BANA Exam 2 Practice - BANA Introduction to Quantitative Analysis Dukedom Chemicals blends its private label orchard spray from four individual

BANA Exam 2 Practice - BANA Introduction to Quantitative...

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BANA: Introduction to Quantitative AnalysisDukedom Chemicalsblends its private label orchard spray from four individual compounds. Management would like to make the blend at as low a cost as possible while maintaining the requirements for chemical structure. The following linear programming problem has been formulated and the computer solution to it is provided.xi= The number of units of compound i to put in the blend, i=1,...,4MIN Z = 25x1+ 36x2+ 16x3+ 4x4S.T.12x1+ 9x2+ 10x3+ 6x4780Units of Chemical 15x1+ 22x2+ 19x3+ 10.2x4900Units of Chemical 2x1, x2, x3, x40Microsoft Excel 9.0 Answer ReportWorksheet: [Book2]Sheet1Report Created: 5/8/2003 12:49:19 PMTarget Cell (Min)CellNameOriginal ValueFinal Value\$B\$18Total Cost Amount0520Adjustable CellsCellNameOriginal ValueFinal Value\$A\$11Compound 100\$B\$11Compound 200\$C\$11Compound 300\$D\$11Compound 40130ConstraintsCellNameCell ValueFormulaStatusSlack\$B\$15Chemical 1 Amount780\$B\$15>=\$F\$5Binding0\$B\$16Chemical 2 Amount1326\$B\$16>=\$F\$6Not Binding426
Microsoft Excel 9.0 Sensitivity ReportWorksheet: [Book2]Sheet1Report Created: 5/8/2003 12:49:19 PMAdjustable CellsFinalReducedObjectiveAllowableAllowableCellNameValueCostCoefficientIncreaseDecrease\$A\$11Compound 1017251E+3017\$B\$11Compound 2030361E+3030\$C\$11Compound 309.333333333161E+309.333333333\$D\$11Compound 4130045.64ConstraintsFinalShadowConstraintAllowableAllowableCellNameValuePriceR.H. SideIncreaseDecrease\$B\$15Chemical 1 Amount7800.6666666677801E+30250.5882353\$B\$16Chemical 2 Amount132609004261E+30a.How much of each compound is to be used?Compound 10Compound 20Compound 30Compound 4130b.What is the total cost to make the blend?Z*=520c.What costs would each of the non-used compounds have to be in order to be used in the blend?Compound 1: 25-17=8; Cost of Compound 1 must be <=8Compound 2: 36-30=6; Cost of Compound 2 must be <=6Compound 3: 16-9.33=6.67; Cost of Compound 3 must be <=6.67d.What would happen to the optimal solution and Z*if compound 1 cost \$30?

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