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