Management Decision Modeling - StudyDaddy

Dr. Jose Garcia‐Rubia

Management Decision ModelingChapter 9. Transportation, Assignment, and Network Models


LEARNING OBJECTIVESAfter completing this chapter, students will be able to:

1. Structure LP problems for the transportation, transshipment, and assignment models.

2. Solve facility location and other application problems with transportation models.

3. Use LP to model shortest-route and maximal-flow problems.

4. Solve minimal-spanning tree problems.

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CHAPTER OUTLINE9.1 Introduction9.2 The Transportation Problem9.3 The Assignment Problem9.4 The Transshipment Problem9.5 Maximal‐Flow Problem 9.6 Shortest‐Route Problem 9.7 Minimal‐Spanning Tree Problem

3


Introduction LP problems modeled as networksHelps visualize and understand problems

Transportation problem Transshipment problem Assignment problem Maximal-flow problem Shortest-route problem Minimal-spanning tree problem

Specialized algorithms available

4


Introduction Common terminology for network

modelsPoints on the network are referred to as

nodes Typically circles

Lines on the network that connect nodes are called arcs

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The Transportation Problem Deals with the distribution of goods from

several points of supply (sources) to a number of points of demand (destinations)Usually given the capacity of goods at each

source and the requirements at each destination

Typically objective is to minimize total transportation and production costs

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Activity 9.1 Linear Program for Transportation Executive Furniture Corporation

transportation problemMinimize transportation costMeet demandNot exceed supply

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Linear Program for Transportation

Let Xij = number of units shipped from source i to destination j

Wherei = 1, 2, 3, with 1 = Des Moines, 2 = Evansville,

and 3 = Fort Lauderdalej = 1, 2, 3, with 1 = Albuquerque, 2 = Boston,

and 3 = Cleveland

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Linear Program for TransportationFIGURE 9.1 – Network Representation of a Transportation Problem

100

300

300

Supply

$5

$4

$3

$8$4

$3

$9

$7$5

Demand

200

200

300

Source

Des Moines(Source 1)

Evansville(Source 2)

Fort Lauderdale(Source 3)

Albuquerque(Destination 1)

Boston(Destination 2)

Cleveland(Destination 3)

Destination

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Linear Program for TransportationMinimize total cost = 5X11 + 4X12 + 3X13 + 8X21 + 4X22 +

3X23 + 9X31 +7X32 + 5X33

Subject to:X11 + X12 + X13≤ 100 (Des Moines supply)X21 + X22 + X23≤ 300 (Evansville supply)X31 + X32 + X33≤ 300 (Fort Lauderdale supply)X11 + X21 + X31= 300 (Albuquerque demand)X12 + X22 + X32= 200 (Boston demand)X13 + X23 + X33= 200 (Cleveland demand)Xij ≥ 0 for all i and j

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Linear Program for Transportation

Optimal solution

100 units from Des Moines to Albuquerque 200 units from Evansville to Boston100 units from Evansville to Cleveland 200 units from Ft. Lauderdale to

Albuquerque 100 units from Ft. Lauderdale to ClevelandTotal cost = $3,900

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Using Excel QMPROGRAM 9.1 – Executive Furniture Corporation Solution in Excel 2013 Using Excel QM

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A General LP Model for Transportation Problems

LetXij = number of units shipped from source i to destination jcij = cost of one unit from source ito destination jsi =supply at source idj =demand at destination j

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Minimize cost cij xiji1

m

j1

n

A General LP Model for Transportation Problems

Subject to:

xij si i 1,2,...,mj1

n

xiji1

m

dj j 1,2...,n

xij 0 for all i and j14


Activity 9.2 Facility Location Analysis Transportation method especially useful New location is major financial importance Several alternative locations evaluated Subjective factors are considered Final decision also involves minimizing total

shipping and production costs Alternative facility locations analyzed within

the framework of one overall distribution system 15


Facility Location Analysis Hardgrave Machine Company produces

computer components in Cincinnati, Salt Lake City, and Pittsburgh

Four warehouses in Detroit, Dallas, New York, and Los Angeles

Two new plant sites being considered –Seattle and Birmingham

Which of the new locations will yield the lowest cost for the firm in combination with the existing plants and warehouses?

