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8/7/2019 ch8_student

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Chap 8.

The Transportation and

Assignment Problem

by Dr. Peitsang Wu

Department of IndustrialEngineering and Management

I-Shou University


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The Transportation Problem

m resources, n destination

sinumber of nit supplied by source i

djnumber of unit required by destinationj

cijtransportation cost per unit shipped from

source i to destinationj


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Objectiveminimize total transportation cost

xijnumber of product shipped from source i

to destinationj


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ost per unit istributed estination

n....321 upply

source

m

/

/

/

/

2

1

mnmmm

n

n

cccc

cccc

cccc

....

/

/

/

/

1/

1/

1/

1/

....

....

321

2232221

1131211

ms

s

s

/

/

/

/

2

1

emand ndddd ....321


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Properties of Transportation Problem

Feasible Solutions Property :

If the total supply { total demand, it meaneithersi ordj represent a bound rather than anexact requirement, in this case introduce

dummy source or dummy destination asthe slack variable.

Integer solutions property: Not onlysi and djmust be integer values. But also all the B.F.S.

have integer values.

! !! !

!!m

i

n

j

ij

m

i

n

j

ji xds

1 11 1


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Example: Production Scheduling

Let xjbe the number of engines to be produced inmonthj.

The problem can be formulated as the general L.P.

model.

MonthScheduled

Installations

Max

Production

Unit cost

of

production

Unit cost

of storage

1 10 25 1.08 0.0152 15 35 1.11 0.015

3 25 30 1.10 0.015

4 20 10 1.13 0.015


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Formulate as Transportation Problem

si: production of jet engines in month i

(i =1.4)

dj: installation of jet engines in monthj(j =1.4)

xij: number of engines produced in month i

for installation in monthjcij: cost associated with each unit ofxij


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Supply is not fixed quantities.

Cost per unit distributed

Destination Supply

1 2 3 4

source

1

2 ?

3 ? ?4 ? ? ?

Demand


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In fact x11+x12+x13+x14e 25

x21+x22+x23+x24e 35

x31+x32+x33+x34e 30

x41+x42+x43+x44e 10

And notice that

i.e. total supply {total demand)+ + + = = supply+ + + = = demand

total supply - total demand =

{ 0demandSupply


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Solution

Use Big-Mmethod and introduce a new dummy

destination d5

Cost per unit distributed

Destination Supply

51 2 3 4

source

1 1.080 1.095 1.110 1.125 0 25

2 M 1.110 1.125 1.140 0 353 M M 1.100 1.115 0 30

4 M M M 1.130 0 10

Demand 10 15 25 20 30


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Example: Distribution of Water Resources

Cost per Acre Foot

Supply

BerdooLos

Devils

San

Go

Holly

glass

Colombo

River16 13 22 17 50

Sacron River 14 13 19 15 60

Calorie River 19 20 23 - 50

Min needed 30 70 0 10

requested 50 70 30 g


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Example: Distribution of Water Resources

Observations:

upper bound for Holly glass:

i.e. Holly glass can get as much as units.

Demand: bounded variables not constant.

regard less the minimum need, the demandis fixed.

now

pexcess demand.


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1 2 3 4 Supply

Colombo 16 13 22 17 50Sacron 14 13 19 15 60

Calorie 19 20 23 M 50

dummy 0 0 0 0 50demand 50 70 30 60


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After considering the minimum needs

1

(min)

2

(extra)

3 4 5 6Supply

1 16 16 13 22 17 17 50

2 14 14 13 19 15 15 60

3 19 19 20 23 M M 50

4 M 0 M 0 M 0 50

demand 30 20 70 30 10 50


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Transportation Simplex Method

estination

1 2 3 4 5 n Supply ui

c11 c12 c1n

1 s1

c21 c22 c212 s2

/

/ //

/

/

cm1 cm2 cmn

source

m

1

1

1

1

1

sm

emand d1 d2 d3 d4 d5 dn

vj

Z


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Procedure

Step 1: Initialization

Step 2: Optimality Test

Step 3: Iteration 3

Determine the entering basic variable

Determine the leaving basic variable

Determine the new B.F.S.


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Initialization

In general L.P. problem, well have exact onebasic variable for each constraint,For transportation problem with m sources

and n destinations, the number of basicvariables is m+n-1.

The reason is that the functional constraintsare equality constraints, this set ofm+n equations with one extra (or redundant)equation that can be deleted without changingthe feasible region i.e. any one of theconstraints are satisfied.


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Procedure for Constructing the Initial B.F.S.

