1. Transportation and Transshipment Models
Chapter 11 Supplement
Transportation and Transshipment Models

2. Just how do you make decisions?
Emotional direction
Intuition
Analytic thinking
Are you an intuit, an analytic, what???
How many of you use models to make decisions??

3. Problems
Arise whenever there is a perceived difference between what is desired and what is in actuality.
Problems serve as motivators for doing something
Problems lead to decisions 42

4. How many of you have used a model before?
Any kind of model??
All of you have!!!
All of the time!!!
Copyright 2011 John Wiley & Sons, Inc.

5. Problem Problem MS Model Mental Model Mental Model Action Action
Decision
Decision
Action
Action

6. Model Classification Criteria
Purpose
Perspective
Use the perspective of the targeted decision-maker
Degree of Abstraction
Content and Form
Decision Environment
{This is what you should start any modeling facilitation meeting with}

7. Purpose
Planning
Forecasting
Training
Behavioral research

8. Perspective Descriptive Prescriptive “Telling it like it is”
Most simulation models are of this type
Prescriptive
“Telling it like it should be”
Most optimization models are of this type

9. Degree of Abstraction
Isomorphic
One-to-one
Homomorphic
One-to-many

10. Content and Form verbal descriptions mathematical constructs
simulations
mental models
physical prototypes

11. Decision Environment Decision Making Under Certainty
TOOL: all of mathematical programming—supplements to Chapters 11 and 14
Decision Making under Risk and Uncertainty
TOOL: Decision analysis--tables, trees, Bayesian revision—supplement to Chapter 1
Decision Making Under Change and Complexity
TOOL: Structural models, simulation models—supplement to Chapter 13

12. Copyright 2011 John Wiley & Sons, Inc.
We will cover parts of….
The supplements to Chapters 11, 14 and 13
In that order
Network programming—suppl to Chap 11 today
Linear programming—suppl to Chap 14 tomorrow
Simulation—suppl to Chap 13 Friday
And test you on this on July 30
Copyright 2011 John Wiley & Sons, Inc.

13. Mathematical Programming
Linear programming
Integer linear programming
some or all of the variables are integer variables
Network programming (produces all integer solutions)
Nonlinear programming
Dynamic programming
Goal programming
The list goes on and on
Geometric Programming

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Network Programming
Transportation model
Transhipment model
Shortest Route model (not covered)
Minimal Spanning Tree (not covered)
Maximal Flow model (not covered)
Assignment model (not covered)
Many other models
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15. A Model of this class
What would we include in it?

16. Management Science Models: A Definition
A QUANTITATIVE REPRESENTATION OF A PROCESS THAT CONSISTS OF THOSE COMPONENTS THAT ARE SIGNIFICANT FOR THE ________ BEING CONSIDERED
Purpose

17. Mathematical programming models covered in Ch 11, Supplement
Transportation Model
Transshipment Model
Not included are:
Shortest Route
Minimal Spanning Tree
Maximal flow
Assignment problem
many others

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Transportation Model
A model formulated for a class of problems with the following characteristics
items are transported from a number of sources to a number of destinations at minimum cost
each source supplies a fixed number of units
each destination has a fixed demand for units
Solution Methods
stepping-stone (by hand—a heuristic algorithm)
modified distribution
Excel’s Solver (uses Dantzig’s Simplex optimization algorithm)
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19. Transportation Method Example
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20. Transportation Method
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21. Problem Formulation with Excel
1. Click on “Data”
Total cost formula for all potato shipments in cell C10
=E5+E6+E7
=C5+D5+E5
2. Solver
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22. Solver Parameters Total cost
Decision variables representing shipment routes
Constraints specifying that supply at the distribution centers equals demand at the plants
Click to “solve”
Click on “Options” to activate “Assume Linear Models”
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Solution
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24. The Underlying Network
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25. Modified Problem Solution
High cost prohibits route C5
Column “H” added for excess supply
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26. Modified Problem Settings
Constraint changed to ≤ to reflect supply > demand
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27. Copyright 2011 John Wiley & Sons, Inc.
OM Tools
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28. Copyright 2011 John Wiley & Sons, Inc.
Transshipment Model
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29. Transshipment Model Solution
=SUM(B6:B7)
=SUM(B6:D6)
=SUM(C13:C15)
=SUM(C13:E13)
=C8-F14
= B8-F13, the amount shipped into KC equals the amount shipped out
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30. Transshipment Settings
Transshipment constraints
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31. For problems in which there is an underlying network:
There are easy (fast) solutions
An exception is the traveling salesman problem
The solutions are always integer ones
{How about solving a 50,000 node problem in less than a minute on a laptop??}

32. CARLTON PHARMACEUTICALS
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.

33. Data Assumptions Unit shipping cost, supply, and demand
Unit shipping cost is constant.
All the shipping occurs simultaneously.
The only transportation considered is between sources and destinations.
Total supply equals total demand.

