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Operational Decision-Making Tools: Transportation and Transshipment Models

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Chapter 11 Supplement

Operational Decision-Making Tools: Transportation and Transshipment Models

Beni AsllaniUniversity of Tennessee at Chattanooga

Operations Management - 6hh Edition

Roberta Russell & Bernard W. Taylor, III

Copyright 2006 John Wiley & Sons, Inc.

Emotional direction

Intuition

Analytic thinking

Are you an intuit, an analytic, what???

How many of you use models to make decisions??

Copyright 2006 John Wiley & Sons, Inc.

- 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

Problem

Problem

MS Model

Mental Model

Mental Model

Decision

Decision

Action

Action

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- 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}

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- Planning
- Forecasting
- Training
- Behavioral research

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- Descriptive
- “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

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- Isomorphic
- One-to-one

- Homomorphic
- One-to-many

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- verbal descriptions
- mathematical constructs
- simulations
- mental models
- physical prototypes

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- Decision Making Under Certainty
- TOOL: all of mathematical programming

- Decision Making under Risk and Uncertainty
- TOOL: Decision analysis--tables, trees, Bayesian revision

- Decision Making Under Change and Complexity
- TOOL: Structural models, simulation models

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- 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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- What would we include in it?

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- A QUANTITATIVE REPRESENTATION OF A PROCESS THAT CONSISTS OF THOSE COMPONENTS THAT ARE SIGNIFICANT FOR THE PURPOSE BEING CONSIDERED

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- Transportation Model
- Transshipment Model

Not included are:

Shortest Route

Minimal Spanning Tree

Maximal flow

Assignment problem

many others

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- A transportation model is formulated for a class of problems with the following characteristics
- a product is 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 product
- each destination has a fixed demand for the product

- Solution (optimization) Algorithms
- stepping-stone
- modified distribution
- Excel’s Solver

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Problem Formulation Using Excel

Total Cost Formula

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Using Solver from Tools Menu

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Solution

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Modified Problem Solution

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- 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??}

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- 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.

Copyright 2006 John Wiley & Sons, Inc.

- Data
- Unit shipping cost, supply, and demand

- 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.

Copyright 2006 John Wiley & Sons, Inc.

Destinations

Boston

Sources

35

Cleveland

30

Richmond

40

S1=1200

32

37

40

Detroit

42

25

S2=1000

Atlanta

35

15

20

St.Louis

Greensboro

28

S3= 800

D1=1100

NETWORK

REPRESENTATION

D2=400

D3=750

D4=750

Copyright 2006 John Wiley & Sons, Inc.

- 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)

Copyright 2006 John Wiley & Sons, Inc.

Supply from Cleveland X11+X12+X13+X14 = 1200

Supply from Detroit X21+X22+X23+X24 = 1000

Supply from Greensboro X31+X32+X33+X34 = 800

X11

Cleveland

X12

X31

S1=1200

X21

X13

X14

X22

X32

Detroit

X23

S2=1000

X24

X33

Greensboro

S3= 800

X34

The supply constraints

Boston

D1=1100

Richmond

D2=400

Atlanta

D3=750

St.Louis

D4=750

Copyright 2006 John Wiley & Sons, Inc.

=

=

=

=

=

=

=

Copyright 2006 John Wiley & Sons, Inc.

Excel Optimal Solution

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Range of optimality

WINQSB Sensitivity Analysis

If this path is used, the total cost will increase by $5 per unit

shipped along it

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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

Copyright 2006 John Wiley & Sons, Inc.

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Transshipment Model: Solution

Copyright 2006 John Wiley & Sons, Inc.

Copyright 2006 John Wiley & Sons, Inc.All rights reserved. Reproduction or translation of this work beyond that permitted in section 117 of the 1976 United States Copyright Act without express permission of the copyright owner is unlawful. Request for further information should be addressed to the Permission Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, or damages caused by the use of these programs or from the use of the information herein.

Copyright 2006 John Wiley & Sons, Inc.

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.

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- 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.

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

- 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)

Network presentation

- Intermediate transshipment nodes: [Total flow out of the node] = [Total flow into the node]
- X34+X35 = X13 (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 = 13 (Node 6)

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- The Complete mathematical model

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WINQSB Input Data

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WINQSB Optimal Solution

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- 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.

