Chapter 4: Linear Programming Lesson Plan Mixture Problems  Combining Resources to Maximize Profit Finding the Optimal Production Policy Why the Corner Point Principle Works  Decreasing-Time-List Algorithm Linear Programming  Life Is Complicated A Transportation Problem  Delivering Perishables Improving on the Current Solution 1 Mathematical Literacy in Today’s World, 8th ed. For All Practical Purposes © 2009, W.H. Freeman and Company

Chapter 4: Linear Programming Mixture Problems 2 Mixture Problem  Limited resources are combined into products in such a way that the profit from selling those products is a maximum. Linear Programming  A management science technique that helps a business allocate the resources it has on hand to make a particular mix of products that will maximize profit.  One of the most frequently used management science techniques.

Chapter 4: Linear Programming Mixture Problems 3 Production Policy  A solution to a linear-programming mixture problem is a production policy that tells us how many units of each product to make.  Optimal Production Policy Has Two Properties First, it is possible; that is, it does not violate any of the limitations under which the manufacturer operates, such as availability of resources. Second, the optimal production policy gives the maximum profit. Common Features of Mixture Problems  Resources – Available in limited, known quantities for time period.  Products – Made by combining, or mixing, the resources.  Recipes – How many units of each resource are needed.  Profits – Each product earns a known profit per unit.  Objectives – To find how much of each product to make to maximize profit without exceeding any of the resource limitations.

4 Chapter 4: Linear Programming Mixture Problems Mixture Problem: Making Skateboards and Dolls  Skateboards require five units of plastic and are sold for $1 profit.  Dolls require two units of plastic and are sold for $0.55 profit.  If 60 units of plastic are available, what numbers of skateboards and/or dolls should be manufactured to maximize the profits? Step 1  Mixture Chart – display the verbal information into a chart that includes the unknown variables (x units of Skateboards, and y units of dolls). Step 2  Translate the mixture chart into mathematical form from the chart. 5x + 2y ≤ 60 (plastic) P = 1x y (profit)

5 Chapter 4: Linear Programming Mixture Problems Feasibility Set or Feasibility Region  Our goal is to find the best mixture of x and y (skateboards and/or dolls) to produce the largest profit — two phases: 1.Find the feasible set for the mixture problem subject to limited resources. Graph line below 5x + 2y  60 (plastic) 2.Determine the mixture that gives rise to the largest profit (next slide). Feasibility set (feasibility region) - A collection of all physically possible solutions, or choices, that can be made. Shade in the feasible region is where all equations are true: 5x + 2y  60, and where x ≥ 0, y ≥ 0 Graph of 5x + 2y = 60

6 Chapter 4: Linear Programming Finding the Optimal Production Policy Feasibility Set or Feasibility Region  Next step is to find the optimal production policy, a point within that region that gives a maximum profit. 1.Find the corner points of the feasible region. 2.Evaluate the profit at each corner point. 3.Choose the corner point with the highest profit as the production policy. Corner point principle – States that in a linear programming problem, the maximum value for the profit formula always corresponds to a corner point of the feasible region. Optimal production policy – Corresponds to a corner point of the feasible region where the profit formula has a maximum value. Optimal production policy would be the point (0,30), which gives the maximum profit of $ (0,0)$1.00(0)+ $0.55(0)= $0.00+ $0.00= $0.00 (0,30)$1.00(0)+$0.55(30)= $0.00+ $16.50= $16.50 (12,0)$1.00(12)+ $0.55(0)= $ $0.00= $12.00 Calculation of the Profit Formula for Skateboards and Dolls Corner Point Value of the Profit Formula: $1.00 x + $0.55 y

7 Chapter 4: Linear Programming Finding the Optimal Production Policy Example: Mixture of Two Fruit Juices  Using the data from the mixture chart (two products, two resources): Determine the profit and constraint equations. Graph the equations and find the feasibility region. Maximize profit formula: 3x + 4y Constraints: Cranberry: 3x + 2y  200 Apple: 1x + 2y  100 Minimums: x ≥ 0 and y ≥ 0 Feasible region

8 Chapter 4: Linear Programming Why the Corner Point Principle Works Example Continued: Mixture of Two Fruit Juices  Using the corner point principle, the highest profit value on a polygonal feasible region is always at a corner point.  Evaluate the profit formula at these corner points: Profit Line: If a profit line corresponding to a certain profit does not touch the feasible region, that profit is not possible. Example 3x + 4y = 360 does not work. However, a profit = 160 is feasible, but we can do better. The line shifts from 160 to 360 until it reaches a corner point (50,25) for maximum profit. The profit line for 360 lies outside the feasible region Finding the Optimal Production Policy for Beverages Corner Point Value of the Profit Formula: 3 x + 4 y cents (0, 0)3(0) +4(0)= 0 cents (0, 50)3(0) +4(50)= 200 cents (50, 25)3(50) +4(25)= 250 cents (66.7, 0)3(66.7) +4(0)= 200 cents (rounded) Optimal production policy is (50, 25) with max profit = 250 cents.

