transportation assignment & transshipment problems-solving

8/12/2019 Transportation Assignment & Transshipment Problems-Solving

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Transportation, Assignment and

Transshipment ProblemsMohon disarikan & di-email dalam bentuk e-

paper dg syarat: pilih satu topik:

1. Masalah Transportatsi dg iterasi solusi North-west

Corner, di halaman 20

2. Masalah Transportatsi dg iterasi solusi Minimum

cost, di halaman 25

3. Masalah Transportatsi dg iterasi solusi dg solusiVogel, di halaman 34

4. Masalah assignememt di halaman 51

5. Masalah transhipment di halaman 55


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Formulating Transportation

ProblemsExample 1: Powerco has three electric

power plants that supply the electric needs

of four cities.

The associated supply of each plant and

demand of each city is given in the table 1.

The cost of sending 1 million kwh ofelectricity from a plant to a city depends on

the distance the electricity must travel.


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Transportation tableau

A transportation problem is specified by

the supply, the demand, and the shipping

costs. So the relevant data can be

summarized in a transportation tableau.The transportation tableau implicitly

expresses the supply and demand

constraints and the shipping cost between

each demand and supply point.


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Table 1. Shipping costs, Supply, and Demand

for Powerco Example

From To

City 1 City 2 City 3 City 4 Supply

(M il lion kwh)

Plant 1 $8 $6 $10 $9 35

Plant 2 $9 $12 $13 $7 50

Plant 3$14 $9 $16 $5 40

Demand

(Mil lion kwh)

45 20 30 30

Transportation Tableau


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Solution

1. Decision Variable:

Since we have to determine how much electricity

is sent from each plant to each city;

Xij = Amount of electricity produced at plant i

and sent to city j

X14= Amount of electricity produced at plant 1

and sent to city 4


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2. Objective function

Since we want to minimize the total cost of shipping

from plants to cities;

Minimize Z = 8X11+6X12+10X13+9X14

+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34


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3. Supply Constraints

Since each supply point has a limited productioncapacity;

X11+X12+X13+X14


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5. Sign Constraints

Since a negative amount of electricity can not be

shipped all Xijs must be non negative;

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)


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LP Formulation of Powercos Problem

Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34

S.T.: X11+X12+X13+X14 = 30

X14+X24+X34 >= 30

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)


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General Description of a Transportation

Problem

1. A set of m supply pointsfrom which a good is

shipped. Supply point ican supply at mostsi

units.

2. A set of n demand pointsto which the good is

shipped. Demand pointjmust receive at least di

units of the shipped good.

3. Each unit produced at supply point iand shippedto demand pointjincurs a variable costof cij.


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Xij = number of units shipped fromsupply point ito

demand point j

),...,2,1;,...,2,1(0

),...,2,1(

),...,2,1(..

min

1

1

1 1

njmiX

njdX

misXts

Xc

ij

mi

i

jij

nj

j

iij

mi

i

nj

j

ijij


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Balanced Transportation Problem

If Total supply equals to total demand, theproblem is said to be a balanced

transportation problem:

nj

j

j

mi

i

i ds11


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Balancing a TP if total supply exceeds total

demand

If total supply exceeds total demand, we

can balance the problem by adding dummy

demand point. Since shipments to thedummy demand point are not real, they are

assigned a cost of zero.


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Balancing a transportation problem if total

supply is less than total demand

If a transportation problem has a total

supply that is strictly less than total

demand the problem has no feasiblesolution. There is no doubt that in such a

case one or more of the demand will be left

unmet. Generally in such situations a

penalty cost is often associated with unmetdemand and as one can guess this time the

total penalty cost is desired to be minimum


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Finding Basic Feasible Solution

for Transportation ProblemUnlike other Linear Programming

problems, a balancedTP with m supply

points and n demand points is easier to

solve, although it has m + n equality

constraints. The reason for that is, if a set

of decision variables (xijs) satisfy all but

one constraint, the values for xijs willsatisfy that remaining constraint

automatically.