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Facility Location Analysis

WAREHOUSE

MONTHLY DEMAND (UNITS)

PRODUCTION PLANT

MONTHLY SUPPLY

COST TO PRODUCE

ONE UNIT ($)Detroit 10,000 Cincinnati 15,000 48

Dallas 12,000 Salt Lake City 6,000 50

New York 15,000 Pittsburgh 14,000 52

Los Angeles 9,000 35,000

46,000

Supply needed from a new plant = 46,000 – 35,000 = 11,000 units per month

TABLE 9.1 – Hardgrave’s Demand and Supply Data

ESTIMATED PRODUCTION COSTPER UNIT AT PROPOSED PLANTS

Seattle $53

Birmingham $4917


Facility Location Analysis

TOFROM DETROIT DALLAS NEW YORK

LOS ANGELES

CINCINNATI $25 $55 $40 $60

SALT LAKE CITY 35 30 50 40

PITTSBURGH 36 45 26 66

SEATTLE 60 38 65 27

BIRMINGHAM 35 30 41 50

TABLE 9.2 – Hardgrave’s Shipping Costs

Solve two transportation problems – one for each combination

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Facility Location AnalysisXij = number of units shipped from source i to

destination j

Wherei = 1, 2, 3, 4 with 1 = Cincinnati, 2 = Salt Lake

City, 3 = Pittsburgh, and 4 = Seattlej = 1, 2, 3, 4 with 1 = Detroit, 2 = Dallas, 3 = New

York, and 4 = Los Angeles

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Facility Location AnalysisMinimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22

+ 100X23 + 90X24 + 88X31 + 97X32 + 78X33+ 118X34 + 113X41 + 91X42 + 118X43 + 80X44

Subject to:X11 + X21 + X31 + X41 = 10,000 Detroit demandX12 + X22 + X32 + X42 = 12,000 Dallas demandX13 + X23 + X33 + X43 = 15,000 New York demandX14 + X24 + X34 + X44 = 9,000 Los Angeles demandX11 + X12 + X13 + X14 ≤ 15,000 Cincinnati supplyX21 + X22 + X23 + X24 ≤ 6,000 Salt Lake City supplyX31 + X32 + X33 + X34 ≤ 14,000 Pittsburgh supplyX41 + X42 + X43 + X44 ≤ 11,000 Seattle supplyAll variables Xij ≥ 0

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Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22+ 100X23 + 90X24 + 88X31 + 97X32 + 78X33+ 118X34 + 113X41 + 91X42 + 118X43 + 80X44


Facility Location AnalysisThe total cost for the Seattle alternative = $3,704,000

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Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22+ 100X23 + 90X24 + 88X31 + 97X32 + 78X33+ 118X34 + 113X41 + 91X42 + 118X43 + 80X44


Facility Location AnalysisThe total cost for the Seattle alternative = $3,704,000

Reformulating the problem for the Birmingham alternative and solving, the total cost = $3,741,000

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Using Excel QMPROGRAM 9.2 – Facility Location (Seattle) Solution in Excel 2013 Using Excel QM

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Using Excel QMPROGRAM 9.3 – Facility Location (Birmingham) Solution in Excel 2013 Using Excel QM

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The Assignment Problem This class of problem determines the most

efficient assignment of people or equipment to particular tasks

Objective is typically to minimize total cost or total task time

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Activity 9.2 Linear Program for Assignment The Fix-it Shop has just received three new

repair projects that must be repaired quickly1. A radio2. A toaster oven3. A coffee table

Three workers with different talents are able to do the jobs

Owner estimates wage cost for workers on projects

Objective – minimize total cost 26


Linear Program for Assignment Example

FIGURE 9.2 – Assignment Problem in a Transportation Network Format

1

1

1

Supply

$11

$14

$6

$8$10

$11

$9

$12$7

1

1

1

Demand

Person

Adams(Source 1)

Brown(Source 2)

Cooper(Source 3)

Project 1(Destination 1)



