Linear Programming, A Geometric Approach Chapter 3 Linear Programming, A Geometric Approach Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Outline 3.1 A Linear Programming Problem 3.2 Linear Programming I 3.3 Linear Programming II Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

3.1 A Linear Programming Problem The Problem Tabulate Data Translate the Constraints The Objective Function Linear Programming Problem Production Schedule No Waste Feasible Set Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

The Problem A furniture manufacturer makes two types of furniture - chairs and sofas. The manufacture of a chair requires 6 hours of carpentry, 1 hour of finishing, and 2 hours of upholstery. Manufacture of a sofa requires 3 hours of carpentry, 1 hour of finishing, and 6 hours of upholstery. Each day the factory has available 96 labor hours for carpentry, 18 labor-hours for finishing, and 72 labor-hours for upholstery. The profit per chair is $80 and per sofa is $70. How many chairs and sofas should be produced each day to maximize the profit? Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Tabulate Data It is helpful to tabulate data given in the problem. Chair Sofa Available time Carpentry Finishing Upholstery Profit 6 hours 3 hours 96 labor-hours 1 hour 18 labor-hours 2 hours 72 labor-hours $80 $70 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Translate the Constraints Translate each of the constraints (restrictions on labor-hours available) into mathematical language. Let x be the number of chairs and y be the number of sofas manufactured each day, respectively. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Translate the Constraints (2) Carpentry: [number of labor-hours per day] = (number of hours required per chair)  (number of chairs per day) + (number of hours required per sofa)  (number of sofas per day) = 6x + 3y [number of labor-hours per day] < [maximum available] 6x + 3y < 96 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Translate the Constraints (3) Similarly, Finishing: x + y < 18 Upholstery: 2x + 6y < 72 Number of chairs and sofas cannot be negative: x > 0, y > 0 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

The Objective Function The objective of the problem is to optimize profit. Translate the profit (objective function) into mathematical language. [profit] = [profit from chairs] + [profit from sofas] = [profit per chair][number of chairs] + [profit per sofa][number of sofas] = 80x + 70y Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Linear Programming Problem The manufacturing problem can now be written as a mathematical problem. Find x and y for which 80x + 70y is as large as possible, and for which the following hold simultaneously: This is called a linear programming problem. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Production Schedule In the manufacturing problem, each pair of numbers (x,y) that satisfies the system of inequalities is called a production schedule. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Production Schedule Which of the following is a production schedule for (11,6)? (6,11)? Yes No Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

No Waste It seems clear that a factory will operate most efficiently when its labor is fully utilized (no waste). This would require x and y to satisfy the system Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example No Waste Solve According to the graph of the three equations, there is no common intersection and therefore no solution. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Feasible Set The set of solutions to the system of inequalities is called the feasible set of the system. This represents all possible production schedules. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Feasible Set Find the feasible set for Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Feasible Set (2) Notice that (0,0) satisfies all the inequalities. Graph the boundaries: y < -2x + 32 y < -x + 18 y < -x/3 + 12 x > 0, y > 0 Feasible Set Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Summary Section 3.1 A linear programming problem asks us to find the point (or points) in the feasible set of a system of linear inequalities at which the value of a linear expression involving the variables, called the objective function, is either maximized or minimized. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

