Chapter 3 Introduction to Linear Programming PART 1 Assoc. Prof. Dr. Arslan M. Örnek http://homes.ieu.edu.tr/~aornek/ise203%20optimization%20i.htm 1
3.1 What Is a Linear Programming Problem? Linear Programming (LP) is a tool for solving optimization problems. 2
Example 1: Giapetto s Woodcarving Giapetto s, Inc., manufactures wooden soldiers and trains. Each soldier built: Sells for $27 and uses $10 worth of raw materials. Increases Giapetto s variable labor/overhead costs by $14. Requires 2 hours of finishing labor. Requires 1 hour of carpentry labor. Each train built: Sells for $21 and uses $9 worth of raw materials. Increases Giapetto s variable labor/overhead costs by $10. Requires 1 hour of finishing labor. Requires 1 hour of carpentry labor. 3
Ex. 1 - continued Each week Giapetto can obtain: All needed raw material. Only 100 finishing hours. Only 80 carpentry hours. Demand for the trains is unlimited. At most 40 soldiers are bought each week. Giapetto wants to maximize weekly profit (revenues costs). Formulate a mathematical model of Giapetto s situation that can be used maximize weekly profit. 4
Ex. 1 - continued Profit = Revenue cost Revenue = Price (TL/unit) * Quantity Produced (unit) (= TL) Cost = [Variable cost (TL/unit) + Overhead cost (TL/unit)] * Quantity Produced (unit) (=TL) 5
Example 1: Solution The Giapetto solution model incorporates the characteristics shared by all linear programming problems. Decision variables should completely describe the decisions to be made. x 1 = number of soldiers produced each week x 2 = number of trains produced each week The decision maker wants to maximize (usually revenue or profit) or minimize (usually costs) some function of the decision variables. This function to be maximized or minimized is called the objective function. 6
Ex. 1 - Solution continued Giapetto s weekly profit can be expressed in terms of the decision variables x 1 and x 2 : Weekly profit = weekly revenue (weekly raw material costs + weekly variable costs) Weekly revenue = 27x 1 + 21x 2 Weekly raw material costs = 10x 1 + 9x 2 Weekly variable costs = 14x 1 + 10x 2 Weekly profit = 3x 1 + 2x 2 7
Ex. 1 - Solution continued Thus, Giapetto s objective is to chose x 1 and x 2 to maximize weekly profit. The variable z denotes the objective function value of any LP. Giapetto s objective function is Maximize Z = 3x 1 + 2x 2 Maximize = Max, Minimize = Min The coefficient of an objective function variable is called an objective function coefficient. 8
Ex. 1 - Solution continued As x 1 and x 2 increase, Giapetto s objective function grows larger. For Giapetto, the values of x 1 and x 2 are limited by the following three restrictions (constraints): Each week, no more than 100 hours of finishing time may be used. Each week, no more than 80 hours of carpentry time may be used. Because of limited demand, at most 40 soldiers should be produced. 9
Back to the parameters Each soldier built x 1 : Requires 2 hours of finishing labor. Requires 1 hour of carpentry labor. Each train built x 2 : Requires 1 hour of finishing labor. Requires 1 hour of carpentry labor. 10
Ex. 1 - Solution continued Constraints: Each week, no more than 100 hours of finishing time may be used: 2 x 1 + x 2 100 Each week, no more than 80 hours of carpentry time may be used: x 1 + x 2 80 Because of limited demand, at most 40 soldiers should be produced: x 1 40 11
The coefficients of the decision variables in the constraints are called the technological coefficients. The number on the right-hand side of each constraint is called the constraint s right-hand side (or rhs). To complete the formulation of a linear programming problem, the following question must be answered for each decision variable. Can the decision variable only assume nonnegative values, or is the decision variable allowed to assume both positive and negative values? If the decision variable can assume only nonnegative values, the sign restriction x i 0 is added. If the variable can assume both positive and negative values, the decision variable x i is unrestricted in sign (often abbreviated urs). 12
Ex. 1 - Solution continued For the Giapetto problem model, combining the sign restrictions x 1 0 and x 2 0 with the objective function and constraints yields the following optimization model: 13
Definitions A function f(x 1, x 2,, x n ) of x 1, x 2,, x n is a linear function if and only if for some set of constants, c 1, c 2,, c n, f(x 1, x 2,, x n ) = c 1 x 1 + c 2 x 2 + + c n x n. Ex. f(x 1, x 2 ) = 2x 1 + x 2 For any linear function f(x 1, x 2,, x n ) and any number b, the inequalities f(x 1, x 2,, x n ) b and f(x 1, x 2,, x n ) b are linear inequalities. 14
Definitions A linear programming problem (LP) is an optimization problem for which we do the following: Attempt to maximize (or minimize) a linear function (called the objective function) of the decision variables. The values of the decision variables must satisfy a set of constraints. Each constraint must be a linear equation or inequality. A sign restriction is associated with each variable. For any variable x i, the sign restriction specifies either that x i must be nonnegative (x i 0) or that x i may be unrestricted in sign. 15
Proportionality and Additivity Assumptions The fact that the objective function for an LP must be a linear function of the decision variables has two implications: 1. The contribution of the objective function from each decision variable is proportional to the value of the decision variable. For example, the contribution to the objective function for 4 soldiers is exactly fours times the contribution of 1 soldier. 2. The contribution to the objective function for any variable is independent of the other decision variables. For example, no matter what the value of x 2, the manufacture of x 1 soldiers will always contribute 3x 1 dollars to the objective function. 16