Project

27



Xij 1 if person i is assigned to project j0 otherwise

Let

where

i = 1, 2, 3, with 1 = Adams, 2 = Brown, and 3 = Cooper

j = 1, 2, 3, with 1 = Project 1, 2 = Project 2, and 3 = Project 3

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Minimize total cost = 11X11 + 14X12 + 6X13 + 8X21+ 10X22 + 11X23 + 9X31 + 12X32 + 7X33

subject toX11 + X12 + X13 = 1X21 + X22 + X23 = 1X31 + X32 + X33 = 1X11 + X21 + X31 = 1X12 + X22 + X32 = 1X13 + X23 + X33 = 1

Xij = 0 or 1 for all i and j29



Minimize total cost = 11X11 + 14X12 + 6X13 + 8X21+ 10X22 + 11X23 + 9X31 + 12X32 + 7X33

subject toX11 + X12 + X13 ≤ 1X21 + X22 + X23 ≤ 1X31 + X32 + X33 ≤ 1X11 + X21 + X31 = 1X12 + X22 + X32 = 1X13 + X23 + X33 = 1Xij = 0 or 1 for all i and j

Solution

X13 = 1, Adams assigned to Project 3X22 = 1, Brown assigned to Project 2X31 = 1, Cooper is assigned to Project 1

Total cost = $25

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Using Excel QMPROGRAM 9.4 – Mr. Fix-It Shop Assignment Solution in Excel 2013 Using Excel QM

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The Transshipment Problem Items are being moved from a source to a

destination through an intermediate point (a transshipment point)

Transshipment problem

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Activity 9.4 The Transshipment Problem Frosty Machines manufactures snow blowers in

Toronto and Detroit Shipped to regional distribution centers in

Chicago and Buffalo Then shipped to supply houses in New York,

Philadelphia, and St. Louis Shipping costs vary by location and destination Snow blowers cannot be shipped directly from

the factories to the supply houses33


The Transshipment Problem

FIGURE 9.3 – Network Representation of Transshipment Example

800

700

Supply

300

350

450

DemandSource

Toronto(Node 1)

Detroit(Node 2)

New York City(Node 5)

Philadelphia(Node 6)

St. Louis(Node 7)

Destination

Chicago(Node 3)

Transshipment Point

Buffalo(Node 4)

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The Transshipment ProblemTABLE 9.3 – Frosty Machine Transshipment Data

TO

FROM CHICAGO BUFFALONEW YORK

CITY PHILADELPHIAST.

LOUIS SUPPLY

Toronto $4 $7 — — — 800

Detroit $5 $7 — — — 700

Chicago — — $6 $4 $5 —

Buffalo — — $2 $3 $4 —

Demand — — 450 350 300

Minimize transportation costs associated with shipping snow blowers subject to demands and supplies

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The Transshipment Problem Minimize cost subject to

1. The number of units shipped from Toronto is not more than 8002. The number of units shipped from Detroit is not more than 7003. The number of units shipped to New York is 4504. The number of units shipped to Philadelphia is 3505. The number of units shipped to St. Louis is 3006. The number of units shipped out of Chicago is equal to the

number of units shipped into Chicago7. The number of units shipped out of Buffalo is equal to the

number of units shipped into Buffalo

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The Transshipment ProblemDecision variables

Xij = number of units shipped from location (node) i to location (node) j

where

i = 1, 2, 3, 4j = 3, 4, 5, 6, 7

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The Transshipment ProblemMinimize cost = 4X13 + 7X14 + 5X23 + 7X24 + 6X35 + 4X36 + 5X37

+ 2X45 + 3X46 + 4X47subject to

X13 + X14 ≤ 800 (Supply at Toronto [node 1])X23 + X24 ≤ 700 (Supply at Detroit [node 2])X35 + X45 = 450 (Demand at New York [node 5])X36 + X46 = 350 (Demand at Philadelphia [node 6])X37 + X47 = 300 (Demand at St. Louis [node 7])X13 + X23 = X35 + X36 + X37 (Shipping through Chicago [node 3])X14 + X24 = X45 + X46 + X47 (Shipping through Buffalo [node 4])

