McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003 7.1 Network Models Chapters 6 and 7

© The McGraw-Hill Companies, Inc., 20037.1McGraw-Hill/Irwin

Network Models

Chapters 6 and 7


Where network flows arise• Transportation

– Transportation of goods over transportation networks

– Scheduling of fleets of airplanes: time/space networks

• Manufacturing

– Scheduling of goods for manufacturing

– Flow of manufactured items within inventory systems

• Communications

– Design and expansion of communication systems

– Flow of information across networks

• Personnel Assignment

– Assignment of crews to airline schedules

– Assignment of drivers to vehicles


Network Optimization Problem Types

Many optimization problems can be represented by a graphical network representation.

Examples:– Distribution problems– Routing problems– Maximum flow problems– Designing computer / phone / road networks– Equipment replacement


Network examples

• Shortest path

• Maximum flow

• Transportation problem (Chapter 6)

• Assignment problem (Chapter 6)

• All are examples of a more general model type:– The Minimum-Cost-Network Flow Model


Advantages of Network Models• They can be solved very quickly with specialized algorithms.

• They have naturally integer solutions.– By recognizing that a problem can be formulated as a network program, it

is possible to solve special types of integer programs without resorting to the ineffective and time consuming integer programming algorithms.

• They are intuitive.– Network models provide a language for talking about problems that is much

more intuitive than the “variables, objective, and constraints” language of linear and integer programming.

• These advantages come with a drawback (of course):

– Network models cannot formulate the wide range of models that linear and integer programs can.

– However, they occur often enough that they form an important tool for real decision making.


Table of ContentsChapter 7 (Network Optimization

Problems)Minimum-Cost Flow Problems (Section 7.1)

A Case Study: The BMZ Maximum Flow Problem (Section 7.2)

Maximum Flow Problems (Section 7.3)

Shortest Path Problems: Littletown Fire Department (Section 7.4)

Shortest Path Problems: General Characteristics (Section 7.4)

Shortest Path Problems: Minimizing Sarah’s Total Cost (Section 7.4)

Shortest Path Problems: Minimizing Quick’s Total Time (Section 7.4)

Minimum Spanning Trees: The Modern Corp. Problem (Section 7.5)


Distribution Unlimited Co. Problem• The Distribution Unlimited Co. has two factories producing a product

that needs to be shipped to two warehouses– Factory 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?


The Distribution Network

F1

DC

F2 W2

W180 unitsproduced

70 units produced

60 unitsneeded

90 units needed


Data for Distribution Network

F1

DC

F2 W2

W180 unitsproduced

70 units produced

60 unitsneeded

90 units needed

$700/unit

$900/unit

$300/unit [50 units max.]




Both transportation cost and arc capacity are considered.


A Network Model

F1

DC

F2 W2

W1$700

$900

[80] [- 60]

[- 90][70]

[0]

$300 [50]

$200 [50]

$400 [50]

$400 [50]


The Optimal Solution

F1

DC

F2 W2

W1(30)

(40)

[80] [- 60]

[- 90][70]

[0](50)

(30)

(30)

(50)


Terminology for Minimum-Cost Flow Problems

1. The model for any minimum-cost flow problem is represented by a network with flow passing through it.

2. The circles in the network are called nodes.

3. Each node where the net amount of flow generated (outflow minus inflow) is a fixed positive number is a supply node.

4. Each node where the net amount of flow generated is a fixed negative number is a demand node.

5. 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.

6. The arrows in the network are called arcs.

7. The maximum amount of flow allowed through an arc is referred to as the capacity of that arc.


Assumptions of a Minimum-Cost Flow Problem

1. At least one of the nodes is a supply node.

2. At least one of the other nodes is a demand node.

3. All the remaining nodes are transshipment nodes.

4. 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.)

5. The network has enough arcs with sufficient capacity to enable all the flow generated at the supply nodes to reach all the demand nodes.

6. The cost of the flow through each arc is proportional to the amount of that flow, where the cost per unit flow is known.

7. 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.)


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.


