optimization lecture 4

8/10/2019 optimization lecture 4

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Lecture 4

Transportation Pro ebl ms

as ca y ea s w t

the roblem which aims to fin

transpor

d the to

tat on pro em

best wa

,demand pfulfill the of usingointsdemand n

.

variable cost of shipping the p oneroduct from

,

asupp y po nt eman poto ont a s m ar

cons

r

should be taken into contrai sident ration.


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) The Definition of the Transportation Model1

transportatwe present the standard definiti ion model

transportation mod

on of the

the direct sense the seeks the deel termination

, .

,

Now

In

of trana spo plan of a single from a number of

to a number

rta

of

tion

.source destinations

com

s

modity

of su l at each and amount of at each

data of the incmo l dedel u

Level

The

demsource

1. and

.destinationunit cost of the from eatransport chation The commod2. sity o eurc

to each .destination


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there is only one can receive, aSince descomm tinationodity

its from one or moreof the model is to determine the amount to be

. The

sourdem cesobjective

and

shipped from each to eachsource destina such that the

total cost istransportation minimized.

tion

basic assumption of the model is that the

cost on a given route

transportation

directly proportionalis to the number of

The

units transportedtransportationdefinition of unit of will vary depending

.The

transportedon the

units of and must be consistent with our

.

The

commodi

supp

ty

l ey d mandtransportedefinition of a nid u t.


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depicts the model as a network with

and

transportation

.

Figure 1

n destinations

m sources

or is represented by

arc joining and represents the

.

The

a destination

a destination

A s a noource

a source

de

route through

which the is

amount of at is the at is

transported.

, .The ji demandsupply source i a destination j b

commodity

unit cost betweentransportat ano d si in .The ijsource desti cnationi

Figure 1


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) Formulating Transportation Problems2

represent the amo transportedunt fromto then the model of

Let ijes ination L

xource i P the problem is gitrans venpo generalrt lyi aa st on

Minimize = =

i ii 1

j jj 1

c xz

., , , = i in

jSubject to x a i 1 m

., , ,

m

i b j 1 nx =

., for all and

=

i 1

ij x 0 ji


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stipulatesthat the sum of the shipmentsn

a

canno exceerom s

that the sum of the shipmentsreq iresu

.=

m

j 1 source supp y

total sup ypl

to must satisfy i st .=1i

i a destination demand

model just described implies that the must atThe total supplyleast following the that isequal ,

i

LP total dema dn

a . nm

b

resulting formulationthe is called a, ==

Whenm

ii

n

11ja b

differs from the model above onl in the fact that all

.

con

It

Balanced Transportation Model

straintsare equations; that is,

== = mn

= ., , , , , ,= =1j i1


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has lants in and

( )

.

TheM Los An eles Detroit New OrleanG Auto Com an s

Standard Transportation ModExample e1 l

major distribution centers are located in and

capacities of

.Its

The

Denver Miami

the three plants during the next quarter are and , ,1000 1500

cars

quarterly at the two distribution centers are and cars

.

.The demands 2300 1400

1200

cost per car per mile itrain transportatio snThe approximately cents

mileage chart between the plants and distribution centers is as follows:

.

The

8


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milea e chart can be translated to cost er car at the rateThe

Example 1 - con

of cents per mile

.8

which represent in the generalijc model:

represent the plants anddistribution centers we let,

represent the number

of cars from totransported

.

ji

destsource i

x

ination jthe sum of the (Since total supply 100 50 1 0 )

ha ens to e ual the thetotal demand 2300 140

0 1200 3700

0 3700

transportation balanresulting model is ced.


