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Intro Management Science 472.21 2 Fall 2011 Bruce Duggan Providence University College

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This Week Review Cases from ch 1 Linear Programming ch 2 formulas & graphs

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Case 1: Clean Clothes Corner A.Current volume? shes just breaking even v= cfcf p-c v v= $1,700.00 $1.10 - $0.25

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Case 1: Clean Clothes Corner B. Increase needed to break even? v= cfcf p-c v v= $16,200.00/12 $1.10 - $0.25

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Case 1: Clean Clothes Corner C. Monthly profit? Z = vp - c f - vc v Z = 4,300.00 $1.10 - ($1,700.00 + $1,350.00 ) - 4,300 $0.25

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Case 1: Clean Clothes Corner D. If lower price? BE? Z?Z? Z = vp - c f - vc v v= cfcf p-c v

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Case 1: Clean Clothes Corner E. Which is the better choice? Z with new equipment? Z without new equipment?

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Case 2: Ocobee Which option is better? make the rafts yourself? buy them from North Carolina?

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ch 2: Linear Programming George Dantzig http://forum.stanford.edu/blog/?p=27

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Linear Programming Jargon Linear programming l.p. figuring stuff out with basic algebra Model formulation Stating our problem in words/math/graphs Sensitivity analysis What happens if…?

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Linear Programming Jargon Why is there jargon? handout

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Applications Kellogg pg 35 Nutrition Coordinating Center pg 46 Soquimich pg 51

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Example: Maximization The St. Adolphe Historical Museum We have a group of older volunteers The St. Adolphe Craft League Theyve offered to make toothpick tchochkes to sell at the gift shop Red River ox carts the first church in St. Adolphe We can sell everything they make

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St. Adolphe Craft League They want to know: How many ox carts? How many churches? Goal To make the most profit possible for the museum

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St. Adolphe Craft League Resource availability 40 hrs of labor 120 boxes of toothpicks Decision variables x 1 = number of ox carts to make x 2 = number of churches to make

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St. Adolphe Craft League Product resource requirements and unit profit: 41 32 Product cart church Profit ($/unit) 40 50 Material (boxes/unit) Labour (hr/unit) Resource Requirements

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St. Adolphe Craft League Objective function Maximize Z = $40 x 1 + $50 x 2 Resource constraints 1 x 1 + 2 x 2 40 hours of labor 4 x 1 + 3 x 2 120 boxes of toothpicks Non-Negativity constraints x 1 0; x 2 0

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Maximize Z = $40 x 1 + $50 x 2 subject to: 1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0 St. Adolphe Craft League Problem definition Complete linear programming model

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Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0 St. Adolphe Craft League Model formulation: l.p. no computers yet

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St. Adolphe Craft League words math graphs Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0

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St. Adolphe Craft League x2x2 0 10203040 10 20 30 40 x1x1 Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0

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St. Adolphe Craft League x2x2 0 10203040 10 20 30 40 x1x1 Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0

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St. Adolphe Craft League x2x2 0 10203040 10 20 30 40 x1x1 Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0

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Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0 St. Adolphe Craft League x2x2 0 10203040 10 20 30 40 x1x1

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St. Adolphe Craft League x2x2 0 10203040 10 20 30 40 x1x1 Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0 x 1 = 0 ox carts x 2 = 20 churches Z = $1,000 x 1 = 30 ox carts x 2 = 0 churches Z = $1,200 x 1 = 24 ox carts x 2 = 8 churches Z = $1,360

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Linear Programming lp has 2 main tools maximization most profit minimization least cost Z means profit Z means cost

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Example: Minimization Friesen Farms section of land needs at least 16 lb nitrogen 24 lb phosphate 2 brands of fertilizer available DeSallaberry Superior Carmen Crop Goal Meet fertilizer needs at minimum cost Problem How much of each brand should you buy? words math graphs

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Friesen Farms Chemical Contributions Product Nitrogen (lb/bag) Phosphate (lb/bag) Cost ($/bag) DeSallaberry Superior Carmen Crop words math graphs

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Friesen Farms Chemical Contributions Product Nitrogen (lb/bag) Phosphate (lb/bag) Cost ($/bag) DeSallaberry Superior 24$6 Carmen Crop43$3 words math graphs

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Friesen Farms Objective function Minimize Z = $6 x 1 + $3 x 2 Decision variables x 1 = bags of DeSallaberry to buy x 2 = bags of Carmen to buy words math graphs

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Friesen Farms Objective function Minimize Z = $6 x 1 + $3 x 2 Model constraints 2 x 1 + 4 x 2 16 (lb) nitrogen constraint 4 x 1 + 3 x 2 24 (lb) phosphate constraint x 1, x 2 0 non-negativity constraint words math graphs

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Min Z = $6 x 1 + $3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0 Friesen Farms Model formulation: l.p. words math graphs

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Min Z = $6 x 1 + 3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0 Friesen Farms x2x2 0 2468 2 4 6 8 x1x1 words math graphs

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Friesen Farms x2x2 0 2468 2 4 6 8 x1x1 Min Z = $6 x 1 + 3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0

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Friesen Farms x2x2 0 2468 2 4 6 8 x1x1 Min Z = $6 x 1 + 3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0

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x2x2 0 2468 2 4 6 8 x1x1 Friesen Farms Min Z = $6 x 1 + 3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0 x 1 = 0 bags of DeSallaberry x 2 = 8 bags of Carmen Z = $24 x 1 = 5 DeSallaberry x 2 = 2 Carmen Z = $36 x 1 = 8 DeSallaberry x 2 = 0 Carmen Z = $48

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x2x2 0 2468 2 4 6 8 x1x1 Friesen Farms Min Z = $6 x 1 + 3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0

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x2x2 0 2468 2 4 6 8 x1x1 Friesen Farms Min Z = $6 x 1 + 3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0

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x2x2 0 2468 2 4 6 8 x1x1 Friesen Farms Min Z = $6 x 1 + 3 x 2 s.t.2 x 1 + 4 x 2 16 4 x 1 + 3 x 2 24 x 1, x 2 0 Surplus Variables whats left over - dont contribute to - slack x 1 = 0 bags of DeSallaberry x 2 = 8 bags of Carmen s 1 = 16 lb of nitrogen s 2 = 0 lb of phosphate Z = $24 00 x 1 = 4.8 DeSallaberry x 2 = 1.6 Carmen s 1 = 0 nitrogen s 2 = 0 phosphate Z = $33 60 x 1 = 8 DeSallaberry x 2 = 0 Carmen s 1 = 0 nitrogen s 2 = 8 phosphate Z = $48 00

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On computer much easier to do goals up to now the idea the formulas

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l.p. usual characteristics & limitations clear goal choice amongst alternatives certainty non-probabilistic constraints exist relationships linear slope constant additivity divisibility for graphical solution 2 variables

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Assignment ch 2 problems in group 2 38 yourself 1 16

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Max Z = $40 x 1 + $50 x 2 s.t.1 x 1 + 2 x 2 40 4 x 1 + 3 x 2 120 x 1, x 2 0 St. Adolphe Craft League Model formulation: l.p.

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Next Week review ch 2 problems ch 3 on the computer sensitivity analysis

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