Spreadsheet Model

34567891011

B C D E F G H I J K LFrom To Ship Capacity Unit Cost Nodes Net Flow Supply/Demand

F1 W1 30 $700 F1 80 = 80F1 DC 50 <= 50 $300 F2 70 = 70DC W1 30 <= 50 $200 DC 0 = 0DC W2 50 <= 50 $400 W1 -60 = -60F2 DC 30 <= 50 $400 W2 -90 = -90F2 W2 40 $900

Total Cost $110,000

345678

JNet Flow

=SUMIF(From,I4,Ship)-SUMIF(To,I4,Ship)=SUMIF(From,I5,Ship)-SUMIF(To,I5,Ship)=SUMIF(From,I6,Ship)-SUMIF(To,I6,Ship)=SUMIF(From,I7,Ship)-SUMIF(To,I7,Ship)=SUMIF(From,I8,Ship)-SUMIF(To,I8,Ship)

F1

DC

F2 W2

W1$700

$900

[80] [- 60]

[- 90][70]

[0]

$300 [50]

$200 [50]

$400 [50]

$400 [50]


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”)


Typical Applications of Minimum-Cost Flow Problems

Kind ofApplication

SupplyNodes

Transshipment Nodes

DemandNodes

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

VendorsIntermediate warehouses

Processing facilities

Coordinating product mixes at plants

PlantsProduction 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


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?


The BMZ Distribution Network

ST

LI

BO

RO

NO

NY

LA

New York

Rotterdam

Stuttgart

LisbonNew Orleans

{40 units max.]

Bordeaux[70 units max.]

Los Angeles

[80 units max.]

[60 units max.]

[50 units max.]

[30 units max.]

[50 units max.]

[40 units max.]

[70 units max]


A Network Model for BMZ

ST

LI

BO

RO

NO

NY

LA[70]

[80]

[70]

[60]

[40]

[50]

[30]

[50]

[40]


Spreadsheet Model for BMZ

34567891011121314

B C D E F G H I J KFrom To Ship Capacity Nodes Net Flow Supply/Demand

Stuttgart Rotterdam 50 <= 50 Stuttgart 150Stuttgart Bordeaux 70 <= 70 Rotterdam 0 = 0Stuttgart Lisbon 30 <= 40 Bordeaux 0 = 0

Rotterdam New York 50 <= 60 Lisbon 0 = 0Bordeaux New York 30 <= 40 New York 0 = 0Bordeaux New Orleans 40 <= 50 New Orleans 0 = 0

Lisbon New Orleans 30 <= 30 Los Angeles -150New York Los Angeles 80 <= 80

New Orleans Los Angeles 70 <= 70

Maximum Flow 150


Assumptions of Maximum Flow Problems

1. 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.)

2. All the remaining nodes are transshipment nodes.

3. 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.

4. 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.


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?


Network Model for the expanded BMZ Problem

ST

BE

BO

NY

LI

NO

HA

BN

RO

LA

SE

[70]

[20]

[20]

[40]

[10]

[80]

[70]

[40]

[30]

[60]

[40]

[50]

[30]

[60]

[50]

[40]


Spreadsheet Model

3456789101112131415161718192021


Stuttgart Rotterdam 40 <= 50 Stuttgart 140Stuttgart Bordeaux 70 <= 70 Berlin 80Stuttgart Lisbon 30 <= 40 Hamburg 0 = 0

Berlin Rotterdam 20 <= 20 Rotterdam 0 = 0Berlin Hamburg 60 <= 60 Bordeaux 0 = 0

Rotterdam New York 60 <= 60 Lisbon 0 = 0Bordeaux New York 30 <= 40 Boston 0 = 0Bordeaux New Orleans 40 <= 50 New York 0 = 0

Lisbon New Orleans 30 <= 30 New Orleans 0 = 0Hamburg New York 30 <= 30 Los Angeles -160Hamburg Boston 30 <= 40 Seattle -60

New Orleans Los Angeles 70 <= 70New York Los Angeles 80 <= 80New York Seattle 40 <= 40

Boston Los Angeles 10 <= 10Boston Seattle 20 <= 20

Maximum Flow 220


Some Applications of Maximum Flow Problems

1. Maximize the flow through a distribution network, as for BMZ.

2. Maximize the flow through a company’s supply network from its vendors to its processing facilities.

3. Maximize the flow of oil through a system of pipelines.

4. Maximize the flow of water through a system of aqueducts.

5. Maximize the flow of vehicles through a transportation network.


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?


