03 networks 2 print

8/13/2019 03 Networks 2 Print

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Network Models II

Shortest PathCross Docking

Enhance Modeling SkillsModeling with AMPL

15.057 Spring 03 Vande Vate 1


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The Shortest Path Model

aFind theshortestpath fromHome to 5

H

H

6

53

4

21

8

3

3

21

6

14

2 1

3

7

1



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Directiona

Two-waystreets a One-way

streets H

H

6

53

4

21

8

3

3

21

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14

2 1

3

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1



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03ShortestPathModel.xls

ConnectivityFrom\To Home Site 1 Site 2 Site 3 Site 4 Site 5 Site 6 Site 7

DistanceFrom\To Home Site 1 Site 2 Site 3 Site 4 Site 5 Site 6 Site 7

Home 1 Home 4 7 8Site 1 1 1 1 Site 1 4 6 1Site 2 1 1 1 Site 2 6 1 2Site 3 1 1 1 Site 3 1 1 1Site 4 1 1 1 1 Site 4 7 1 3 2Site 5 1 1 1 Site 5 2 3 3Site 6 1 1 1 Site 6 3 3Site 7 1 1 1 Site 7 8 2 1

RouteFrom\To Home Site 1 Site 2 Site 3 Site 4 Site 5 Site 6 Site 7

TotalFrom

TotalDistance Home Site 1 Site 2 Site 3 Site 4 Site 5 Site 6 Site 7

TotalFrom

Home 0 Home 0 0 0Site 1 0 Site 1 0 0 0Site 2 0 Site 2 0 0 0Site 3 0 Site 3 0 0 0

Site 4 0 Site 4 0 0 0Site 5 0 Site 5 0 0 0Site 6 0 Site 6 0 0 0Site 7 0 Site 7 0 0 0

Total To 0 Total To 0 0 0Total From -

Total To 0Net Required 1 -1 0

Shortest Path Model

11

1

00 000000 000000 000000 0000

00 000000 000000 000000 0000

0000000 00 0000

00000000000 0



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Challengea

Build a Solver model



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A Solver Model

aThe Objective: Minimize $U$21a The Variables: $C$13:$J$20

a The Constraints:Only travel on existing edges

$C$13:$J$20


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Flow Conservation

aNumber From - Number To = Net Requireda Number of times - Number of = ?we leave times we enter

a +1 at Home (we leave once)a -1 at Site 5 (we arrive once)a 0 everywhere else

each time we arrive (if ever), we leave



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Compare with Assignment Modela Assignment Model

Sum across each row = 1Sum down each column = 1

Each variable appears in 2 constraintsa Shortest Path Model

Sum across a row - Sum down the column = 0Trips out of a site - Trips into the siteEach variable appears in ? constraints



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Network Flow Problems

a Each variable appears in at most two constraints

At most one constraint as - the variable At most one constraint at + the variable

a AssignmentSum across each row = 1Sum down each column = 1

aShortest PathSum across the a row - sum down the col = #



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Bounds

a Variables can also have boundse.g., in the Shortest Path Model:

Number of times we use each variableLower bound: >= 0Upper bound:


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Properties of Network Flowsa If the bounds and RHS are integral, the

solution will be integrala It the costs are integral, the reduced costs

and marginal values will be integrala Can be solved very quicklya Limited demands on memory



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Crossdocking

a 2 customersa Minimize shipping costs

a 3 plantsa 2 distribution centers



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A Network Model

15.057 Spring 03 Vande Vate

Unit Shipping Costs Arc Capacities

Plant toDC DC 1 DC 2 Costs Plant to DC DC 1 DC 2Plant 1 5.0$ 5.0$ Capacities Costs Plant 1 200 200Plant 2 1.0$ 1.0$ Flows Flows Plant 2 200 200Plant 3 1.0$ 0.5$ Payments Payments Plant 3 200 200

DC toCustomer DC 1 DC 2

DC toCustomer DC 1 DC 2

Customer 1 2.0$ 2.0$ Customer 1 200 200Customer 2 12.0$ 12.0$ Customer 2 200 200

Shipments Payments

Plant toDC DC 1 DC 2 Total Out Supply Plant to DC DC 1 DC 2 Total OutPlant 1 - - - 200 Plant 1 -$ -$ -$Plant 2 - - - 300 Costs Plant 2 -$ -$ -$

Plant 3 - - - 100 Capacities Plant 3 -$ -$ -$Total In - - Flows Total In -$ -$ -$Costs

CapacitiesDC to

Customer DC 1 DC 2 Total In DemandDC to

Customer DC 1 DC 2 Total OutPayments Customer 1 - - - 400 Customer 1 -$ -$ -$

Customer 2 - - - 180 Customer 2 -$ -$ -$

Total Out - - Total In -$ -$ -$

Net Flow DC 1 DC 2- - Total Shipping Cost -$

Minimum Cost Network Flow Problem

Transportation Costs ($ 000/Ton) Transportation Capacities


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Challengea Build a Solver Model



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A Solver Model

a Objective: Minimize $K$28a Variables: $C$17:$D$19, $C$23:$D$24 a Constraints:

Do not exceed supply at the plants$E$17:$E$19 = $F$23:$F$24

Do not exceed shipping capacity$C$17:$D$19


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And...

a Flow conservation at the DCs$C$28:$D$28 = 0

a Supply and Demand like Autopowera Flow conservation at DCs like Shortest Path



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Network Flows: Good News a Lots of applications a Simple Models a Optimal Solutions Quickly a Integral Data, Integral Answers



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Network Flows: Bad News a Underlying Assumptions

Single Homogenous ProductLinear Costs

No conversions or losses...



