Download - Bba 3274 qm week 9 transportation and assignment models

Transportation & Transportation & Assignment ModelsAssignment ModelsTransportation & Transportation &

Assignment ModelsAssignment Models

BBA3274 / DBS1084 QUANTITATIVE METHODS for BUSINESSBBA3274 / DBS1084 QUANTITATIVE METHODS for BUSINESS

byStephen Ong

Visiting Fellow, Birmingham City University Business School, UK

Visiting Professor, Shenzhen University

Today’s Overview Today’s Overview

Learning ObjectivesLearning Objectives

1.1. Structure LP problems using the Structure LP problems using the transportation, transshipment and transportation, transshipment and assignment models.assignment models.

2.2. Use the northwest corner and Use the northwest corner and stepping-stone methods.stepping-stone methods.

3.3. Solve facility location and other Solve facility location and other application problems with application problems with transportation models.transportation models.

4.4. Solve assignment problems with the Solve assignment problems with the Hungarian (matrix reduction) Hungarian (matrix reduction) method.method.

After completing this lecture, students will be able to:After completing this lecture, students will be able to:

OutlineOutline

9.19.1 IntroductionIntroduction

9.29.2The Transportation ProblemThe Transportation Problem

9.39.3The Assignment ProblemThe Assignment Problem

9.49.4The Transshipment ProblemThe Transshipment Problem

9.59.5The Transportation AlgorithmThe Transportation Algorithm

9.69.6Special Situations with the Transportation Special Situations with the Transportation AlgorithmAlgorithm

9.79.7Facility Location AnalysisFacility Location Analysis

9.89.8 The Assignment Algorithm The Assignment Algorithm

9.99.9 Special Situations with the Assignment Special Situations with the Assignment AlgorithmAlgorithm

IntroductionIntroduction

We will explore three special We will explore three special linear programming models:linear programming models: The transportation problem.The transportation problem. The assignment problem.The assignment problem. The transshipment problem.The transshipment problem.

These problems are members of These problems are members of a category of LP techniques a category of LP techniques called called network flow problems.network flow problems.

The Transportation ProblemThe Transportation Problem

The The transportation problemtransportation problem deals with deals with the distribution of goods from several the distribution of goods from several points of supply (points of supply (sourcessources) to a number ) to a number of points of demand (of points of demand (destinationsdestinations).).

Usually we are given the capacity of Usually we are given the capacity of goods at each source and the goods at each source and the requirements at each destination.requirements at each destination.

Typically the objective is to minimize Typically the objective is to minimize total transportation and production total transportation and production costs.costs.

The Transportation ProblemThe Transportation Problem The Executive Furniture The Executive Furniture

Corporation manufactures office Corporation manufactures office desks at three locations: Des desks at three locations: Des Moines, Evansville, and Fort Moines, Evansville, and Fort Lauderdale.Lauderdale.

The firm distributes the desks The firm distributes the desks through regional warehouses through regional warehouses located in Boston, Albuquerque, located in Boston, Albuquerque, and Cleveland.and Cleveland.

9-8

The Transportation ProblemThe Transportation Problem

Network ReNetwork Representation of a Transportation Problem, esentation of a Transportation Problem, with Costs, Demands and Supplieswith Costs, Demands and Supplies

100 Units100 Units

300 Units300 Units

300 Units300 Units 200 Units200 Units

200 Units200 Units

300 Units300 Units

FactoriFactories es

(Sourc(Sources)es)Des MoinesDes Moines

EvansvilleEvansville

Fort LauderdaleFort Lauderdale

WarehouseWarehouses s

(Destinatio(Destinations)ns)AlbuquerqueAlbuquerque

BostonBoston

ClevelandCleveland

Figure 9.1

Executive Furniture CompanyExecutive Furniture Company

$5$5$4$4

$3$3$8$8

$4$4$3$3

$9$9$7$7

$5$5

SupplySupply DemandDemand

Linear Programming for the Linear Programming for the Transportation ExampleTransportation Example

Let XLet Xijij = number of units = number of units

shipped from source shipped from source ii to to destination destination jj,,Where:Where:

ii = 1, 2, 3, with 1 = Des Moines, 2 = = 1, 2, 3, with 1 = Des Moines, 2 = Evansville, and 3 = Fort LauderdaleEvansville, and 3 = Fort Lauderdale

jj = 1, 2, 3, with 1 = Albuquerque, 2 = = 1, 2, 3, with 1 = Albuquerque, 2 = Boston, and 3 = Cleveland.Boston, and 3 = Cleveland.

Linear Programming for the Linear Programming for the Transportation ExampleTransportation Example

Minimize total cost = 5XMinimize total cost = 5X1111 + 4X + 4X1212 + 3X + 3X1313 + +

8X8X2121 + 4X + 4X2222 + 3X + 3X2323

+ 9X+ 9X3131 +7X +7X3232 + 5X + 5X3333

Subject to:Subject to: XX1111 + X + X1212 + X + X1313 ≤ 100 (Des Moines supply) ≤ 100 (Des Moines supply) XX2121 + X + X2222 + X + X2323 ≤ 300 (Evansville supply) ≤ 300 (Evansville supply) XX3131 + X + X3232 + X + X3333 ≤ 300 (Fort Lauderdale supply) ≤ 300 (Fort Lauderdale supply) XX1111 + X + X2121 + X + X3131 = 300 (Albuquerque demand) = 300 (Albuquerque demand) XX1212 + X + X2222 + X + X3232 = 200 (Boston demand) = 200 (Boston demand) XX1313 + X + X2323 + X + X3333 = 200 (Cleveland demand) = 200 (Cleveland demand) XXijij ≥ 0 for all ≥ 0 for all ii and and jj..

Executive Furniture Corporation Executive Furniture Corporation Solution in Excel 2010Solution in Excel 2010

Program 9.1

A General LP Model for A General LP Model for Transportation ProblemsTransportation Problems

Let:Let:XXijij = number of units shipped = number of units shipped

from source from source ii to destination to destination jj..ccijij = cost of one unit from = cost of one unit from

source source ii to destination to destination jj..ssii = supply at source = supply at source ii..

ddjj = demand at destination = demand at destination jj..

A General LP Model for A General LP Model for Transportation ProblemsTransportation Problems

Minimize cost = Minimize cost =

Subject to:Subject to:

i = 1, 2,…, m.i = 1, 2,…, m.

j = 1, 2, …, nj = 1, 2, …, n..

xxijij ≥ 0 ≥ 0 for all i and jfor all i and j..

The Assignment ProblemThe Assignment Problem

This type of problem This type of problem determines the most determines the most efficient assignment of efficient assignment of people to particular tasks, people to particular tasks, etc.etc.

Objective is typically to Objective is typically to minimize total cost or total minimize total cost or total task time.task time.

Linear Program for Linear Program for Assignment ExampleAssignment Example

The Fix-it Shop has just received three The Fix-it Shop has just received three new repair projects that must be new repair projects that must be repaired quickly: a radio, a toaster repaired quickly: a radio, a toaster oven, and a coffee table.oven, and a coffee table.

Three workers with different talents are Three workers with different talents are able to do the jobs.able to do the jobs.

The owner estimates the cost in wages The owner estimates the cost in wages if the workers are assigned to each of if the workers are assigned to each of the three jobs.the three jobs.

Objective: Objective: minimize total cost.minimize total cost.

Example of an Assignment Problem Example of an Assignment Problem in a Transportation Network Formatin a Transportation Network Format

Figure 9.2


Let:Let: XXijij = = 1 if person 1 if person ii is assigned is assigned

to project to project jj, or 0 otherwise., or 0 otherwise.

WhereWhere ii = 1,2,3 with 1 = Adams, = 1,2,3 with 1 = Adams,

2=Brown, and 3 = Cooper2=Brown, and 3 = Cooper jj = 1,2,3, with 1 = Project 1, = 1,2,3, with 1 = Project 1,

2=Project 2, and 3 = Project 3.2=Project 2, and 3 = Project 3.


Minimize total cost = 11XMinimize total cost = 11X1111 + 14X + 14X1212 + + 6X6X1313 + 8X + 8X2121 + 10X + 10X2222 + 11X + 11X2323 + 9X + 9X31 31 + 12X+ 12X3232 + + 7X7X3333

Subject to:Subject to: XX1111 + X + X1212 + X + X1313 ≤ 1 ≤ 1 XX2121 + X + X2222 + X + X2323 ≤ 1 ≤ 1 XX3131 + X + X3232 + X + X3333 ≤ 1 ≤ 1 XX1111 + X + X2121 + X + X3131 = 1 = 1 XX1212 + X + X2222 + X + X3232 = 1 = 1 XX1313 + X + X2323 + X + X3333 = 1 = 1 XXijij = 0 or 1 for all = 0 or 1 for all ii and and jj

Fix-it Shop Solution in Excel Fix-it Shop Solution in Excel 20102010

Program 9.2


XX1313 = 1, so Adams is assigned = 1, so Adams is assigned

to project 3.to project 3. XX2222 = 1, so Brown is assigned to = 1, so Brown is assigned to

project 2.project 2. XX3131 = 1, so Cooper is assigned = 1, so Cooper is assigned

to project 3.to project 3. Total cost of the repairs is $25.Total cost of the repairs is $25.

