transportation + assignment models (1)

7/27/2019 Transportation + Assignment Models (1)

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Transportation and

Assignment Models


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Specialized Problems

Transportation Problem Distribution of items from several sources

to several destinations. Supply capacities

and destination requirements known. Assignment Problem

One-to-one assignment of people to jobs,etc.

Special ized algo ri thm s

save t ime!


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Transportation Problem

Goal ~To determine the delivery routes tobe used & units delivered so as tominimise total transportation cost

Initial sol. technique ~ North west cornerrule, Vogels approx., Lowest cost,

Final sol. ~ Stepping Stone, MODI,

Involves step by step process


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Example of transportation problem

Masjid Tanah(D)

Banting(E)

Taiping(F)

Kuantan(A)

Johor Bahru(B)

Gombak(C)

Factories

(Sources)

Retail-shops

(destinations)

Capacities requirementsroutes

100units

300units

300units

300units

200units

200units


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Transportation Problem

Des Moines

(100 units)

capacity

Cleveland

(200 units)

required

Boston(200 units)

requiredEvansville

(300 units)

capacity

Ft. Lauderdale

(300 units)

capacity

Albuquerque(300 units)

required

D

A

E

C B

F


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Transportation Costs

From

(Sources)

To(Destinations)

Albuquerque(A)Boston(B)

Cleveland(C)

Des Moines(D)

Evansville(E)

FortLauderdale(F)

$5

$8

$9

$4

$4

$7

$3

$3

$5


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Unit Shipping Cost:1Unit, Factory toWarehouse

Des Moines

(D)

Evansville

(E)

Fort

Lauderdale (F)

Warehouse

Req.

Albuquerque

(A) Boston (B)Cleveland

(C) Factory

Capacity5 4 3

57

48

9

3


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Total Demand and Total Supply

Des Moines

(D)

Evansville

(E)

Fort

Lauderdale(F)Warehouse

Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

100


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ranspor a on a e or xecu veFurniture Corp.

Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9


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Initial Solution Using the NorthwestCorner Rule

Start in the upper left-hand cell andallocate units to shipping routes asfollows:

Exhaust the supply (factory capacity) ofeach row before moving down to the nextrow.

Exhaust the demand (warehouse)requirements of each column beforemoving to the next column to the right.

Check that all supply and demand

requirements are met.


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Initial SolutionNorth West Corner Rule

Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

100

200 100

100 200

Total Transportation cost = 100x5 + 200x8 + 100x4 + 100x7 + 200x5= $4200


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1. Find the difference between the two lowest unittransportation costs for each row & column

2. Determine the row or column with the greatest

difference3. Assign as many units as possible to the cost

square in the row or column selected

4. Eliminate any row or column that has just beencompletely satisfied by the assignment made

5. Repeat steps 1-4

Initial SolutionVogels Approximation steps (refer p 424)


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Vogels Approximation1. For each row/column of table, find

difference between two lowest costs.(Opportunity cost)

2. Find greatest opportunity cost.

3. Assign as many units as possible tolowest cost square in row/column with

greatest opportunity cost.

4. Eliminate row or column which has beencompletely satisfied.

4. Begin again, omitting eliminated

rows/columns.

I iti l S l ti


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Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

Total Transportation cost = $3900

Initial SolutionVogels Approximation (refer p 427)

100

200

200 100

100


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Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9


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Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9


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Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9


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The Stepping-Stone Method

1. Select any unused square to evaluate.

2. Begin at this square. Trace a closed path back tothe original square via squares that are currentlybeing used (only horizontal or vertical movesallowed).

3. Place + in unused square; alternate - and + oneach corner square of the closed path.

4. Calculate improvement index: add together theunit cost figures found in each square containing a

+; subtract the unit cost figure in each squarecontaining a -.

5. Repeat steps 1 - 4 for each unused square.

If all indices computed > = 0, optimal sol.


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Stepping-Stone Method - The DesMoines-to-Cleveland Route

Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

200

100

100

100 200

- +

-

+

+

-

Start


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Stepping-Stone MethodAn Improved Solution

Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

100

100

200

200100


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Third and Final Solution

Des Moines

(D)

Evansville

(E)

Ft Lauderdale

(F)

Warehouse

Req.

Albuquerque(A)

Boston(B)

Cleveland(C)

FactoryCapacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

100

200

100200

100

I iti l S l ti


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Des Moines

(D)

Evansville

(E)

Fort


Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

Using stepping stone:Each improvement index in the unoccupied cell >=0, optimal solutionTotal Transportation cost = $3900

Initial SolutionVogels Approximation (refer p 427)

100

200

200 100

100

+2 +2

+1

+1

MODI Method: 5 Steps


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MODI Method: 5 Steps1. Compute the values for each row and column: set Ri+ Kj= Ci jfor

those squares current ly used o r occu pied.2. After writing all equations, set R1 = 0.

