if1= 1 and 2= 1:

x2 = b1 =b1+ 1a

13+ 2a

14

= 1 + (1)(0 2) + (1)( 0 4) = 0 8


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= 1 + (1)(0.2) + (1)(0.4) = 0.8

x1 = b2 =b2+ 1a23+ 2a24

= 4 + (1)(0.4) + (1)(0.2) = 4.6

x5 = b3 =b3+ 1a

33+ 2a

34

= 13 + (1)(0.8) + (1)(1.4) = 15.2 (H)

Chen CL 184

Ranges for Resource Limits: =, type

Max: z =x1+ 4x2

t + 2 5


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s.t. x1+ 2x2 5

2x1+ x2= 4

x1 x2 1, x1, x2 0

Chen CL 185

Basic z x1 x2 x3 x4 x5 x6 b bi/aic

x3 0 0 0 1 1 1 1 2

x2 0 0 1 0 2

313

23

23


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2 3 3 3 3

x1 0 1 0 0 13 13 13 53z 1 0 0 0 73

53

73

133

x3: slack (1st constraint), x4: surplus (3rd)

x5, x6: artificial variables (2nd and 3rd constraints)

Chen CL 186

2 1 (j = 3)

3 b1


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max2/31/3, 5/31/3 2 min 21 (j = 5)2 b2 6

max

21, 5/31/3

3 min

2/32/3

(j = 6)

1 b3 2

Chen CL 187

New solution if2x1+ x2= 45

x3 = b1 =b1+ 2a

15

= 2 + (1)(1) = 1


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= 2 + (1)( 1) = 1

x2 = b2 =b2+ 2a25

= 23+ (1)(13) = 1

x1 = b3 =b3+ 2a

35

=

5

3+ (1)(

1

3) = 2

Chen CL 188

New solution ifx1 x2 12

x3 = b1 =b1+ 3a

16

= 2 + (1)(1) = 3


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= 2 + (1)(1) = 3

x2 = b2 =b2+ 3a26

= 23+ (1)(23) = 0

x1 = b3 =b3+ 3a

36

=

5

3+ (1)(

1

3) = 2 (C)

Chen CL 189

Ranging cost Coefficients

ckck + ck : admissible range on ck such that the optimum


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ckck+ ck : admissible range on ck such that the optimum

design variables are not changed

Limits on change ck depend on whether or not xk is a basic

variable at optimum

Note: feasible region for the problem does not change

Chen CL 190

Theorem: range for cost coefficient of NBsLetck be such that xk is not a basic variable.

Ifck ck+ ck, ck ck ,then the optimum solution (design variables and cost function)


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( g )

does not change(ck is the cost coefficient corresponding to xk in final tableau)

Chen CL 191

Theorem: range for cost coefficient of BVsLetck be such that x

k is a basic variable and x

k=b

r.

Then the range on changeck inck for which the optimum design

variables do not change is given as


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maxj

cj

arj; arj 0

The new cost function is f + ckxk

arj : element in rth row and jth column of final tableau.

Index r is determined by the row that determines xkIndex j corresponds to each of the nonbasic columns excluding

artificial columns

if no arj >0 no upper limit,if no arj


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Max: z = 5x1 2x2 (= f)

s.t. 2x1+ x2 9

x1 2x2 2

3x1+ 2x2 3, x1, x2 0


x2 0 0 1 0.2 0.4 0 1

x1 0 1 0 0.4 0.2 0 4

x5 0 0 0 0.8 1.4 1 13

z 1 0 0 1.6 1.8 0 18c1 c

2 c

3 c

4 c

5

Chen CL 193

Max: z= 5x1 2x2 (c1= 5, c2= 2)

2nd row determines BV x1r= 23rd, 4th columns are NBsj = 3, 4


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c1 min

1.60.4,

1.80.2

= 4

c1 1

znew = z + c1x

1 (c1= 54)

= 18 + (1)(4) = 14

Chen CL 194

1rd row determines BV x2r= 13rd, 4th columns are NBsj = 3, 4

max 1.80.4 = 4.5 c2 min

1.60.2 = 8


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0.4 0.22.5 c2 10znew = z

+ c2x2 (c2= 2 3)

= 18 + (1)(1) = 17

Chen CL 195

Ranges for Resource Limits: =, type

Max: z = x1 + 4x2


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Max: z =x1+ 4x2

s.t. x1+ 2x2 5

2x1+ x2= 4

x1 x2 1, x1, x2 0

Basic z x1 x2 x3 x4 x5 x6 b bi/aic

x3 0 0 0 1 1 1 1 2

x2 0 0 1 0 2

313

23

23

x1 0 1 0 0 13

13

13

53

z 1 0 0 0 7353

73

133

c1 c2 c

3 c

4 c

5 c

6

Chen CL 196

3rd row determines x1 as a basic (r= 3, j = 4)

max

7/31/3

= 7 c1

8 c1


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8 c1

znew = z + c1x

1 (c1= 12)

= 133 + (1)(53) = 6

2nd row determines x2 as a basic (r= 2, j = 4)

c2 3.5

c2 0.5

znew = z

+ c2x2 (c2= 43)

= 133 + (1)(23) =

113

Chen CL 197

Changes in the Coefficient Matrix

aijaij+ aij feasible region changes


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j j j

Q: find znew with minor computations

Case 1: when the change is associated with a nonbasic variable Case 2: when the change is associated with a basic variable

Theorem: change associated with a NBLet j in aij be such that xj is not a BV, and k be the column index for theslack or artificial variable associated with the ith row.Define a vector

cB =

cB1 cB2 cBm

T

where cBi = cj if x

j = b

i , i = 1 m (index i corresponds to ith row thatdetermines optimum value of variable xj). Also define a scalar

R=

mr=1

cBrark

Chen CL 198

With this notation, if aij satisfies the following inequalities

ij

cjR

cjR

when R

0


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ij when R

0

then the optimum solution (design variables and cost function) do not change

whenaij aij+ aij.

Also, ifR= 0, then the solution does not change for any value ofaij

Note: re-solve the problem if the inequalities are not satisfied

Chen CL 199

Theorem: change associated with a BVLet j in aij be such that xj is a BV and let x

j = b

t (t is the row index

determining optimum value ofxj). Let index k and scalar R be defined as inprevious Theorem. Let aij satisfy the following inequalities:


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maxr=t

brAr , Ar 0withAr =b

tark b

ratk, r= 1 m; r=t, and

maxq cq

Bq , Bq >0 aij minr=t cq

Bq , Bq 0

The upper and lower limits on aij do not exist if the corresponding

denominators do not exist. If aij satisfies the above inequalities, then the

optimum solution of the changed problem can be obtained without any further

iterations of the Simplex method.

Chen CL 200

Homework

1. Solve the LP problems and verify the solutions graphically


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(1) Max: z=x1+ 4x2s.t. x1+ 2x2 5

x1+ x2= 4

x1 x2 3

x1, x2 0

(2) Max: z= 4x1+ 2x2

s.t. 2x1+ x2 4x1+ 2x2 2

x1, x2 0


202/202

4 Linear Programming.pdf

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