Transshipment ApplicationsTransshipment Applications

When the items are being moved from a source to When the items are being moved from a source to a destination through an intermediate point (a a destination through an intermediate point (a transshipment pointtransshipment point), the problem is called a ), the problem is called a transshipment problem.transshipment problem.Distribution CentersDistribution Centers

Frosty Machines manufactures snow blowers in Frosty Machines manufactures snow blowers in Toronto and Detroit.Toronto and Detroit.

These are shipped to regional distribution centers in These are shipped to regional distribution centers in Chicago and Buffalo.Chicago and Buffalo.

From there they are shipped to supply houses in From there they are shipped to supply houses in New York, Philadelphia, and St Louis.New York, Philadelphia, and St Louis.

Shipping costs vary by location and destination.Shipping costs vary by location and destination. Snow blowers cannot be shipped directly from the Snow blowers cannot be shipped directly from the

factories to the supply houses.factories to the supply houses.

Network Representation of Network Representation of Transshipment ExampleTransshipment Example

Figure 9.3


Frosty Machines Transshipment DataFrosty Machines Transshipment Data

TOTO

FROMFROM CHICAGOCHICAGO BUFFALOBUFFALONEW YORK NEW YORK CITYCITY PHILADELPHIAPHILADELPHIA ST LOUISST LOUIS SUPPLYSUPPLY

TorontoToronto $4$4 $7$7 —— —— —— 800800

DetroitDetroit $5$5 $7$7 —— —— —— 700700

ChicagoChicago —— —— $6$6 $4$4 $5$5 ——

BuffaloBuffalo —— —— $2$2 $3$3 $4$4 ——

DemandDemand —— —— 450450 350350 300300

Frosty would like to minimize the Frosty would like to minimize the transportation costs associated with transportation costs associated with shipping snow blowers to meet the shipping snow blowers to meet the demands at the supply centers given the demands at the supply centers given the supplies available.supplies available.

Table 9.1


A description of the problem would be to A description of the problem would be to minimize cost subject to:minimize cost subject to:

1.1. The number of units shipped from Toronto is not The number of units shipped from Toronto is not more than 800.more than 800.

2.2. The number of units shipped from Detroit is not The number of units shipped from Detroit is not more than 700.more than 700.

3.3. The number of units shipped to New York is 450.The number of units shipped to New York is 450.4.4. The number of units shipped to Philadelphia is 350.The number of units shipped to Philadelphia is 350.5.5. The number of units shipped to St Louis is 300.The number of units shipped to St Louis is 300.6.6. The number of units shipped out of Chicago is The number of units shipped out of Chicago is

equal to the number of units shipped into Chicago.equal to the number of units shipped into Chicago.7.7. The number of units shipped out of Buffalo is equal The number of units shipped out of Buffalo is equal

to the number of units shipped into Buffalo.to the number of units shipped into Buffalo.


The decision variables should represent the number of The decision variables should represent the number of units shipped from each source to the transshipment units shipped from each source to the transshipment points and from there to the final destinations.points and from there to the final destinations.

XX1313 = the number of units shipped from Toronto to = the number of units shipped from Toronto to

ChicagoChicagoXX1414 = the number of units shipped from Toronto to = the number of units shipped from Toronto to

BuffaloBuffaloXX2323 = the number of units shipped from Detroit to = the number of units shipped from Detroit to

ChicagoChicagoXX2424 = the number of units shipped from Detroit to = the number of units shipped from Detroit to

BuffaloBuffaloXX3535 = the number of units shipped from Chicago to = the number of units shipped from Chicago to

New YorkNew YorkXX3636 = the number of units shipped from Chicago to = the number of units shipped from Chicago to

PhiladelphiaPhiladelphiaXX3737 = the number of units shipped from Chicago to = the number of units shipped from Chicago to

St LouisSt LouisXX4545 = the number of units shipped from Buffalo to = the number of units shipped from Buffalo to

New YorkNew YorkXX4646 = the number of units shipped from Buffalo to = the number of units shipped from Buffalo to

PhiladelphiaPhiladelphiaXX4747 = the number of units shipped from Buffalo to = the number of units shipped from Buffalo to

St LouisSt Louis


The linear program is:The linear program is:

Minimize cost = 4XMinimize cost = 4X1313 + 7X + 7X1414 + + 5X5X2323 + 7X + 7X2424 + 6X + 6X3535 + 4X + 4X3636 + 5X + 5X3737 + + 2X2X4545 + 3X + 3X4646 + 4 + 4XX4747 subject tosubject to

XX1313 + X + X1414 ≤ 800≤ 800 (supply at Toronto)(supply at Toronto)XX2323 + + XX2424 ≤ 700≤ 700 (supply at Detroit)(supply at Detroit)XX3535 + X + X4545 = 450= 450 (demand at New York)(demand at New York)XX3636 + X + X4646 = 350= 350 (demand at Philadelphia)(demand at Philadelphia)XX37 37 + X+ X4747 = 300= 300 (demand at St Louis)(demand at St Louis)XX1313 + X + X2323 = X= X3535 + X + X3636 + X + X37 37 (shipping through (shipping through

Chicago)Chicago)XX1414 + X + X2424 = X= X4545 + X + X4646 + X + X47 47 (shipping through (shipping through

Buffalo)Buffalo) XXijij ≥ 0≥ 0for all for all ii and and jj (nonnegativity) (nonnegativity)

Solution to Frosty Machines Solution to Frosty Machines Transshipment ProblemTransshipment Problem

Program 9.3

The Transportation AlgorithmThe Transportation Algorithm

This is an iterative procedure in This is an iterative procedure in which a solution to a transportation which a solution to a transportation problem is found and evaluated problem is found and evaluated using a special procedure to using a special procedure to determine whether the solution is determine whether the solution is optimal.optimal. When the solution is optimal, the When the solution is optimal, the

process stops.process stops. If not, then a new solution is generated.If not, then a new solution is generated.

Transportation Table for Transportation Table for Executive Furniture CorporationExecutive Furniture Corporation

TOTO

FROM FROM

WAREHOUSE WAREHOUSE AT AT

ALBUQUERQUEALBUQUERQUE


BOSTONBOSTON


CLEVELANDCLEVELANDFACTORY FACTORY CAPACITYCAPACITY

DES MOINES DES MOINES FACTORYFACTORY

$5$5 $4$4 $3$3100100

EVANSVILLE EVANSVILLE FACTORYFACTORY

$8$8 $4$4 $3$3300300

FORT LAUDERDALE FORT LAUDERDALE FACTORYFACTORY

$9$9 $7$7 $5$5300300

WAREHOUSE WAREHOUSE REQUIREMENTSREQUIREMENTS 300300 200200 200200 700700

Table 9.2

Des Moines Des Moines capacity capacity constrainttconstraintt

Cell representing a Cell representing a source-to-source-to-destination destination (Evansville to (Evansville to Cleveland) shipping Cleveland) shipping assignment that assignment that could be madecould be made

Total Total supply supply and and demanddemand

ClevelaCleveland nd warehowarehouse use demanddemand

Cost of shipping 1 Cost of shipping 1 unit from Fort unit from Fort Lauderdale factory to Lauderdale factory to Boston warehouseBoston warehouse

Developing an Initial Solution: Developing an Initial Solution: Northwest Corner RuleNorthwest Corner Rule

Once we have arranged the data in a table, we must Once we have arranged the data in a table, we must establish an initial feasible solution.establish an initial feasible solution.

One systematic approach is known as the One systematic approach is known as the northwest northwest corner rule.corner rule.

Start in the upper left-hand cell and allocate units to Start in the upper left-hand cell and allocate units to shipping routes as follows:shipping routes as follows:

1.1. Exhaust the supply (factory capacity) of each row before Exhaust the supply (factory capacity) of each row before moving down to the next row.moving down to the next row.

2.2. Exhaust the demand (warehouse) requirements of each Exhaust the demand (warehouse) requirements of each column before moving to the right to the next column. column before moving to the right to the next column.

3.3. Check that all supply and demand requirements are met.Check that all supply and demand requirements are met. This problem takes five steps to make the initial This problem takes five steps to make the initial

shipping assignments.shipping assignments.

9-31


1.1. Beginning in the upper left hand corner, we assign Beginning in the upper left hand corner, we assign 100 units from Des Moines to Albuquerque. This 100 units from Des Moines to Albuquerque. This exhaust the supply from Des Moines but leaves exhaust the supply from Des Moines but leaves Albuquerque 200 desks short. We move to the second Albuquerque 200 desks short. We move to the second row in the same column.row in the same column.