The Littletown Road System

Fire Station

H

G

F

E

D

C

B

A

3

6

4

2

1

7

5

4

6

8

6

4

3

4

6

7

52

3

Farming Community


The Network Representation

T

H

G

F

E

D

B

C

A

O (Destination)(Origin)

3

6

1

2

6

4

3

4

7

8

6

5

42

34

6

75


Spreadsheet Model

3456789

1011121314151617181920212223242526272829

B C D E F G H I J KFrom To On Route Distance Nodes Net Flow Supply/Demand

Fire St. A 1 3 Fire St. 1 = 1Fire St. B 0 6 A 0 = 0Fire St. C 0 4 B 0 = 0

A B 1 1 C 0 = 0A D 0 6 D 0 = 0B A 0 1 E 0 = 0B C 0 2 F 0 = 0B D 0 4 G 0 = 0B E 1 5 H 0 = 0C B 0 2 Farm Com. -1 = -1C E 0 7D E 0 3D F 0 8E D 0 3E F 1 6E G 0 5E H 0 4F G 0 3F Farm Com. 1 4G F 0 3G H 0 2G Farm Com. 0 6H G 0 2H Farm Com. 0 7

Total Distance 19


Assumptions of a Shortest Path Problem

1. 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.

2. 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.

3. 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.)

4. The objective is to find the shortest path (the path with the minimum total length) from the origin to the destination.


Applications of Shortest Path Problems

1. Minimize the total distance traveled.

2. Minimize the total cost of a sequence of activities.

3. Minimize the total time of a sequence of activities.


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?


Sarah’s Cost Data

Operating and Maintenance Costsfor Ownership Year

Trade-in Value at Endof Ownership Year

PurchasePrice 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


Shortest Path Formulation

(Origin) (Destination)4321

17,000

10,50010,500

5,500 5,500 5,500 5,500

25,000

17,000

10,500

0


Spreadsheet Model

34567891011121314151617181920212223

B C D E F G H I JOperating & Trade-in Value PurchaseMaint. Cost at End of Year Price

Year 1 $2,000 $8,500 $12,000Year 2 $3,000 $6,500Year 3 $4,500 $4,500Year 4 $6,500 $3,000

From To On Route Cost Nodes Net Flow Supply/DemandYear 0 Year 1 0 $5,500 Year 0 1 = 1Year 0 Year 2 1 $10,500 Year 1 0 = 0Year 0 Year 3 0 $17,000 Year 2 0 = 0Year 0 Year 4 0 $25,000 Year 3 0 = 0Year 1 Year 2 0 $5,500 Year 4 -1 = -1Year 1 Year 3 0 $10,500Year 1 Year 4 0 $17,000Year 2 Year 3 0 $5,500Year 2 Year 4 1 $10,500Year 3 Year 4 0 $5,500

Total Cost $21,000


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?


Time and Cost of the Four Phases

LevelRemainingResearch Development

Design ofMfg. System

Initiate Productionand Distribution

Normal 5 months — — —

Priority 4 months 3 months 5 months 2 months

Crash 2 months 2 months 3 months 1 month

LevelRemainingResearch Development

Design ofMfg. System

Initiate Productionand Distribution

Normal $3 million — — —

Priority 6 million $6 million $9 million $3 million

Crash 9 million 9 million 12 million 6 million


Shortest Path Formulation

(Nor

mal)

(Crash)

(Crash)

(Prio

rity) (Crash)

(Crash)

(Crash)

(Priority

)(Crash)

(Crash)

(Crash)

(Crash)

5

3

3

2

3 1

132

(Prio

rity)

(Crash)1

2

3

3

2

,

T

3, 6

3, 9

3, 12

2, 12

2, 15

2, 18

2, 21

1, 21

1, 240, 30

1, 27

3, 3

4, 9

4, 6

4, 3

4, 0

(Origin) (Destination)(Nor

mal)

(Priority)(Crash)

(Crash)

(Prio

rity)

(Priority) (Priority)

(Crash)

(Crash)

(Crash)

(Priority

) (Priority) (Priority)(Crash)

(Crash)

(Crash)

(Priority) ((Priority)

(Crash)

5

35 2

0

0

0

0

25

3

42

3 (Priority) (Priority)5

1

132

(Prio

rity)

