Transportation, Transshipment
and Assignment Models
Learning Objectives
• Structure special LP network flow models.
• Set up and solve transportation models
• Extend basic transportation model to include
transshipment points.
• Set up and solve facility location and other application
problems as transportation models.
• Set up and solve assignment models
Overview
Part of a larger class of linear programming problems
are known as network flow models.
They possess special mathematical features that enabled
the development of very efficient, unique solution
methods.
Transportation Model
 Transportation problem deals with the distribution of goods
from several points of supply to a number of points of
demand. They arise when a cost-effective pattern is needed
to ship items from origins that have limited supply to
destinations that have demand for the goods.
 Resources to be optimally allocated usually involve a given
capacity of goods at each source and a given requirement
for the goods at each destination.
 Most common objective of the transportation problem is to
schedule shipments from sources to destinations so that
total production and transportation costs are minimized
Transshipment Model
 An extension of transportation problems is called
transshipment problem in which a point can have
shipments that both arrive as well as leave.
 Example would be a warehouse where shipments
arrive from factories and then leave for retail outlets
 It may be possible for a firm to achieve cost savings
(economies of scale) by consolidating shipments from
several factories at a warehouse and then sending
them together to retail outlets.
Assignment Model
 Assignment problem refers to a class of LP problems that
involve determining most efficient assignment of:
People to projects,
Salespeople to territories,
Contracts to bidders,
Jobs to machines, and so on
 Objective is to minimize total cost or total time of
performing tasks at hand, although a maximization
objective is also possible.
Transportation Model
Transportation Model
Problem definition
– There are m sources. Source i has a supply capacity of Si.
– There are n destinations.The demand at destination j is D j.
– Objective:
To minimize the total shipping cost of supplying the
destinations with the required demand from the
available supplies at the sources.
The Transportation Model
Characteristics
A product is to be transported from a number of sources to
a number of destinations at the minimum possible cost.
Each source is able to supply a fixed number of units of
the product, and each destination has a fixed demand for
the product.
The linear programming model has constraints for supply
at each source and demand at each destination.
All constraints are equalities in a balanced transportation
model where supply equals demand.
Constraints contain inequalities in unbalanced models
where supply is not equal to demand.
Transportation Model- Example 1
Executive Furniture Corporation
Transportation Model Example 1
Executive Furniture Corporation
Transportation Costs Per Desk
Transportation Model Example 1
Executive Furniture Corporation:LP Transportation Model Formulation
Objective: minimize total shipping costs =
5 XDA + 4 XDB + 3 XDC + 3 XEA + 2 XEB +
1 XEC + 9 XFA + 7 XFB + 5 XFC
Where: Xij = number of desks shipped from factory i to warehouse j
i = D (for Des Moines),
E (for Evansville), or
F (for Fort Lauderdale).
j = A (for Albuquerque),
B (for Boston), or
C (for Cleveland).
Transportation Model Example 1
Executive Furniture Corporation: Supply Constraints
Net flow at Des Moines = (Total flow in) - (Total flow out)
= (0) - (XDA + XDB + XDC)
Net flow at Des Moines =
-XDA - XDB - XDC = -100
(Des Moines capacity) and
-XEA - XEB - XEC = -300
(Evansville capacity)
-XFA - XFB - XFC = -300
(Fort Lauderdale capacity)
Multiply each constraint by -1 and rewrite as:
XDA + XDB + XDC = 100
(Des Moines capacity)
XEA + XEB + XEC = 300
(Evansville capacity)
XFA + XFB + XFC = 300
(Fort Lauderdale capacity)
Transportation Model Example 1
Executive Furniture Corporation: Demand Constraints
Net flow at Albuquerque = (Total flow in) - (Total flow out)
= (XDA + XEA + XFA) - (0)
Net flow at Albuquerque =
XDA + XEA + XFA = 300
(Albuquerque demand) and
XDB + XEB + XFB = 200
(Boston demand)
XDC + XEC + XFC = 200
(Cleveland demand)
Transportation Model Example 1:
The Optimum Solution
SHIP:
100 desks from Des Moines to Albuquerque,
200 desks from Evansville to Albuquerque,
100 desks from Evansville to Boston,
100 desks from Fort Lauderdale to Boston,
and 200 desks from Fort Lauderdale to Cleveland.
