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Table of Contents Chapter 6 (Network Optimization Problems) Table of Contents Chapter 6 (Network Optimization Problems)

Table of Contents Chapter 6 (Network Optimization Problems) - PowerPoint Presentation

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Table of Contents Chapter 6 (Network Optimization Problems) - PPT Presentation

MinimumCost Flow Problems Section 61 62612 A Case Study The BMZ Maximum Flow Problem Section 62 613616 Maximum Flow Problems Section 63 617621 Shortest Path Problems Littletown Fire Department Section 64 622625 ID: 667777

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Slide1

Table of ContentsChapter 6 (Network Optimization Problems)

Minimum-Cost Flow Problems (Section 6.1) 6.2–6.12A Case Study: The BMZ Maximum Flow Problem (Section 6.2) 6.13–6.16Maximum Flow Problems (Section 6.3) 6.17–6.21Shortest Path Problems: Littletown Fire Department (Section 6.4) 6.22–6.25Shortest Path Problems: General Characteristics (Section 6.4) 6.26–6.27Shortest Path Problems: Minimizing Sarah’s Total Cost (Section 6.4) 6.28–6.31Shortest Path Problems: Minimizing Quick’s Total Time (Section 6.4) 6.32–6.36

© 2014 by McGraw-Hill Education.  This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner.  This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.  Slide2

Distribution Unlimited Co. Problem

The Distribution Unlimited Co. has two factories producing a product that needs to be shipped to two warehousesFactory 1 produces 80 units.Factory 2 produces 70 units.Warehouse 1 needs 60 units.Warehouse 2 needs 90 units.There are rail links directly from Factory 1 to Warehouse 1 and Factory 2 to Warehouse 2.Independent truckers are available to ship up to 50 units from each factory to the distribution center, and then 50 units from the distribution center to each warehouse.Question: How many units (truckloads) should be shipped along each shipping lane?

6-

2Slide3

The Distribution Network

6-3Slide4

Data for Distribution Network

F1

DC

F2

W2

W1

80 units

produced

70 units

produced

60 units

needed

90 units

needed

$700/unit

$1,000/unit

$300/unit

[50 units max.]

$500/unit

[50 units max.]

$200/unit

[50 units max.]

$400/unit

[50 units max.]

6-

4Slide5

A Network Model

F1

DC

F2

W2

W1

$700

$1,000

[80]

[- 60]

[- 90]

[70]

[0]

$300

[50]

$200

[50]

$500

[50]

$400

[50]

6-

5Slide6

The Optimal Solution

6-6Slide7

Terminology for Minimum-Cost Flow Problems

The model for any minimum-cost flow problem is represented by a network with flow passing through it.The circles in the network are called nodes.Each node where the net amount of flow generated (outflow minus inflow) is a fixed positive number is a supply node.Each node where the net amount of flow generated is a fixed negative number is a demand node.

Any node where the net amount of flow generated is fixed at

zero

is a

transshipment node

. Having the amount of flow out of the node equal the amount of flow into the node is referred to as

conservation of flow

.

The arrows in the network are called

arcs

.

The maximum amount of flow allowed through an arc is referred to as the

capacity

of that arc.

6-

7Slide8

Assumptions of a Minimum-Cost Flow Problem

At least one of the nodes is a supply node.At least one of the other nodes is a demand node.All the remaining nodes are transshipment nodes.Flow through an arc is only allowed in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. (If flow can occur in both directions, this would be represented by a pair of arcs pointing in opposite directions.)The network has enough arcs with sufficient capacity to enable all the flow generated at the

supply nodes

to reach all the

demand nodes

.

The cost of the flow through each arc is

proportional

to the amount of that flow, where the cost per unit flow is known.

The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand. (An alternative objective is to maximize the total profit from doing this.)

6-

8Slide9

Properties of Minimum-Cost Flow Problems

The Feasible Solutions Property: Under the previous assumptions, a minimum-cost flow problem will have feasible solutions if and only if the sum of the supplies from its supply nodes equals the sum of the demands at its demand nodes.The Integer Solutions Property: As long as all the supplies, demands, and arc capacities have integer values, any minimum-cost flow problem with feasible solutions is guaranteed to have an optimal solution with integer values for all its flow quantities.6-9Slide10

Spreadsheet Model

6-

10Slide11

The SUMIF Function

The SUMIF formula can be used to simplify the node flow constraints.=SUMIF(Range A, x, Range B)For each quantity in (Range A) that equals x, SUMIF sums the corresponding entries in (Range B).The net outflow (flow out – flow in) from node x is then=SUMIF(“From labels”, x, “Flow”) – SUMIF(“To labels”, x, “Flow”)

6-

11Slide12

Typical Applications of Minimum-Cost Flow Problems

Kind ofApplication

Supply

Nodes

Transshipment Nodes

Demand

Nodes

Operation of a distribution network

Sources of goods

Intermediate storage facilities

Customers

Solid waste management

Sources of solid waste

Processing facilities

Landfill locations

Operation of a supply network

Vendors

Intermediate warehouses

Processing facilities

Coordinating product mixes at plants

Plants

Production of a specific product

Market for a specific product

Cash flow management

Sources of cash at a specific time

Short-term investment options

Needs for cash at a specific time

6-

12Slide13

The BMZ Maximum Flow Problem

The BMZ Company is a European manufacturer of luxury automobiles. Its exports to the United States are particularly important.BMZ cars are becoming especially popular in California, so it is particularly important to keep the Los Angeles center well supplied with replacement parts for repairing these cars.BMZ needs to execute a plan quickly for shipping as much as possible from the main factory in Stuttgart, Germany to the distribution center in Los Angeles over the next month.The limiting factor on how much can be shipped is the limited capacity of the company’s distribution network.Question: How many units should be sent through each shipping lane to maximize the total units flowing from Stuttgart to Los Angeles?6-