3.2 Linear Programming I Vertex Fundamental Theorem of Linear Programming Linear Programming Steps Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Vertex The boundary of the feasible set is composed of line segments. The line segments intersect in points, each of which is a corner of the feasible set. Such a corner is called a vertex. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Vertex Find the vertices of (0,12) (9,9) (14,4) (0,0) (16,0) y < -2x + 32 y < -x + 18 y < -x/3 + 12 x > 0, y > 0. (0,12) (9,9) (14,4) (0,0) (16,0) Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Fundamental Theorem of Linear Programming Fundamental Theorem of Linear Programming The maximum (or minimum) value of the objective function in a linear programming problem is achieved at one of the vertices of the feasible set. The point that yields the maximum (or minimum) value of the objective function is called an optimal point. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Optimal Point Profit = 80x + 70y Find the point which maximizes Profit = 80x + 70y for the feasible set with vertices (0,0), (0,12), (9,9), (14,4) and (16,0). Vertex Profit = 80x + 70y (0,0) 80(0) + 70(0) = 0 (0,12) 80(0) + 70(12) = 840 (9,9) 80(9) + 70(9) = 1350 (14,4) 80(14) + 70(4) = 1400 (16,0) 80(16) + 70(0) = 1280 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Linear Programming Steps - Step 1 Translate the problem into mathematical language. Organize the data. Identify the unknown quantities and define corresponding variables. Translate the restrictions into linear inequalities. Form the objective function. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Linear Programming Steps - Step 2 Graph the feasible set. Put the inequalities in standard form. Graph the straight line corresponding to each inequality. Determine the side of the line belonging to the graph of each inequality. Cross out the other side. The remaining region is the feasible set. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Linear Programming Steps - Steps 3 & 4 Determine the vertices of the feasible set. Step 4 Evaluate the objective function at each vertex. Determine the optimal point. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Linear Programming Steps Rice and soybeans are to be part of a staple diet. One cup of uncooked rice costs 21 cents and contains 15 g of protein, 810 calories, and 1/9 mg of B2 (riboflavin). One cup of uncooked soybeans costs 14 cents and contains 22.5 g of protein, 270 calories, and 1/3 mg of B2. The minimum daily requirements are 90 g of protein, 1620 calories and 1 mg of B2. Find the optimal point that will minimize cost. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 1A Organize the data. Rice Soybeans Required level per day Protein (g/cup) 15 22.5 90 Calories (per cup) 810 270 1620 B2 (mg/cup) 1/9 1/3 1 Cost (cents/cup) 21 14 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 1B B. Identify the unknown quantities and define corresponding variables. x = number of cups of rice per day y = number of cups of soybeans per day Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 1C C. Translate the restrictions into linear inequalities. Protein: 15x + 22.5y > 90 Calories: 810x + 270y > 1620 B2: (1/9)x + (1/3)y > 1 Nonnegative: x > 0, y > 0 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 1D D. Form the objective function. Minimize the cost in cents: [Cost] = 21x + 14y Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 2A A. Put the inequalities in standard form. Protein: y > (-2/3)x + 4 Calories: y > -3x + 6 B2: y > (-1/3)x + 3 Nonnegative: x > 0 y > 0 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 2B 4 B. Graph the straight line corresponding to each inequality. 1. y = (-2/3)x + 4 2. y = -3x + 6 3. y = (-1/3)x + 3 4. x = 0 5. y = 0 2 1 3 5 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 2C C. Determine the side of the line. y > (-2/3)x + 4 feasible set Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 3 Determine the vertices of the feasible set. x = 0 & y = -3x + 6: (0,6) y = -3x + 6 & y = (-2/3)x + 4: (6/7,24/7) y = (-2/3)x + 4 & y = (-1/3)x + 3: (3,2) y = (-1/3)x + 3 & y = 0: (9,0) (0,6) (6/7,24/7) (3,2) (9,0) Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Step 4 Determine the objective function at each vertex. Determine the optimal point. Vertex Cost = 21x + 14y (0,6) 21(0) + 14(6) = 84 (6/7,24/7) 21(6/7) + 14(24/7) = 66 (3,2) 21(3) + 14(2) = 91 (9,0) 21(9) + 14(0) = 189 The minimum cost is 66 cents for 6/7 cups of rice and 24/7 cups of soybeans. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Summary Section 3.2 - Part 1 The fundamental theorem of linear programming states that the optimal value of the objective function for a linear programming problem occurs at a vertex of the feasible set. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Summary Section 3.2 - Part 2 To solve a linear programming word problem, assign variables to the unknown quantities, translate the restrictions into a system of linear inequalities involving no more than two variables, form a function for the quantity to be optimized, graph the feasible set, evaluate the objective function at each vertex, and identify the vertex that gives the optimal value. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

3.3 Linear Programming II Number of Variables Transportation Example Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Number of Variables On the surface some problems may appear to have more than two variables. However, sometimes they can be translated into mathematical language so that only two variables are required. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Example Transportation A TV dealer has stores in city A and B and warehouses in cities W and V. The cost of shipping a TV from W to A is $6, from V to A is $3, from W to B is $9 and from V to B is $5. Store in A orders 25 TV sets and store in B orders 30 sets. The W warehouse has a stock of 45 sets and V warehouse has 40. What is the most economical way to supply the two stores the requested TV sets? Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 1A V Stock: 40 B A $9 $5 Needs: 30 Needs: 25 W Stock: 45 $6 $3 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 1B The number of variables can be reduced by observing that what is not shipped from the warehouse in W must be shipped from the warehouse in V. Let x be the number of TVs shipped from the W warehouse to the store in A and y be the number of TVs shipped from W to B. Then 30 - x is going from V to A and 25 - y from V to B. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 1A&B V Stock: 40 30 - x 25 - y B A $9 $5 Needs: 30 Needs: 25 W $6 $3 x y Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 1C Warehouse W: x + y < 45 Warehouse V: Nonnegative restrictions: 0 < x and 30 - x > 0 0 < y and 25 - y > 0 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 1C - Simplified x + y < 45 x + y > 15 x < 30 y < 25 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

[cost] = 3x + 6y + 5(30 - x) + 9(25 - y) Step 1D The cost of transporting the TVs is to be minimized. [cost] = 3x + 6y + 5(30 - x) + 9(25 - y) [cost] = 375 - 2x - 3y Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 2 1. y < 45 - x 2. y > 15 - x 3. x < 30, 0 < x 4. y < 25, 0 < y 4 3 1 2 3 4 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 3 y < 45 - x y > 15 - x x < 30 0 < x y < 25 (0,25) (20,25) (0,15) (30,15) (15,0) (30,0) Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 4 Vertex Cost = 375 - 2x - 3y (0,15) 375 - 2(0) - 3(15) = 330 (0,25) 375 - 2(0) - 3(25) = 300 (20,25) 375 - 2(20) - 3(25) = 260 (30,15) 375 - 2(30) - 3(15) = 270 (30,0) 375 - 2(30) - 3(0) = 315 (15,0) 375 - 2(15) - 3(0) = 345 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Step 4 - Solution Optimum Point (20,25) V Stock: 40 10 B A $9 $5 Needs: 30 Needs: 25 W Stock: 45 10 $5 $9 $3 $6 20 25 Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e

Summary Section 3.3 Sometimes it is necessary to use algebra to reduce the number of variables. Once the number of variables is reduced to two, the steps for solving a linear programming problem are followed. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e