34. NETWORK REPRESENTATION Destinations Boston Sources Cleveland Richmond
Atlanta
St.Louis
Destinations
Sources
Cleveland
Detroit
Greensboro
D1=1100 37 40 42 32 35 30 25 15 20 28
S1=1200
S2=1000
S3= 800
D2=400
D3=750
D4=750

35. The Associated Linear Programming 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)

36. The supply constraints
Cleveland
S1=1200
X11
X12
X13
X14
Supply from Cleveland X11+X12+X13+X14 =
The supply constraints
Detroit
S2=1000
X21
X22
X23
X24
Supply from Detroit X21+X22+X23+X =
Boston
Greensboro
S3= 800
X31
X32
X33
X34
Supply from Greensboro X31+X32+X33+X34 =
D1=1100
Richmond
D2=400
Atlanta
D3=750
St.Louis
D4=750

37. The complete mathematical programming model=

38. Excel Optimal Solution

39. WINQSB Sensitivity Analysis
If this path is used, the total cost will increase by $5 per unit
shipped along it
Range of optimality

40. Range of feasibility
Shadow prices for warehouses - the cost resulting from 1 extra case of vaccine
demanded at the warehouse
Shadow prices for plants - the savings incurred for each extra case of vaccine available at
the plant

41. Transshipment Model

42. Transshipment Model: Solution

43. A General Network Problem
DEPOT MAX
A General Network Problem
Depot Max has six stores.
Stores 5 and 6 are running low on the model 65A Arcadia workstation, and need a total of 25 additional units.
Stores 1 and 2 are ordered to ship a total of 25 units to stores 5 and 6.
Stores 3 and 4 are transshipment nodes with no demand or supply of their own.

44. Other restrictions
There is a maximum limit for quantities shipped on various routes.
There are different unit transportation costs for different routes.
Depot Max wishes to transport the available workstations at minimum total cost.

45. Copyright 2006 John Wiley & Sons, Inc.
DATA: 20 10 7 1 3 5
Arcs: Upper bound and lower bound constraints: 6 5 12 11 7 2 4 6 15 15
Network presentation
Supply nodes: Net flow out of the node] = [Supply at the node]
X12 + X13 + X15 - X21 = 10 (Node 1) X21 + X24 - X12 = 15 (Node 2)
Intermediate transshipment nodes: [Total flow out of the node] = [Total flow into the node]
X34+X35 = X (Node 3) X46 = X24 + X34 (Node 4)
Transportation
unit cost
Demand nodes: [Net flow into the node] = [Demand for the node]
X15 + X35 +X65 - X56 = 12 (Node 5) X46 +X56 - X65 = (Node 6)
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46. The Complete mathematical model

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WINQSB Input Data
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48. Copyright 2006 John Wiley & Sons, Inc.
WINQSB Optimal Solution
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49. 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-of-quarter inventory requirement.
Management would like to schedule production to minimize its costs for the quarter.

50. Data: Initial inventory = 200 pairs
Ending inventory required =1200 pairs
Production capacity for the next quarter = 400 pairs in regular time.
= 200 pairs in overtime.
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)

51. Analysis of Unit costs Analysis of demand: Analysis of Supplies:
Net demand to satisfy in July = = 200 pairs
Net demand in August = 600
Net demand in September = = 2200 pairs
Analysis of Supplies:
Production capacities are thought of as supplies.
There are two sets of “supplies”:
Set 1- Regular time supply (production capacity)
Set 2 - Overtime supply
Initial inventory
Analysis of Unit costs
Unit cost = [Unit production cost] +
[Unit holding cost per month][the number of months stays in inventory]
Example: A unit produced in July in Regular time and sold in September costs 25+ (3%)(25)(2 months) = $26.50
Forecasted demand
In house inventory

52. Network representation
Production
Month/period
Network representation
Month
sold
July
R/T
July
R/T 25
25.75
26.50
1000
200
July
July
O/T
500 30
30.90
31.80 +M 26
26.78 +M 37 +M 29
Aug.
R/T
800 +M 32
32.96
Aug.
600
Demand
Production Capacity
Aug.
O/T
400
Sept.
2200
Sept.
R/T
400
Dummy
300
Sept.
O/T
200

53. Copyright 2006 John Wiley & Sons, Inc.
Source: July production in R/T
Destination: July‘s demand.
Source: Aug. production in O/T
Destination: Sept.’s demand
Unit cost= $25 (production)
32+(.03)(32)=$32.96
Unit cost =Production+one month holding cost
Copyright 2006 John Wiley & Sons, Inc.

54. Copyright 2006 John Wiley & Sons, Inc.

55. Summary of the optimal solution
In July produce at capacity (1000 pairs in R/T, and 500 pairs in O/T). Store = 1300 at the end of July.
In August, produce 800 pairs in R/T, and 300 in O/T. Store additional = 500 pairs.
In September, produce 400 pairs (clearly in R/T). With 1000 pairs retail demand, there will be
( ) = 1200 pairs available for shipment to Ski Chalet.
Inventory +
Production -
Demand

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Problem 4-25
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59. Copyright 2006 John Wiley & Sons, Inc.