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Homogenous Product


Must be able to interchange positions of product anywhere


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Linear Costs a No Fixed Chargesa No Volume Discountsa No Economies of Scale



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Summarya Network FlowsSimple Formulation

Flow Out (sum across a row) = DemandFlow In - Flow Out = Constant

Limited byHomogenous ProductLinear Costsetc.

Integer Data give Integral Solutions



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Modeling with AMPLa Problems with Excel Solver

Integration of Model and DataExample:

Change the time horizon of our Inventory Model

Excel is a limited database toola Algebraic Modeling Languages

Separate the Model from the Data Keep the data in databases



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How they work

Conceptual Model

Data

Algebraic Modeling LanguageAMPL/OPL/GAMS/XPress/...

OptimizerCPLEXOSLXPress



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Our Use of AMPL

a Pseudo AMPL to discuss modelsIn classIn exams

a Need to be precise aboutWhats a parameter, variable, Indexing: relationships between variables,data, constraints

a Challenges and Project



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Is this necessary/valuable?a AMPL is very detailed

Expect 1 or 2 per team to masterRest to read and understand

a Brings out the real issuesPractical implementation -- you can oversee

Data issues -- the real challengea Valuable tool



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The Transportation Modela set ORIG; a set DEST; a param supply {ORIG}; a param demand {DEST}; a param cost {ORIG, DEST}; a var Trans {ORIG, DEST} >= 0;



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Transportation Modelminimize Total_Cost:

sum{o in ORIG, d in DEST}cost[o,d]*Trans[o,d];

s.t. Supply {o in ORIG}:sum{d in DEST} Trans[o,d] = demand[d];



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The Dataa An Access Database called TransportationData.mdba Tables in the database

Origins: Supply information

Destinations: Demand information

Origin Supply Amsterdam 500

Antwerp 700The Hague 800

Destination DemandLeipzig 400Liege 200Nancy 900Tilburg 500



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The Costs

a Cost: Unittransportationcosts

origin destination cost Amsterdam Leipzig 120 Amsterdam Liege 41 Amsterdam Nancy 130 Amsterdam Tilburg 59.5 Antwerp Leipzig 61 Antwerp Liege 100 Antwerp Nancy 40 Antwerp Tilburg 110

The Hague Leipzig 102.5The Hague Liege 122The Hague Nancy 90The Hague Tilburg 42



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Reading Data

table OriginTable IN "ODBC""D:\Personal\15057\TransportationData.mdb" "Origins": ORIG


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Explanation

a "D:\Personal\15057\TransportationData.mdb is thepath to the database. Alternatively you can create aDSN (data source name) for this file, say TransportData,and use the command DSN=TransportData.

a "Origins is the name of the table in the database.

Alternatively we can use an SQL command like SQL=SELECT * FROM Originsa The : is syntax. What follows is the mapping of the data

we read to AMPL objects that will hold it.a The brackets [] around Origin mean that this field in the

database indexes the data, e.g., 500 is the supply for Amsterdam.



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Explanation Continueda ORIG


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Explanation

a "D:\Personal\15057\TransportationData.mdb is thepath to the database. Alternatively you can create aDSN (data source name) for this file, say TransportData,and use the command DSN=TransportData.

a Destinations is the name of the table in the database.

Alternatively we can use an SQL command like SQL=SELECT * FROM Destinationsa The : is syntax. What follows is the mapping of the data

we read to AMPL objects that will hold it.a The brackets [] around Destination mean that this field

in the database indexes the data, e.g., 400 is thedemand for Leipzig.



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Explanation Continueda DEST


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Reading Costtable CostTable IN "ODBC"

"D:\Personal\15057\TransportationData.mdb" "Cost": [origin, destination], cost;

Explanation:a table is a keyword that says we will read or write data a CostTable is a name we made up. No other AMPL

model entity can have this namea IN is a key word that says we are reading data.a ODBC says we are using ODBC to read the data



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Explanation Continueda We dont have an


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Running AMPLa model d:\15057\TransportationModel.mod; a option solver cplex; # use cplex to solve a solve; a display Trans;



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Writing Output

table TransOutTable OUT "ODBC""D:\Personal\15057\TransportationData.mdb"

"TransOut": {origin in ORIG, destination in DEST:

Trans[origin, destination] > 0}

-> [origin, destination], Trans[origin,destination]~Trans;write table TransOutTable;Explanation:

a table is a keyword that says we will read or write dataa TransOutTable is a name we made up. No other AMPL

model entity can have this name



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Explanationa OUT is a key word that says we are writing data.a ODBC says we are using ODBC to write the dataa "D:\Personal\15057\TransportationData.mdb is the

path to the database. Or you can use DSN=a "TransOut is the name of the table to create. AMPL

drops and writes this table. Any data currently in thetable is lost.

a : is syntax. It separates the description of thedestination from the definition of the data and themapping of the columns



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More Explanation

a {origin in ORIG, destination in DEST:Trans[origin, destination] > 0} defines the index set

that will control the data to write out. This says to onlyreport on origin-destination pairs where we actuallysend a positive flow.

a -> is syntax. It separates the indexing from the datadefinition and mapping to fields of the output table.

a [origin, destination] indicates that the records of theoutput table are indexed by the origin-destination pairs.

AMPL will write a new record for each pair.a Trans[origin,destination]~Trans says to create a field

called Trans in the table and to populate it with thevalues of the Trans variable.



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03 networks 2 print

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