TOTOFROMFROM

ALBUQUERQUE ALBUQUERQUE ((AA))

BOSTON BOSTON ((BB))

CLEVELAND CLEVELAND ((CC))

FACTORY FACTORY CAPACITYCAPACITY

DES MOINES DES MOINES ((DD)) 100100

$5$5 $4$4 $3$3100100

EVANSVILLE EVANSVILLE ((EE))

$8$8 $4$4 $3$3300300

FORT LAUDERDALE FORT LAUDERDALE ((FF))

$9$9 $7$7 $5$5300300



2.2. Assign 200 units from Evansville to Albuquerque. Assign 200 units from Evansville to Albuquerque. This meets Albuquerque’s demand. Evansville has This meets Albuquerque’s demand. Evansville has 100 units remaining so we move to the right to the 100 units remaining so we move to the right to the next column of the second row.next column of the second row.

TOTOFROMFROM






$5$5 $4$4 $3$3100100

EVANSVILLE EVANSVILLE ((EE)) 200200

$8$8 $4$4 $3$3300300


$9$9 $7$7 $5$5300300



3.3. Assign 100 units from Evansville to Boston. The Assign 100 units from Evansville to Boston. The Evansville supply has now been exhausted but Evansville supply has now been exhausted but Boston is still 100 units short. We move down Boston is still 100 units short. We move down vertically to the next row in the Boston column.vertically to the next row in the Boston column.

TOTOFROMFROM






$5$5 $4$4 $3$3100100


$8$8100100

$4$4 $3$3300300


$9$9 $7$7 $5$5300300



4.4. Assign 100 units from Fort Lauderdale to Assign 100 units from Fort Lauderdale to Boston. This fulfills Boston’s demand and Boston. This fulfills Boston’s demand and Fort Lauderdale still has 200 units available.Fort Lauderdale still has 200 units available.

TOTOFROMFROM






$5$5 $4$4 $3$3100100


$8$8100100

$4$4 $3$3300300


$9$9100100

$7$7 $5$5300300



5.5. Assign 200 units from Fort Lauderdale to Assign 200 units from Fort Lauderdale to Cleveland. This exhausts Fort Lauderdale’s Cleveland. This exhausts Fort Lauderdale’s supply and Cleveland’s demand. The initial supply and Cleveland’s demand. The initial shipment schedule is now complete.shipment schedule is now complete.

TOFROM

ALBUQUERQUE (A)

BOSTON (B)

CLEVELAND (C)

FACTORY CAPACITY

DES MOINES (D) 100

$5 $4 $3100

EVANSVILLE (E) 200

$8100

$4 $3300

FORT LAUDERDALE (F)

$9100

$7200200

$5300

WAREHOUSE REQUIREMENTS 300 200 200 700

Table 9.3


The cost of this shipping assignment:The cost of this shipping assignment:ROUTEROUTE

UNITS UNITS SHIPPEDSHIPPED xx

PER UNIT PER UNIT COST ($)COST ($) ==

TOTAL TOTAL COST ($)COST ($)FROMFROM TOTO

DD AA 100100 55 500500

EE AA 200200 88 1,6001,600

EE BB 100100 44 400400

FF BB 100100 77 700700

FF CC 200200 55 1,0001,000

4,2004,200

This solution is feasible but This solution is feasible but we need to check to see if it we need to check to see if it is optimal.is optimal.

Stepping-Stone Method: Stepping-Stone Method: Finding a Least Cost SolutionFinding a Least Cost Solution

The The stepping-stone methodstepping-stone method is an iterative is an iterative technique for moving from an initial feasible technique for moving from an initial feasible solution to an optimal feasible solution.solution to an optimal feasible solution.

There are two distinct parts to the process:There are two distinct parts to the process: Testing the current solution to determine if Testing the current solution to determine if

improvement is possible.improvement is possible. Making changes to the current solution to Making changes to the current solution to

obtain an improved solution.obtain an improved solution. This process continues until the optimal This process continues until the optimal

solution is reached.solution is reached.

Stepping-Stone Method: Finding a Stepping-Stone Method: Finding a Least Cost SolutionLeast Cost Solution

There is one very important rule: There is one very important rule: The number of The number of occupied routes (or squares) must always be occupied routes (or squares) must always be equal to one less than the sum of the number of equal to one less than the sum of the number of rows plus the number of columnsrows plus the number of columns In the Executive Furniture problem this means the initial In the Executive Furniture problem this means the initial

solution must have 3 + 3 – 1 = 5 squares used.solution must have 3 + 3 – 1 = 5 squares used.

Occupied Occupied shipping routes shipping routes

(squares)(squares)

NumbNumber of er of rowsrows

Number Number of of

columnscolumns

= + = + –– 1 1

When the number of occupied rows is When the number of occupied rows is less than this, the solution is called less than this, the solution is called degenerate.degenerate.

Testing the Solution for Testing the Solution for Possible ImprovementPossible Improvement

The stepping-stone method works The stepping-stone method works by testing each unused square in the by testing each unused square in the transportation table to see what transportation table to see what would happen to total shipping costs would happen to total shipping costs if one unit of the product were if one unit of the product were tentatively shipped on an unused tentatively shipped on an unused route.route.

There are five steps in the process.There are five steps in the process.

Five Steps to Test Unused Squares Five Steps to Test Unused Squares with the Stepping-Stone Methodwith the Stepping-Stone Method

1.1. Select an unused square to evaluate.Select an unused square to evaluate.

2.2. Beginning at this square, trace a closed Beginning at this square, trace a closed path back to the original square via path back to the original square via squares that are currently being used squares that are currently being used with only horizontal or vertical moves with only horizontal or vertical moves allowed.allowed.

3.3. Beginning with a plus (+) sign at the Beginning with a plus (+) sign at the unused square, place alternate minus unused square, place alternate minus ((––) signs and plus signs on each corner ) signs and plus signs on each corner square of the closed path just traced.square of the closed path just traced.


4.4. Calculate an Calculate an improvement indeximprovement index by adding by adding together the unit cost figures found in each together the unit cost figures found in each square containing a plus sign and then square containing a plus sign and then subtracting the unit costs in each square subtracting the unit costs in each square containing a minus sign.containing a minus sign.

5.5. Repeat steps 1 to 4 until an improvement Repeat steps 1 to 4 until an improvement index has been calculated for all unused index has been calculated for all unused squares. If all indices computed are greater squares. If all indices computed are greater than or equal to zero, an optimal solution has than or equal to zero, an optimal solution has been reached. If not, it is possible to improve been reached. If not, it is possible to improve the current solution and decrease total the current solution and decrease total shipping costs.shipping costs.


For the Executive Furniture Corporation data:For the Executive Furniture Corporation data:

Steps 1 and 2Steps 1 and 2. Beginning with Des Moines. Beginning with Des Moines––Boston route we trace a closed path using Boston route we trace a closed path using only currently occupied squares, alternately only currently occupied squares, alternately placing plus and minus signs in the corners placing plus and minus signs in the corners of the path.of the path. In a In a closed pathclosed path, only squares , only squares

currently used for shipping can be currently used for shipping can be used in turning corners.used in turning corners.

Only oneOnly one closed route is possible for closed route is possible for each square we wish to test.each square we wish to test.


Step 3Step 3. Test the cost-effectiveness of the Des . Test the cost-effectiveness of the Des MoinesMoines––Boston shipping route by pretending that Boston shipping route by pretending that we are shipping one desk from Des Moines to we are shipping one desk from Des Moines to Boston. Put a plus in that box.Boston. Put a plus in that box.

But if we ship one But if we ship one moremore unit out of Des Moines unit out of Des Moines we will be sending out 101 units.we will be sending out 101 units.

Since the Des Moines factory capacity is only Since the Des Moines factory capacity is only 100, we must ship 100, we must ship fewerfewer desks from Des Moines desks from Des Moines to Albuquerque so place a minus sign in that box.to Albuquerque so place a minus sign in that box.

But that leaves Albuquerque one unit short so But that leaves Albuquerque one unit short so increase the shipment from Evansville increase the shipment from Evansville tto o Albuquerque by one unit and so on until the Albuquerque by one unit and so on until the entire closed path is completed.entire closed path is completed.