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Example 1- con

s mo e as a eq cons ra n su :a yMinimize

1 2 1 2 1 21 1 2 2 3 380 215 100 108 102 68

Subject to

x x x x x x

11 12 1000x x

3 31 2

1200x x

1 2 3

1

1 1 1

2 32 2 2

Dem

1400

x x x

x x x

and Constraints

., for all andijx i 0 j


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Example 1- con

equalitythis model has all constraints:

Hence 11

1

2

xx

LP

T

12

2

2

80 215 100 108 102 68Minx

C Xx

z z

3

3

1

2Subj ect to

xx

10001500 211 1 0 0 0 00 0 1 1 0 0 10001500x

x

2300

=

2

1

2

3

23001 0 1 0 1 0x

x

32x

.,ij


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.Example 1- con

o t stan ar s:s

The1000 1000

LPua

1 2 3 4 5

T

1500 1500

1200 1200y y y y yMax w wY

2300 2300

1400 1400

1 2 4 53 2300 1401000y 1500y 120 y 00y yw

1

2

1 0 0 0 1 215y

yT A Y

3

4

0 1 0 0 1 108 y

y

C

0 50 1 0 1 68y


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transportation model

transportation

more compact method for representing the is to

use what we call the .

A

tableau

is a matrix form with its rows representing the and its columnstheIt ourc

dees

stina .tions

model can thus be summarized as shown in

, .

.The

ij

TablG e 1M


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( )Balanced TransportatioExample 2 n Model

suppose that the plant capacity is cars

,Ino a emano a s pup y

1300Example 1

( )

situation

instead of .

The

1500

is said to be unbalanced because the totalsupply( ) is strictly than the total ( )less .demand 33500 700

manner that will distribu

te the shortage quantity ( -3700 3500

exceeds a fictitious or dummy

.

,Since

deman supply sourd ce

can e a e w t ts capac ty equa top ant cars

is allowed under normal conditions to ship

.

, ,Thedummy plant

its production to all distribution ce snter .


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( )total demtotal su ly andppExample 2 - con

s mo e ecomes as:Minimize

1 2 11 1 2 2 3 32 2180 215 100 108 102 68

Subject to

x x x x x x

11 12xx 1000

3 31 2

1200x x

1 2 3

1

1 1 1

2 22

x x x

x x x

23

emand Constraints

1400., for all andji x i j0


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Example 2 - con

of the

the does not exist no physical shipping will occur and the

.

,Since

Detroitplan

pla

tnt

corres un tpon ng s zero

( ) has a capacity

transportat on c

of car

os

s

t .

shown by a tint overlay .Thedummy pla t 2n 00

Table 2


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the exceeds the we can add a destination

.

,If su demand dumml

Example 2 - con

that will absorb the differenceexample, that the at drops to cars

.

.For demand 1suppose 900Denver

cars shippAny

surplus

ed from a plant to a added distribution center

represent a quantity at that plant

.

dummy

un ranspor a on cosassoc a e s zero .e

Table 3


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Exercise 1

to p balanced transportationroduce a ? [ ]model

destina

Ans.tion No.

the to and are and respectively

[ ]

, .

orders at and will be and cars short.

Denver Miami

Ans. T

150 50

150he Denver Mia

dummy pla

5

nt

mi 0

each of the following ca( ) Inc ses indicate whether a or

should be added to the modelbalance

,

. destination

sourcedummy

( ) , , , and , , , . 1 2 3 41 2 3 4 b 10 ba 10 a 5 a 51 4 a 6 b 7 9b

( ) , and , , .

1 2 1 2 3b 25 b 30 b 102 a 30 a 4 4

. .


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) Basic feasible solution fortransportation model3

other problems awith supply points and demand points is easier to solve

balanced transportation model ,,

Unlike LPm n

although it h equalitasm n y constraints

basic methods balancedto find the for a

.

