optimization lecture 4
TRANSCRIPT
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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.
From To Albuquerque Boston Cleveland Supply
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
From To Albuquerque Boston Cleveland Supply
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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) Least - Cost Method2 - con.
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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) Least - Cost Method2 - con.
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
rqueBoston Cleveland Total
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
rqueBoston 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
, 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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