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Graphical Linear Programming

Graphical Linear Programming

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    Management Science Management Science Presentation Transcript

    • Spreadsheet Modeling & Decision Analysis Chapter 2 Introduction to Optimization and Linear Programming
    • Characteristics of Optimization Problems
      • Decisions – that can be represented by decision variables (DVs). Typically represented as (X 1 , X 2 , X 3 , …. X n ); a set of n decision variables
      • Constraints – one or more functions of the DVs that must be satisfied. Typically should be thought of as restrictions (constraints) on the resources available to achieve the objective.
      • Objective - some function of the DVs the decision maker wants to MAXimize of MINimize.
    • General Form of an Optimization Problem
      • MAX (or MIN): f 0 (X 1 , X 2 , …, X n )
      • Subject to: f 1 (X 1 , X 2 , …, X n )<=b 1
      • f k (X 1 , X 2 , …, X n )>=b k
      • f m (X 1 , X 2 , …, X n ) = b m
      • As in, MAX or MIN some objective subject to a specific set of constraints based on the decision problem.
      • The 3 general forms of constraint relationships are represented here.
      • Note: If all the functions in an optimization problem are linear (the objective functions and the constraints) , then the problem is a Linear Programming (LP) problem
    • Linear Programming (LP) Problems
      • MAX (or MIN): c 1 X 1 + c 2 X 2 + … + c n X n
      • Subject to: a 11 X 1 + a 12 X 2 + … + a 1 n X n <= b 1
      • :
      • a k 1 X 1 + a k 2 X 2 + … + a kn X n >=b k
      • :
      • a m 1 X 1 + a m 2 X 2 + … + a mn X n = b m
      • So, this general form on an LP specifies up to n decision variables and up to m constraints.
      • c 1, c 2, c 3 … and so on are the coefficients in the objective function for DVs X 1, X 2, X 3 … etc.
      • a ij is the coefficient for variable X j in the i th constraint.
    • 5 Steps In Formulating LP Models:
      • 1. Understand the problem.
      • 2. Identify the decision variables.
      • 3. State the objective function as a linear combination of the decision variables.
      • 4. State the constraints as linear combinations of the decision variables.
      • 5. Identify any upper or lower bounds on the decision variables.
    • Solving LP Problems: A Graphical Approach
      • The constraints of an LP problem defines its feasible region.
      • Feasible solutions (region) of an LP model is the set of points (values for the DVs) that simultaneously satisfy all the constraints in the problem.
      • The best point in the feasible region is the optimal solution to the problem.
      • For LP problems with 2 variables, it is possible to plot the feasible region and find the optimal solution graphically.
    • Summary of Graphical Solution to LP Problems
      • 1. Plot the boundary line of each constraint in the model.
      • 2. Identify the feasible region; i.e. the set of all points on the graph that simultaneously satisfy all the constraints.
      • 3. Locate the optimal solution by either:
      • a. Plotting level curves for the objective function to identify the value of the coordinates for the DVs that satisfy the objective function (will occur at extreme point in the feasible region).
      • b. Enumerate all the extreme points and calculate the value of the objective function for each of these points.
    • Special Conditions in LP Models
        • Alternate Optimal Solutions ; instances where more than one point in the feasible region MAX (or MIN) the objective function
        • Redundant Constraints ; can occur if a constraint plays no role in determining the feasible region.
        • Unbounded Solutions ; the objective function can be made infinitely large (for MAX …small for MIN) and still satisfy the constraints. problem
        • Infeasibility ; instances where there is no way to satisfy all the constraints simultaneously.
    • An Example LP Problem Blue Ridge Hot Tubs produces two types of hot tubs: Aqua-Spas & Hydro-Luxes. There are 200 pumps, 1566 hours of labor, and 2880 feet of tubing available. Aqua-Spa Hydro-Lux Pumps 1 1 Labor 9 hours 6 hours Tubing 12 feet 16 feet Unit Profit $350 $300