Three s TP

)

1 Northwest Corner Method

) Least - ost Met o

3 Vogel s Method


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manufactures office TheExecutive Furniture Corporation

Example 3

desks at three locations: , ,Des Moines Evansville Fort Lauderdale

firm distributes the desks through regiona The l

located in:

warehouses

of the monthly production of each factory

, ,

Estimates supply

costs are the same at the three factories so the only Production

ar constante no matter the quantity shipped

. Costs

the number of desks on each route tominimize total transportation


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.Example 3 - con

Cleveland00 units

Locations of factories and warehousesExecutive Furniture

(100 units)

supply

demand

Albuquerque

(200 units)demandEvansville(300 units)

un sdemand

Ft. Lauderdale(300 units)

supply

Figure 1


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.Example 3 - con

From To lbuquerque Boston Cleveland

es o nes

Evansville $8 $4 $3

Ft. Lauderdale $9 $7 $5


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.Example 3 - con

Minimize

1 1 1 2 2 2 3 3 31 2 3 1 2 3 1 2 3

Subject to

x x x x x x x x x

11 1

x x

2 31 100x

2 3

1 2 3

2 2

3 3 3

300x x x

1 2 31 1 1

Demand Constraints

300x x x

2 21 22 3 200

x x x

200x x x

., for all andji jix 0


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andTotal demandExample 3 - con. Total supply

ToFrom

Albuquerque Boston Cleveland FactorySupply

Des Moines

Factory

x11 $5 x12 $4 x13 $3

100Evansville

Factor

x21 $8 x22 $4 x23 $3300

Ft LauderdaleFactor

x31 $9 x32 $7 x33 $5 300

Warehouse

Demand300 200 200 700

Balanced

TransportationTable 5

Model


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) the byFind bfs Northwest Corner Method1

in and allocate units to shippingroutes as follows:

northwest cornerBegin

the of each row before moving down toExhaust supply

the

.

Exhaust of each column before moving todemand

the next column.

it take initials to make the

.

,ForExam ive stele 3 sshipping assignme tsn .


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in setting as as possible.large: Begin 11xStep 1

ass gn un s rom o . exhaust the from but leaves

eThis

11 uquerquelbuquerque

es o nessupply Dx

es Moines

es s s or .

move

We to the in thesecon samed ro co .mnw lu

From To Albuquerque Boston Cleveland Supply

Des Moinesx11 x12 x13

100100

Evansvillex21 $8 x22 $4 x23 $3

300

Ft Lauderdalex31 $9 x32 $7 x33 $5 300

Demand 300 200 200 700


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to the in the .second row same column: MovingSt ep 2

ass gn un s rom omeets

. .This

12 uqvansv e uerqueAlbuquerque s dem d

xan

ha units remaining so we move tosEvansville 100

next column

the right to

the o secondf the row.


Des Moinesx11 x12 x13

100100

Evansvillex21 $8 x22 $4 x23 $3

300200

Ft Lauderdalex31 $9 x32 $7 x33 $5 300

Demand 300 200 200 700


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to the in the .samsecond col e rumn ow: MovingStep 3

has now been exhaustedbut s

.

iThe22

Evansv Bille supp ostol ny

move

.

We next rodown vertically to the in columthe nw .Boston


Des Moines11 12 13

100100

Evansville 21 22 23 300200 100

Ft Lauderdale x31 32 x33 300

eman


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vertically to the in the .columthird row n: Down BostonStep 4

ass gn un s rom o fulfills and still has

.

eThis

22 or au er a eFort Lauderda

x os onBo leston s demand

units available.

uquerque os on eve an upp y

x11 $5 x12 $4 x13 $3

100x21 $8 x22 $4 x23 $3

200 100

x31 $9 x32 $7 x33 $5 au er a e100


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:Step 5

.

and .

sexhaust s s This32

Fort Lauder Cleveland demanddale su ly

pp

.The initial shipment schedule is now complete

uquerque os on eve an upp y

X11 $5 x12 $4 x13 $3

100

x21 $8 x22 $4 x23 $3

200 100

x31 $9 x32 $7 x33 $5 au er a e100 200

ill h th f ll i hi h i) Fi ll bf1 con


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we will have the following which is:) , ,Finally bfs1 - con.

and .