Total shipping cost is $3,000.
Transportation Model Example 2
Problem Definition and Data
Problem: How many tons of wheat to transport from each grain
elevator to each mill on a monthly basis in order to minimize the
total cost of transportation?
Data:
Grain Elevator
Supply
Mill
Demand
1. Kansas City
150
A. Chicago
200
2. Omaha
175
B. St. Louis
100
3. Des Moines
275
C. Cincinnati
300
Total
600 tons
Total
600 tons
Grain Elevator
1. Kansas City
2. Omaha
3. Des Moines
Transport Cost from Grain Elevator to Mill ($/ton)
A. Chicago
B. St. Louis
C. Cincinnati
$ 6
$ 8
$10
7
11
11
4
5
12
Transportation Model Example 2
Model Formulation
Transportation Model Example 2
Model Formulation
Minimize Z = $6x1A + 8x1B + 10x1C + 7x2A + 11x2B + 11x2C +
4x3A + 5x3B + 12x3C
subject to:
x1A + x1B + x1C = 150
x2A + x2B + x2C = 175
x3A + x3B + x3C = 275
x1A + x2A + x3A = 200
x1B + x2B + x3B = 100
x1C + x2C + x3C = 300
xij  0
xij = tons of wheat from each grain elevator, i, i = 1, 2, 3, to each
mill j, j = A,B,C
Transportation Model- Example 3
Carlton Pharmateuticals
• Carlton Pharmaceuticals supplies drugs and other
medical supplies.
• It has three plants in: Cleveland, Detroit, Greensboro.
• It has four distribution centers in:
Boston, Richmond, Atlanta, St. Louis.
• Management at Carlton would like to ship cases of a
certain vaccine as economically as possible.
• Data
– Unit shipping cost, supply, and demand
To
From
Cleveland
Detroit
Greensboro
Demand
Boston
$35
37
40
1100
Richmond
30
40
15
400
Atlanta
40
42
20
750
St. Louis
32
25
28
750
• Assumptions
– Unit shipping cost is constant.
– All the shipping occurs simultaneously.
– The only transportation considered is between
sources and destinations.
– Total supply equals total demand.
Supply
1200
1000
800
NETWORK
REPRESENTATION
Sources
Destinations
D1=1100
Boston
Cleveland
S1=1200
Richmond
D2=400
Detroit
S2=1000
Atlanta
D3=750
Greensboro
S3= 800
St.Louis
D4=750
• The Mathematical Model
– The structure of the model is:
Minimize <Total Shipping Cost>
ST
[Amount shipped from a source] = [Supply at that source]
[Amount received at a destination] = [Demand at that destination]
– Decision variables
Xij = amount shipped from source i to destination j.