1.From the rows and columns still under

consideration, select the next basic variable

(allocation) according to some criterion.2.Make that allocation large enough to

exactly use up the remaining supply in its

row or the remaining demand in its column(which ever is smaller) .


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Procedure for Constructing the Initial B.F.S.

3.Eliminate that row or column from furtherconsideration. (If the row and the column havethe sane remaining supply and demand, select

arbitrary row as the one to be eliminated. Thecolumn will be use later to provide a degeneratebasic variable, i.e. a circled allocation with zero)

4.If only one row or column remains underconsideration, then the procedure is completed byselecting every remaining variable associatedwith that row or column to be basic with the onlyfeasible allocation. Otherwise return to step 1.


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Alternative Criteria for Initialization

Northwest corner rule.

Vogels approximation method.

Russells approximation method.


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Northwest Corner Rule

Begin by selecting x11

(the northwestcorner).

Thereafter, if

xij was the last variableselected, then next select xi , j+1 (that is,

move one column to the right) if source ihas any supply remaining.

Otherwise, select next xi+1 , j (that is, moveone row down) .


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Example

Destination

1 2 3 4 5 Supply u i

1 1 1 2 11 5 0

1 1 1 1 12 6 0

1 1 2 2 M

3 5 0

M 0 M 0 0

source

4 (D ) 5 0

D e m a n d 3 0 2 0 7 0 3 0 6 0

vj=


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Vogels Approximation Method

For each row and column remaining underconsideration, calculate its difference,which is defined as the arithmeticdifference between the smallest and thenext-to-the-smallest unit cost cij stillremaining in that row or column.

In that row or column has the largestdifference, select the variable having thesmallest remaining unit cost. (If tie, breaksarbitrarily)


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Example

estination

1 2 3 4 5 Supply di erence

1 16 16 13 22 17 50

2 14 14 13 19 15 60

3 19 19 20 23 50Source

4 0 0 0 50emand 20 70 30 60 Select x

di erence eliminate column


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Example

Destination

1 2 3 4 5 Supply difference

1 16 16 13 17 50

2 14 14 13 15 60

3 19 19 20 M 50Source

4 M 0 M 0 20

Demand 30 20 70 60 Select x =

difference eliminate row


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Example

estination

1 2 3 5 Supply di erence

1 16 16 13 17 50

2 14 14 13 15 60Source

3 19 19 20 50

emand 30 20 70 40 Select x

di erence eliminate ro


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Example

1 2 3 5 Supply difference

2 14 14 13 15 60Source3 19 19 20 M 50

Demand 30 20 70 40 Select x =

difference eliminate column


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Example

(not column 3,use or ourth iteration in degeneracy)

1 2 3 Supply di erence

2 14 14 13 20S

ource 3 19 19 20 50

emand 30 20 20 Select x

di erence eliminate ro


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Example

1 2 3 Supply

Source 3 19 19 20 50

emand 30 20 0

Select x31

x32

x33

z


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Russells Approximation Method

For each source row i remaining underconsideration, determine its, which is thelargest unit cost cij still remaining in that row.

For each destination columnj remaining underconsideration, determine its which is thelargest unit cost cij still in that column.

For each xij not previously selected in these rows

and columns, calculate . Select the variable having the largest negative

value of (Tie breaks arbitrarily).

jv

jiijij vuc!(

ij(


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Example

Iteration1u 2u 3u 4u 1v 2v 3v 4v 5v ij( llocation

1 x =

2 x =

3 x =

4 x =

5 x =

6 x =

x =

x =

Z =


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Destination

1 2 3 4 5 Supply iu

1 16 16 13 22 17 50

2 14 14 13 19 15 60

3 19 19 20 23 M 50Source

4 M 0 M 0 0 50Demand 30 20 70 30 60

jv


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Comparison of Three Methods

Northwest corner : quick and easy. But itpays no attention to cij, the solution will befar from optimal.

Vogels approximation : popular, easy toimplement by hand.

Russells approximation : excellent

criterion, quick implement is computer (butnot hand) better solution than the other twomethods.


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Optimality Test

A basic feasible solution is optimal if and

only if for every ( i ,j )

such that xij is non-basic.

0u jiij vuc


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Note

Since cij ui vj = 0 ifxij is a basic variable, cij = ui +vj

There are

m+n-1 basic variables.

m+n-1 equations,,but m+n unknowns.

A convenient choice is to select ui that hasthe largest number of allocations in

its row (tie broken arbitrarily), and toassign it to be zero.