9 Chapter 4: Linear Programming Linear Programming: Life is Complicated Two Ways the Feasible Region Can Be More Complex 1.It becomes difficult when there are many corners to calculate. 2.It may be difficult to visualize the feasible region as part of two- dimensional space when there are more than two products.  The Simplex Method The oldest and most common method to solve typically large linear programming problems (devised by George Dantzig, ~1940). How it works: We start at any corner and evaluate the neighboring corners to see which is closer to optimal (like the temperature hints “getting warmer”), then move to the warmer corner and re-evaluate the new neighbors.  Alternative to the Simplex Method This method finds the optimal corner points in fewer steps than the simplex algorithm by making use of search routes through the interior of the feasible region (by Narendra Karmarkar, 1984).

10 Chapter 4: Linear Programming A Transportation Problem: Delivering Perishables Transportation Problem: Delivering Bread  A supermarket chain gets bread deliveries from a bakery chain that does its baking in different places. Each supermarket store needs a certain number of loaves/day. The supplier bakes enough bread to exactly meet the demands. How many loaves of bread should be shipped for each locale to each of the stores to stay within the demands and to minimize the cost? Delivering Bread Example  Three stores require 3 dozen, 7 dozen, and 1 dozen loaves of bread, respectively.  Three bakeries can supply 8 dozen, 1 dozen, and 2 dozen loaves, respectively.  Graph shows supply-and- demand transportation problem.

11 A representation of a specific problem involving meeting the demands of the three stores for breads from the supplies available at three bakeries. Chapter 4: Linear Programming A Transportation Problem: Delivering Perishables Tableau  Table showing costs and rim conditions for transportation problem. The rim conditions:  The number of breads available (supply) on the outside right.  The number of breads required (demand) on the bottom. Each table entry is known as a cell (denoted by row and column). Aggregate costs (the cost of shipping and time consideration, traffic etc.) are in the upper right corner of each cell. Matrix representing the costs and conditions of a bread problem

12 Chapter 4: Linear Programming A Transportation Problem: Delivering Perishables Tableau to Solve the Linear-Programming Problem 1.Find a solution that is feasible (does not violate any constraints). 2.If the current solution is not optimal, move to a better, improved solution (until you have found an optimal solution). Total cost of this “shipment plan”: 2(8) + 6(9) + 1(15) = = 93 A possible solution for the bread transportation problem  Circled number, such as 6 in row I, column 2, means we plan to ship six breads to store 2 from bakery I.  These circled numbers must add up to the rim conditions across the rows and also down the column.

13 Chapter 4: Linear Programming A Transportation Problem: Delivering Perishables Northwest Corner Rule  An easier approach that is a simple rule since it is based on the geometry of the table and does not look at the costs (initially). Northwest Corner Rule Algorithm 1.Locate the top far-left-hand cell and ship this cell along with the costs. Ship via this cell with the smaller of the two rim cells (call the value s) and put a circle around the entry in the tableau. 2.Cross out the row or column that had rim value s and reduce the other rim value for this cell by s. 3.When a single cell remains, there will be a tie for the rim conditions of both the row and column involved, and this amount is entered into the cell and circled.

14 Chapter 4: Linear Programming A Transportation Problem: Delivering Perishables Example Using the Northwest Corner Rule:  The diagrams show the construction of an initial solution to the bread transportation problem.  To get the total cost of this “shipment plan” we sum all the shipped amounts by the associated costs to get: 3(8) + 5(9) + 1(1) + 1(3) + 1(5) = 78 This is a better solution. However, it may be unlikely that it is the cheapest solution since it did not make use of the costs on the cells.

15 Chapter 4: Linear Programming Improving on the Current Solution Stepping Stone Method  Consists of improving an existing feasible solution of a transportation problem, if it is not optimal, by shipping an additional amount using a cell with a negative indicator value. Improves the solution to the Northwest Corner Rule (if not optimal). Compute the indicator values of the cells to improve the current solution by shipping using a cell with a negative indicator value. Continue to find better solutions (cheaper ways of meeting the demands from the supplies) by using a cell with a negative indicator value (optimal solution when cells have positive value, or zero). Indicator value = − 1; minimum circled number with negative label is 2; optimal solution cost = 68 Indicator value + 3 − − 5 = − 8; solution cost = 70