where: i=1 (Cleveland), 2 (Detroit), 3 (Greensboro)
j=1 (Boston), 2 (Richmond), 3 (Atlanta), 4(St.Louis)
Supply from Cleveland X11+X12+X13+X14 = 1200
Supply from Detroit X21+X22+X23+X24
= 1000
Supply from Greensboro X31+X32+X33+X34 = 800
The supply constraints
Boston
D1=1100
X11
Cleveland
S1=1200
X12
X13
X21
X31
Richmond
X14
X22
Detroit
S2=1000
D2=400
X32
X23
X24
Atlanta
X33
St.Louis
Greensboro
S3= 800
D3=750
X34
D4=750
•
The complete mathematical model
Minimize 35X11+30X12+40X13+ 32X14 +37X21+40X22+42X23+25X24+
40X31+15X32+20X33+38X34
ST
Supply constrraints:
X11+ X12+ X13+ X14
X21+ X22+ X23+ X24
X31+ X32+ X33+ X34
Demand constraints:
X11+
X12+
X13+
X21+
X31
X22+
X32
X23+
X14+
All Xij are nonnegative
X33
X24+
X34
= 1200
= 1000
= 800
= 1000
= 400
= 750
= 750
Excel Optimal Solution
CARLTON PHARMACEUTICALS
UNIT COSTS
BOSTON RICHMOND ATLANTA ST.LOUIS
CLEVELAND
$
35.00 $
30.00 $
40.00 $
32.00
DETROIT
$
37.00 $
40.00 $
42.00 $
25.00
GREENSBORO $
40.00 $
15.00 $
20.00 $
28.00
DEMANDS
1100
400
750
750
SHIPMENTS (CASES)
BOSTON RICHMOND ATLANTA ST.LOUIS
CLEVELAND
850
350
0
0
DETROIT
250
0
0
750
GREENSBORO
0
50
750
0
TOTAL
1100
400
SUPPLIES
1200
1000
800
750
TOTAL
1200
1000
800
750
TOTAL COST =
84000
WINQSB Sensitivity Analysis
If this path is used, the total cost
will increase by $5 per unit
shipped along it
Shadow prices for plants - the cost savings realized for each extra case of vaccine
available at the plant
Shadow prices for warehouses - the cost incurred from having 1 extra case of vaccine
demanded at the warehouse
Interpreting sensitivity analysis results
– Reduced costs
• The amount of transportation cost reduction per unit that
makes a given route economically attractive.
• If the route is forced to be used under the current cost
structure, for each item shipped along it, the total cost
increases by an amount equal to the reduced cost.
– Shadow prices
• For the plants, shadow prices convey the cost savings
realized for each extra case of vaccine available at plant.
• For the warehouses, shadow prices convey the cost
incurred from having an extra case demanded at the
warehouse.
Transportation Model- Example 4
Montpelier Ski Company: Using a Transportation model for
production scheduling
– Montpelier is planning its production of skis for the months of
July, August, and September.
– Production capacity and unit production cost will change from
month to month.
– The company can use both regular time and overtime to produce
skis.
– Production levels should meet both demand forecasts and end-ofquarter inventory requirement.
– Management would like to schedule production to minimize its
costs for the quarter.
• Data:
– Initial inventory = 200 pairs
– Ending inventory required =1200 pairs
– Production capacity for the next quarter is shown on the table
– Holding cost rate is 3% per month per ski.
– Production capacity, and forecasted demand for this quarter
(in pairs of skis), and production cost per unit (by months)
Month
July
August
September
Forecasted
Demand
400
600
1000
Production
Production Costs
Capacity Regular Time Overtime
1000
25
30
800
26
32
400
29
37
• Analysis of demand:
– Net demand to satisfy in July = 400 - 200 = 200 pairs
Initial inventory
– Net demand in August = 600
– Net demand in September = 1000 + 1200 = 2200 pairs
• Analysis of Unit costs
Forecasted demand
In house inventory
• Analysis
Supplies:
Unitofcost
= [Unit production cost] +
– Production capacities are thought of as supplies.
[Unit holding cost per month][the number of months
– There are two sets of “supplies”:
stays in
• Set 1- Regular time supply (production capacity)
inventory]
• Set 2 - Overtime supply
Example:
A unit produced in July in Regular time and
sold in
September costs 25+ (3%)(25)(2 months) = $26.50
Production
Month/period
1000
800
July
O/T
Aug.
R/T
25
25.75
26.50
0
30
30.90
31.80 +M
0
26
26.78
400
Aug.
O/T
Month
sold
July
+M
+M
32.96
200
Sept.
R/T
Sept.
O/T
0
0
Aug.
600
Sept.
2200
Dummy
300
+M
0
29
400
+M
+M
32
200
Demand
Production Capacity
500
July
July
R/T
R/T
Network representation
37
0
Source: July production in R/T
Destination: July‘s demand.
Unit cost= $25 (production)
Source: Aug. production in O/T
Destination: Sept.’s demand
32+(.03)(32)=$32.96
Unit cost =Production+one month holding cost
• Summary of the optimal solution
– In July produce at capacity (1000 pairs in R/T, and 500 pairs in
O/T). Store 1500-200 = 1300 at the end of July.