(Crash)2

1

2

3

3

2

5

2


Spreadsheet Model

34567891011121314151617181920212223242526272829303132

B C D E F G H I J KFrom To On Route Time Nodes Net Flow Supply/Demand(0, 30) (1, 27) 0 5 (0, 30) 1 = 1(0, 30) (1, 24) 0 4 (1, 27) 0 = 0(0, 30) (1, 21) 1 2 (1, 24) 0 = 0(1, 27) (2, 21) 0 3 (1, 21) 0 = 0(1, 27) (2, 18) 0 2 (2, 21) 0 = 0(1, 24) (2, 18) 0 3 (2, 18) 0 = 0(1, 24) (2, 15) 0 2 (2, 15) 0 = 0(1, 21) (2, 15) 1 3 (2, 12) 0 = 0(1, 21) (2, 12) 0 2 (3, 12) 0 = 0(2, 21) (3, 12) 0 5 (3, 9) 0 = 0(2, 21) (3, 9) 0 3 (3, 6) 0 = 0(2, 18) (3, 9) 0 5 (3, 3) 0 = 0(2, 18) (3, 6) 0 3 (4, 9) 0 = 0(2, 15) (3, 6) 0 5 (4, 6) 0 = 0(2, 15) (3, 3) 1 3 (4, 3) 0 = 0(2, 12) (3, 3) 0 5 (4, 0) 0 = 0(3, 12) (4, 9) 0 2 (T) -1 = -1(3, 12) (4, 6) 0 1(3, 9) (4, 6) 0 2(3, 9) (4, 3) 0 1(3, 6) (4, 3) 0 2(3, 6) (4, 0) 0 1(3, 3) (4, 0) 1 2(4, 9) (T) 0 0(4, 6) (T) 0 0(4, 3) (T) 0 0(4, 0) (T) 1 0

Total Time 10



Phase Level Time 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


Minimum Spanning Trees:The Modern Corp. Problem

• Modern Corporation has decided to have a state-of-the-art fiber-optic network installed to provide high-speed communication (data, voice, and video) between its major centers.

• Any pair of centers do not need to have a cable directly connecting them in order to take advantage of the technology. All that is necessary is to have a series of cables that connect the centers.

Question: Which cables should be installed to provide high-speed communications between every pair of centers.


Modern Corporation’s Major Centers

B

A

D

C

F

E

G2 2

1 3 1

5

7

5

4

4

7

4



B

A

D

C

F

E

G2 2

1 3 1

5


Assumptions of a Minimum-Spanning Tree Problem

1. You are given the nodes of a network but not the links. Instead, you are given the potential links and the positive cost (or a similar measure) for each if it is inserted into the network.

2. You wish to design the network by inserting enough links to satisfy the requirement that there be a path between every pair of nodes.

3. The objective is to satisfy this requirement in a way that minimizes the total cost of doing so.


Algorithm for a Minimum-Spanning-Tree Problem

1. Choice of the first link: Select the cheapest potential link.

2. Choice of the next link: Select the cheapest potential link between a node that already is touched by a link and a node that does not yet have such a link.

3. Repeat step 2 over and over until every node is touched by a link (perhaps more than one). At that point, an optimal solution (a minimum spanning tree) has been obtained.

(Ties for the cheapest potential link at each step may be broken arbitrarily.)


Application of Algorithm to Modern Corp.: First Link

B

A

D

C

F

E

G2 2

1 3 1

5

7

45

7

4

4


Application of Algorithm to Modern Corp.: Second Link

B

A

D

C

F

E

G2 2

1 3 1

5

7

45

7

4

4


Application of Algorithm to Modern Corp.: Third Link

B

A

D

C

F

E

G2 2

1 3 1

5

7

45

7

4

4


Application of Algorithm to Modern Corp.: Fourth Link

B

A

D

C

F

E

G2 2

1 3 1

5

7

45

7

4

4


Application of Algorithm to Modern Corp.: Fifth Link

B

A

D

C

F

E

G2 2

1 3 1

5

7

45

7

4

4


Application of Algorithm to Modern Corp.: Final Link

B

A

D

C

F

E

G2 2

1 3 1

5

7

45

7

4

4


Applications of Minimum-Spanning-Tree Problems

1. Design of telecommunication networks (computer networks, lease-line telephone networks, cable television networks, etc.)

