linear programming example 5 transportation problem
TRANSCRIPT
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