Evaluating the unused Des Evaluating the unused Des MoinesMoines––Boston shipping Boston shipping routeroute

TOTOFROMFROM

ALBUQUERQUEALBUQUERQUE BOSTONBOSTON CLEVELANDCLEVELAND FACTORY FACTORY CAPACITYCAPACITY

DES MOINESDES MOINES 100100$5$5 $4$4 $3$3

100100

EVANSVILLEEVANSVILLE 200200$8$8

100100$4$4 $3$3

300300

FORT LAUDERDALEFORT LAUDERDALE$9$9

100100$7$7

200200$5$5

300300

WAREHOUSE WAREHOUSE REQUIREMENTSREQUIREMENTS 300300 200200 200200 700700 Table 9.3

Warehouse Warehouse BB$4$4

FactorFactoryyDD

Warehouse Warehouse AA$5$5

100100

FactorFactoryyEE

$8$8

200200

$4$4

100100

+

– +

–



TOTOFROMFROM



100100


100100$4$4 $3$3

300300


100100$7$7

200200$5$5

300300


Warehouse Warehouse AA

FactoryD

$5Warehouse Warehouse BB

$4

FactorFactoryyEE

$8 $4

100

200 100

201

991

+

– +

–99



TOTOFROMFROM



100100


100100$4$4 $3$3

300300


100100$7$7

200200$5$5

300300


Warehouse Warehouse AA

FactorFactoryyDD

$5Warehouse Warehouse BB

$4

FactorFactoryyEE

$8 $4

100

991

201

200 100

99+

– +

–

Result of Result of Proposed Shift Proposed Shift in Allocationin Allocation= 1 x $4= 1 x $4–– 1 x $51 x $5+ 1 x $8+ 1 x $8–– 1 x $4 = +$31 x $4 = +$3


Step 4Step 4. Now compute an . Now compute an improvement indeximprovement index ((IIijij) for the Des Moines) for the Des Moines––Boston route.Boston route.

Add the costs in the squares with Add the costs in the squares with plus signs and subtract the costs in plus signs and subtract the costs in the squares with minus signs:the squares with minus signs:

This means for every desk shipped via This means for every desk shipped via the Des Moinesthe Des Moines––Boston route, total Boston route, total transportation cost will transportation cost will increaseincrease by $3 by $3 over their current level.over their current level.


Step 5Step 5. Now examine the Des Moines. Now examine the Des Moines––Cleveland unusedCleveland unused route which is slightly route which is slightly more difficult to draw.more difficult to draw. Again, only turn corners at Again, only turn corners at

squares that represent squares that represent existing routes.existing routes.

Pass through the Pass through the EvansvilleEvansville–Cleveland square –Cleveland square but we can not turn there but we can not turn there or put a + or – sign.or put a + or – sign.

The closed path we will use The closed path we will use is:is:+ + DCDC – – DADA + + EAEA – – EBEB + + FBFB – –

FCFC


Evaluating the Des MoinesEvaluating the Des Moines––Cleveland Shipping RouteCleveland Shipping Route

TOTOFROMFROM



100100


100100$4$4 $3$3

300300


100100$7$7

200200$5$5

300300


Table 9.5

Start+

+ –

–

+ –


Opening the Des MoinesOpening the Des Moines–Cleveland route will not lower our –Cleveland route will not lower our total shipping costs.total shipping costs.Evaluating the other two routes we find:Evaluating the other two routes we find:

The closed path is + The closed path is + ECEC – – EBEB + + FBFB – – FCFC

The closed path is + The closed path is + FAFA – – FBFB + + EBEB – – EAEA

Opening the Fort Lauderdale-Albuquerque route Opening the Fort Lauderdale-Albuquerque route willwill lower lower our total transportation costs.our total transportation costs.

Evansville-Evansville-Cleveland indexCleveland index= = IIECEC = + $3 = + $3 –– $4 + $7 $4 + $7 –– $5 = + $1 $5 = + $1

Fort Fort LauderdaleLauderdale––Albuquerque Albuquerque indexindex

= = IIFAFA = + $9 = + $9 –– $7 + $4 $7 + $4 –– $8 = $8 = –– $2 $2

Obtaining an Improved SolutionObtaining an Improved Solution In the Executive Furniture problem there is In the Executive Furniture problem there is

only one unused route with a negative index only one unused route with a negative index (Fort Lauderdale-Albuquerque).(Fort Lauderdale-Albuquerque). If there was more than one route with a negative If there was more than one route with a negative

index, we would choose the one with the largest index, we would choose the one with the largest improvementimprovement

We now want to ship the maximum allowable We now want to ship the maximum allowable number of units on the new routenumber of units on the new route

The quantity to ship is found by referring to the The quantity to ship is found by referring to the closed path of plus and minus signs for the closed path of plus and minus signs for the new route and selecting the new route and selecting the smallest numbersmallest number found in those squares containing minus found in those squares containing minus signs.signs.

Obtaining an Improved Obtaining an Improved SolutionSolution

To obtain a new solution, that number is added To obtain a new solution, that number is added to all squares on the closed path with plus signs to all squares on the closed path with plus signs and subtracted from all squares the closed path and subtracted from all squares the closed path with minus signs.with minus signs.

All other squares are unchanged.All other squares are unchanged. In this case, the maximum number that can be In this case, the maximum number that can be

shipped is 100 desks as this is the smallest shipped is 100 desks as this is the smallest value in a box with a negative sign (value in a box with a negative sign (FBFB route). route).

We add 100 units to the We add 100 units to the FAFA and and EBEB routes and routes and subtract 100 from subtract 100 from FBFB and and EAEA routes. routes.

This leaves balanced rows and columns and an This leaves balanced rows and columns and an improved solution.improved solution.

Obtaining an Improved SolutionObtaining an Improved Solution

Stepping-Stone Path Used to Stepping-Stone Path Used to Evaluate Route Evaluate Route F-AF-A

TOTOFROMFROM

AA BB CC FACTORY FACTORY CAPACITYCAPACITY

DD 100100$5$5 $4$4 $3$3

100100

EE 200200$8$8

100100$4$4 $3$3

300300

FF$9$9

100100$7$7

200200$5$5

300300WAREHOUSE WAREHOUSE

REQUIREMENTSREQUIREMENTS 300300 200200 200200 700700

Table 9.6

+

+ –

–

Obtaining an Improved SolutionObtaining an Improved SolutionSecond Solution to the Executive Second Solution to the Executive Furniture ProblemFurniture Problem

TOTOFROMFROM


DD 100100$5$5 $4$4 $3$3

100100

EE 100100$8$8

200200$4$4 $3$3

300300

FF 100100$9$9 $7$7

200200$5$5

300300

WAREHOUSE WAREHOUSE REQUIREMENTSREQUIREMENTS 300300 200200 200200 700700Table 9.7

Total shipping costs have been Total shipping costs have been reduced by (100 units) x ($2 reduced by (100 units) x ($2 saved per unit) and now equals saved per unit) and now equals $4,000.$4,000.

Obtaining an Improved SolutionObtaining an Improved Solution This second solution may or may not be optimal.This second solution may or may not be optimal. To determine whether further improvement is To determine whether further improvement is

possible, we return to the first five steps to test each possible, we return to the first five steps to test each square that is square that is nownow unused. unused.

The four new improvement indices are:The four new improvement indices are:

DD to to BB = = IIDBDB = + $4 = + $4 –– $5 + $8 $5 + $8 –– $4 = + $3 $4 = + $3(closed path: + (closed path: + DBDB – – DADA + + EAEA – – EBEB))

DD to to CC = = IIDCDC = + $3 = + $3 –– $5 + $9 $5 + $9 –– $5 = + $2 $5 = + $2(closed path: + (closed path: + DCDC – – DADA + + FAFA – – FCFC))

EE to to CC = = IIECEC = + $3 = + $3 –– $8 + $9 $8 + $9 –– $5 = $5 = –– $1 $1(closed path: + (closed path: + ECEC – – EAEA + + FAFA – – FCFC))

FF to to BB = = IIFBFB = + $7 = + $7 –– $4 + $8 $4 + $8 –– $9 = + $2 $9 = + $2(closed path: + (closed path: + FBFB – – EBEB + + EAEA – – FAFA))

Obtaining an Improved SolutionObtaining an Improved Solution

An improvement can be made by An improvement can be made by shipping the maximum allowable number shipping the maximum allowable number of units from of units from EE to to C.C.

TOTOFROMFROM


DD 100100$5$5 $4$4 $3$3

100100

EE 100100$8$8

200200$4$4 $3$3

300300

FF 100100$9$9 $7$7

200200$5$5

300300


Table 9.8

Path to Evaluate the Path to Evaluate the E-E-C C RouteRoute

Start+

+ –

–


Total cost of third solution:Total cost of third solution:ROUTEROUTE

DESKS DESKS SHIPPEDSHIPPED xx

PER UNIT PER UNIT COST ($)COST ($) ==

TOTAL TOTAL COST ($)COST ($)FROMFROM TOTO

DD AA 100100 55 500500

EE BB 200200 44 800800

EE CC 100100 33 300300

FF AA 200200 99 1,8001,800

FF CC 100100 55 500500

3,9003,900


TOTOFROMFROM


DD 100100$5$5 $4$4 $3$3

100100

EE$8$8

200200$4$4

100100$3$3

300300

FF 200200$9$9 $7$7

100100$5$5

300300WAREHOUSE WAREHOUSE

REQUIREMENTSREQUIREMENTS 300300 200200 200200 700700

Table 9.9

Third and optimal solution:Third and optimal solution:

Obtaining an Improved SolutionObtaining an Improved SolutionThis solution is optimal as the improvement This solution is optimal as the improvement indices that can be computed are all greater indices that can be computed are all greater than or equal to zero.than or equal to zero.