3. Solve the system of equations forRiand Kjvalues.4. Compute the improvement index for each unused square by the

formula improvement index:

Ii j = Ci j- Ri- Kj5. Select the largest negative index and proceed to solve the problem

as you did using the stepping-stone method.

1. Set up equation for each occupied celllet R1=0 then solve for K1=5, R2=3, K2=1, R3=6, and K3= -1

2. Compute improvement index for each unoccupied cellI_EA = 8 - 3 - 5 = 0I_DB = 4 - 0 - 1 = 3I_DC = 3 - 0 - (-1) = 4I_FB = 7 - 6 - 1 = 0


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Des Moines

(D)

Evansville

(E)

Fort

Lauderdale(F)

Warehouse

Req.

Albuquerque

(A)Boston

(B)

Cleveland

(C)Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

200

100

100

100 200

Kj K1 K2 K3

Ri

R1

R2

R3

3 4

1

-2

Each improvement index Iij in the unoccupied cell >=0, optimal solutionI_FA isve (not optimal yet). Total Transportation cost = $4200

To

From

+

-

-

+


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Special Problems in TransportationMethod

Unbalanced Problem

Demand Less than Supply

Demand Greater than Supply

Degeneracy (one or more Iij =0)

More Than One Optimal Solution


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Unbalanced ProblemDemand Less than Supply

Factory 1

Factory 2

Factory 3

Customer

Requirements

Customer1 Customer 2

DummyFactory

Capacity

150 80 150 380

80

130

1708 5 0

0

09

1015

3


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Unbalanced ProblemSupply Less than Demand

Factory 1

Factory 2

Dummy

Customer

Requirements

Customer1

Customer2

Customer3

FactoryCapacity

150 80 150 380

80

130

1708 5 16

7

00

1015

0


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Degeneracy

Factory 1

Factory 2

Factory 3

Customer

Requirements

Customer1

Customer2

Customer3

FactoryCapacity

100 100 100 300

80

120

1005 4 3

3

57

48

9

100

100

80

20


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Degeneracy - Coming Up!

Factory 1

Factory 2

Factory 3

Customer

Requirements

Customer1

Customer2

Customer3

FactoryCapacity

150 80 50 280

80

130

708 5

7

9

16

10

1015

3

70

80

50

50

30


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Des Moines

(D)

Evansville

(E)

Fort

Lauderdale(F)

Warehouse

Req.

Albuquerque(A)

Boston

(B)

Cleveland(C)

Factory

Capacity

300 200 200 700

300

300

1005 4 3

3

57

48

9

Start

200

100

100

100 200

- +

-

+

+

-


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The Assignment Method

1. subtract the smallest number in each row fromevery number in that row

subtract the smallest number in eachcolumn from every number in that column

2. draw the minimum number of vertical andhorizontal straight lines necessary to coverzeros in the table

if the number of lines equals the number ofrows or columns, then one can make anoptimal assignment (step 4)


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The Assignment Method - continued

3.if the number of lines does not equal thenumber of rows or columns

subtract the smallest number not covered bya line from every other uncovered number

add the same number to any number lying atthe intersection of any two lines

return to step 2

4. make optimal assignments at locations ofzeros within the table


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The Assignment Problem

Project

Person 1 2 3

Adams $11 $14 $6

Brown $8 $10 $11

Cooper $9 $12 $7


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Hungarian Method

Initial TablePerson Project

1 2 3

Adams 11 14 6

Brown 8 10 11

Cooper 9 12 7


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Hungarian Method

Row ReductionPerson Project

1 2 3

Adams

Brown

Cooper

5 8 0

0 2 3

2 5 0


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Hungarian Method

Column Reduction

Person Project

1 2 3

Adams 5 6 0

Brown 0 0 3

Cooper 2 3 0


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Hungarian Method

Person Project

1 2 3

Adams

Brown

Cooper

5 6 0

0 0 3

2 3 0

Testing CoveringLine 2

CoveringLine

1


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Hungarian Method

Person Project

1 2 3

Adams 3 4 0

Brown 0 0 5

Cooper 0 1 0

Revised Opportunity Cost Table


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Hungarian Method

Person Project

1 2 3

Adams 3 4 0

Brown0 0 5

Cooper0 1 0

TestingCovering Line1

Covering

Line 2

Covering Line 3


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Hungarian Method

AssignmentsPerson Project

1 2 3

Adams 6

Brown 10

Cooper 9


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Maximization Assignment Problem

Project

1 2 3 Dummy

Adams $11 $14 $6 $0

Brown $8 $10 $11 $0

Cooper $9 $12 $7 $0

Davis $10 $13 $8 $0


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Maximization Assignment Problem

Project

1 2 3 Dummy

Adams $32 $0 $8 $14

Brown $6 $4 $3 $14

Cooper $5 $2 $7 $14

Davis $4 $1 $6 $14


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Part A ~ 14 October 2004


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