Chapter 3 Introduction to optimization models

Linear Programming The PCTech company makes and sells two models for computers, Basic and XP. Profits for Basic is $80/unit and for XP is $129/unit. Sales estimate is 600 Basics and 1200 XPs Making the computers involves two operations: Assembly:Basic requires 5 hours and XP requires 6 hours Testing:Basic requires 1 hour and XP requires 2 hours Available labor hours: Assembly:10000 hours Testing:3000 hours

Linear Programming PC Tech wants to know how many of each model it should produce (assemble and test) to maximize its net profit, but it cannot use more labor hours than are available, and it does not want to produce more than it can sell. The problem objective: – Use LP to find the best mix of computer models that maximizes profit – Stay within the company’s labor availability – Don’t produce more than what can be sold

Graphical Method x 1 = Number of basic computer model x 2 = Number of XP computer model Net profit = 80x x 2 x1x1 x2x2

x 1 = Number of basic computer model x 2 = Number of XP computer model If x 1 = 1290, x 2 = 0, Net profit = 103,200 If x 1 = 0, x 2 = 800, Net profit = 103,200 Net profit = $103,200 x2x2 x1x1 Graphical Method Net profit = 80x x 2

Net profit = 80x x 2 x 1 = Number of basic computer model x 2 = Number of XP computer model If x 1 = 1290, x 2 = 0, Net profit = 103,200 If x 1 = 0, x 2 = 800, Net profit = 103,200 Net profit = $103,200 x2x2 x1x1 Net profit = $130,00 Net profit = $140,00 Graphical Method Iso-profit line

Constraints Basic ModelXP ModelHours available Assembly labor5 hours/unit6 hours/unit10,000 hours Testing labor1 hour/unit2 hours/unit3,000 hours Labor hours constraints Basic ModelXP Model Maximum sales/month Sales constraints

x 1 = Number of basic computer model x 2 = Number of XP computer model Assembly hours constraint: 5x 1 + 6x 2 <= 10,000 If we make no XP model at all 5(2000) + 6(0) = 10,000 If we make no Basic model at all 5(0) + 6( ) = 10,000 X 1 = 2000, x 2 = 0 Assembly Hours Constraints X 1 = 0, x 2 = x1x1 x2x2

x 1 = Number of basic computer model x 2 = Number of XP computer model Testing hours constraint: x 1 + 2x 2 <= 3,000 If we make no XP model at all (3000) + 2(0) = 3,000 If we make no Basic model at all (0) + 2(1500) = 3,000 Testing Hours Constraints x2x2 x1x1 X 1 = 3000, x 2 = 0 X 1 = 0, x 2 = 1500

x 1 = Number of basic computer model x 2 = Number of XP computer model Maximum sales Constraints x2x2 x1x1 Maximum sales for basic model: x 1 <= 600 X 1 = 600, x 2 = 0

x 1 = Number of basic computer model x 2 = Number of XP computer model x2x2 x1x1 Maximum sales for XP model: x 2 <= 1200 X 1 = 0, x 2 = 1200 Maximum sales Constraints

x 1 = Number of basic computer model x 2 = Number of XP computer model Feasible region x2x2 x1x1 x 2 = 1200 x 1 = 600 5x 1 + 6x 2 =10000 Redundant constraint x 1 + 2x 2 <= 3000

x 1 = Number of basic computer model x 2 = Number of XP computer model Optimum solution x2x2 x1x1 Redundant constraint x 1 + 2x 2 <= 3000 Iso-profit line

x 1 = Number of basic computer model x 2 = Number of XP computer model Optimum solution x2x2 x1x1

x 1 = Number of basic computer model x 2 = Number of XP computer model Optimum solution x2x2 x1x1 Optimum Solution is the intersection between: x 2 = 1200, and 5x 1 + 6x 2 = Solve and x 1 = 560 and x 2 = 1200 Profit = 80(560) + 129(1200) = $199,600 5x 1 + 6x 2 =10000 x 2 = 1200

The algebraic model Maximize 80x x 2 subject to: 5x 1 + 6x 2 < x 1 + 2x 2 < 3000 x 1 < 600 x 2 < 1200 x 1, x 2 > 0

Elements of LP model Decision variables – The variable whose values must be determined Objective function – A linear function of decision variables – The value of this function is to be optimized – minimized or maximized Constraints – Linear functions of the variables – Represents limited resources or minimum requirements

LP requirements Proportionality of variables Additivity of resources Divisibility of variables Non-negativity Linear objective function Linear constraints

Scaling in LP Poorly scaled model – model contains some very large numbers (e.g. 100,000 or more) and some very small numbers (e.g or less) – Solver may erroneously give an error that the linearity conditions are not satisfied Three remedies for poorly scaled model – Use Automatic Scaling option in Solver/Options – Redefine the units in the model – Change the Precision setting in Solver's Options dialog box to a larger number, such or (The default has five zeros.)

Solutions to LP problem Feasible solution Feasible region Optimal solution – Unique – Multiple – Unbounded

x 1 = Number of basic computer model x 2 = Number of XP computer model Multiple Optimum solution x2x2 x1x1

x 1 = Number of basic computer model x 2 = Number of XP computer model Multiple Optimum solution x2x2 x1x1 Iso-profit line

x 1 = Number of basic computer model x 2 = Number of XP computer model x2x2 x1x1 Unbounded Solution Constraint 1 Constraint 2

x 1 = Number of basic computer model x 2 = Number of XP computer model x2x2 x1x1 Unbounded Solution Constraint 1 Constraint 2

x 1 = Number of basic computer model x 2 = Number of XP computer model x2x2 x1x1 Infeasible Solution Constraint 1 Constraint 2

Summary An LP model may result in – an unique optimum solution – multiple optimum solutions – unbounded feasible region – infeasible region