– In August, produce 800 pairs in R/T, and 300 in O/T. Store
additional 800 + 300 - 600 = 500 pairs.
– In September, produce 400 pairs (clearly in R/T). With 1000
pairs
retail demand, there will be
(1300 + 500) + 400 - 1000 = 1200 pairs available for shipment
Inventory + Production - Demand
Unbalanced Transportation Problems
•
If supplies are not equal to demands, an unbalanced
transportation model exists.
• In an unbalanced transportation model, supply or
demand constraints need to be modified.
• There are two possible scenarios:
(1) Total supply exceeds total requirement.
(2) Total supply is less than total requirement.
Total Supply Exceeds Total Requirement
Total flow out of Des Moines ( XDA + XDB + XDC) should be
permitted to be smaller than total supply (100).
The constraint should be written as
-XDA - XDB - XDC >= -100
-XEA - XEB - XEC >= -300
-XFA - XFB - XFC >= -300
XDA + XDB + XDC <= 100
XEA + XEB + XEC <= 100
XFA + XFB + XFC <= 100
(Des Moines capacity)
(Evansville capacity)
(Fort Lauderdale capacity)
Total Supply Less Than Total Requirement
Total flow in to Albuquerque (that is, XDA + XEA + XFA) should
be permitted to be smaller than total demand (namely, 300).
This warehouse should be written as:
XDA + XEA + XFA <= 300
(Albuquerque demand)
XDB + XEB + XFB <= 200
(Boston demand)
XDC + XEC + XFC <= 200
(Cleveland demand)
Transportation Example 5: Formulation
Develop the linear programming model and solve using Excel:
Plant
1
2
3
Demand (tons)
A
$ 8
15
3
150
Construction site
B
C
$ 5
$ 6
10
12
9
10
70
100
Supply (tons)
120
80
80
Transportation Example 5: Formulation
Minimize Z = $8x1A + 5x1B + 6x1C + 15x2A + 10x2B + 12x2C +
3x3A + 9x3B + 10x3C
subject to:
x1A + x1B + x1C = 120
x2A + x2B + x2C = 80
x3A + x3B + x3C = 80
x1A + x2A + x3A  150
x1B + x2B + x3B  70
x1C + x2C + x3C  100
xij  0
Transportation Model-Example 6
Hardgrave Machine Company - New Factory Location
• Produces computer components at its plants in Cincinnati,
Kansas City, and Pittsburgh.
• Plants not able to keep up with demand for orders at four
warehouses in Detroit, Houston, New York, and Los Angeles.
• Firm has decided to build a new plant to expand its
productive capacity.
• Two sites being considered:
– Seattle, Washington and
– Birmingham, Alabama.
• Both cities attractive in terms: labor supply, municipal
services, and ease of factory financing.
Transportation Model-Example 6
Hardgrave Machine Company: Demand Supply Data and Production Costs
Transportation Model-Example 6
Hardgrave Machine Company: Shipping Costs
Transshipment Model
Transshipment Model
In a Transshipment Problem flows can occur both out of and into the
same node in three ways:
1. If total flow into a node is less than total flow out from node,
node represents a net creator of goods (a supply point).
- Flow balance equation will have a negative right hand
side (RHS) value.
2. If total flow into a node exceeds total flow out from node,
node represents a net consumer of goods, (a demand point).
- Flow balance equation will have a positive RHS value.
3. If total flow into a node is equal to total flow out from node,
node represents a pure transshipment point.
- Flow balance equation will have a zero RHS value.
The Transshipment Model
Characteristics
 It is an extension of the transportation model.
 Intermediate transshipment points are added between the sources
and destinations.
 Items may be transported from:
 Sources through transshipment points to destinations
 One source to another
 One transshipment point to another
 One destination to another
 Directly from sources to to destinations
 Some combination of these
Executive Furniture Corporation – Revisited
 Assume it is possible for Executive Furniture to ship
desks from Evansville factory to its three warehouses
at very low unit shipping costs.
 Consider shipping all desks produced at other two
factories (Des Moines and Fort Lauderdale) to
Evansville.