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Methods to find the bfs for a balanced TP

There are three basic methods:

1. Northwest Corner Method

2. Minimum Cost Method

3. Vogels Method


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1. Northwest Corner Method

To find the bfs by the NWC method:

Begin in the upper left (northwest) corner of the

transportation tableau and set x11as large as

possible (here the limitations for setting x11to a

larger number, will be the demand of demandpoint 1 and the supply of supply point 1. Your

x11value can not be greater than minimum of

this 2 values).


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According to the explanations in the previous slide

we can set x11=3 (meaning demand of demand

point 1 is satisfied by supply point 1).5

6

2

3 5 2 3

3 2

6

2

X 5 2 3


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After we check the east and south cells, we saw that

we can go east (meaning supply point 1 still has

capacity to fulfill some demand).

3 2 X

6

2

X 3 2 3

3 2 X

3 3

2

X X 2 3


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After applying the same procedure, we saw that we

can go south this time (meaning demand point 2

needs more supply by supply point 2).

3 2 X

3 2 1

2

X X X 3

3 2 X

3 2 1 X

2

X X X 2


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

x11

=3, x12

=2, x22

=3, x23

=2, x24

=1, x34

=2

3 2 X

3 2 1 X

2 X

X X X X


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2. Minimum Cost Method

The Northwest Corner Method dos not utilize shippingcosts. It can yield an initial bfs easily but the total

shipping cost may be very high. The minimum cost

method uses shipping costs in order come up with a

bfs that has a lower cost. To begin the minimum costmethod, first we find the decision variable with the

smallest shipping cost (Xij). Then assignXijits largest

possible value, which is the minimum ofsiand dj


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After that, as in the Northwest Corner Method we

should cross out row i and column j and reduce the

supply or demand of the noncrossed-out row orcolumn by the value of Xij. Then we will choose the

cell with the minimum cost of shipping from the

cells that do not lie in a crossed-out row or column

and we will repeat the procedure.


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An example for Minimum Cost Method

Step 1: Select the cell with minimum cost.

2 3 5 6

2 1 3 5

3 8 4 6

5

10

15

12 8 4 6


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Step 2: Cross-out column 2

2 3 5 6

2 1 3 5

8

3 8 4 6

12 X 4 6

5

2

15


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Step 3: Find the new cell with minimum shipping

cost and cross-out row 2

2 3 5 6

2 1 3 5

2 8

3 8 4 6

5

X

15

10 X 4 6


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cost and cross-out row 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

X

X

15

5 X 4 6


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cost and cross-out column 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5

X

X

10

X X 4 6


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cost and cross-out column 3

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4

X

X

6

X X X 6


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Step 7: Finally assign 6 to last cell. The bfs is found

as: X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4 6

X

X

X

X X X X


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3. Vogels Method

Begin with computing each row and column a penalty.The penalty will be equal to the difference between

the two smallest shipping costs in the row or column.

Identify the row or column with the largest penalty.

Find the first basic variable which has the smallestshipping cost in that row or column. Then assign the

highest possible value to that variable, and cross-out

the row or column as in the previous methods.

Compute new penalties and use the same procedure.


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An example for Vogels Method

Step 1: Compute the penalties.

Supply Row Penalty

6 7 8

15 80 78

Demand

Column Penalty 15-6=9 80-7=73 78-8=70

7-6=1

78-15=63

15 5 5

10

15


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Step 2: Identify the largest penalty and assign the

highest possible value to the variable.

Supply Row Penalty

6 7 8

5

15 80 78

Demand

Column Penalty 15-6=9 _ 78-8=70

8-6=2

78-15=63

15 X 5

5

15


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Supply Row Penalty

6 7 8

5 5

15 80 78

Demand

Column Penalty 15-6=9 _ _

_

_

15 X X

0

15


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Supply Row Penalty

6 7 8

0 5 5

15 80 78

Demand

Column Penalty _ _ _

_

_

15 X X

X

15


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Step 5: Finally the bfs is found as X11=0, X12=5,

X13=5, and X21=15

Supply Row Penalty

6 7 8

0 5 5

15 80 78

15

Demand

Column Penalty _ _ _

_

_

X X X

X

X


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The Transportation Proble as a

Simplex Method

In this section we will explain how the simplex

algorithm is used to solve a transportation problem.


