CHAPTER 7:TRANSPORTATION, ASSIGNMENT AND TRANSSHIPMENT PROBLEMS

NETWORK FLOW MODEL

Consists of nodes representing a set of origins and a set of destinations.

An arc is used to represent the route from each origins to each destinations.

Each origin has a supply and each destination has a demand.

Objective: To determine the optimal amount to ship from each origin to each destination

Network flow problems Transportation

Problem

Assignment Problem

Transshipment Problem

TRANSPORTATION PROBLEM Transportation Problem Problems of distributing goods and

services from several supply location to several demand locations

Supply locations are called as Origin Demand locations are called as

Destination Quantity of goods at origin are limited Quantity of goods at destinations are

known

TRANSPORTATION PROBLEM Transportation Problem Each origin and destinations are represented

by Circles called as nodes Each origin and destinations are connected

by arc Each node requires one constraint Each arc requires one variable The sum of variables corresponding to arcs

from an origin node must be less than or equal to origin supply. (Rule 3)

The sum of variables corresponding to arcs into an destination node must equal to destination ‘s demand (Rule 4)

TRANSPORTATION PROBLEM

The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply si) to n destinations (each with a demand dj), when the unit shipping cost from an origin, i, to a destination, j, is cij.

The network representation for a transportation problem with two sources and three destinations is given on the next slide.

TRANSPORTATION PROBLEM Network Representation 1

2

3

1

2

c1

1 c12

c13c21 c22

c23

d1

d2

d3

s1

s2

SOURCES DESTINATIONS

TRANSPORTATION PROBLEM

LP FormulationThe LP formulation in terms of the amounts shipped from the origins to the destinations, xij , can be written as:

Min cijxij i j

s.t. xij < si for each origin i j

xij = dj for each destination j i

xij > 0 for all i and j

TRANSPORTATION PROBLEM Distribution of goods from three plants

to four distributionsSupply

Origin Plant Three Months Production Capacity (units)

1 Cleveland 50002 Bedford 60003 York 2500,Total=13,500

Destination Distribution Centre

Three-Months Demand Forecast (Units)

1 Boston 60002 Chicago 40003 St. Louis 20004 Lexington 1500,Total=13,500

Demands

TRANSPORTATION PROBLEMObjectives To determine the routes to be used and

quantity to be shipped from each origin to demand route that will provide minimum total transportation cost.

Construct a network graph Connect each origin with the destination

with arcs representing the routes between origin and the destinations.

12 Possible Routes

TRANSPORTATION NETWORK

7

Cleveland

Bedford

York

Boston

Chicago

S.Louis

Lexington

3

2

6

7

5

2

3

2

5

4

5

Transportation Cost per unit

DemandsSupplies

FORMULATING THE PROBLEMX ijnumber of units shipped from origin I to

destination jX11 number of units shipped from origin

(Cleveland) to destination 1 (boston)X12 number of units shipped from origin

(Cleveland) to destination 2 (Chicago)Cost per unit Origin Boston Chicago

St.Louis

Lexington

Cleveland

3 2 7 6

Bedford 7 5 2 3York 2 5 4 5Objective Function=sum of cost from each source to destinations

Transportation cost shipped from Cleveland 3x11+2x12+7x13+6x14

Transportation cost shipped from Bedford 7x21+5x22+2x23+3x24

Transportation cost shipped from York 2x31+5x32+4x33+5x34

How many total variables and constraint?? What are supply and demand constraints???

SUPPLY CONSTRAINT (RULE 3) Transportation shipped from

Cleveland X11+x12+x13+x14<= 5000

Transportation shipped from Bedford

X21+x22+x23+x24<=6000 Transportation shipped from York

X31+x32+x33+x34<=2500

DEMAND CONSTRAINT (RULE 4) Transportation To boston

X11+x21+x31= 6000 Transportation to Chicago

X12+x22+x32=4000 Transportation to St.Louis

X13+x23+x33=2500 Transportation to Lexington

X14+x24+x34=1500Objective function????