Xij ≥ 0 for all i and j (nonnegativity)

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Minimize cost = 4X13 + 7X14 + 5X23 + 7X24 + 6X35 + 4X36 + 5X37+ 2X45 + 3X46 + 4X47

subject toX13 + X14 ≤ 800 (Supply at Toronto [node 1])X23 + X24 ≤ 700 (Supply at Detroit [node 2])X35 + X45 = 450 (Demand at New York [node 5])X36 + X46 = 350 (Demand at Philadelphia [node 6])X37 + X47 = 300 (Demand at St. Louis [node 7])X13 + X23 = X35 + X36 + X37 (Shipping through Chicago [node 3])X14 + X24 = X45 + X46 + X47 (Shipping through Buffalo [node 4])

Xij ≥ 0 for all i and j (nonnegativity)

The Transshipment Problem

Ship 650 units from Toronto to ChicagoShip 150 units from Toronto to BuffaloShip 300 units from Detroit to BuffaloShip 350 units from Chicago to PhiladelphiaShip 300 units form Chicago to St. LouisShip 450 units from Buffalo to New York

Total cost = $9,550

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Using Excel QMPROGRAM 9.5 – Excel QM Solution to Frosty Machine Transshipment Problem

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Maximal-Flow Problem Determining the maximum amount of

material that can flow from one point (the source) to another (the sink) in a network

Two common methodsLinear programmingMaximal-flow technique

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Activity 9.5 Maximal-Flow Problem Determine maximum number of cars from

east to west for Waukesha WI road system

1

2

3

4

5

6

West Point

East Point

Capacity in Hundreds of Cars per Hour

32

10

1 2

1

0 1 1

0 32

21

0

16

1

FIGURE 9.4 – Road Network for Waukesha Maximal-Flow Example 42


Maximal-Flow ProblemVariables

Xij = flow from node i to node jwhere

i = 1, 2, 3, 4, 5, 6j = 1, 2, 3, 4, 5, 6

Constraints necessary for Capacity of each arc Equal flows into and out of each arc

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Maximal-Flow ProblemMaximize flow = X61

subject to

(X21 + X61) – (X12 + X13 + X14) = 0 Flows into = flows out of node 1

(X12 + X42 + X62) – (X21 + X24 + X26) = 0 Flows into = flows out of node 2

(X13 + X43 + X53) – (X34 + X35) = 0 Flows into = flows out of node 3

(X14 + X24 + X34 + X64) – (X42 + X43 + X46) = 0 Flows into = flows out of node 4

(X35) – (X56 + X53) = 0 Flows into = flows out of node 5

(X26 + X46 + X56) – (X61 + X62 + X64) = 0 Flows into = flows out of node 6

Xij ≥ 0

X12 ≤ 3 X13 ≤ 10 X14 ≤ 2 Capacities for arcs from node 1


X34 ≤ 3 X35 ≤ 2 Capacities for arcs from node 3




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Using Excel QMPROGRAM 9.6 –Waukesha Maximal-Flow Solution

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Shortest-Route Problem Find the shortest distance from one

location to another Can be modeled asA linear programming problem with 0-1

variables A special type of transshipment problem

Using specialized algorithm

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Activity 9.6 Shortest-Route Problem Ray Design transports beds, chairs, and other

furniture items from the factory to the warehouse Travel through several cities No direct interstate highways

Find the route with the shortest distance

Plant

Warehouse

1

2

3

4

5

650

40

150

200

FIGURE 9.5 –Roads from Ray’s Plant to Warehouse

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Shortest-Route ProblemVariables

Xij = 1 if arc from node i to node j is selected and Xij = 0 otherwise

wherei = 1, 2, 3, 4, 5j = 2, 3, 4, 5, 6

Constraints specify the number of units going into a node must equal the number of units going out of the node

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Shortest-Route ProblemOrigin point must ship one unit