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Example

30

0 20

40

30

30

10

50

estination

1 2 3 4 5 Supplyui

16 16 13 22 17

1 50

14 14 13 19 152 60

19 19 20 233

-2250

0 0 0

Source

4+3 +4

50

emand 30 20 70 30 60

vjZ =


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An Iteration

STEP 1:Select the one with the largest negative value

ofcij - ui - vj to be the entering basic variable.

STEP 2: Increase the entering basic variable from 0 sets

off a chain reaction of compensating changesin other basic variables.

In order to satisfy the supply and demandconstraints. The first basic variable to bedecreased to 0 then becomes the leaving basicvariable.


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An Iteration

STEP 3:

The new B.F.S. is identified by adding value

of the leaving basic variable to the allocationfor each recipient cell and subtracting this

same amount from the allocation for each

donor cell.


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Example (STEP 2)

3 4 5 Supply

1

22 171 50

1

19 152 60

emand 70 30 60

Step 3: (Z=10(15-17+13-13)=10(-2)=10(c25-u2-v5)

40

30

10

+

+

+4

+1

-

-

-2


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Summary

Initialization:

Construct an initial B.F.S. by any of the threemethods, go to optimality test.

Optimality Test: Derives ui and vjby selecting the row having the

largest number of allocations, setting it ui = 0 ,

and then solving the set of equations cij = ui +vj

for each ( i ,j ) such that xij is basic. If for every ( i ,j ) such that xij is

non-basic, then the current solution is optimal.pSTOP.

Otherwise, go to an iteration.

0u jiij vuc


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Iteration

STEP 3:

Determine the new B.F.S. : Add the value of

the leaving basic variable to the allocation foreach recipient cell. Subtract this value from

the allocation for each donor cell. Return to

Optimality test.


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Transportation Simplex Tableaux

estinationIteration

0 1 2 3 4 5Supply ui

16 16 13 22 171 50

14 14 13 19 152 60

19 19 20 233 50

0 0 0

Source

4( ) 50

emand 30 20 70 30 60

vj

Z =

40 10

30

0 20

30

30

50


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estinationIteration

1 1 2 3 4 5Supply

ui

16 16 13 22 171 50

14 14 13 19 152 60

19 19 20 233 50

0 0 0

Source

4( ) 50

emand 30 20 70 30 60

vjZ =

50

30

0 20 30

50

20 10


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DestinationIteration

2 1 2 3 4 5 Supply ui

16 16 13 22 171 50

14 14 13 19 152 60

19 19 20 23 M3 50

M 0 M 0 0

Source

4(D) 50

Demand 30 20 70 30 60

vj=

50

20 40

0

30 20

30 20


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Complete set of transportation simplex

tableaux (P330,331 Table 8.23)

DestinationIteration

3 1 2 3 4 5 Supply ui

16 16 13 22 171 50

14 14 13 19 152

+260

19 19 20 233 50

0 0 0

Source

4(D) 50

Demand 30 20 70 30 60

vjZ =

50

20 40

30 20 0

30 20


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Changing Coefficient of a Nonbasic Variable

The change of coefficient of a non basicvariable xij will leave the r.h.s. of theoptimal tableau unchanged. Thus thecurrent basis will still be feasible.

is not changed, ui , vj s remainunchanged.

In row 0, only the coefficient ofxijchanged, thus as long as the coefficient ofxij in the optimal row 0 is non negativethe current basis remains optimal.

1T

B

Bc


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Assure the Optimal Tableau

Destination

1 2 3 4 Supply ui

8 6 10 91

+2 +735 0

9 12 13 72

+3 +250 3

14 9 16 5

Source

3+5 +3

40 3

Demand 45 20 30 30

vj 6 6 10 2Z =1,020

10 25

45 5

10


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Ex: Change c11

from 8 to 8+

Change c11

from 8 to 8+ , for what

values of will the current basis remains

optimal?

ifc11u 6 the current basis remains optimal.

!d 1111 vuc


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Changing Coefficient of a Basic Variable

Since we are changing . willchange too. The coefficient of each nonbasic variable in row 0 may change.

To determine whether the current basisremains optimal, we must find the new uis,and vj s and use these values to price out

all non basic variables.The current basis remains optimal, as long

as all non basic variables are non negative.

Tc 1T

B

Bc


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Ex:Change c13

from 10 to 10+

,

,,,

,

,,0

010..0

32

1432

4332

2123

31121

313113

!!

!!!!

!!

