Network Models with Excel

Simple StructureIntuition into solverNumerous applicationsIntegral data means integral

solutions

PROTRAC Engine Distribution

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Germany

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The Hague

Amsterdam

Antwerp

Nancy

Liege

Tilburg

Leipzig

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Transportation Costs

To DestinationFrom Origin Leipzig Nancy Liege TilburgAmsterdam 120 130 41 62Antwerp 61 40 100 110The Hague 102.5 90 122 42

Unit transportation costs from harbors to plants

Minimize the transportation costs involved in

moving the engines from the harbors to the

plants

A Transportation Model

PROTRAC Transportation ModelUnit Cost From/To Leipzig Nancy Liege Tilburg

Amsterdam 120.0$ 130.0$ 41.0$ 62.0$ Antwerp 61.0$ 40.0$ 100.0$ 110.0$ The Hague 102.5$ 90.0$ 122.0$ 42.0$

Shipments From/To Leipzig Nancy Liege Tilburg Total Available

Amsterdam - - - - - 500Antwerp - - - - - 700The Hague - - - - - 800Total - - - - - Required 400 900 200 500

Total Cost From/To Leipzig Nancy Liege Tilburg Total

Amsterdam -$ -$ -$ -$ -$ Antwerp -$ -$ -$ -$ -$ The Hague -$ -$ -$ -$ -$ Total -$ -$ -$ -$ -$

Model ComponentsAdjustables or Variables

By changing cells selection ranges separated by commas

Objective Target Cell Min or Max

Constraints LHS is a cell reference >=, <=, = (others for later) RHS is a cell reference or number.

How the Solver works

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A Basic Feasible Solution

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Finding an Entering Variable

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Finding an Entering Variable

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Computing Reduced Cost

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$122

Computing Reduced Cost

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Computing Reduced Cost

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Computing Reduced Cost

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Costs$122 $ 40 $162

Saves$100 $ 90 $190

Net $28

Finding a Leaving Variable

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Red flows decrease.

Green flowsincrease.

Leaving variableis first to reach 0

New Basic Feasible Solution

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Tilburg

New Basic Feasible Solution

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Quantity Discounts

Minimize Cost

Shipment Size

Tot

al C

ost

$4

$3

Crossdocks and Warehouses

PlantSupply

(Tons/Year) CustomerDemand (Tons/Year)

Plant 1 200 Customer 1 400Plant 2 300 Customer 2 180Plant 3 100

Inter Plant Plant 1 Plant 2 Plant 3Plant to DC DC 1 DC 2

Plant 1 -$ 5.0$ 3.0$ Plant 1 5.0$ 5.0$ Plant 2 9.0$ -$ 9.0$ Plant 2 1.0$ 1.0$ Plant 3 4.0$ 8.0$ -$ Plant 3 1.0$ 0.5$

Plant to Customer Customer 1 Customer 2

DC to DC DC 1 DC 2

Plant 1 20.0$ 20.0$ DC 1 -$ 1.2$ Plant 2 8.0$ 15.0$ DC 2 0.8$ -$ Plant 3 10.0$ 12.0$

Customer to

Customer Customer 1 Customer 2Customer 1 -$ 1.0$ Customer 2 7.0$ -$

The RedBrand Company Example 15.4

A Network Flow Model with Transshipments

Transportation Costs ($ 000/Ton)

Flow Balance

At the DCs Flow into the DC - Flow out of the DC = 0

At the Plants Flow out of Plant - Flow into the Plant Supply

At the Customers Flow into the Cust. - Flow out of the Cust. Demand

A Solver ModelRedBrand Minimum Cost Network Flow Problem

Unit Shipping Costs

Inter Plant Plant 1 Plant 2 Plant 3Plant to

DC DC 1 DC 2Plant 1 -$ 5.0$ 3.0$ Plant 1 5.0$ 5.0$ Plant 2 9.0$ -$ 9.0$ Plant 2 1.0$ 1.0$ Plant 3 0.4$ 8.0$ -$ Plant 3 1.0$ 0.5$