MODEL FORMALATION Objective functionMin

3x11+2x12+7x13+6x14+7x21+5x22+2x23+3x24+2x31+5x32+4x33+5x34

S.TX11+x12+x13+x14 <= 5000

X21+x22+x23+x24 <=6000X31+x32+x33+x34<=2500

X11 +x21 +x31 = 6000X12 +x22 +x32 =4000

X13 +x23 +x33 =2500 X14 +x24 +x34 =1500

SOLUTION

Variable

Value Reduced Cost

X11 3500.00 0.0X12 1500.00 0.0X13 0.0 8.0X14 0.0 6.0X21 2500.00 1.0X22 2000.00 0.0X23 1500.00 0.0X24 2500.00 0.0X31 0.00 0.0X32 0.00 4.0X33 0.00 6.0X34 0.00 6.0

Minimum total transportation cost?

Units shipped=3500;cost per unit=3Total cost from Cleveland to Boston??

GENERAL LINEAR PROGRAMMING MODEL OF TRANSPORTATION PROBLEM

0

....3,2,1,

sup.....3,2,1,

.

1

1

11

ij

m

i

i

n

jij

ij

n

jij

m

i

x

Demandnjdjxij

plymisx

ts

xcMin

Xij=number of units shipped from origin I to destination jCij = cost per unit shipping from origin I to destination jSi= supply or capacity in units at origin IDj= demand in units at destinations j

GENERALIZED ASSIGNMENT PROBLEM

0

,........3,2,;1

,.....3,2,1,1

..

1

1

1 1

xij

Tasksnix

Agentsmix

ts

xcMin

m

jij

n

jij

ij

m

i

n

iij

• Assigning jobs to machine• Agents to tasks• Sales personnel to sales territory• One to one assignment, i.e. one agent is assigned to

one and only one task

ASSIGNMENT PROBLEM An assignment problem seeks to minimize the

total cost assignment of m workers to m jobs, given that the cost of worker i performing job j is cij.

It assumes all workers are assigned and each job is performed.

An assignment problem is a special case of a transportation problem in which all supplies and all demands are equal to 1; hence assignment problems may be solved as linear programs.

ASSIGNMENT PROBLEM

Network Representation

2

3

1

2

3

1c11

c12c13

c21 c22

c23

c31 c32

c33

AGENTS TASKS

ASSIGNMENT PROBLEM LP Formulation

Min cijxij i j

s.t. xij = 1 for each agent i j

xij = 1 for each task j i

xij = 0 or 1 for all i and j

Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than 1 job.

LP Formulation Special Cases• Number of agents exceeds the number of

tasks:

xij < 1 for each agent i j • Number of tasks exceeds the number of agents:

Add enough dummy agents to equalize the number of agents and the number of tasks. The objective function coefficients for these new variable would be zero.

Assignment Problem

AGENT TASK PROBLEM

Project Leader

Client 1 Client 2 Client 3

Terry 10 15 9

Carle 9 18 5

Jack 6 14 3

Estimated project completion Time

ASSIGNMENT PROBLEM NETWORK

7

Terry

Carle

Jack

Client 1

Client 2

Client 3

10

15

6

9

185

614

3

Completion Time in Days

Demands

Supplies

Number of Constraints=6;Number of variables=9

PROBLEM FORMULATION

Days requires for Terry’s assignment 10x11+15x12+9x13

Days required for Carle ‘s assignment 9x21+18x22+5x23

Days required for Jack ‘s assignment 6x31+14x32+3x33

Objective Function Min 10x11+15x12+9x13 +10x11+15x12+9x13+

6x31+14x32+3x33

CONSTRAINT Each project leader can be assigned to at

most one client. X11+x12+x13 <=1 ;Terry ‘s assignment X21+x21+x23<=1; Carle ‘s assignment X31+X21 +x31<=1 ;Jack assignment Each Client must have at least one leader X11+X21+X31=1; Client 1 X12+X22+X32=1; Client 2 X13+X23+X33=1 ; Client 3

SOLUTIONVariable

Value Reduced Cost

X11 0.00 0.00X12 1.00 0.00X13 0.00 3.00X21 0.00 0.00X22 0.00 4.00X23 1.00 0.00X31 1.00 0.00X32 0.00 3.0X33 0.00 1.00

Terry is assigned to client2;x12=1Carle is assigned to client3;x23=1Jack is assigned to client 1 x31=1Total completion time required is 26 days

A contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects.

Projects Subcontractor A B C Westside 50 36 16

Federated 28 30 18

Goliath 35 32 20

Universal 25 25 14How should the contractors be assigned to minimize total costs?

Example: Hungry Owner

Example: Hungry Owner Network Representation

5036

1628

301835 32

2025 25

14

West.

C

B

A

Univ.

Gol.

