lp examples solid waste management

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LP Examples Solid Waste Management

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LP Examples Solid Waste Management. A SOLID WASTE PROBLEM. A city generates 200 tons/day of solid wastes and must dispose it to three landfills . The data about the cost is given in the table. For environmental reason all the three landfills has to be utilized. - PowerPoint PPT Presentation

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Page 1: LP Examples Solid Waste Management

LP ExamplesSolid Waste Management

Page 2: LP Examples Solid Waste Management

A SOLID WASTE PROBLEM

LandfillMaximum capacity

(tons/day)

Cost of transfer to landfill ($/ton)

Cost of disposal at the

landfill ($/ton)

L1 120 4 3L2 100 3 5L3 50 2 4

A city generates 200 tons/day of solid wastes and must dispose it to three landfills. The data about the cost is given in the table. For environmental reason all the three landfills has to be utilized.Develop the needed equations (objective function and constraints) for Linear programming model to find the optimal distribution of the waste to the three landfills.

Page 3: LP Examples Solid Waste Management

MIN 7 X1 + 8 X2 + 6 X3SUBJECT TO

X1 + X2 + X3 = 200X1 <= 120X2 <= 100X3 <= 50X1 >= 0X2 >= 0X3 >= 0

END

MODEL FORMULATION

Page 4: LP Examples Solid Waste Management

SOLID WASTE SOLUTION using LINDO

Objective function Value = $1380X1 = 120; X2 = 30; X3 = 50

Page 5: LP Examples Solid Waste Management

LP Example of

Soil Stability

Page 6: LP Examples Solid Waste Management

SOIL STABILITY PROBLEMIn order to assure adequate stability under load repetition, a soil mixture for base and sub-base courses in the construction of a certain highway must have a liquid limit, 21=<LL<=28, and a Plasticity Index, 4=<PI<=6.

Two materials, A and B, are available as follows:

Properties A BLL 35 20PI 8 3.5Cost ($/cu. m ) $.35 $.65

Assume that the LL and the PI are linear functions of the combinations of the two materials A and B and determine the optimal proportion of base and sub-base.

Page 7: LP Examples Solid Waste Management

MODEL FORMULATION

MIN 0.35 XA + 0.65 XBSUBJECT TO

L.L. 35 XA + 20 XB >= 21L.L. 35 XA + 20 XB <= 28P.I. 8 XA + 3.5 XB >= 4P.I. 8 XA + 3.5 XB <= 6Proportionality XA + XB = 1

END

Page 8: LP Examples Solid Waste Management

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Case 2: LINDO OUTPUTSOLUTION:LP OPTIMUM FOUND AT STEP 4OBJECTIVE FUNCTION VALUE1) .4900000VARIABLE VALUE REDUCED COSTXA .533333 .000000XB .466667 .000000ROW SLACK OR SURPLUS DUAL PRICES2) 7.000000 .0000003) .000000 .0200004) 1.900000 .0000005) .100000 .0000006) .000000 -1.050000NO. ITERATIONS= 4RANGE(SENSITIVITY) ANALYSIS:Y ?:RANGES IN WHICH THE BASIS IS UNCHANGEDOBJ COEFFICIENT RANGESVARIABLE CURRENT ALLOWABLE ALOWABLECOEF INCREASE DECREASEXA .350000 .300000 INFINITYXB .650000 INFINITY .300000RIGHTHAND SIDE RANGESROW CURRENT ALLOWABLE ALOWABLERHS INCREASE DECREASE2 21.000000 7.000000 INFINITY3 28.000000 .333333 6.3333334 4.000000 1.900000 INFINITY5 6.000000 INFINITY .1000006 1.000000 .400000 .040000