Plant to Customer Customer 1 Customer 2 DC to DC DC 1 DC 2

Plant 1 20.0$ 20.0$ DC 1 -$ 1.2$ Plant 2 8.0$ 15.0$ DC 2 0.8$ -$ Plant 3 10.0$ 12.0$

Customer to

Customer Customer 1 Customer 2DC to

Customer Customer 1 Customer 2Customer 1 -$ 1.0$ DC 1 2.0$ 12.0$ Customer 2 7.0$ -$ DC 2 2.0$ 12.0$

Shipments

Inter Plant Plant 1 Plant 2 Plant 3 Total OutPlant to

DC DC 1 DC 2 Total OutPlant 1 - - - - Plant 1 - - - Plant 2 - - - - Plant 2 - - - Plant 3 - - - - Plant 3 - - -

Total In - - - Total In - -

Plant to Customer Customer 1 Customer 2 Total Out DC to DC DC 1 DC 2 Total Out

Plant 1 - - - DC 1 - - - Plant 2 - - - DC 2 - - - Plant 3 - - - Total In - -

Total In - -

Customer to

Customer Customer 1 Customer 2 Total OutDC to

Customer Customer 1 Customer 2 Total OutCustomer 1 - - - DC 1 - - - Customer 2 - - - DC 2 - - - Total In - - Total In - -

Net FlowsNet Flow

Out Supply Net Flow In Demand Net FlowPlant 1 - 200 Customer 1 - 400 DC 1 - Plant 2 - 300 Customer 2 - 180 DC 2 - Plant 3 - 100

Transportation Costs ($ 000/Ton)

Arc Capacities

Inter Plant Plant 1 Plant 2 Plant 3Plant to

DC DC 1 DC 2Plant 1 200 200 200 Plant 1 200 200Plant 2 200 200 200 Plant 2 200 200Plant 3 200 200 200 Plant 3 200 200

Plant to Customer Customer 1 Customer 2 DC to DC DC 1 DC 2

Plant 1 200 200 DC 1 200 200Plant 2 200 200 DC 2 200 200Plant 3 200 200

Customer to

Customer Customer 1 Customer 2DC to

Customer Customer 1 Customer 2Customer 1 200 200 DC 1 200 200Customer 2 200 200 DC 2 200 200

Incurred Costs

Inter Plant Plant 1 Plant 2 Plant 3 Total OutPlant to

DC DC 1 DC 2 Total OutPlant 1 -$ -$ -$ -$ Plant 1 -$ -$ -$ Plant 2 -$ -$ -$ -$ Plant 2 -$ -$ -$ Plant 3 -$ -$ -$ -$ Plant 3 -$ -$ -$

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

Plant to Customer Customer 1 Customer 2 Total Out DC to DC DC 1 DC 2 Total Out

Plant 1 -$ -$ -$ DC 1 -$ -$ -$ Plant 2 -$ -$ -$ DC 2 -$ -$ -$ Plant 3 -$ -$ -$ Total In -$ -$ -$

Total In -$ -$ -$

Customer to

Customer Customer 1 Customer 2 Total Out DC to

Customer Customer 1 Customer 2 Total OutCustomer 1 -$ -$ -$ DC 1 -$ -$ -$ Customer 2 -$ -$ -$ DC 2 -$ -$ -$ Total In -$ -$ -$ Total In -$ -$ -$

Total Shipping Cost -$

Transportation Capacities (Tons)

Network Flow Models

Variables are flows of a single homogenous commodity

Constraints are Net flow Supply/Demand Lower Bound Flow on arc Upper Bound

Theorem: If the data are integral, any solution solver finds will be integral as well.

An Important Special Case

One unit available at one plantOne unit required at one customerMinimizing the cost of shipping is....