, , , ,

1 1 2 32 3 3 1 0

x x x x

o s s pp ng aso a s gnmeno ssc

1 21 1 22

5 8 4 7x x x x z

.2 33 35 $4200x

RouteUnitsShipped

Per unitCost ($) =

TotalCost ($)From To

D A 100 5 500,

E B100 4 400

F B 100 7 700

4,200


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( )) the byFind Lowest Costbfs Least - cost Met hod2

the cell with the lowest cost .IdentifyStep 1.

exceeding or

.

dsupply ema ; then cross out the or

that is exhausted b this assi nmentcolum

r

n

ow

or both .

nd

the cell with thelowest co froms the remainingt

FindStep 3.

a

.

RepeatStep 4. step 2 nd until all units have been allocated.3

) L t C t M th d2


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) Least - Cost Method2 - con.

s e ce so s p un s roowes cos cross mto and the first asrowoff,rs ClDes Moines eveland

s sat s e .

Des Mo sine

From To Albuquerque Boston Cleveland SupplyX $5 x $4 x $3

es o nes100

x 1 $8 x $4 x 3 $3vansv e

x31 $9 x32 $7 x33 $5 au er a e

emand

) h d2


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is again the cell so ship ucolumn

lowest cos nitsfrom to and

tcross o ff

,SecondEvan Cleveland 3sv

$3 100ille

as is satisfi d eCleveland

From To Albuquerque Boston Cleveland SupplyX11 $5 x12 $4 x13 $3

100x21 $8 x22 $4 x23 $3

100

x31

$9 x32

$7 x33

$5

Demand 300 200 200 700

) Least Cost Method2 con


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s e ce so s p un s romto andowes coscross off rcolumn ow, andrEvansville Boston 2

as an are sat s eEvansvi e Boston

rom o uquerque oston eve an upp y

es Moinesx11 $5 x12 $4 x13 $3 100

Evansvillex21 $8 x22 $4 x23 $3 300

Ft Lauderdalex31 $9 x32 $7 x33 $5

300

Demand 300 200 200 700

) L t C t M t dh2


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) Least - Cost Met doh2 - con.

s p un ts rom toas this is the only remaining cell to complete the allocations,na y u ort auq eruerqu ae e

$3 100 $4 200 $3 100 $9 3 00 $4100

rom o uquerque oston eve an upp y

es Moinesx11 $5 x12 $4 x13 $3 100

Evansvillex21 $8 x22 $4 x23 $3 300

Ft Lauderdalex31 $9 x32 $7 x33 $5

300

) th b Find St 13 V l M th dbf


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=

) the by .Find Step 13 Vogel s Methodb fs

smallest costs the

between the and in the or .next smallest cost columnrow

penalty 3 = 8-5 0 =4-4 0 =3-3 cost

To Albu ue Total RowFrom rque Supply penalty$5 $4 $3

$8 $4 $3

$9 $7 $5 =

Total Demand 300 200 200 700

greatestthe or with thecolumnrowIdentify


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greatest

opportunity

the or with the

( ).cost

columnrow)

.

in this example

IdentifyStep 2

column 13 - con.

as as possible to the lowest - costmany units.

AssignStep 3

.

Column

penalty 3 = 8-5 0 =4-4 0 =3-3

Opportunity

costTo

FromAlbuquer

queBoston Cleveland Total

SupplyRow

penalty

Des Moines $5 $4 $3 100 1 =4-3100

Evansville$8 $4 $3

300 1 =4-3

tLauderdale

$9 $7 $5300 2 = 7-5

Total Demand 300 200 200 700

) S 43


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or which has beencolumrow n) EliminateStep 43 - con. .

penalty 3 = 8-5 0 =4-4 0 =3-3 cost

To Albu ue Total RowFrom rque Supply penalty$5

$4

$3=

$8 $4 $3

t $9 $7 $5 = -Lauderdale

Total Demand 300 200 200 700


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t itti) R tSt 5 St 1 St 43 con


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to omitting, r wo s) . RepeatStep 5 Step 1 Step 43 - con. .

