Thank you and welcome Linear Programming (LP) Modeling Application in manufacturing And marketing By M. Dadfar, PhD

First example The problem: A toy manufacturing company needs to know how to optimize weekly profit of making two different toys: wooden Soldier and Train. Information at hand: Each soldier built: Sells for $27 and uses $10 worth of raw materials. Increases variable labor/overhead costs by $14. Requires 2 hours of finishing labor. Requires 1 hour of carpentry labor. At most 40 soldiers are bought each week. Each train built: Sells for $21 and used $9 worth of raw materials. Increases variable labor/overhead costs by $10. Requires 1 hour of finishing labor. Requires 1 hour of carpentry labor. Demand for trains are unlimited. Each week the company can obtain: All needed raw material. Only 100 finishing hours. Only 80 carpentry hours. A mathematical model to maximize weekly profit (revenues – expenses) is needed

Calculations x 1 = number of soldiers produced each week x 2 = number of trains produced each week Profit = Revenue - Raw material costs - Variable costs Weekly revenue = 27x x 2 Weekly raw material costs = 10x 1 + 9x 2 Weekly variable costs = 14x x 2 Weekly profit = (27x x 2 ) – (10x 1 + 9x 2 ) – (14x x 2 ) = 3x 1 + 2x 2

Constrains The objective is to chose x 1 and x 2 to maximize 3x 1 + 2x 2. But this can not be achieved by simply increasing X1 and X2. The values of x 1 and x 2 are limited by the following three constraints: Constraint 1 Each week, no more than 100 hours of finishing time may be used. Constraint 2 Each week, no more than 80 hours of carpentry time may be used. Constraint 3 Because of limited demand, at most 40 soldiers should be produced. These three constraints can be expressed mathematically by the following equations: Constraint 1: 2 x 1 + x 2 ≤ 100 Constraint 2: x 1 + x 2 ≤ 80 Constraint 3: x 1 ≤ 40

The Model This is the Optimization model of the problem Max, z = 3x 1 + 2x 2 (objective function) Subject to (constraints): 2 x 1 + x 2 ≤ 100 (finishing constraint) x 1 + x 2 ≤ 80 (carpentry constraint) x 1 ≤ 40 (constraint on demand for soldiers) x 1 ≥ 0 (can’t be negative) x 2 ≥ 0(can’t be negative)

Finding the Feasible Solution Since this example of LP has two variables, it may be solved graphically. The feasible region is the set of all points satisfying the constraints. For example x 1 = 40 and x 2 = 20 are in the feasible region but x 1 = 15, x 2 = 70 is not in the feasible region since is not less than or equal to 80. Constraints 2 x 1 + x 2 ≤ 100 (finishing constraint) x 1 + x 2 ≤ 80 (carpentry constraint) x 1 ≤ 40 (demand constraint) x 1 ≥ 0 (can’t be negative) x 2 ≥ 0 (can’t be negative) A graph of the constraints and feasible region is shown on the next slide.

The set of points satisfying the LP is bounded by the five sided polygon DGFEH. Any point on or in the interior of this polygon is in the feasible region. LP Graph

The optimal solution must be an extreme point of the feasible region. For this example, Maximum weekly profit occurs at point G where: x 1 = 20, x 2 = 60 Max z = $180. Solution

Second example, a marketing problem * The company believes that its most likely customers are high-income women and men. * To reach these groups, the CEO has embarked on an ambitious TV advertising campaign and has decide to purchase one minute commercial spots on two type of programs: comedy shows and football games. How to maximize viewers with minimized cost?

 Each comedy commercial is seen by 7 million high income women and 2 million high-income men.  Each football game is seen by 2 million high-income women and 12 million high-income men.  A 1-minute comedy ad costs $50,000 and a 1-minute football ad costs $100,000. The company likes for commercials to be seen by at least 28 million high-income women and 24 million high-income men. We use LP to determine how they can meet its advertising requirements at minimum cost. Marketing problem Data

The decision variables are: x 1 = number of 1-minute comedy ads x 2 = number of 1-minute football ads Problem Formulation Min Cost; z = 50x x 2 (objective function in $1,000 ) 7x 1 + 2x 2 ≥ 28(high-income women) 2x x 2 ≥ 24(high-income men) x 1, x 2 ≥ 0(non-negativity constraints) A graph of the feasible area is show on the next slide.

Like the first example LP, This LP has a convex feasible region. However, it contains points for which the value of at least one variable can assume arbitrarily large values. The feasible region is not bounded. LP Graph

Since the company wants to minimize total advertising costs, the optimal solution to the problem is the point in the feasible region with the smallest z-value; point E and it is the optimal solution: x 1 = 3.6 x 2 = 1.4 Optimal Solution

If only 1-minute commercials are available, it is unreasonable to say 3.6 comedy and 1.4 football commercials should be purchased. So, the Divisibility Assumption has been violated, and the Dorian LP should, in reality, be considered as an integer programming model problem. Since there is no way of knowing with certainty of how many viewers are added with each type of commercial, the Certainty Assumption is also violated. Resolving violations is not in the scope of this presentation. We have used similar models to help companies determine their optimal media mix. Violations?

We all use calculators and computers without knowing what is going on inside these devices, we don’t need to know, all we need is to “Use” them. Conclusion The decision makers in the above two examples may not have any idea of what Linear Programming (LP) Modeling is –they don’t need to. It is the CTO’s job to solve the problem and provide them with results, and that is what we do. Ideal CTO, 2012www.idealcto.com