Fed.

ProjectsSubcontractors

Example: Hungry Owner Linear Programming Formulation

Min 50x11+36x12+16x13+28x21+30x22+18x23

+35x31+32x32+20x33+25x41+25x42+14x43

s.t. x11+x12+x13 < 1 x21+x22+x23 < 1 x31+x32+x33 < 1 x41+x42+x43 < 1 x11+x21+x31+x41 = 1 x12+x22+x32+x42 = 1 x13+x23+x33+x43 = 1 xij = 0 or 1 for all i and j

Agents

Tasks

HUNGARIAN METHOD Step 1: For each row, subtract the minimum

number in that row from all numbers in that row. Step 2: For each column, subtract the minimum

number in that column from all numbers in that column.

Step 3: Draw the minimum number of lines to cover all zeroes. If this number = m, STOP -- an assignment can be made.

Step 4: Subtract d (the minimum uncovered number) from uncovered numbers. Add d to numbers covered by two lines. Numbers covered by one line remain the same. Then, GO TO STEP 3.

HUNGARIAN METHOD Finding the Minimum Number of Lines and

Determining the Optimal Solution Step 1: Find a row or column with only one

unlined zero and circle it. (If all rows/columns have two or more unlined zeroes choose an arbitrary zero.)

Step 2: If the circle is in a row with one zero, draw a line through its column. If the circle is in a column with one zero, draw a line through its row. One approach, when all rows and columns have two or more zeroes, is to draw a line through one with the most zeroes, breaking ties arbitrarily.

Step 3: Repeat step 2 until all circles are lined. If this minimum number of lines equals m, the circles provide the optimal assignment.

EXAMPLE: HUNGRY OWNER Initial Tableau Setup

Since the Hungarian algorithm requires that there be the same number of rows as columns, add a Dummy column so that the first tableau is:

A B C Dummy Westside 50 36 16 0

Federated 28 30 18 0 Goliath 35 32 20 0 Universal 25 25 14 0

EXAMPLE: HUNGRY OWNER Step 1: Subtract minimum number in each row from all

numbers in that row. Since each row has a zero, we would simply generate the same matrix above.

Step 2: Subtract the minimum number in each column from all numbers in the column. For A it is 25, for B it is 25, for C it is 14, for Dummy it is 0. This yields:

A B C Dummy Westside 25 11 2 0 Federated 3 5 4 0 Goliath 10 7 6 0 Universal 0 0 0 0

EXAMPLE: HUNGRY OWNER Step 3: Draw the minimum number of lines to

cover all zeroes. Although one can "eyeball" this minimum, use the following algorithm. If a "remaining" row has only one zero, draw a line through the column. If a remaining column has only one zero in it, draw a line through the row.

A B C Dummy Westside 25 11 2 0

Federated 3 5 4 0 Goliath 10 7 6 0 Universal 0 0 0 0

EXAMPLE: HUNGRY OWNER Step 3: Draw the minimum number of lines

to cover all zeroes.

A B C Dummy Westside 23 9 0 0

Federated 1 3 2 0 Goliath 8 5 4 0 Universal 0 0 0 2

EXAMPLE: HUNGRY OWNER Step 4: The minimum uncovered

number is 1. Subtract 1 from uncovered numbers. Add 1 to numbers covered by two lines. This gives:

A B C Dummy

Westside 23 9 0 1

Federated 0 2 1 0

Goliath 7 4 3 0

Universal 0 0 0 3

EXAMPLE: HUNGRY OWNER Step 3: The minimum number of lines to cover all

0's is four. Thus, there is a minimum-cost assignment of 0's with this tableau. The optimal assignment is:

Subcontractor Project Distance Westside C 16

Federated A 28Goliath (unassigned) Universal B 25 Total Distance = 69 miles

TRANSSHIPMENT PROBLEM Transshipment problems are transportation problems in

which a shipment may move through intermediate nodes (transshipment nodes)before reaching a particular destination node.

Transshipment problems can be converted to larger transportation problems and solved by a special transportation program.

Transshipment problems can also be solved by general purpose linear programming codes.

The network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations is shown on the next slide.

TRANSSHIPMENT PROBLEM Network Representation

2

3

4

5

6

7

1c13

c14

c23

c24

c25

c15

s1

c36

c37

c46

c47

c56

c57

d1

d2

INTERMEDIATE NODES

SOURCES DESTINATIONS

s2

DemandSupply

TRANSHIPMENT PROBLEM Contains three types of nodes:

origin ,transhipment node and destination nodes

For origin nodes sum of shipments out minus The sum of shipment in must be less than or equal to origin supply.