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How to Pivot a Transportation Problem

Based on the transportation tableau, the followingsteps should be performed.

Step 1.Determine (by a criterion to be developed

shortly, for example northwest corner method) the

variable that should enter the basis.

Step 2.Find the loop (it can be shown that there is

only one loop) involving the entering variable and

some of the basic variables.Step 3.Counting the cells in the loop, label them as

even cells or odd cells.


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Step 4.Find the odd cells whose variable assumes the

smallest value. Call this value . The variable

corresponding to this odd cell will leave the basis. To

perform the pivot, decrease the value of each odd cell

by and increase the value of each even cell by . The

variables that are not in the loop remain unchanged.The pivot is now complete. If =0, the entering

variable will equal 0, and an odd variable that has a

current value of 0 will leave the basis. In this case a

degenerate bfs existed before and will result after thepivot. If more than one odd cell in the loop equals ,

you may arbitrarily choose one of these odd cells to

leave the basis; again a degenerate bfs will result


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Assignment Problems

Example: Machineco has four jobs to be completed.

Each machine must be assigned to complete one job.

The time required to setup each machine for completing

each job is shown in the table below. Machinco wants tominimize the total setup time needed to complete the

four jobs.


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Setup times

(Also called the cost matrix)Time (Hours)

Job1 Job2 Job3 Job4

Machine 1 14 5 8 7

Machine 2 2 12 6 5

Machine 3 7 8 3 9

Machine 4 2 4 6 10


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The Model

According to the setup table Machincos problem can be

formulated as follows (for i,j=1,2,3,4):

10

1

1

1

1

1

1

1

1..

10629387

5612278514min

44342414

43332313

42322212

41312111

44434241

34333231

24232221

14131211

4443424134333231

2423222114131211

ijij orXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXXXXXXts

XXXXXXXX

XXXXXXXXZ


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For the model on the previous page note that:

Xij=1 if machine iis assigned to meet the demands of

jobj

Xij=0 if machine iis not assigned to meet the demands

of jobj

In general an assignment problem is balancedtransportation problem in which all supplies and

demands are equal to 1.


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The Assignment Problem

In general the LP formulation is given as

Minimize 1 1

1

1

1 1

1 1

0

, , ,

, , ,

or 1,

n n

i j i j

i j

n

i j

j

n

i j

i

i j

c x

x i n

x j n

x i j

Each supply is 1

Each demand is 1

Comments on the Assignment


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Comments on the Assignment

Problem The Air Force has used this for assigning

thousands of people to jobs.

This is a classical problem. Research on the

assignment problem predates research on LPs.

Very efficient special purpose solution techniques

exist.

10 years ago, Yusin Lee and J. Orlin solved a problem

with 2 million nodes and 40 million arcs in hour.

Al h h h i i l b


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Although the transportation simplex appears to be very

efficient, there is a certain class of transportation

problems, called assignment problems, for which the

transportation simplex is often very inefficient. For thatreason there is an other method called The Hungarian

Method. The steps of The Hungarian Method are as

listed below:

Step1.Find a bfs. Find the minimum element in each row

of the mxmcost matrix. Construct a new matrix by

subtracting from each cost the minimum cost in its row.

For this new matrix, find the minimum cost in eachcolumn. Construct a new matrix (reduced cost matrix) by

subtracting from each cost the minimum cost in its

column.


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Step2.Draw the minimum number of lines (horizontal

and/or vertical) that are needed to cover all zeros in the

reduced cost matrix. If m lines are required , an optimalsolution is available among the covered zeros in the

matrix. If fewer than m lines are required, proceed to step

3.