D to B = IDB = + $4 – $5 + $9 – $5 + $3 – $4 = + $2(closed path: + DB – DA + FA – FC + EC – EB)

D to C = IDC = + $3 – $5 + $9 – $5 = + $2(closed path: + DC – DA + FA – FC)

E to A = IEA = + $8 – $9 + $5 – $3 = + $1(closed path: + EA – FA + FC – EC)

F to B = IFB = + $7 – $5 + $3 – $4 = + $1(closed path: + FB – FC + EC – EB)

9-60

Summary of Steps in Transportation Summary of Steps in Transportation Algorithm (Minimization)Algorithm (Minimization)

1.1. Set up a balanced transportation table.Set up a balanced transportation table.

2.2. Develop initial solution using the northwest Develop initial solution using the northwest corner method method.corner method method.

3.3. Calculate an improvement index for each empty Calculate an improvement index for each empty cell using the stepping-stone method. If cell using the stepping-stone method. If improvement indices are all nonnegative, stop as improvement indices are all nonnegative, stop as the optimal solution has been found. If any index the optimal solution has been found. If any index is negative, continue to step 4.is negative, continue to step 4.

4.4. Select the cell with the improvement index Select the cell with the improvement index indicating the greatest decrease in cost. Fill this indicating the greatest decrease in cost. Fill this cell using the stepping-stone path and go to step cell using the stepping-stone path and go to step 3.3.

Unbalanced Transportation Unbalanced Transportation ProblemsProblems

In real-life problems, total demand is frequently In real-life problems, total demand is frequently not equal to total supply.not equal to total supply.

These These unbalanced problemsunbalanced problems can be handled can be handled easily by introducing easily by introducing dummy sourcesdummy sources or or dummy dummy destinations.destinations.

If total supply is greater than total demand, a If total supply is greater than total demand, a dummy destination (warehouse), with demand dummy destination (warehouse), with demand exactly equal to the surplus, is created. exactly equal to the surplus, is created.

If total demand is greater than total supply, we If total demand is greater than total supply, we introduce a dummy source (factory) with a introduce a dummy source (factory) with a supply equal to the excess of demand over supply equal to the excess of demand over supply.supply.

Special Situations with the Special Situations with the Transportation AlgorithmTransportation Algorithm

Unbalanced Transportation ProblemsUnbalanced Transportation Problems In either case, shipping cost In either case, shipping cost

coefficients of zero are assigned to coefficients of zero are assigned to each dummy location or route as no each dummy location or route as no goods will actually be shipped.goods will actually be shipped.

Any units assigned to a dummy Any units assigned to a dummy destination represent excess capacity.destination represent excess capacity.

Any units assigned to a dummy source Any units assigned to a dummy source represent unmet demand.represent unmet demand.

Demand Less Than SupplyDemand Less Than Supply Suppose that the Des Moines factory increases its Suppose that the Des Moines factory increases its

rate of production from 100 to 250 desks.rate of production from 100 to 250 desks. The firm is now able to supply a total of 850 desks The firm is now able to supply a total of 850 desks

each period.each period. Warehouse requirements remain the same (700) Warehouse requirements remain the same (700)

so the row and column totals do not balance.so the row and column totals do not balance. We add a dummy column that will represent a fake We add a dummy column that will represent a fake

warehouse requiring 150 desks.warehouse requiring 150 desks. This is somewhat analogous to adding a slack This is somewhat analogous to adding a slack

variable.variable. We use the stepping-stone method to find the We use the stepping-stone method to find the

optimal solution.optimal solution.

Demand Less Than SupplyDemand Less Than Supply

Initial Solution to an Unbalanced Problem Where Initial Solution to an Unbalanced Problem Where Demand is Less Than SupplyDemand is Less Than Supply

TOTOFROMFROM AA BB CC DUMMY DUMMY

WAREHOUSEWAREHOUSETOTAL TOTAL AVAILABLEAVAILABLE

DD 250250$5$5 $4$4 $3$3 00

250250

EE 5050$8$8

200200$4$4

5050$3$3 00

300300

FF$9$9 $7$7

150150$5$5

15015000

300300

WAREHOUSE WAREHOUSE REQUIREMENTSREQUIREMENTS 300300 200200 200200 150150 850850

Table 9.10New Des New Des Moines Moines capacitycapacityTotal cost = 250($5) + 50($8) + 200($4) + 50($3) + 150($5) + Total cost = 250($5) + 50($8) + 200($4) + 50($3) + 150($5) +

150(0) = $3,350150(0) = $3,350

Demand Greater than SupplyDemand Greater than Supply The second type of unbalanced The second type of unbalanced

condition occurs when total condition occurs when total demand is greater than total demand is greater than total supply.supply.

In this case we need to add a In this case we need to add a dummy row representing a dummy row representing a fake factory.fake factory.

The new factory will have a The new factory will have a supply exactly equal to the supply exactly equal to the difference between total difference between total demand and total real supply.demand and total real supply.

The shipping costs from the The shipping costs from the dummy factory to each dummy factory to each destination will be zero.destination will be zero.

Demand Greater than SupplyDemand Greater than Supply

Unbalanced Transportation Table for Happy Sound Stereo Company

TOTOFROMFROM

WAREHOUSE WAREHOUSE AA

WAREHOUSE WAREHOUSE BB

WAREHOUSE WAREHOUSE CC

PLANT PLANT SUPPLYSUPPLY

PLANT PLANT WW$6$6 $4$4 $9$9

200200

PLANT PLANT XX$10$10 $5$5 $8$8

175175

PLANT PLANT YY$12$12 $7$7 $6$6

7575

WAREHOUSE WAREHOUSE DEMANDDEMAND 250250 100100 150150

450450500500

Table 9.11

Totals Totals do not do not

balancebalance

Demand Greater than SupplyDemand Greater than SupplyInitial Solution to an Initial Solution to an

Unbalanced Problem in Which Unbalanced Problem in Which Demand is Greater Than SupplyDemand is Greater Than SupplyTOTO

FROMFROMWAREHOUSE WAREHOUSE

AAWAREHOUSE WAREHOUSE

BBWAREHOUSE WAREHOUSE

CCPLANT PLANT

SUPPLYSUPPLY

PLANT PLANT WW 200200$6$6 $4$4 $9$9

200200

PLANT PLANT XX 5050$10$10

100100$5$5

2525$8$8

175175

PLANT PLANT YY$12$12 $7$7

7575$6$6

7575

PLANT PLANT YY00 00

505000

5050

WAREHOUSE WAREHOUSE DEMANDDEMAND 250250 100100 150150 500500

Table 9.12Total cost of initial solution =Total cost of initial solution = 200($6) + 200($6) +

50($10) + 100($5) + 25($8) + 50($10) + 100($5) + 25($8) + 75($6) + $50(0) = $2,85075($6) + $50(0) = $2,850

Degeneracy in Transportation Degeneracy in Transportation ProblemsProblems

DegeneracyDegeneracy occurs when the number of occurs when the number of occupied squares or routes in a occupied squares or routes in a transportation table solution is less than transportation table solution is less than the number of rows plus the number of the number of rows plus the number of columns minus 1.columns minus 1.

Such a situation may arise in the initial Such a situation may arise in the initial solution or in any subsequent solution.solution or in any subsequent solution.

Degeneracy requires a special procedure to Degeneracy requires a special procedure to correct the problem since there are not correct the problem since there are not enough occupied squares to trace a closed enough occupied squares to trace a closed path for each unused route and it would be path for each unused route and it would be impossible to apply the stepping-stone impossible to apply the stepping-stone method.method.

Degeneracy in Transportation Degeneracy in Transportation ProblemsProblems

To handle degenerate problems, create an To handle degenerate problems, create an artificially occupied cell.artificially occupied cell.

That is, place a zero (representing a fake That is, place a zero (representing a fake shipment) in one of the unused squares and shipment) in one of the unused squares and then treat that square as if it were occupied.then treat that square as if it were occupied.

The square chosen must be in such a position The square chosen must be in such a position as to allow all stepping-stone paths to be as to allow all stepping-stone paths to be closed.closed.

There is usually a good deal of flexibility in There is usually a good deal of flexibility in selecting the unused square that will receive selecting the unused square that will receive the zero.the zero.

Degeneracy in an Initial Degeneracy in an Initial SolutionSolution

Initial Solution of a Degenerate ProblemInitial Solution of a Degenerate Problem

TOFROM

CUSTOMER 1 CUSTOMER 2 CUSTOMER 3 WAREHOUSE SUPPLY

WAREHOUSE 1 100$8 $2 $6

100

WAREHOUSE 2$10

100$9

20$9

120

WAREHOUSE 3$7 $10

80$7

80

CUSTOMER DEMAND 100 100 100 300

Table 9.13

00

00

Possible choices of Possible choices of cells to address the cells to address the degenerate solutiondegenerate solution

Degeneracy in an Initial Degeneracy in an Initial SolutionSolution

The Martin Shipping Company example The Martin Shipping Company example illustrates degeneracy in an initial illustrates degeneracy in an initial solution.solution.