X12 + X13 = 1

Final destination must have one unit shipped into it

X46 + X56 = 1

Intermediate nodes must have same amounts entering and leaving

X12 + X32 = X23 + X24 + X25

or

X12 + X32 – X23 – X24 – X25 = 049


Shortest-Route ProblemMinimize distance = 100X12 + 200X13 + 50X23 + 50X32

+ 200X24 + 200X42 + 100X25+ 100X52 + 40X35 + 40X53 + 150X45+ 150X54 + 100X46 + 100X56

subject toX12 + X13 = 1 Node 1

X12 + X32 – X23 – X24 – X25 = 0 Node 2X13 + X23 – X32 – X35 = 0 Node 3

X24 + X54 – X42 – X45 – X46 = 0 Node 4X25 + X35 + X45 – X52 – X53 – X54 – X56 = 0 Node 5

X46 + X56 = 1 Node 6All variables = 0 or 1

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Using Excel QMPROGRAM 9.7 –Ray Designs, Inc.Solution

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Using Excel QMPROGRAM 9.7 –Ray Designs, Inc.Solution

Solution

X12 = X23 = X35 = X56 = 1

Route is City 1 to City 2 to City 3 to City 5 to City 6

Total distance traveled = 290 miles

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Minimal‐Spanning Tree Problem Connecting all points of a network together

while minimizing the total distance of the connections

Linear programming can be used but is complex

Minimal-spanning tree technique is quite easy

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Minimal‐Spanning Tree ProblemSteps for the Minimal-Spanning Tree Technique

1. Select any node in the network.2. Connect this node to the nearest node that

minimizes the total distance.3. Considering all of the nodes that are now connected,

find and connect the nearest node that is not connected. If there is a tie for the nearest node, select one arbitrarily. A tie suggests there may be more than one optimal solution.

4. Repeat the third step until all nodes are connected.

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Activity 9.7 Minimal‐Spanning Tree Problem Lauderdale Construction

Housing project in Panama City Beach

Determine the least expensive way to provide water and power to each house

1

2

4

6

Gulf

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

FIGURE 9.6 – Network for Lauderdale Construction

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Minimal‐Spanning Tree ProblemStep 1 – Arbitrarily select

node 1Step 2 – Connect node 1

to node 3 (nearest)

1

2

4

6

Gulf

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

FIGURE 9.7 – First Iteration

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Minimal‐Spanning Tree ProblemStep 3 – Connect next nearest unconnected

node, node 4Continue for other unconnected nodes

1

2

4

6

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

FIGURE 9.8 – Second and Third Iterations

1

2

4

6

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

(a) Second Iteration (b) Third Iteration57


Minimal‐Spanning Tree ProblemStep 4 – Repeat the process

1

2

4

6

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

FIGURE 9.9 – Last Four Iterations

1

2

4

6

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

(a) Fourth Iteration (b) Fifth Iteration

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Step 4 – Repeat the processMinimal‐Spanning Tree Problem

1

2

4

6

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

FIGURE 9.9 – Last Four Iterations

1

2

4

6

3

5

7

8

3

27

5

6

1

43

3

3

2

2

5

(c) Sixth Iteration (d) Seventh Iteration

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Minimal‐Spanning Tree ProblemTABLE 9.4 – Summary of Steps in Lauderdale Construction Minimal-Spanning Tree Problem

STEPCONNECTED

NODESUNCONNECTED

NODES

CLOSE UNCONNECTED

NODESARC

SELECTEDARC

LENGTHTOTAL

DISTANCE

1 1 2, 3, 4, 5, 6, 7, 8 3 1–3 2 2

2 1, 3 2, 4, 5, 6, 7, 8 4 3–4 2 4

3 1, 3, 4 2, 5, 6, 7, 8 2 or 6 3–2 3 7

4 1, 2, 3, 4 5, 6, 7, 8 5 or 6 2–5 3 10

5 1, 2, 3, 4, 5 6, 7, 8 6 3–6 3 13

6 1, 2, 3, 4, 5, 6 7, 8 8 6–8 1 14

7 1, 2, 3, 4, 5, 6, 8 7 7 8–7 2 16

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Using Excel QMPROGRAM 9.8 –Lauderdale Construction Minimal-Spanning Tree Example

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