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61. 6.3 The Assignment Problem
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 job so that each worker is assigned a job, and each job is performed.

62. 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.

63. Data: Travel time in minutes from assembly lines to inspection areas.

64. NETWORK REPRESENTATION
Assembly Line
Inspection Areas
S1=1
S2=1
S3=1
S4=1
S5=1
D1=1 1 A 2 B
D2=1 3 C
D3=1
D4=1 4 D
D5=1 5 E

65. Assumptions and restrictions
The number of workers equals 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.

66. Computer solutions Special cases
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.

67. 6.5 The Shortest Path Problem
For a given network find the path of minimum distance, time, or cost from a starting point, the start node, to a destination, the terminal node.
Problem definition
There are n nodes, beginning with start node 1 and ending with terminal node n.
Bi-directional arcs connect connected nodes i and j with nonnegative distances, d i j.
Find the path of minimum total distance that connects node 1 to node n.

68. Fairway Van Lines
Determine the shortest route from Seattle to El Paso over the following network highways.

69. Seattle Butte 1 2 Boise 3 4 Cheyenne Salt Lake City Portland Reno 7 8
599 2
497
Boise
691
180
420 3 4
Cheyenne
345
Salt Lake City
432
Portland
440
Reno 7 8
526 6
138
102 5
432
621
Sac.
291
Denver 9
Las Vegas 11
280 10
108
Bakersfield
Kingman
452
155
Barstow
114
469 15
207 12 14 13
Albuque.
Phoenix
Los Angeles
386 16
403
118 19 17 18
San Diego
425
314
Tucson
El Paso

70. Objective = Minimize S dijXij
Solution - a linear programming approach
Decision variables
Objective = Minimize S dijXij

71. 2 7 Subject to the following constraints: Salt Lake City 1 3 4 Seattle
Boise
Portland
599
497
180
432
345
Butte
[The number of highways traveled out of Seattle (the start node)] = 1 X12 + X13 + X14 = 1
In a similar manner:
[The number of highways traveled into El Paso (terminal node)] = 1 X12,19 + X16,19 + X18,19 = 1
[The number of highways used to travel into a city] = [The number of highways traveled leaving the city]. For example, in Boise (City 4):
X14 + X34 +X74 = X41 + X43 + X47.
Nonnegativity constraints

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WINQSB Optimal Solution
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73. Solution - a network approach The Dijkstra’s algorithm:
Find the shortest distance from the “START” node to every other node in the network, in the order of the closet nodes to the “START”.
Once the shortest route to the m closest node is determined, the shortest route to the (m+1) closest node can be easily determined.
This algorithm finds the shortest route from the start to all the nodes in the network.

74. An illustration of the Dijkstra’s algorithm
SLC.
SLC.
SLC
SLC
420
CHY.
691 + =
1119
1290
BUT
599
POR
180
497
BOI
BUT.
599
180
497
345
SLC + =
842
BOI
BOI.
SEA. +
BOI
432
SAC
602 =
612
782
… and so on until the
whole network
is covered.
POR.
SAC.

75. 6.6 The Minimal Spanning Tree
This problem arises when all the nodes of a given network must be connected to one another, without any loop.
The minimal spanning tree approach is appropriate for problems for which redundancy is expensive, or the flow along the arcs is considered instantaneous.

76. THE METROPOLITAN TRANSIT DISTRICT
The City of Vancouver is planning the development of a new light rail transportation system.
The system should link 8 residential and commercial centers.
The Metropolitan transit district needs to select the set of lines that will connect all the centers at a minimum total cost.
The network describes:
feasible lines that have been drafted,
minimum possible cost for taxpayers per line.

77. SPANNING TREE NETWORK PRESENTATION 55 North Side University 3 50 5 30
Business
District 39 38 33 4 34
West Side 45 1 32 8 28 43 35 2 6
East Side
City
Center
Shopping
Center 41 40 37 44 36 7
South Side

78. Solution - a network approach
The algorithm that solves this problem is a very easy (“trivial”) procedure.
It belongs to a class of “greedy” algorithms.
The algorithm:
Start by selecting the arc with the smallest arc length.
At each iteration, add the next smallest arc length to the set of arcs already selected (provided no loop is constructed).
Finish when all nodes are connected.
Computer solution
Input consists of the number of nodes, the arc length, and the network description.

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WINQSB Optimal Solution
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80. Loop OPTIMAL SOLUTION NETWORK REPRESENTATION Total Cost = $236 million55
University 3 50 5
North Side 30
Business
District 39 38 33 4 34
West Side 45
Loop 32 1 8 28 43 35 2 6
East Side
City
Center
Shopping
Center 41 40 37 44 36
Total Cost = $236 million 7
South Side

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