It has three warehouses which supply It has three warehouses which supply three major retail customers.three major retail customers.

Applying the northwest corner rule the Applying the northwest corner rule the initial solution has only four occupied initial solution has only four occupied squaressquares

To correct this problem, place a zero in To correct this problem, place a zero in an unused square, typically one adjacent an unused square, typically one adjacent to the last filled cell.to the last filled cell.

Degeneracy During Later Degeneracy During Later Solution StagesSolution Stages

A transportation problem can become A transportation problem can become degenerate after the initial solution degenerate after the initial solution stage if the filling of an empty square stage if the filling of an empty square results in two or more cells becoming results in two or more cells becoming empty simultaneously.empty simultaneously.

This problem can occur when two or This problem can occur when two or more cells with minus signs tie for the more cells with minus signs tie for the lowest quantity.lowest quantity.

To correct this problem, place a zero in To correct this problem, place a zero in one of the previously filled cells so that one of the previously filled cells so that only one cell becomes empty.only one cell becomes empty.


Bagwell Paint ExampleBagwell Paint Example After one iteration, the cost analysis at After one iteration, the cost analysis at

Bagwell Paint produced a transportation Bagwell Paint produced a transportation table that was not degenerate but was not table that was not degenerate but was not optimal.optimal.

The improvement indices are:The improvement indices are:

factory factory AA – warehouse 2 index = +2 – warehouse 2 index = +2factory factory AA – warehouse 3 index = +1 – warehouse 3 index = +1factory factory BB – warehouse 3 index = – warehouse 3 index = –15–15factory factory CC – warehouse 2 index = +11 – warehouse 2 index = +11

Only route Only route with a with a

negative negative indexindex


Bagwell Paint Transportation TableBagwell Paint Transportation Table

TOTOFROMFROM

WAREHOUSE WAREHOUSE 11

WAREHOUSE WAREHOUSE 22

WAREHOUSE WAREHOUSE 33 FACTORY FACTORY

CAPACITYCAPACITY

FACTORY FACTORY AA 7070$8$8 $5$5 $16$16

7070

FACTORY FACTORY BB 5050$15$15

8080$10$10 $7$7

130130

FACTORY FACTORY CC 3030$3$3 $9$9

5050$10$10

8080

WAREHOUSE WAREHOUSE REQUIREMENTREQUIREMENT 150150 8080 5050 280280

Table 9.14


Tracing a Closed Path for the Factory Tracing a Closed Path for the Factory B – Warehouse 3 RouteB – Warehouse 3 Route

TOTOFROMFROM

WAREHOUSWAREHOUSE 1E 1

WAREHOUSWAREHOUSE 3E 3

FACTORY FACTORY BB 5050$15$15 $7$7

FACTORY FACTORY CC 3030$3$3

5050$10$10

Table 9.15

+

+ –

–

This would cause two cells to drop This would cause two cells to drop to zero.to zero.

We need to place an artificial zero We need to place an artificial zero in one of these cells to avoid in one of these cells to avoid degeneracy.degeneracy.

More Than One Optimal More Than One Optimal SolutionSolution

It is possible for a transportation problem to have It is possible for a transportation problem to have multiple optimal solutions.multiple optimal solutions.

This happens when one or more of the improvement This happens when one or more of the improvement indices is zero in the optimal solution.indices is zero in the optimal solution. This means that it is possible to design alternative shipping This means that it is possible to design alternative shipping

routes with the same total shipping cost.routes with the same total shipping cost. The alternate optimal solution can be found by shipping the The alternate optimal solution can be found by shipping the

most to this unused square using a stepping-stone path.most to this unused square using a stepping-stone path.

In the real world, alternate optimal solutions provide In the real world, alternate optimal solutions provide management with greater flexibility in selecting and management with greater flexibility in selecting and using resources.using resources.

Maximization Transportation Maximization Transportation ProblemsProblems

If the objective in a transportation problem If the objective in a transportation problem is to maximize profit, a minor change is is to maximize profit, a minor change is required in the transportation algorithm.required in the transportation algorithm.

Now the optimal solution is reached when Now the optimal solution is reached when all the improvement indices are negative or all the improvement indices are negative or zero.zero.

The cell with the largest positive The cell with the largest positive improvement index is selected to be filled improvement index is selected to be filled using a stepping-stone path.using a stepping-stone path.

This new solution is evaluated and the This new solution is evaluated and the process continues until there are no process continues until there are no positive improvement indices.positive improvement indices.

Unacceptable Or Prohibited Unacceptable Or Prohibited RoutesRoutes

At times there are transportation problems in At times there are transportation problems in which one of the sources is unable to ship to which one of the sources is unable to ship to one or more of the destinations.one or more of the destinations. The problem is said to have an The problem is said to have an unacceptableunacceptable or or

prohibited route.prohibited route. In a minimization problem, such a prohibited In a minimization problem, such a prohibited

route is assigned a very high cost to prevent route is assigned a very high cost to prevent this route from ever being used in the this route from ever being used in the optimal solution.optimal solution.

In a maximization problem, the very high In a maximization problem, the very high cost used in minimization problems is given cost used in minimization problems is given a negative sign, turning it into a very bad a negative sign, turning it into a very bad profit.profit.

Facility Location AnalysisFacility Location Analysis

The transportation method is especially The transportation method is especially useful in helping a firm to decide where useful in helping a firm to decide where to locate a new factory or warehouse.to locate a new factory or warehouse.

Each alternative location should be Each alternative location should be analyzed within the framework of one analyzed within the framework of one overalloverall distribution system. distribution system.

The new location that yields the The new location that yields the minimum cost for the minimum cost for the entire systementire system is is the one that should be chosen.the one that should be chosen.

Locating a New Factory for Locating a New Factory for Hardgrave Machine CompanyHardgrave Machine Company

Hardgrave Machine produces computer Hardgrave Machine produces computer components at three plants and ships to four components at three plants and ships to four warehouses.warehouses.

The plants have not been able to keep up The plants have not been able to keep up with demand so the firm wants to build a new with demand so the firm wants to build a new plant.plant.

Two sites are being considered, Seattle and Two sites are being considered, Seattle and Birmingham.Birmingham.

Data has been collected for each possible Data has been collected for each possible location. Which new location will yield the location. Which new location will yield the lowest cost for the firm in combination with lowest cost for the firm in combination with the existing plants and warehouses?the existing plants and warehouses?

Locating a New Factor for y Locating a New Factor for y Hardgrave Machine CompanyHardgrave Machine Company

Hardgrave’s Demand and Supply DataHardgrave’s Demand and Supply Data

WAREHOUSEWAREHOUSE

MONTHLY MONTHLY DEMAND DEMAND (UNITS)(UNITS)

PRODUCTIOPRODUCTION PLANTN PLANT

MONTHLY MONTHLY SUPPLYSUPPLY

COST TO COST TO PRODUCE ONE PRODUCE ONE UNIT ($)UNIT ($)

DetroitDetroit 10,00010,000 CincinnatiCincinnati 15,00015,000 4848

DallasDallas 12,00012,000 Salt LakeSalt Lake 6,0006,000 5050

New YorkNew York 15,00015,000 PittsburghPittsburgh 14,00014,000 5252

Los AngelesLos Angeles 9,0009,000 35,00035,000

46,00046,000

Supply needed from new plant = 46,000 – 35,000 = 11,000 units per monthSupply needed from new plant = 46,000 – 35,000 = 11,000 units per month

ESTIMATED PRODUCTION ESTIMATED PRODUCTION COST PER UNIT AT COST PER UNIT AT PROPOSED PLANTSPROPOSED PLANTS

SeattleSeattle $53$53

BirminghamBirmingham $49$49

Table 9.16


Hardgrave’s Shipping CostsHardgrave’s Shipping Costs

TOTOFROMFROM DETROITDETROIT DALLASDALLAS

NEW NEW YORKYORK

LOS LOS ANGELESANGELES

CINCINNATICINCINNATI $25$25 $55$55 $40$40 $60$60

SALT LAKESALT LAKE 3535 3030 5050 4040

PITTSBURGHPITTSBURGH 3636 4545 2626 6666

SEATTLESEATTLE 6060 3838 6565 2727

BIRMINGHAMBIRMINGHAM 3535 3030 4141 5050

Table 9.17

Locating a New Factory for Locating a New Factory for Hardgrave Machine CompanyHardgrave Machine CompanyBirmingham Plant Optimal Solution: Total Birmingham Plant Optimal Solution: Total

Hardgrave Cost is $3,741,000Hardgrave Cost is $3,741,000

TOTOFROMFROM DETROITDETROIT DALLASDALLAS NEW YORKNEW YORK



CINCINNATICINCINNATI 10,00010,0007373 103103

1,0001,0008888

4,0004,000108108

15,00015,000

SALT LAKESALT LAKE8585

1,0001,0008080 100100

5,0005,0009090

6,0006,000

PITTSBURGHPITTSBURGH8888 9797

14,00014,0007878 118118

14,00014,000

BIRMINGHAMBIRMINGHAM8484

11,00011,0007979 9090 9999

11,00011,000

WAREHOUSE WAREHOUSE REQUIREMENTREQUIREMENT 10,00010,000 12,00012,000 15,00015,000 9,0009,000 46,00046,000

Table 9.18

Locating a New Factory for Locating a New Factory for Hardgrave Machine CompanyHardgrave Machine CompanySeattle Plant Optimal Solution: Total Hardgrave Cost is Seattle Plant Optimal Solution: Total Hardgrave Cost is $3,704,000.$3,704,000.