Column= - = - = -

Opportunity

To

From

Albuquer ue Boston Cleveland

Total

Su l

Row

enalt

Des Moines 100$5

$4

$3100

Evansville $8 $4 $3 300 1 =4-3200

Ft

Lauderdale

$9 $7 $5300 2 = 7-5

Total Demand 300 200 200 700

, of Second Run Step 2 Step 3

t) R t ittiSt 5 St 1 St 43


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to o sr w) . ,Repeat omittingStep 5 Step 1 Step 43 - con .

or e m na e n e co umns prev ous s ep. wo

Columnenalt 1 =9-8 = 7-4 2 =5-3

Opportunitycost

To

From

Albuque

rqueBoston Cleveland Total

Supply

Row

penalty

Des Moines 100$5

$4

$3100

Evansville $8 200 $4 $3 300 1 =4-3

FtLauderdale

$9

$7 $5300 2 = 7-5

Total Demand 300 200 200 700

of Second Run Step 4

t) R t ittiSt 5 St 1 St 43 con


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to or rows) . ,Repeat omittingStep 5 Step 1 Step 43 - con. . ,

Columnenalt 1 =9-8 2 =5-3

Opportunitycost

To

From

Albuque


Supply

Row

penalty

Des Moines 100$5

$4

$3100

Evansville $8 200 $4 $3 300 5 = 8-3

FtLauderdale

$9

$7 $5300 4 =9-5

Total Demand 300 200 200 700

o r un tep

to orrows) Repeat omittingStep 5 Step 1 Step 43 - con


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to orrows) . ,Repeat omittingStep 5 Step 1 Step 43 - con. . ,

Column1 =9-8

=5-Opportunity

To

From

Albuque


Supply

Row

penalty

Des Moines 100$5

$4

$3100

Evansville $8 200 $4 $3 300 5 = 8-3100

FtLauderdale

$9

$7 $5300 4 =9-5

Total Demand 300 200 200 700

, of Third Run Step 2 Step3

to orrows) Repeat omittingStep 5 Step 1 Step 43 - con


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to orrows) . ,Repeat omittingStep 5 Step 1 Step 43 con. . ,

Columnenalt 1 =9-8 2 =5-3

Opportunitycost

To

From

Albuque

rque

Boston Cleveland Total

Supply

Row

penalty

Des Moines 100$5

$4

$3100

Evansville $8 200 $4 $3 300 5 = 8-3100

FtLauderdale

$9

$7 $5300 4 =9-5

Total Demand 300 200 200 700

Step 4of Th ird Run

to orwro s) Repeat omittingStep 5 Step 1 Step 43 - con.


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to or wro s) . ,Repeat omittingStep 5 Step 1 Step 43 con. . ,, ,

Columnpenalty

Opportunitycost

ToFrom

Albuque

rque

Boston Cleveland Total

Supply

Row

penalty

Des Moines 1005

4

3100

Evansville $8

200$4

10$3

300

Ft $9

$7 $5300 4 =9-5

Total Demand 300 200 200 700

) Vogel s Meth3 - con . od


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onl is not eliminated we finall assi nmentsrow to,Since 3

bala and as following:nce row supplcolumn demand yColumn O ortunitpenalty cost

To Albuque Total Rowrom rque Supply penalty$5

$4

$3100

$8 $4 10 $3

Ft00

$9

$7 10 $500 au er a e

Total Demand 300 200 200 700

)3 con


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)3 - con.

we will have the following which is:, ,Finally bfs

and

, ,1 3 3 2 11 2 3 2 3

.2 31 1 2 31 2 0x x x x

tota cost sThe

i

.

This

s the of the threbest solutio e solun tions.


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optimization lecture 4

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