For destination nodes sum of shipments out minus The sum of shipment in must be equal to demand

For transhipment nodes the sum of shipments out must equal to sum of shipments in.

ASSIGNMENT PROBLEM NETWORK1Denver

2Atlan

ta

4Louisvill

e

5Detro

it

6Miami

7Dalla

s

10

Distribution Routes DemandsSupplies

Number of Constraints=8;Number of variables=12

3Kansas

8NewOrlea

ns

600

400

200

150

350

300

CONSTRAINTS Origin Nodes ?? X13+X14 <=600 (Denver) X23+x24 <=400 (Atlanta) For transhipment Nodes x35+x36+x37+x38=x13+ x23 (Node 3;units in=units out) X45+X46+X47+X48=X14+X24(Node 4;units in= units out) For Destination nodes X35+x45=200 X36+x46=150 X37+x47=350 X38+x48=300

TRASPORTION COST PER UNIT Ware House

Plant Kansas city (3)

Lousville (4)

Denver (1)

2 3

Atlanta (2)

3 1Retail Outlet

Warehouse Detroit (5)

Miami (6)

Dallas (7)

New Orleans (8)

Kansas City

2 6 3 6

Louisville 4 4 6 5

2x13+3x14+3x23+x24+2x35+6x36+3x37+6x38+4x45+4x46+6x47+5x48+4x28+x78

Obj function

SOLUTION

Variable

Value Reduced Costs

X13 550.00

0.00

X14 50.00 0.00X23 0.00 3.00X24 400.0

00.00

X35 200.00

0.00

X36 0.00 1.00X37 350.0

00.00

X38 0.00 0.00X45 0.00 3.00X46 150.0

00.00

X47 0.00 4.00X48 300.0

00.00

Value of objective function?

EXAMPLE: TRANSSHIPPINGThomas Industries and Washburn Corporation supply three firms (Zrox, Hewes, Rockwright) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc. Currently weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockwright. Both Arnold and Supershelf can supply at most 75 units to its customers. Additional data is shown on the next slide.

Example: Transshipping

Because of long standing contracts based on past orders, unit costs from the manufacturers to the suppliers are:

Thomas Washburn Arnold 5 8 Supershelf 7 4

The costs to install the shelving at the various locations are:

Zrox Hewes Rockwright

Thomas 1 5 8 Washburn 3 4 4

EXAMPLE: TRANSSHIPPING Network Representation

ARNOLD

WASHBURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

15

8

34

4

Arnold

SuperShelf

Hewes

Zrox

Thomas

Wash-Burn

Rock-Wright

EXAMPLE: TRANSSHIPPING Linear Programming Formulation

Decision Variables Defined xij = amount shipped from manufacturer i to supplier j

xjk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) j = 3 (Thomas), 4 (Washburn) k = 5 (Zrox), 6 (Hewes), 7 (Rockwright)

Objective Function DefinedMinimize Overall Shipping Costs:

Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37 + 3x45 + 4x46 + 4x47

EXAMPLE: TRANSSHIPPING Constraints Defined

Amount Out of Arnold: x13 + x14 < 75Amount Out of Supershelf: x23 + x24 < 75Amount Through Thomas: x13 + x23 - x35 - x36 - x37 = 0Amount Through Washburn: x14 + x24 - x45 - x46 - x47 = 0Amount Into Zrox: x35 + x45 = 50Amount Into Hewes: x36 + x46 = 60Amount Into Rockwright: x37 + x47 = 40Non-negativity of Variables: xij > 0, for all i and j.

Example: Transshipping Optimal Solution (from The Management

Scientist ) Objective Function Value =

1150.000 Variable Value Reduced

Costs X13 75.000

0.000 X14 0.000

2.000 X23 0.000

4.000 X24 75.000

0.000 X35 50.000

0.000 X36 25.000

0.000 X37 0.000

3.000 X45 0.000

3.000 X46 35.000

0.000 X47 40.000

0.000

Example: Transshipping Optimal Solution

ARNOLD

WASHBURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

15

8

3 4

4

Arnold

SuperShelf

Hewes

Zrox

Thomas

Wash-Burn

Rock-Wright

75

75

50

25

35

40