!!!!!

d

!d

d

(!d

uu

vvvv

vuvu

vuvuvuvuulet

vueivuc


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Ex:Change c13

from 10 to 10+

Price out each non basic variable

d

d

d

d

d

d

:

:

:

:

:

:

33

31

24

22

14

11

c

c

c

c

c

c


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Increasing bothsi and djby

This change maintains a balance transportation

problem.

Since bothsi

and dj

increase, this will cancel

out the supply demand balance.

Now.

New value = old -value +ui +vj

e.g. increase source 1 and destination 2 by 1 unit.

New cost = 1020+1*0+1*(6) = 1026


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Increasing bothsi and djby

We may also find new values of the decisionvariables as follows:

(i) Ifxij is a basic variable in the optimal solution,

increase xijby(ii) Ifxij is a non basic variable in the optimal

solution, find the loop involving xij and someof the basic variables. Find an odd cell in the

loop that is in row i. Increase the value of thisodd cell by and go around the loop,alternatively increasing and then decreasingcurrent basic variables in the loop by.


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Ex: Increases1, d

2by 2

x12

is a basic variable in the optimal solution.

Now


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Ex: Increases1, d

1by 1

x11 is a non basic variable in the optimal solution.

Destination

1 2 3 4 Supply ui

1 35 0

2 50 3Source

3 40 3

Demand 45 20 30 30

vj 6 6 10 2

Z=1020+u1+v1

=

10 26

46 4

10 30


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The Assignment Problem

Assumptions:

The number of assignees and the number of tasksare the same. (denote by n)

Each assignee is to be assigned to exactly on task.

Each task is to be performed by exactly oneassignee.

There is a cost cij

associated with assignee iperforming taskj ( i ,j = 1..n )

The objective is to determine how all nassignments should be made in order to minimizethe total cost.


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Example

ocation

n =4

ocation

n =4

1 2 3 4 1 2 3 41 13161211 1 13161211

2 15 - 1320 2 15 1220

achine

n =3

3 5 7 10 6 3 5 7 10 6

achine

n =4

4

(D)0 0 0 0


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The Mathematical Model

( actually xij n_a )

jix

x

xts

xcZMin

ij

n

i

ij

n

j

ij

n

j

n

i

ijij

,,0

1

1..

1

1

1 1

u

!

!

!

!

!

! !

.

.0

1..

jtaskperformsi

not

assignee

if

ifxei ij

!


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The Mathematical Model

Assignment problem is a special case of

transportation problem:

Number of sources m = number ofdestination n

Every supply si = 1

Every demand dj = 1


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The Two Properties

Feasible solution property

Integer solution property holds here.

More over, the constraints prevent anyvariable from being greater than 1, and the

non-negativity constraints prevent values

less than 0. So it automatically satisfiesbinary restriction.


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Our example

1 2 3 4 Supply

1 13 16 12 11 1

2 15 M 13 20 1

3 5 7 10 6 1

4(D) 0 0 0 0 1

Demand 1 1 1 1


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Hungarian Method:

STEP 2:

Draw the minimum number of lines

(horizontal and/or vertical) that are needed to

cover all the zeros in the reduced cost matrix.

Ifn lines are required to cover all the zeros, an

optimal solution is available among the cover

zeros in the matrix. If fewer than n lines are needed to cover all the

zeros, GOTO STEP 3.


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Hungarian Method:

STEP 3:

Find the smallest nonzero element (call itsvalue k) in the reduced cost matrix that is not

covered by lines draw n in STEP2.Now subtract kfrom each uncovered element

of the reduced cost matrix and add kto eachelement of the reduced cost matrix that is

covered by two lines.Return to STEP 2.


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Example

row min

14 5 8 7

2 12 6 5

7 8 3 9

2 4 6 10

Column

min


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Example

9 0 3 0

0 10 4 1 p

4 5 0 4Smallest

Uncovered:1

0 2 4 6


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Transshipment Problem

A supply node is a node that can send goods to

another node but cannot receive goods from any

other node.

A demand node is a node that can receive goods

from other nodes but cannot send goods to any

other node.

A transshipment node is a node that can bothreceive goods from other nodes and send goods

to other nodes.


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Solution Procedure

Step 1: If necessary, add a dummy demand tobalance the problem.s =total available supply.

Step 2: Construct a transportation tableau as

follow:A row consists of supply node and

transshipment node.

A column consists of demand node and

transshipment node.The transshipment node will have a supply

equal to (nodes original supply)+s, and ademand equal to (nodes original demand)+s.


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Example

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