Presentation on theme: "Chapter 19 – Linear Programming"— Presentation transcript:
1Chapter 19 – Linear Programming BUAD306Chapter 19 – Linear Programming
2Optimization QUESTION: Have you ever been limited to what you can get done because you don’thave enough ________?Examples of how we optimize in daily decisions?Why do we strive for optimization?Factors that limit optimization?
3What is Linear Programming A model consisting of linear relationships that represent a firm’s objective and resource constraintsProblems are typically referred to as constrained optimization problems
4LP Objectives Maximize profits Maximize outputs Minimize costs Determine combinations of outputs to meet/surpass goals
6LP AssumptionsLinearity - The impact of decision variables is linear in constraints and objective functionDivisibility – non-integer values of decision variables are acceptableCertainty – values of parameters are known and constantNon-Negativity – negative values of decision variables are unacceptable
7LP Concepts Objective - The goal of the LP model Decision Variables - Amounts of either inputs or outputsConstraints - Limitations of available alternatives: Equal to, Greater than, Less thanFeasible Solution Space - The set of all feasible combinations of decision variables as defined by the constraints
8Constraints Restrictions on the company’s resources TimeLaborEnergyMaterialsMoneyRestriction guidelinesrecipe for making food productsengineering specifications
9LP Steps Step 1: Set up LP model Step 2: Plot the constraints Step 3: Identify the feasible solutionspaceStep 4: Plot the objective functionStep 5: Determine the optimal solution
10LP Model Example X1 = Quantity of Product 1 to produce X2 = Quantity of Product 2 to produce Decision VariablesMaximize: 5x1 + 8x2 (profit) Objective FunctionSubject to:Labor: x1 + 4x2 <= hoursMaterial: x1 + 6X2 <= pounds ConstraintsProduct 1: X = unitsx2 >= Non-negativity constraint
11LP ExampleA pottery company employs artisans to produce clay bowls and mugs. The two primary resources used by the company are special pottery clay and skilled labor. Given these limited resources, the company would like to determine how many bowls and mugs to produce each day to maximize profit.
12Formulate this problem as an LP model. The two products have the following resource requirements for production and selling price per item produced:There are 40 hours of labor and 60 pounds of clay available each day.Formulate this problem as an LP model.
13There are 40 hours of labor and 60 pounds of clay available each day There are 40 hours of labor and 60 pounds of clay available each day. What are the decision variables, objective function & constraints?Decision variables: x1 = # of bowls to producex2 = # of mugs to produceObjective: Maximize Z = 20x1 + 20x2Constraints: Labor: x1 + 2x2 <= 40Clay: 3x1 + 2x2 <= 60Non- Neg: x1, x2 >=0
14Plot ConstraintsStep 1: For each constraint, set x1 = 0, get value for x2Step 2: For each constraint, set x2 = 0, get value for x1Step 3: Plot these as coordinates and then drawconstraint lines
18Identify Feasible Solution Space This is the area that satisfies all of theLP model’s constraints simultaneously.Includes all of the points on the bordersas well.504540353025x220151051020304050x1
19Plot the Objective Function In order to locate the point in the feasible solution space that will maximize revenue, we now need to plot the Objective Function:Maximize Z = $20x1 + 20x2We will select and arbitrary level of revenue, such as $240, in order to plot the objective function.240 = $20x1 + 20x2if x1 = 0, then x2 = 12 (0, 12)if x2 = 0, then x1 = 12 (12, 0)
20Identify Feasible Solution Space Observation: Every point on thisLine in the feasible solution areaWill result in a revenue of $240.(Every combination of X1 and X2on this line in this space will yieldan $240 revenue total.)504540353025x220151051020304050x1
21Identify Feasible Solution Space Shift the objective function line to locatethe optimal solution point.504540353025x220151051020304050x1
22Finding the Optimal Solution As you increase the revenue from $240, the line will move up and to the right in a parallel manner.The point where it moves thru the boundary of the feasible space contains the points farthest from the origin– which correspond to the points yielding the greatest revenue.Note: Once it passes outside the feasible solution area, it’s no longer a feasible revenue line because of constraint limitations.
23Finding the Optimal Solution As we shift the objective function line away from the origin, we search for the farthest point that is still in/on the feasible solution.As been proven mathematically, the solution point will not only be on the boundary of the feasible solution space, it will be at one of the corners of the boundary where two constraints intersect– the most you can use of those constraints. This is called an extreme point.
24Finding the Optimal Solution In our example, by shifting the objective function line away from the origin, we reach the farthest point that is still in/on the feasible solution space at the point where the constraints for labor and clay intersect.To determine what quantities of Q1 and Q2 are valid for that combination, we set the two constraints equal to each other.