2. Design of a lightly-used transportation network to minimize the total cost of providing the links (rail lines, roads, etc.)

3. Design of a network of high-voltage electrical power transmission lines.

4. Design of a network of wiring on electrical equipment (e.g., a digital computer system) to minimize the total length of the wire.

5. Design of a network of pipelines to connect a number of locations.


Network Optimization Problems

Many optimization problems can be represented by a graphical network representation.

Examples:– Distribution problems– Routing problems– Maximum flow problems– Designing computer / phone / road networks– Equipment replacement


Components of a Minimum-Cost-Flow Model

• Nodes– can represent a location, point in time, or state

– supply node (flow is generated)

– demand node (flow is consumed)

– transshipment node (flow in = flow out)

• Arcs– can represent potential flow (e.g., a shipping lane) or a

transition from state to state.

– directional (one-way)

• if both ways, use two arcs

– cost (assumed proportional to flow)

– may have capacity limitations


Minimum-Cost-Flow Model• Objective: Minimize the total cost of all flow, while sending supply,

subject to constraints, through the network to satisfy demand.

• Integer Solutions Property: If supplies, demands, and arc capacities are integer, then the optimal flow will also be integer.

• Network Simplex Method: A streamlined version of the simplex method.– extremely efficient

– computer software may have graphical interface (with nodes and arcs)

• Excel uses the standard simplex method. However, the minimum-cost-flow model is a useful tool for modeling a problem:– visual

– intuitive

– easy to set up

– transforms easily to a spreadsheet model


Minimum-Cost-Flow Model• Consider a directed network with n nodes. The decision variables

are xij, the flow through arc (i, j). The given information includes:

– cij: cost per unit of flow from i to j (may be negative),

– uij: capacity (or upper bound) on flow from i to j,

– bi: net flow generated at i.

• This last value has a sign convention: – bi > 0 if i is a supply node,

– bi < 0 if i is a demand node,

– bi = 0 if i is a transshipment node.

• The objective is to minimize the total cost of sending the supply through the network to satisfy the demand.


Minimum-Cost-Flow Model

• Linear programming formulation for this model is…


Minimum-Cost-Flow Model• Things you can do with this model…

– Lower bounds on arcs. If a variable xij has a lower bound of lij, upper

bound of uij, and cost of cij, change the problem as follows:

• Replace the upper bound with uij - lij,

• Replace the supply at i with bi - lij,

• Replace the supply at j with bi + lij,

– Now this is a minimum cost flow problem. Add cijlij to the objective after solving and

lij to the flow on arc (i, j) to obtain a solution of the original problem.

– Upper bounds on flow through a node. Replace the node i with nodes i' and i''. Create an arc from i' to i'' with the appropriate capacity, and cost 0. Replace every arc (j, i) with one from j to i' and every arc (i, j) with one from i'' to j. Lower bounds can also be handled this way.

– Convex, piecewise linear costs on arc flows (for minimization). This is handled by introducing multiple arcs between the nodes, one for each portion of the piecewise linear function. The convexity will assure that costs are handled correctly in an optimal solution.


Multi-Echelon DistributionConsider a multi-echelon distribution problem. Product must be distributed from a pair of factories to three warehouses. Product is then shipped to five distribution centers. A private trucking fleet is used for all shipping. Some shipping lanes are currently capacitated due to a limited number of trucks.

[2000]

[3000]

[-800]

[-700]

[-1500]

[-900]

[-1100]

$4.00

$3.75

$2.50

$5.25

$6.00

$6.75

$8.25

$7.50

$6.50

$8.75

$7.75

[300]

[500][1200]

[900]

F1

F2

WH1

WH2

WH3

DC1

DC2

DC3

DC4

DC5

Question: How many units should be shipped along each shipping lane?