 Consider using a new shipping company to move
desks from Evansville to all its warehouses.
Executive Furniture Corporation - Revisited
• Revised unit shipping costs are shown here.
• Note Evansville factory shows up in both the “From” and
“To” entries.
Executive Furniture Corporation Revisited
LP Model for theTransshipment Problem
 Two new additional decision variables for new shipping
routes are to be added.
XDE= number of desks shipped from Des Moines to Evansville
XFE = number of desks shipped from Fort Lauderdale to Evansville
Objective Function: minimize total shipping costs =
5XDA + 4XDB + 3XDC + 2XDE + 3XEA + 2XEB +
+1XEC + 9XFA + 7XFB + 5XFC + 3XFE
Executive Furniture Corporation Revisited
LP Model for theTransshipment Problem
• Relevant flow balance equations written as:
(0) - (XDA + XDB + XDC + XDE) = -100
(Des Moines capacity)
(0) - (XFA + XFB + XFC + XFE) = -300
(Fort Lauderdale capacity)
• Supplies have been expressed as negative numbers in the RHS.
Net flow at Evansville = (Total flow in) - (Total flow out)
= (XDE + XFE) - (XEA + XEB + XEC)
• Net flow equals total number of desks produced (the supply) at
Evansville.
Net flow at Evansville = (XDE + XFE) - (XEA + XEB + XEC) = -300
• No change in demand constraints for warehouse requirements:
XDA + XEA + XFA = 300
(Albuquerque demand)
XDB + XEB + XFB = 200
(Boston demand)
XDC + XEC + XFC = 200
(Cleveland demand)
Transshipment Model Example 2
Problem Definition and Data
Extension of the transportation model in which intermediate
transshipment points are added between sources and destinations.
An example of a transshipment point is a distribution center or
warehouse located between plants and stores
Data:
Farm
1. Nebrasca
2. Colorado
3. Kansas City
$16
15
Warehouses 6. Chicago
3. Kansas
$6
4. Omaha
7
5. Des Moines
4
Grain Elevator
4. Omaha
10
14
Stores
7. St. Louis
8
11
5
5. Des Moines
12
17
8. Cincinnati
10
11
12
Transshipment Model Example 2
Problem Definition and Data
Transshipment Model Example 2
Model Formulation
Minimize Z = $16x13 + 10x14 + 12x15 + 15x23 + 14x24 +
17x25 + 6x36 + 8x37 + 10x38 + 7x46 + 11x47 +
11x48 + 4x56 + 5x57 + x58
subject to:
x13 + x14 + x15 = 300
x23+ x24 + x25 = 300
x36 + x46 + x56 = 200
x37+ x47 + x57 = 100
x38 + x48 + x58 = 300
x13 + x23 - x36 - x37 - x38 = 0
x14 + x24 - x46 - x47 - x48 = 0
x15 + x25 - x56 - x57 - x58 = 0
xij  0
Assignment Model
The Assignment Model
Problem definition
– m workers are to be assigned to m jobs
– A unit cost (or profit) Cij is associated with
worker i performing job j.
– Minimize the total cost (or maximize the total
profit) of assigning workers to jobs so that each
worker is assigned a job, and each job is
performed.
The Assignment Model
Characteristics
It is a special form of linear programming models
similar to the transportation model.
Supply at each source and demand at each destination is
limited to one unit.
In a balanced model supply equals demand.
In an unbalanced model supply is not equal to demand.
The Assignment Model Assumptions
– The number of workers is equal to the number of jobs.
– Given a balanced problem, each worker is assigned
exactly once, and each job is performed by exactly one
worker.
– For an unbalanced problem “dummy” workers (in case
there are more jobs than workers), or “dummy” jobs
(in case there are more workers than jobs) are added to
balance the problem.
Assignment Model Example 1
Fix-It Shop Example
Received three new rush projects to repair:
(1) a radio,
(2) a toaster oven, and
(3) a broken coffee table.
Three workers (each has different talents and
abilities).
Estimated costs to assign each worker to each of the
three projects.
Assignment Model Example 1
Fix-It Shop
• Rows denote people or objects to be assigned, and columns
denote tasks or jobs assigned.
