Linear Programming Example 5 Transportation Problem.

Linear ProgrammingLinear Programming

Example 5

Transportation

Problem

Transportation ProblemTransportation Problem• The transportation problem seeks to

minimize the total shipping costs of transporting goods from m origins (each with a supply si) to n destinations (each with a demand dj), when the unit shipping cost from an origin, i, to a destination, j, is cij.

• The network representation for a transportation problem with two sources and three destinations is given on the next slide.

Network RepresentationNetwork Representation

22

cc11

11cc1212

cc1313

cc2121

cc2222

cc2323

dd11

dd22

dd33

ss11

s2

SourcesSources DestinationsDestinations

33

22

11

11

LP FormulationLP FormulationThe LP formulation in terms of the amounts shipped from the origins to the destinations, xij , can be written as:

Min cijxij i j

s.t. xij < si for each origin i j

xij = dj for each destination j i

xij > 0 for all i and j

Example: Acme Block Co.Example: Acme Block Co.

Acme Block Company has orders for 80 tons ofAcme Block Company has orders for 80 tons of concrete blocks at three suburban locationsconcrete blocks at three suburban locations as as follows: follows: Northwood -- 25 Northwood -- 25 tons,tons,

Westwood -- 45 tons, andWestwood -- 45 tons, and

Eastwood -- 10 tons. Eastwood -- 10 tons. AcmeAcme has two plants, has two plants, eacheach of which of which can produce can produce

50 tons per week.50 tons per week. Delivery cost per ton from each plantDelivery cost per ton from each plant to each to each

suburban location is shown on the next slide.suburban location is shown on the next slide.

How should end of week shipments be made to How should end of week shipments be made to fillfill thethe above orders?above orders?

Delivery Cost Per TonDelivery Cost Per Ton

NorthwoodNorthwood WestwoodWestwood EastwoodEastwood

Plant 1 24 Plant 1 24 30 30 40 40

Plant 2 Plant 2 30 40 30 40 42 42

Delivery CostDelivery Cost

Decision VariableDecision VariableXX1111: Tons of Concrete shipped from Plan 1 to : Tons of Concrete shipped from Plan 1 to

NorthwoodNorthwood

XX1212: Tons of Concrete shipped from Plan 1 to : Tons of Concrete shipped from Plan 1 to WestwoodWestwood

XX1313: Tons of Concrete shipped from Plan 1 to Eastwood: Tons of Concrete shipped from Plan 1 to Eastwood

XX2121: Tons of Concrete shipped from Plan 2 to Northwood: Tons of Concrete shipped from Plan 2 to Northwood

XX2222: Tons of Concrete shipped from Plan 2 to Westwood: Tons of Concrete shipped from Plan 2 to Westwood

XX1313: Tons of Concrete shipped from Plan 2 to Eastwood: Tons of Concrete shipped from Plan 2 to Eastwood

Objective FunctionObjective FunctionMin. 24XMin. 24X1111+30X+30X1212+40X+40X1313+30X+30X2121+40X+40X2222+42X+42X2323

LP ModelLP Model

ConstraintsConstraintsXX1111 + X + X12 12 ++ XX1313 <= 50 <= 50 Plant 1 Plant 1

XX2121 + X + X2222 + X + X2323 <= 50 <= 50 Plant 2Plant 2

XX1111+X+X2121 = 25= 25 NorthwoodNorthwood

XX1212+X+X2222 = 45 = 45 WestwoodWestwood

XX1313+X+X2323 = 10 = 10 EastwoodEastwood

Non-NegativityNon-NegativityXX1111, X, X1212, X, X1313, X, X2121, X, X22, 22, XX23 23 >=0>=0

LP ModelLP Model

ExExcel Inputcel Input

Northwood Westwood Eastwood Total Shipped SupplyPlant 1 0 50Plant 2 0 50Total Received 0 0 0 0Demand 25 45 10

Northwood Westwood EastwoodPlant 1 24 30 40Plant 2 30 40 42

Transportation Problem

Solver SolutionSolver Solution

Northwood Westwood Eastwood Total Shipped SupplyPlant 1 5 45 0 50 50Plant 2 20 0 10 30 50Total Received 25 45 10 2490Demand 25 45 10

Northwood Westwood EastwoodPlant 1 24 30 40Plant 2 30 40 42

Transportation Problem

Optimal SolutionOptimal Solution

FromFrom ToTo AmountAmount CostCost

Plant 1 Northwood 5 Plant 1 Northwood 5 120120

Plant 1 Westwood 45 Plant 1 Westwood 45 1,3501,350

Plant 2 Northwood 20 Plant 2 Northwood 20 600600

Plant 2 Eastwood 10 Plant 2 Eastwood 10 420420

Total Cost = $2,490Total Cost = $2,490

SolutionSolution

Partial Sensitivity Report (first half)Partial Sensitivity Report (first half)

Sensitivity ReportSensitivity Report

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$C$4Plant 1 Northwood 5 0 24 4 4$D$4Plant 1 Westwood 45 0 30 4 1E+30$E$4 Plant 1 Eastwood 0 4 40 1E+30 4$C$5Plant 2 Northwood 20 0 30 4 4$D$5Plant 2 Westwood 0 4 40 1E+30 4$E$5 Plant 2 Eastwood 10 0 42 4 1E+30

Partial Sensitivity Report (second half)Partial Sensitivity Report (second half)

Sensitivity ReportSensitivity Report

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$F$4 Plant 1 Total Shipped 50 -6 50 20 5$F$5 Plant 2 Total Shipped 30 0 50 1E+30 20$C$6Total Received Northwood 25 30 25 20 20$D$6Total Received Westwood 45 36 45 5 20$E$6 Total Received Eastwood 10 42 10 20 10