Spreadsheet Model

345678910111213141516

B C D E F G H I J K LFrom To Ship Capacity Unit Cost Nodes Net Flow Supply/Demand

F1 WH1 900 <= 900 $4.00 F1 2000 = 2000F1 WH2 1100 $3.75 F2 3000 = 3000F2 WH2 1200 <= 1200 $2.50 WH1 0 = 0F2 WH3 1800 $5.25 WH2 0 = 0

WH1 DC1 800 $6.00 WH3 0 = 0WH1 DC2 100 <= 300 $6.75 DC1 -800 = -800WH2 DC2 600 $8.25 DC2 -700 = -700WH2 DC3 1500 $7.50 DC3 -1500 = -1500WH2 DC4 200 <= 500 $6.50 DC4 -900 = -900WH3 DC4 700 $8.75 DC5 -1100 = -1100WH3 DC5 1100 $7.75

Total Cost $57,800

345678910111213

JNet Flow

=D4+D5=D6+D7=D8+D9-D4=D10+D11+D12-D5-D6=D13+D14-D7=-D8=-D9-D10=-D11=-D12-D13=-D14


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”)


Spreadsheet Model using SUMIF

345678910111213141516

B C D E F G H I J K LFrom To Flow Capacity Unit Cost Nodes Net Flow Supply/Demand

F1 WH1 900 <= 900 $4.00 F1 2000 = 2000F1 WH2 1100 $3.75 F2 3000 = 3000F2 WH2 1200 <= 1200 $2.50 WH1 0 = 0F2 WH3 1800 $5.25 WH2 0 = 0

WH1 DC1 800 $6.00 WH3 0 = 0WH1 DC2 100 <= 300 $6.75 DC1 -800 = -800WH2 DC2 600 $8.25 DC2 -700 = -700WH2 DC3 1500 $7.50 DC3 -1500 = -1500WH2 DC4 200 <= 500 $6.50 DC4 -900 = -900WH3 DC4 700 $8.75 DC5 -1100 = -1100WH3 DC5 1100 $7.75

Total Cost $57,800

345678910111213

JNet Flow

=SUMIF($B$4:$B$14,I4,$D$4:$D$14)-SUMIF($C$4:$C$14,I4,$D$4:$D$14)=SUMIF($B$4:$B$14,I5,$D$4:$D$14)-SUMIF($C$4:$C$14,I5,$D$4:$D$14)=SUMIF($B$4:$B$14,I6,$D$4:$D$14)-SUMIF($C$4:$C$14,I6,$D$4:$D$14)=SUMIF($B$4:$B$14,I7,$D$4:$D$14)-SUMIF($C$4:$C$14,I7,$D$4:$D$14)=SUMIF($B$4:$B$14,I8,$D$4:$D$14)-SUMIF($C$4:$C$14,I8,$D$4:$D$14)=SUMIF($B$4:$B$14,I9,$D$4:$D$14)-SUMIF($C$4:$C$14,I9,$D$4:$D$14)=SUMIF($B$4:$B$14,I10,$D$4:$D$14)-SUMIF($C$4:$C$14,I10,$D$4:$D$14)=SUMIF($B$4:$B$14,I11,$D$4:$D$14)-SUMIF($C$4:$C$14,I11,$D$4:$D$14)=SUMIF($B$4:$B$14,I12,$D$4:$D$14)-SUMIF($C$4:$C$14,I12,$D$4:$D$14)=SUMIF($B$4:$B$14,I13,$D$4:$D$14)-SUMIF($C$4:$C$14,I13,$D$4:$D$14)


The Minimum-Cost-Flow Model is an LP

• Any minimum cost flow model consists of a set of nodes:– Supply node(s), with supply si

– Demand node(s), with demand di

– Transshipment nodes• A set of arcs from node i to node j

– with cost cij

– some with limited capacity kij

• LP Formulation:Let xij = flow from i to jMinimize Cost = ∑ ij cij xij

subject toFlow: ∑all j flowing out of i xij – ∑all j flowing into i xji = (si, di, or 0)

Capacity: xij ≤ kij

and xij ≥ 0.


Maximum Flow ProblemAn oil company has the following pipeline network, where each pipeline is labeled with its maximum flow rate (in thousands of gallons per hour).

Question: What is the maximum possible flow rate from A to G?