• Numbers in table are costs associated with each particular
assignment.
Assignment Model Example 1
Fix-It Shop: Assignment Alternatives and Costs
• Owner's objective is to assign three projects to workers in
a way that result is lowest total cost.
Assignment Model Example 1
Fix-It Shop
• Owner's objective is to assign three projects to workers
in a way that results in lowest total cost.
Assignment Model Example 1
Fix-It Shop
Formulate LP model Xij = “Flow” on arc from node denoting worker i to node denoting
project j.
Solution value will equal 1 if worker i is assigned to project j :
i = A (for Adams), B (for Brown), or C (for Cooper)
j = 1 (for project 1), 2 (for project 2), or 3 (for project 3)
Objective Function: minimize total assignment cost =
11XA1 + 14XA2 + 6XA3 + 8XB1 + 10XB2 + 11XB3 +
+ 9XC1 + 12XC2 + 7XC3
Assignment Model Example 1
Fix-It Shop
Constraints expressed using standard flow balance equations
are as follows:
-XA1 - XA2 - XA3 = -1
(Adams availability)
-XB1 - XB2 - XB3 = -1
(Brown availability)
-XC1 - XC2 - XC3 = -1
(Cooper availability)
XA1 + XB1 + XC1 = 1
(Project 1 requirement)
XA2 + XB2 + XC2 = 1
(Project 2 requirement)
XA3 + XB3 + XC3 = 1
(Project 3 requirement)
Assignment Model- Example 2
Ballston Electronics
• Five different electrical devices produced on five
production lines, are needed to be inspected.
• The travel time of finished goods to inspection areas
depends on both the production line and the inspection
area.
• Management wishes to designate a separate inspection
area to inspect the products such that the total travel
time is minimized.
Assignment Model- Example 2
• Data: Travel time in minutes from assembly
lines to inspection areas.
Assembly
Assembly
Lines
Lines
11
22
33
44
55
AA
10
10
11
11
13
13
14
14
19
19
BB
44
77
88
16
16
17
17
Inspection
Inspection Area
Area
CC
66
77
12
12
13
13
11
11
DD
10
10
99
14
14
17
17
20
20
EE
12
12
14
14
15
15
17
17
19
19
Assignment Model Example 2: Network Representation
(3 of 3)
Assembly Line
Inspection Areas
S1=1
A D1=1
1
S2=1
2
B
D2=1
S3=1
3
C D3=1
S4=1
4
D
D4=1
S5=1
5
E
D5=1
• Computer solutions
– A complete enumeration is not efficient even
for moderately large problems (with m=8, m! >
40,000 is the number of assignments to
enumerate).
– The Hungarian method provides an efficient
solution procedure.
• Special cases
– A worker is unable to perform a particular job.
– A worker can be assigned to more than one job.
– A maximization assignment problem.
Assignment Model Example 3
Problem Definition and Data
Problem: Assign four teams of officials to four games in a way
that will minimize total distance traveled by the officials. Supply
is always one team of officials, demand is for only one team of
officials at each game.
Data:
Assignment Model Example 3
Model Formulation
Minimize Z = 210xAR + 90xAA + 180xAD + 160xAC + 100xBR +
70xBA + 130xBD + 200xBC + 175xCR + 105xCA +
140xCD + 170xCC + 80xDR + 65xDA + 105xDD +
120xDC
subject to:
xAR + xAA + xAD + xAC = 1
xBR + xBA + xBD + xBC = 1
xCR + xCA + xCD + xCC = 1
xDR + xDA + xDD + xDC = 1
xAR + xBR + xCR + xDR = 1
xAA + xBA + xCA + xDA = 1
xAD + xBD + xCD + xDD = 1
xAC + xBC + xCC + xDC = 1
xij  0
Summary
Three network flow models have been presented:
1. Transportation model deals with distribution of
goods from several supplier to a number of demand
points.
2. Transshipment model includes points that permit
goods to flow both in and out of them.
3. Assignment model deals with determining the most
efficient assignment of issues such as people to
projects.