TOTOFROMFROM DETROITDETROIT DALLASDALLAS NEW YORKNEW YORK



CINCINNATICINCINNATI 10,00010,0007373

4,0004,000103103

1,0001,0008888 108108

15,00015,000

SALT LAKESALT LAKE8585

6,0006,0008080 100100 9090

6,0006,000

PITTSBURGHPITTSBURGH8888 9797

14,00014,0007878 118118

14,00014,000

SEATTLESEATTLE113113

2,0002,0009191 118118

9,0009,0008080

11,00011,000

WAREHOUSE WAREHOUSE REQUIREMENTREQUIREMENT 10,00010,000 12,00012,000 15,00015,000 9,0009,000 46,00046,000

Table 9.19


By comparing the total system costs of the By comparing the total system costs of the two alternatives, Hardgrave can select the two alternatives, Hardgrave can select the lowest cost option:lowest cost option: The Birmingham location yields a total system The Birmingham location yields a total system

cost of $3,741,000.cost of $3,741,000. The Seattle location yields a total system cost The Seattle location yields a total system cost

of $3,704,000.of $3,704,000. With the lower total system cost, the Seattle With the lower total system cost, the Seattle

location is favored.location is favored. Excel QM can also be used as a solution Excel QM can also be used as a solution

tool.tool.

Excel QM Solution for Facility Excel QM Solution for Facility Location ExampleLocation Example

Program 9.4

The Assignment AlgorithmThe Assignment Algorithm The second special-purpose LP algorithm is the The second special-purpose LP algorithm is the

assignment method.assignment method. Each assignment problem has associated with it a Each assignment problem has associated with it a

table, or matrix.table, or matrix. Generally, the rows contain the objects or people we Generally, the rows contain the objects or people we

wish to assign, and the columns comprise the tasks or wish to assign, and the columns comprise the tasks or things to which we want them assigned.things to which we want them assigned.

The numbers in the table are the costs associated with The numbers in the table are the costs associated with each particular assignment.each particular assignment.

An assignment problem can be viewed as a An assignment problem can be viewed as a transportation problem in which the capacity from transportation problem in which the capacity from each source is 1 and the demand at each destination is each source is 1 and the demand at each destination is 1.1.

Assignment Model ApproachAssignment Model Approach

The Fix-It Shop has three rush projects to The Fix-It Shop has three rush projects to repair.repair.

The shop has three repair persons with The shop has three repair persons with different talents and abilities.different talents and abilities.

The owner has estimates of wage costs The owner has estimates of wage costs for each worker for each project.for each worker for each project.

The owner’s objective is to assign the The owner’s objective is to assign the three project to the workers in a way that three project to the workers in a way that will result in the lowest cost to the shop.will result in the lowest cost to the shop.

Each project will be assigned exclusively Each project will be assigned exclusively to one worker.to one worker.


Estimated Project Repair Costs for the Fix-It Estimated Project Repair Costs for the Fix-It Shop Assignment ProblemShop Assignment Problem

PROJECTPROJECT

PERSONPERSON 11 22 33

AdamsAdams $11$11 $14$14 $6$6

BrownBrown 88 1010 1111

CooperCooper 99 1212 77

Table 9.20


Summary of Fix-It Shop Assignment Alternatives and Summary of Fix-It Shop Assignment Alternatives and CostsCosts

PRODUCT ASSIGNMENTPRODUCT ASSIGNMENT

11 22 33 LABOUR LABOUR COSTS ($)COSTS ($)

TOTAL TOTAL COSTS ($)COSTS ($)

AdamsAdams BrownBrown CooperCooper 11 + 10 + 711 + 10 + 7 2828

AdamsAdams CooperCooper BrownBrown 11 + 12 + 11 + 12 + 1111 3434

BrownBrown AdamsAdams CooperCooper 8 + 14 + 78 + 14 + 7 2929

BrownBrown Cooper Cooper AdamsAdams 8 + 12 + 68 + 12 + 6 2626

CooperCooper AdamsAdams BrownBrown 9 + 14 + 119 + 14 + 11 3434

CooperCooper BrownBrown AdamsAdams 9 + 10 + 69 + 10 + 6 2525

Table 9.21

The Hungarian Method The Hungarian Method (Flood’s Technique)(Flood’s Technique)

The The Hungarian methodHungarian method is an efficient method of is an efficient method of finding the optimal solution to an assignment finding the optimal solution to an assignment problem without having to make direct comparisons problem without having to make direct comparisons of every option.of every option.

It operates on the principle of It operates on the principle of matrix reduction.matrix reduction. By subtracting and adding appropriate numbers in By subtracting and adding appropriate numbers in

the cost table or matrix, we can reduce the problem the cost table or matrix, we can reduce the problem to a matrix of to a matrix of opportunity costs.opportunity costs.

Opportunity costs show the relative penalty Opportunity costs show the relative penalty associated with assigning any person to a project as associated with assigning any person to a project as opposed to making the opposed to making the bestbest assignment. assignment.

We want to make assignment so that the opportunity We want to make assignment so that the opportunity cost for each assignment is zero.cost for each assignment is zero.

Three Steps of the Three Steps of the Assignment MethodAssignment Method

1.1. Find the opportunity cost table byFind the opportunity cost table by::

(a)(a) Subtracting the smallest number in each row of Subtracting the smallest number in each row of the original cost table or matrix from every the original cost table or matrix from every number in that row.number in that row.

(b)(b) Then subtracting the smallest number in each Then subtracting the smallest number in each column of the table obtained in part (a) from column of the table obtained in part (a) from every number in that column.every number in that column.

2.2. Test the table resulting from step 1 to see whether Test the table resulting from step 1 to see whether an optimal assignment can be madean optimal assignment can be made by drawing the by drawing the minimum number of vertical and horizontal straight minimum number of vertical and horizontal straight lines necessary to cover all the zeros in the table. If lines necessary to cover all the zeros in the table. If the number of lines is less than the number of rows the number of lines is less than the number of rows or columns, proceed to step 3.or columns, proceed to step 3.

Three Steps of the Three Steps of the Assignment MethodAssignment Method

3.3. Revise the opportunity cost tableRevise the opportunity cost table by by subtractingsubtracting the smallest number the smallest number not covered by a line from all not covered by a line from all numbers not covered by a straight numbers not covered by a straight line. This same number is also line. This same number is also added to every number lying at the added to every number lying at the intersection of any two lines. intersection of any two lines. Return to step 2 and continue the Return to step 2 and continue the cycle until an optimal assignment cycle until an optimal assignment is possible.is possible.

Figure 9.4

Steps in the Steps in the Assignment MethodAssignment Method

The Hungarian Method (Flood’s The Hungarian Method (Flood’s Technique)Technique)

Step 1: Find the opportunity cost table. We can compute row opportunity

costs and column opportunity costs. What we need is the total opportunity

cost. We derive this by taking the row

opportunity costs and subtract the smallest number in that column from each number in that column.

The Hungarian Method (Flood’s The Hungarian Method (Flood’s Technique)Technique)

Cost of Each Person-Cost of Each Person-Project Assignment for Project Assignment for the Fix-it Shop Problemthe Fix-it Shop Problem

Row Opportunity Row Opportunity Cost Table for Cost Table for the Fix-it Shop the Fix-it Shop Step 1, Part (a)Step 1, Part (a)PROJECTPROJECT


AdamsAdams $11$11 $14$14 $6$6

BrownBrown 88 1010 1111


Tables 9.22-9.23

PROJECTPROJECT


AdamsAdams $5$5 $8$8 $0$0

BrownBrown 00 22 33


The opportunity cost of The opportunity cost of assigning Cooper to project assigning Cooper to project 2 is $12 2 is $12 –– $7 = $5. $7 = $5.


Derive the total opportunity costs by taking the costs Derive the total opportunity costs by taking the costs in Table 9.23 and subtract the smallest number in in Table 9.23 and subtract the smallest number in each column from each number in that column.each column from each number in that column.

Table 9.24

Total Opportunity Cost Total Opportunity Cost Table for the Fix-it Shop Table for the Fix-it Shop

Step 1, Part (b)Step 1, Part (b)PROJECTPROJECT



BrownBrown 00 00 33



Step 2: Test for the optimal assignment.Step 2: Test for the optimal assignment. We want to assign workers to projects in such We want to assign workers to projects in such

a way that the total labor costs are at a a way that the total labor costs are at a minimum.minimum.

