# Chapter 19 – Linear Programming

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Chapter 19 – Linear Programming
BUAD306 Chapter 19 – Linear Programming

Optimization QUESTION: Have you ever been limited to
what you can get done because you don’t have enough ________? Examples of how we optimize in daily decisions? Why do we strive for optimization? Factors that limit optimization?

What is Linear Programming
A model consisting of linear relationships that represent a firm’s objective and resource constraints Problems are typically referred to as constrained optimization problems

LP Objectives Maximize profits Maximize outputs Minimize costs
Determine combinations of outputs to meet/surpass goals

Business Examples of LP
Product Planning - Landscaping Portfolio Selection - Investments Menu Planning – Restaurants Route Planning / Pricing - Airlines

LP Assumptions Linearity - The impact of decision variables is linear in constraints and objective function Divisibility – non-integer values of decision variables are acceptable Certainty – values of parameters are known and constant Non-Negativity – negative values of decision variables are unacceptable

LP Concepts Objective - The goal of the LP model
Decision Variables - Amounts of either inputs or outputs Constraints - Limitations of available alternatives: Equal to, Greater than, Less than Feasible Solution Space - The set of all feasible combinations of decision variables as defined by the constraints

Constraints Restrictions on the company’s resources
Time Labor Energy Materials Money Restriction guidelines recipe for making food products engineering specifications

LP Steps Step 1: Set up LP model Step 2: Plot the constraints
Step 3: Identify the feasible solution space Step 4: Plot the objective function Step 5: Determine the optimal solution

LP Model Example X1 = Quantity of Product 1 to produce
X2 = Quantity of Product 2 to produce Decision Variables Maximize: 5x1 + 8x2 (profit) Objective Function Subject to: Labor: x1 + 4x2 <= hours Material: x1 + 6X2 <= pounds Constraints Product 1: X = units x2 >= Non-negativity constraint

LP Example A 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.

Formulate 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.

There 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 produce x2 = # of mugs to produce Objective: Maximize Z = 20x1 + 20x2 Constraints: Labor: x1 + 2x2 <= 40 Clay: 3x1 + 2x2 <= 60 Non- Neg: x1, x2 >=0

Plot Constraints Step 1: For each constraint, set x1 = 0, get value for x2 Step 2: For each constraint, set x2 = 0, get value for x1 Step 3: Plot these as coordinates and then draw constraint lines

Plot Constraints – Identify Coordinates
Labor Constraint: x1 + 2x2 <= 40 hrs

Plot Constraints – Identify Coordinates
Clay Constraint: 3x1 + 2x2 <= 60 lbs.

Plot Constraints Clay Labor 50 45 40 35 30 25 x2 20 15 10 5 10 20 30
10 20 30 40 50 x1

Identify Feasible Solution Space
This is the area that satisfies all of the LP model’s constraints simultaneously. Includes all of the points on the borders as well. 50 45 40 35 30 25 x2 20 15 10 5 10 20 30 40 50 x1

Plot 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 + 20x2 We will select and arbitrary level of revenue, such as \$240, in order to plot the objective function. 240 = \$20x1 + 20x2 if x1 = 0, then x2 = 12 (0, 12) if x2 = 0, then x1 = 12 (12, 0)

Identify Feasible Solution Space
Observation: Every point on this Line in the feasible solution area Will result in a revenue of \$240. (Every combination of X1 and X2 on this line in this space will yield an \$240 revenue total.) 50 45 40 35 30 25 x2 20 15 10 5 10 20 30 40 50 x1

Identify Feasible Solution Space
Shift the objective function line to locate the optimal solution point. 50 45 40 35 30 25 x2 20 15 10 5 10 20 30 40 50 x1

Finding 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.

Finding 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.

Finding 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.

Solving 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:

Solving for the Optimal Solution
Thus, the optimal solution is to produce ___ bowls and ____ mugs for a maximized revenue of \$___________. Objective Function: Maximize: 20x1 + 20x2 Z =

Testing 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.)

Confirm Answer (0,0) = (0,20) = (20,0) = (10,15) = 50 45 40 35 30 25
x2 20 15 10 5 10 20 30 40 50 x1

Total Minutes Available
Example #1 Delstate 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. Product Cutting Polishing Gem 1 14 7 Gem 2 6 12 Total Minutes Available 84

Slack and Surplus A 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.

Slack 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; <=40 Clay: 3(10) + 2(15) <= <= 60 We can conclude that there is no slack of resources given our optimal solution What if our solution was (20, 5)??? Do we have slack??? What does that tell us since this is a 2 constraint problem??

Example #2 A 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.

In Class Example max 5x1 + 6x2 st A 2x1+3x2<=60 B 4x1+x2<=80
C x1+x2<=50 N/N x1, x2 >=0

Binding Constraints Redundant Constraints Multiple Optimal Solutions Slack & Surplus

Binding Constraint When a constraint forms the optimal corner point of the feasible solution space, it is BINDING.

Redundant Constraint A constraint that does not form a unique boundary of the feasible solution space. Its removal would not alter the feasible solution space.

Multiple Optimal Solutions
In some instances, you may get an objective function which is parallel to a constraint line This ends up NOT being tangent to the feasible solution space!

Sensitivity Analysis Assessing the impact of potential changes to the numerical values of an LP model Objective function coefficients Right-hand values of constraints Constraint coefficients Max 20x1 + 20x2 S.T. L: 1x1 + 2x2 <= 40 C: 3x1 + 2x2 <= 60 N/N: x1, x2 >=0

LINDO Computer-based modeling system for solving linear programming problems All PCs in the Purnell computer lab have LINDO installed

LINDO Steps Designate functions within LINDO Max/Min Constraints
Solve with Sensitivity Analysis

LINDO Answers Notice that the results provide the same answers
we obtained in our graphical approach.

Objective 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 solution Revenue coefficient for Mugs can range between 20 and 60 without impacting solution

RHS Ranges With 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) Price RHS for Labor constraint can range between and 100 RHS for Clay constraint can range between 20 and 60

Shadow (Dual) Price The amount by which the value of the objective function would change if there is a one unit change in the RHS of that constraint

Shadow (Dual) Pricing If constraint has slack, shadow price = 0
Increasing or decreasing RHS has no effect - would only increase slack/surplus If constraint is binding, shadow price = the amount objective function will change for each one unit change in the RHS

Review of LP Handout #3

Homework #5 – PIES! Max Z: 1.5x X2 Subject to: Flour: 3x1 + 3x2 <= 2100 Sugar: 1.5x1 + 2x2 <= 1200 Time: 6x1 + 3x2 <= 3600 N/N: X1, X2 >= 0

Homework #5

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