    • 5 Steps In Formulating LP Models (Blue Ridge example):
      • 1. Understand the problem.
      • 2. Identify the decision variables.
      • X 1 =number of Aqua-Spas to produce
      • X 2 =number of Hydro-Luxes to produce
      • 3. State the objective function as a linear combination of the decision variables.
      • MAX: 350X 1 + 300X 2
    • 5 Steps In Formulating LP Models ((Blue Ridge example …. continued)
      • 4. State the constraints as linear combinations of the decision variables.
      • 1X 1 + 1X 2 <= 200 } pumps
      • 9X 1 + 6X 2 <= 1566 } labor
      • 12X 1 + 16X 2 <= 2880 } tubing
      • 5. Identify any upper or lower bounds on the decision variables.
      • X 1 >= 0
      • X 2 >= 0
    • So, the formulation of the LP Model for Blue Ridge Hot Tubs is; MAX: 350X 1 + 300X 2 S.T.: 1X 1 + 1X 2 <= 200 9X 1 + 6X 2 <= 1566 12X 1 + 16X 2 <= 2880 X 1 >= 0 X 2 >= 0
    • Solving LP Problems: An Intuitive Approach
      • Idea: Each Aqua-Spa (X 1 ) generates the highest unit profit ($350), so let’s make as many of them as possible!
      • How many would that be?
        • Let X 2 = 0
          • 1st constraint: 1X 1 <= 200
          • 2nd constraint: 9X 1 <=1566 or X 1 <=174
          • 3rd constraint: 12X 1 <= 2880 or X 1 <= 240
      • If X 2 =0, the maximum value of X 1 is 174 and the total profit is $350*174 + $300*0 = $60,900
      • This solution is feasible , but is it optimal?
      • No!
    • Graphical Solution to the Blue Ridge Hot Tub example …
    • Plotting the First Constraint X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 (0, 200) (200, 0) boundary line of pump constraint X 1 + X 2 = 200
    • Plotting the Second Constraint X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 (0, 261) (174, 0) boundary line of labor constraint 9X 1 + 6X 2 = 1566
    • Plotting the Third Constraint X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 (0, 180) (240, 0) boundary line of tubing constraint 12X 1 + 16X 2 = 2880 Feasible Region
    • X 2 Plotting A Level Curve of the Objective Function X 1 250 200 150 100 50 0 0 50 100 150 200 250 (0, 116.67) (100, 0) objective function 350X 1 + 300X 2 = 35000
    • A Second Level Curve of the Objective Function X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 (0, 175) (150, 0) objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500
    • Using A Level Curve to Locate the Optimal Solution X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500 optimal solution
    • Calculating the Optimal Solution
      • The optimal solution occurs where the “pumps” and “labor” constraints intersect.
      • This occurs where:
        • X 1 + X 2 = 200 (1)
        • and 9X 1 + 6X 2 = 1566 (2)
      • From (1) we have, X 2 = 200 -X 1 (3)
      • Substituting (3) for X 2 in (2) we have,
        • 9X 1 + 6 (200 -X 1 ) = 1566
        • which reduces to X 1 = 122
      • So the optimal solution is,
        • X 1 =122, X 2 =200-X 1 =78
        • Total Profit = $350*122 + $300*78 = $66,100
    • Enumerating The Corner Points Note: This technique will not work if the solution is unbounded. X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 (0, 180) (174, 0) (122, 78) (80, 120) (0, 0) obj. value = $54,000 obj. value = $64,000 obj. value = $66,100 obj. value = $60,900 obj. value = $0
    • Example of Alternate Optimal Solutions X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 450X 1 + 300X 2 = 78300 objective function level curve alternate optimal solutions
    • Example of a Redundant Constraint X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 boundary line of tubing constraint Feasible Region boundary line of pump constraint boundary line of labor constraint
    • Example of an Unbounded Solution X 2 X 1 1000 800 600 400 200 0 0 200 400 600 800 1000 X 1 + X 2 = 400 X 1 + X 2 = 600 objective function X 1 + X 2 = 800 objective function -X 1 + 2X 2 = 400
    • Example of Infeasibility X 2 X 1 250 200 150 100 50 0 0 50 100 150 200 250 X 1 + X 2 = 200 X 1 + X 2 = 150 feasible region for second constraint feasible region for first constraint