25Solving for the Optimal Solution Given our constraint equations:x1 + 2x2 = (Labor)3x1 + 2x2 = 60 (Clay)We set them equal to each other and solve for the optimal quantities of X1 and X2:
26Solving for the Optimal Solution Thus, the optimal solution is to produce ___ bowls and ____ mugs for a maximized revenue of $___________.Objective Function: Maximize: 20x1 + 20x2Z =
27Testing the Optimal Solution You can test to make sure this is the optimal solution by solving for your revenue at the other extreme points (which in this case, intersect with the x1 and x2 axis.)
29Total Minutes Available Example #1Delstate Jewelers manufacturers two types of semi-precious gems for sale to its wholesale customers, Gem 1 and Gem 2. Delstate realizes revenue of $4.00 for each Gem 1 sold and revenue of $3 per gem for each Gem 2 sold. To produce the gems, the company must cut and polish each gem. The company is limited to a certain number of minutes for each machine that is used in the cutting and polishing process as shown in the table below. Given this information, determine the optimal quantities of Gem 1 and Gem 2 that maximize revenue.ProductCuttingPolishingGem 1147Gem 2612Total Minutes Available84
30Slack and SurplusA slack variable is a variable representing unused resources added to a <= constraint to make it an equality.A surplus variable is a variable representing an excess above a resource requirement that is subtracted from a >= constraint to make it an equality.
31Slack and Surplus Back to Mugs and Bowls… We’ve got (10, 15) as optimal solution and the following constraints:Labor x1 + 2x2 <= 40 hrs.Clay 3x1 + 2x2 <= 60 lbs.So by substituting our values for x1 and x2:Labor: (15) <= 40; <=40Clay: 3(10) + 2(15) <= <= 60We can conclude that there is no slack of resources given our optimal solutionWhat if our solution was (20, 5)???Do we have slack??? What does that tell us since this is a 2 constraint problem??
32Example #2A leather shop makes custom, hand-tooled briefcases and luggage. The shop makes a $400 profit from each briefcase and $200 profit from each piece of luggage. The shop has a contract to provide a local store with exactly 30 items each month. A tannery supplies the shop with at least 80 square yards of leather per month. The shop must at least use this amount, but can use more if needed. Each briefcase requires 2 square yards of leather, each piece of luggage requires 8 square yards of leather. From past performance, the shop owner knows that they cannot make more than 20 briefcases per month.Determine the number of briefcases and luggage to produce each month in order to maximize revenue. Formulate an LP model and solve graphically. Determine the optimal quantities of briefcases and luggage and the maximized revenue amount. Also determine if there is any slack or surplus.
33In Class Example max 5x1 + 6x2 st A 2x1+3x2<=60 B 4x1+x2<=80 C x1+x2<=50N/N x1, x2 >=0
35Binding ConstraintWhen a constraint forms the optimal corner point of the feasible solution space, it is BINDING.
36Redundant ConstraintA constraint that does not form a unique boundary of the feasible solution space.Its removal would not alter the feasible solution space.
37Multiple Optimal Solutions In some instances, you may get an objective function which is parallel to a constraint lineThis ends up NOT being tangent to the feasible solution space!
38Sensitivity AnalysisAssessing the impact of potential changes to the numerical values of an LP modelObjective function coefficientsRight-hand values of constraintsConstraint coefficientsMax 20x1 + 20x2S.T.L: 1x1 + 2x2 <= 40C: 3x1 + 2x2 <= 60N/N: x1, x2 >=0
39LINDOComputer-based modeling system for solving linear programming problemsAll PCs in the Purnell computer lab have LINDO installed
40LINDO Steps Designate functions within LINDO Max/Min Constraints Solve with Sensitivity Analysis
41LINDO Answers Notice that the results provide the same answers we obtained in our graphicalapproach.
42Objective Function Ranges With this report, we can determine the range in which the objective coefficients can exist before the optimal solution changes:Revenue coefficient for Bowls can range between 10 and 30 without impacting solutionRevenue coefficient for Mugs can range between 20 and 60 without impacting solution
43RHS RangesWith this report, we can determine the range in which the RHS of the constraints can change while yielding the same savings/cost indicated by the Shadow (Dual) PriceRHS for Labor constraint can range between and 100RHS for Clay constraint can range between 20 and 60
44Shadow (Dual) PriceThe amount by which the value of the objective function would change if there is a one unit change in the RHS of that constraint
45Shadow (Dual) Pricing If constraint has slack, shadow price = 0 Increasing or decreasing RHS has no effect - would only increase slack/surplusIf constraint is binding, shadow price = the amount objective function will change for each one unit change in the RHS