Copyright 2006 John Wiley & Sons, Inc.

- 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)

Copyright 2006 John Wiley & Sons, Inc.

Initial inventory

- Analysis of Unit costs
- Unit cost = [Unit production cost] +

- Analysis of demand:
- Net demand to satisfy in July = 400 - 200 = 200 pairs
- Net demand in August = 600
- Net demand in September = 1000 + 1200 = 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

Forecasted demand

In house inventory

Copyright 2006 John Wiley & Sons, Inc.

Network representation

Production

Month/period

Month

sold

July

R/T

July

R/T

25

25.75

26.50

0

1000

200

July

July

O/T

30

30.90

31.80

0

500

+M

26

26.78

0

+M

+M

37

0

+M

+M

29

0

Aug.

R/T

600

+M

32

32.96

0

Aug.

800

Demand

Production Capacity

Aug.

O/T

400

2200

Sept.

Sept.

R/T

400

Dummy

300

Sept.

O/T

200

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.

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- 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 to Ski Chalet.

Inventory +

Production -

Demand

Copyright 2006 John Wiley & Sons, Inc.

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- 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.

Copyright 2006 John Wiley & Sons, Inc.

- 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 thatthe total travel time is minimized.

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

Copyright 2006 John Wiley & Sons, Inc.

S1=1

S2=1

S3=1

S4=1

S5=1

Assembly Line

Inspection Areas

D1=1

1

A

2

B

D2=1

3

C

D3=1

D4=1

4

D

D5=1

5

E

Copyright 2006 John Wiley & Sons, Inc.

- 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.

Copyright 2006 John Wiley & Sons, Inc.

- 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.

Copyright 2006 John Wiley & Sons, Inc.

- 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 jwith nonnegative distances, d i j.
- Find the path of minimum total distance that connects node 1 to node n.

Copyright 2006 John Wiley & Sons, Inc.

Fairway Van Lines

Determine the shortest route from Seattle to El Paso over the following network highways.

Copyright 2006 John Wiley & Sons, Inc.

Seattle

Butte

599

1

2

497

691

Boise

180

420

3

4

Cheyenne

345

Salt Lake City

432

Portland

440

Reno

7

8

526

6

138

102

432

5

621

Sac.

291

Denver

9

Las Vegas

11

280

10

108

452

Bakersfield

Kingman

155

Barstow

114

469

15

207

12

14

13

Albuque.

Phoenix

Los Angeles

386

403

16

118

19

17

18

San Diego

425

314

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Tucson

El Paso

- Solution - a linear programming approach
Decision variables

- Objective = Minimize S dijXij

Copyright 2006 John Wiley & Sons, Inc.

Butte

Seattle

2

599

1

497

Boise

180

3

4

345

Salt Lake City

432

Portland

7

Subject to the following constraints:

- [The number of highways traveled out of Seattle (the start node)] = 1X12 + X13 + X14 = 1

In a similar manner:

[The number of highways traveled into El Paso (terminal node)] = 1X12,19 + 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.

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Nonnegativity constraints

WINQSB Optimal Solution

Copyright 2006 John Wiley & Sons, Inc.

- 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.

Copyright 2006 John Wiley & Sons, Inc.

420

+

=

SLC.

1119

SLC

599

BUT.

599

BUT

+

=

691

1290

CHY.

345

+

SLC

SLC.

=

SLC

497

497

BOI.

BOI

BOI

+

=

432

612

BOI

BOI

180

POR.

180

POR

+

=

602

SAC.

782

SAC

An illustration of the Dijkstra’s algorithm

842

SEA.

… and so on until the

whole network

is covered.

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- 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.

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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 commercialcenters.
- 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.

Copyright 2006 John Wiley & Sons, Inc.

SPANNING TREE

NETWORK

PRESENTATION

55

North Side

University

50

3

5

30

Business

District

39

38

4

33

34

West Side

45

32

1

8

28

43

35

2

6

East Side

City

Center

Shopping

Center

41

40

37

44

36

7

South Side

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- 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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OPTIMAL SOLUTION

NETWORK

REPRESENTATION

55

University

50

3

5

30

North Side

Business

District

39

38

4

33

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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