We would like to have a total assigned We would like to have a total assigned opportunity cost of zero.opportunity cost of zero.

The test to determine if we have reached an The test to determine if we have reached an optimal solution is simple.optimal solution is simple.

We find the We find the minimumminimum number of straight lines number of straight lines necessary to cover all the zeros in the table.necessary to cover all the zeros in the table.

If the number of lines equals the number of If the number of lines equals the number of rows or columns, an optimal solution has rows or columns, an optimal solution has been reached.been reached.


Test for Optimal Solution to Fix-it Test for Optimal Solution to Fix-it Shop ProblemShop Problem

PROJECTPROJECT



BrownBrown 00 00 33


Table 9.25

Covering line 1Covering line 1


This requires only two lines This requires only two lines to cover the zeros so the to cover the zeros so the solution is not optimal.solution is not optimal.


Step 3: Revise the opportunity-Step 3: Revise the opportunity-cost table.cost table. We We subtractsubtract the smallest number not the smallest number not

covered by a line from all numbers covered by a line from all numbers not covered by a straight line.not covered by a straight line.

The same number is added to every The same number is added to every number lying at the intersection of number lying at the intersection of any two lines.any two lines.

We then return to step 2 to test this We then return to step 2 to test this new table.new table.


Revised Opportunity Cost Revised Opportunity Cost Table for the Fix-it Shop Table for the Fix-it Shop

ProblemProblemPROJECTPROJECT

PERSOPERSONN 11 22 33


BrownBrown 00 00 55


Table 9.26

9-102


Optimality Test on the Optimality Test on the Revised Fix-it Shop Revised Fix-it Shop

Opportunity Cost TableOpportunity Cost TablePROJECTPROJECT



BrownBrown 00 00 55


Table 9.27



This requires three lines This requires three lines to cover the zeros so the to cover the zeros so the solution is optimal.solution is optimal.


Making the Final AssignmentMaking the Final Assignment

The optimal assignment is Adams The optimal assignment is Adams to project 3, Brown to project 2, to project 3, Brown to project 2, and Cooper to project 1.and Cooper to project 1.

For larger problems one approach For larger problems one approach to making the final assignment is to making the final assignment is to select a row or column that to select a row or column that contains only one zero.contains only one zero. Make the assignment to that cell Make the assignment to that cell

and rule out its row and column.and rule out its row and column. Follow this same approach for all Follow this same approach for all

the remaining cells.the remaining cells.


Total labour costs of this assignment are:Total labour costs of this assignment are:

ASSIGNMENTASSIGNMENT COST ($)COST ($)

Adams to project 3Adams to project 3 66

Brown to project 2Brown to project 2 1010

Cooper to project 1Cooper to project 1 99

Total costTotal cost 2525


Making the Final Fix-it Shop AssignmentsMaking the Final Fix-it Shop Assignments

(A) FIRST (A) FIRST ASSIGNMENTASSIGNMENT

(B) SECOND (B) SECOND ASSIGNMENTASSIGNMENT

(C) THIRD (C) THIRD ASSIGNMENTASSIGNMENT

11 22 33 11 22 33 11 22 33

AdamsAdams 33 44 00 AdamsAdams 33 44 00 AdamsAdams 33 44 00

Brown Brown 00 00 55 Brown Brown 00 00 55 Brown Brown 00 00 55

CooperCooper 00 11 00 CooperCooper 00 11 00 CooperCooper 00 11 00

Table 9.28

Excel QM Solution for Fix-It Excel QM Solution for Fix-It Shop Assignment ProblemShop Assignment Problem

Program 9.5

Unbalanced Assignment Unbalanced Assignment ProblemsProblems

Often the number of people or objects to be Often the number of people or objects to be assigned does not equal the number of tasks or assigned does not equal the number of tasks or clients or machines listed in the columns, and the clients or machines listed in the columns, and the problem is problem is unbalanced.unbalanced.

When this occurs, and there are more rows than When this occurs, and there are more rows than columns, simply add a columns, simply add a dummy columndummy column or task. or task.

If the number of tasks exceeds the number of people If the number of tasks exceeds the number of people available, we add a available, we add a dummy rowdummy row..

Since the dummy task or person is nonexistent, we Since the dummy task or person is nonexistent, we enter zeros in its row or column as the cost or time enter zeros in its row or column as the cost or time estimate.estimate.

Unbalanced Assignment Unbalanced Assignment ProblemsProblems

Suppose the Fix-It Shop has another worker Suppose the Fix-It Shop has another worker available.available.

The shop owner still has the same basic problem The shop owner still has the same basic problem of assigning workers to projects, but the problem of assigning workers to projects, but the problem now needs a dummy column to balance the four now needs a dummy column to balance the four workers and three projects.workers and three projects.

PROJECTPROJECT

PERSONPERSON 11 22 33 DUMMYDUMMY

AdamsAdams $11$11 $14$14 $6$6 $0$0

BrownBrown 88 1010 1111 00

CooperCooper 99 1212 77 00

DavisDavis 1010 1313 88 00Table 9.29

Maximization Assignment Maximization Assignment ProblemsProblems

Some assignment problems are phrased in terms Some assignment problems are phrased in terms of maximizing the payoff, profit, or effectiveness.of maximizing the payoff, profit, or effectiveness.

It is easy to obtain an equivalent minimization It is easy to obtain an equivalent minimization problem by converting all numbers in the table to problem by converting all numbers in the table to opportunity costs.opportunity costs.

This is brought about by subtracting every This is brought about by subtracting every number in the original payoff table from the number in the original payoff table from the largest single number in that table.largest single number in that table.

Transformed entries represent opportunity costs.Transformed entries represent opportunity costs. Once the optimal assignment has been found, the Once the optimal assignment has been found, the

total payoff is found by adding the original payoffs total payoff is found by adding the original payoffs of those cells that are in the optimal assignment.of those cells that are in the optimal assignment.


The British navy wishes to The British navy wishes to assign four ships to patrol four assign four ships to patrol four sectors of the North Sea.sectors of the North Sea.

Ships are rated for their probable Ships are rated for their probable efficiency in each sector.efficiency in each sector.

The commander wants to The commander wants to determine patrol assignments determine patrol assignments producing the greatest overall producing the greatest overall efficiencies.efficiencies.


Efficiencies of British Ships in Patrol SectorsEfficiencies of British Ships in Patrol Sectors

SECTORSECTOR

SHIPSHIP AA BB CC DD11 2020 6060 5050 555522 6060 3030 8080 757533 8080 100100 9090 808044 6565 8080 7575 7070

Table 9.30


Opportunity Costs of British ShipsOpportunity Costs of British Ships

SECTORSECTORSHISHIPP AA BB CC DD

11 8080 4040 5050 454522 4040 7070 2020 252533 2020 00 1010 202044 3535 2020 2525 3030

Table 9.31


Convert the maximization efficiency table into a Convert the maximization efficiency table into a minimizing opportunity cost table by subtracting minimizing opportunity cost table by subtracting each rating from 100, the largest rating in the each rating from 100, the largest rating in the whole table.whole table.

The smallest number in each row is subtracted The smallest number in each row is subtracted from every number in that row and the smallest from every number in that row and the smallest number in each column is subtracted from every number in each column is subtracted from every number in that column.number in that column.

The minimum number of lines needed to cover The minimum number of lines needed to cover the zeros in the table is four, so this represents the zeros in the table is four, so this represents an optimal solution.an optimal solution.


Row Opportunity Costs for the British Navy ProblemRow Opportunity Costs for the British Navy Problem

SECTORSECTOR


Table 9.32


Total Opportunity Costs for the Total Opportunity Costs for the British Navy ProblemBritish Navy Problem

SECTORSECTOR


Table 9.33


The overall efficiencyThe overall efficiency

ASSIGNMENTASSIGNMENT EFFICIENCYEFFICIENCY

Ship 1 to sector Ship 1 to sector DD 5555

Ship 2 to sector Ship 2 to sector CC 8080

Ship 3 to sector Ship 3 to sector BB 100100

Ship 4 to sector Ship 4 to sector AA 6565

Total efficiencyTotal efficiency 300300

TutorialTutorial

Lab Practical : Spreadsheet Lab Practical : Spreadsheet

1 - 117

Further ReadingFurther Reading

Render, B., Stair Jr.,R.M. & Hanna, M.E. (2013) Quantitative Analysis for Management, Pearson, 11th Edition

Waters, Donald (2007) Quantitative Methods for Business, Prentice Hall, 4th Edition.

Anderson D, Sweeney D, & Williams T. (2006) Quantitative Methods For Business Thompson Higher Education, 10th Ed.

QUESTIONS?QUESTIONS?

Download - Bba 3274 qm week 9 transportation and assignment models

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