Step3.Find the smallest nonzero element (call its value

k) in the reduced cost matrix that is uncovered by the

lines drawn in step 2. Now subtract k from eachuncovered element of the reduced cost matrix and add k

to each element that is covered by two lines. Return to

step2.


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Transshipment Problems

A transportation problem allows only shipments that godirectly from supply points to demand points. In many

situations, shipments are allowed between supply points

or between demand points. Sometimes there may also

be points (called transshipment points) through whichgoods can be transshipped on their journey from a

supply point to a demand point. Fortunately, the optimal

solution to a transshipment problem can be found by

solving a transportation problem.

Transshipment Problem


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Transshipment Problem

An extension of a transportation problem

More general than the transportation problem in that in thisproblem there are intermediate transshipment points. In

addition, shipments may be allowed between supply points

and/or between demand points

LP Formulation Supply point: it can send goods to another point but cannot

receive goods from any other point

Demand point It can receive goods from other points but

cannot send goods to any other point Transshipment point: It can both receive goods from other

points send goods to other points


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Each supply point will have a supply equal to its

original supply, and each demand point will have a

demand to its original demand. Let s= total available

supply. Then each transshipment point will have a supply

equal to (points original supply)+s and a demand equal

to (points original demand)+s. This ensures that anytransshipment point that is a net supplier will have a net

outflow equal to points original supply and a net

demander will have a net inflow equal to points original

demand. Although we dont know how much will beshipped through each transshipment point, we can be

sure that the total amount will not exceed s.

Transshipment Example


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Transshipment Example Example 5: Widgetco manufactures widgets at two

factories, one in Memphis and one in Denver. The

Memphis factory can produce as 150 widgets, and theDenver factory can produce as many as 200 widgetsper day. Widgets are shipped by air to customers inLA and Boston. The customers in each city require130 widgets per day. Because of the deregulation of

airfares, Widgetco believes that it may be cheaperfirst fly some widgets to NY or Chicago and then flythem to their final destinations. The cost of flying awidget are shown next. Widgetco wants to minimizethe total cost of shipping the required widgets tocustomers.

Transportation Tableau Associated


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Transportation Tableau Associatedwith the Transshipment Example

NY Chicago LA Boston Dummy Supply Memphis $8 $13 $25 $28 $0 150 Denver $15 $12 $26 $25 $0 200

NY $0 $6 $16 $17 $0 350

Chicago $6 $0 $14 $16 $0 350 Demand 350 350 130 130 90

Supply points: Memphis, Denver

Demand Points: LA Boston

Transshipment Points: NY, Chicago

The problem can be solved using the transportation simplex

method

Limitations of Transportation Problem


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One commodity ONLY: any one product supplied

and demanded at multiple locations Merchandise Electricity, water

Invalid for multiple commodities: (UNLESStransporting any one of the multiple commodities iscompletely independent of transporting any othercommodity and hence can be treated by itself alone) Example: transporting product 1 and product 2 from the

supply points to the demand points where the total amount(of the two products) transported on a link is subject to acapacity constraint

Example: where economy of scale can be achieved bytransporting the two products on the same link at a largertotal volume and at a lower unit cost of transportation



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Difficult to generalize the technique to accommodate

(these are generic difficulty for mathematical

programming, including linear and non-linearprogramming

Economy of scale the per-unit cost of transportation on a link

decreasing with the volume (nonlinear and concave; there is a trick

to convert a non-linear program with a piecewise linear but

convex objective function to a linear program; no such tricks exists

for a piecewise linear but concave objective function)

Fixed-cost: transportation usually involves fixed charges. For

example, the cost of truck rental (or cost of trucking in general)

consists of a fixed charge that is independent of the mileage and a

mileage charge that is proportional to the total mileage driven.

Such fixed charges render the objective function NON-LINEAR

and CONCAVE and make the problem much more difficult to

solve

transportation assignment & transshipment problems-solving

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