A

B

C

D

F

E

G

10

10

8

61

4

12

2

3

7

8

2


Spreadsheet Model

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101112131415161718


A B 7 <= 10 A 17A C 10 <= 10 B 0 = 0B C 0 <= 1 C 0 = 0B D 7 <= 8 D 0 = 0B F 0 <= 6 E 0 = 0C B 0 <= 1 F 0 = 0C E 8 <= 12 G -17C F 2 <= 4D F 0 <= 3D G 7 <= 7E F 0 <= 2E G 8 <= 8F G 2 <= 2

Maximum Flow 17


Seattle

Portland

Boise

Salt LakeCity

GrandJunction

Denver

Cheyenne

Butte9

3

7

6

10

77

7

51

4

7

14

(I90)

(I84)

(I82,I84)

(H93)

(I15)

(I84)

(I80)

(I90)

(H30)

(I25)

(I70)

(H6)

(I5)

Billings4

(I25)12

(H287)

Shortest Path ProblemThe travel times along various routes in the Pacific Northwest is shown below.

Question: What is the quickest route from Seattle to Denver?


Spreadsheet Model

34567891011121314151617181920

B C D E F G H I JFrom To On Route Time (hr) Nodes Net Flow Supply/Demand

Seattle Portland 0 3 Seattle 1 = 1Seattle Boise 0 9 Portland 0 = 0Seattle Butte 1 10 Boise 0 = 0

Portland Boise 0 7 Butte 0 = 0Boise Butte 0 7 Billings 0 = 0Boise Cheyenne 0 14 Salt Lake City 0 = 0Boise Salt Lake City 0 6 Cheyenne 0 = 0Butte Billings 1 4 Grand Junction 0 = 0Butte Cheyenne 0 12 Denver -1 = -1Butte Salt Lake City 0 7Butte Boise 0 7

Billings Cheyenne 1 7Cheyenne Denver 1 1

Salt Lake City Grand Junction 0 5Grand Junction Denver 0 4

Total Time 22


Equipment ReplacementA production department needs to purchase a new machine. As the machine ages, it requires additional maintenance and also has a higher defect rate. The production department plans to replace the machine every few years. The purchase price of a new machine is $10,000. The maintenance cost and cost of defective product is given below. A used machine has no resale value.

Age of machine Maintenance Cost Cost of Defects

First Year $3,000 $2,000

Second Year $4,000 $4,000

Third Year $6,000 $7,000

Fourth Year $10,000 $11,000

Fifth Year $20,000 $24,000

Question: What is the best replacement policy over the next five years?


Spreadsheet Solution

3456789

10111213141516171819202122232425262728

B C D E F G H I JAge of Maintenance Cost of PurchaseMachine Cost Defects CostFirst Year $3,000 $2,000 $10,000Second Year $4,000 $4,000Third Year $6,000 $7,000Fourth Year $10,000 $11,000Fifth Year $20,000 $24,000

From To On Route Cost Year Net Flow Supply/DemandYear 0 Year 1 0 $15,000 Year 0 1 = 1Year 0 Year 2 0 $23,000 Year 1 0 = 0Year 0 Year 3 1 $36,000 Year 2 0 = 0Year 0 Year 4 0 $57,000 Year 3 0 = 0Year 0 Year 5 0 $101,000 Year 4 0 = 0Year 1 Year 2 0 $15,000 Year 5 -1 = -1Year 1 Year 3 0 $23,000Year 1 Year 4 0 $36,000Year 1 Year 5 0 $57,000Year 2 Year 3 0 $15,000Year 2 Year 4 0 $23,000Year 2 Year 5 0 $36,000Year 3 Year 4 0 $15,000Year 3 Year 5 1 $23,000Year 4 Year 5 0 $15,000

Total Cost $59,000


Planning Vehicle Replacement at Phillips Petroleum

• Phillips Petroleum had a fleet of 1,500 cars and 3,800 trucks.

• Modeled replacement strategy as shortest path model (20-year time horizon)—solved model once for each class of vehicle.

• Could keep, purchase (replace), or lease, at 3-month intervals.

• Costs considered included:– Maintenance and operating costs (fuel, oil, repair),

– Leasing cost for leased vehicles,

– Purchasing cost for purchased vehicles,

– State license fees and road taxes,

– Tax effects (investment tax credits, depreciation)

• First used to make lease-or-buy decision, then vehicle-replacement strategy, and more recently for other equipment (non-vehicle).

For more details, see Waddell (1983) Jul-Aug Interfaces article, “A Model for Equipment Replacement Decisions and Policies”, downloadable at www.mhhe.com/hillier2e/articles.

Documents

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003 7.1 Network Models Chapters 6 and 7