13Slide14

The BMZ Distribution Network

6-14Slide15

A Network Model for BMZ

6-15Slide16

Spreadsheet Model for BMZ

6-

16Slide17

Assumptions of Maximum Flow Problems

All flow through the network originates at one node, called the source, and terminates at one other node, called the sink. (The source and sink in the BMZ problem are the factory and the distribution center, respectively.)All the remaining nodes are transshipment nodes.Flow through an arc is only allowed in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. At the source, all arcs point away from the node. At the sink, all arcs point into the node.The objective is to maximize the total amount of flow from the source to the sink. This amount is measured in either of two equivalent ways, namely, either the amount leaving the source or the amount

entering the sink.

6-

17Slide18

BMZ with Multiple Supply and Demand Points

BMZ has a second, smaller factory in Berlin.The distribution center in Seattle has the capability of supplying parts to the customers of the distribution center in Los Angeles when shortages occur at the latter center.Question: How many units should be sent through each shipping lane to maximize the total units flowing from Stuttgart and Berlin to Los Angeles and Seattle?6-18Slide19

Network Model for The Expanded BMZ Problem

6-19Slide20

Spreadsheet Model

6-

20Slide21

Some Applications of Maximum Flow Problems

Maximize the flow through a distribution network, as for BMZ.Maximize the flow through a company’s supply network from its vendors to its processing facilities.Maximize the flow of oil through a system of pipelines.Maximize the flow of water through a system of aqueducts.Maximize the flow of vehicles through a transportation network.

6-

21Slide22

Littletown Fire Department

Littletown is a small town in a rural area.Its fire department serves a relatively large geographical area that includes many farming communities.Since there are numerous roads throughout the area, many possible routes may be available for traveling to any given farming community.Question: Which route from the fire station to a certain farming community minimizes the total number of miles?6-22Slide23

The Littletown Road System

6-23Slide24

The Network Representation

6-24Slide25

Spreadsheet Model

6-

25Slide26

Assumptions of a Shortest Path Problem

You need to choose a path through the network that starts at a certain node, called the origin, and ends at another certain node, called the destination.The lines connecting certain pairs of nodes commonly are links (which allow travel in either direction), although arcs (which only permit travel in one direction) also are allowed.Associated with each link (or arc) is a nonnegative number called its length. (Be aware that the drawing of each link in the network typically makes no effort to show its true length other than giving the correct number next to the link.)The objective is to find the shortest path (the path with the minimum total length) from the origin to the destination.

6-

26Slide27

Applications of Shortest Path Problems

Minimize the total distance traveled.Minimize the total cost of a sequence of activities.Minimize the total time of a sequence of activities.6-27Slide28

Minimizing Total Cost: Sarah’s Car Fund

Sarah has just graduated from high school.As a graduation present, her parents have given her a car fund of $21,000 to help purchase and maintain a three-year-old used car for college.Since operating and maintenance costs go up rapidly as the car ages, Sarah may trade in her car on another three-year-old car one or more times during the next three summers if it will minimize her total net cost. (At the end of the four years of college, her parents will trade in the current used car on a new car for Sarah.)Question: When should Sarah trade in her car (if at all) during the next three summers?6-

28Slide29

Sarah’s Cost Data

Operating and Maintenance Costs

for Ownership Year

Trade-in Value at End

of Ownership Year

Purchase

Price

1

2

3

4

1

2

3

4

$12,000

$2,000

$3,000

$4,500

$6,500

$8,500

$6,500

$4,500

$3,000

6-

29Slide30

Shortest Path Formulation

6-30Slide31

Spreadsheet Model

6-

31Slide32

Minimizing Total Time: Quick Company

The Quick Company has learned that a competitor is planning to come out with a new kind of product with great sales potential.Quick has been working on a similar product that had been scheduled to come to market in 20 months.Quick’s management wishes to rush the product out to meet the competition.Each of four remaining phases can be conducted at a normal pace, at a priority pace, or at crash level to expedite completion. However, the normal pace has been ruled out as too slow for the last three phases.$30 million is available for all four phases.Question: At what pace should each of the four phases be conducted?6-

32Slide33

Time and Cost of the Four Phases

LevelRemaining

Research

Development

Design of

Mfg. System

Initiate Production

and Distribution

Normal

5 months

Priority

4 months

3 months

5 months

2 months

Crash

2 months

2 months

3 months

1 month

Level

Remaining

Research

Development

Design of

Mfg. System

Initiate Production

and Distribution

Normal

$3 million

Priority

6 million

$6 million

$9 million

$3 million

Crash

9 million

9 million

12 million

6 million

6-

33Slide34

Shortest Path Formulation

6-34Slide35

Spreadsheet Model

6-

35Slide36

The Optimal Solution

PhaseLevel

Ti

me

Cost

Remaining research

Crash

2 months

$9 million

Development

Priority

3 months

6 million

Design of manufacturing system

Crash

3 months

12 million

Initiate production and distribution

Priority

2 months

3 million

Total

10 months

$30 million

6-

36