Proportionality and Additivity Assumptions Analogously, the fact that each LP constraint must be a linear inequality or linear equation has two implications: 1. The contribution of each variable to the left-hand side of each constraint is proportional to the value of the variable. For example, it takes exactly 3 times as many finishing hours to manufacture 3 soldiers as it does 1 soldier. 2. The contribution of a variable to the left-hand side of each constraint is independent of the values of the variable. For example, no matter what the value of x 1, the manufacture of x 2 trains uses x 2 finishing hours and x 2 carpentry hours 17
Four Assumptions of an LP The first item in each list is called the Proportionality Assumption of Linear Programming. The second item in each list is called the Additivity Assumption of Linear Programming. The divisibility assumption requires that each decision variable be permitted to assume fractional values. The certainty assumption is that each parameter (objective function coefficients, right-hand side, and technological coefficients) are known with certainty. 18
Definitions The feasible region of an LP is the set of all points satisfying all the LP s constraints and sign restrictions. For a maximization problem, an optimal solution to an LP is a point in the feasible region with the largest objective function value. Similarly, for a minimization problem, an optimal solution is a point in the feasible region with the smallest objective function value. 19
Feasible Region and Optimal Solution x 1 = 40 and x 2 = 20 are in the feasible region since they satisfy all the Giapetto constraints. On the other hand, x 1 = 15, x 2 = 70 is not in the feasible region because this point does not satisfy the carpentry constraint [15 + 70 is > 80]. Giapetto Constraints 2 x 1 + x 2 100 (finishing constraint) x 1 + x 2 80 (carpentry constraint) x 1 x 1 40 (demand constraint) 0 (sign restriction) x 2 0 (sign restriction) 20
Feasible Region and Optimal Solution Most LPs have only one optimal solution. However, some LPs have no optimal solution, and some LPs have an infinite number of solutions. Section 3.2 shows that the optimal solution to the Giapetto LP is x 1 = 20 and x 2 = 60. This solution yields an objective function value of: z = 3x 1 + 2x 2 = 3(20) + 2(60) = $180 When we say x 1 = 20 and x 2 = 60 is the optimal solution, we are saying that no point in the feasible region has an objective function value (profit) exceeding 180. 21
3.2 The Graphical Solution to a Two-Variable LP Problem X2 Any LP with only two variables can be solved graphically. The variables are always labeled x 1 and x 2 and the coordinate axes the x 1 and x 2 axes. -1 4 3 2 1 2x 1 + 3x 2 6 1 2 3 4 X1-1 22
Since the Giapetto LP has two variables, it may be solved graphically. The feasible region is the set of all points satisfying the constraints 2 x 1 + x 2 100 (finishing constraint) x 1 + x 2 80 (carpentry constraint) x 1 40 x 1 0 x 2 0 (demand constraint) (sign restriction) (sign restriction) 23
100 80 60 40 20 The set of points satisfying the Giapetto LP is bounded by the five sided polygon DGFEH. Any point on or in the interior of this polygon (the shade area) is in the feasible region. X2 B finishing constraint Feasible Region D demand constraint G z = 100 carpentry constraint F z = 180 z = 60 E A C H 10 20 40 50 60 80 X1 24
Having identified the feasible region for the Giapetto LP, a search can begin for the optimal solution which will be the point in the feasible region with the largest z-value. To find the optimal solution, graph a line on which the points have the same z-value. In a max problem, such a line is called an isoprofit line while in a min problem, this is called the isocost line. 25
100 80 60 40 20 X2 B finishing constraint Feasible Region D demand constraint z = 3x 1 + 2x 2 The figure shows the isoprofit lines for z = 60, z = 100, and z = 180 H G z = 100 carpentry constraint F z = 180 z = 60 E A C 10 20 40 50 60 80 X1 26
100 80 60 40 20 The last isoprofit intersecting (touching) the feasible region indicates the optimal solution for the LP. For the Giapetto problem, this occurs at point G (x 1 = 20, x 2 = 60, z = 180). X2 B finishing constraint Feasible Region D demand constraint G z = 100 carpentry constraint F z = 180 z = 60 E A C H 10 20 40 50 60 80 X1 27
Binding and Nonbinding Constraints A constraint is binding if the left-hand and right-hand side of the constraint are equal when the optimal values of the decision variables are substituted into the constraint. In the Giapetto LP, the finishing and carpentry constraints are binding: (x 1 = 20, x 2 = 60, z = 180) 2 x 1 + x 2 100 (finishing constraint) 2*20 + 60 = 100 x 1 + x 2 80 (carpentry constraint) 20 + 60 = 80 28
Binding and Nonbinding Constraints A constraint is nonbinding if the left-hand side and the right-hand side of the constraint are unequal when the optimal values of the decision variables are substituted into the constraint. In the Giapetto LP, the demand constraint for wooden soldiers is nonbinding: (x 1 = 20, x 2 = 60, z = 180) x 1 40 20 < 40. (demand constraint) 29
Convex Sets, Extreme Points, and LP A set of points S is a convex set if the line segment joining any two pairs of points in S is wholly contained in S. For any convex set S, a point P in S is an extreme point if each line segment that lies completely in S and contains the point P as an endpoint of the line segment. Extreme points are sometimes called corner points, because if the set S is a polygon, the extreme points will be the vertices, or corners, of the polygon. The feasible region for the Giapetto LP is a convex set. 30
100 80 60 40 20 The feasible region for the Giapetto LP is a convex set. X2 B CONVEX finishing constraint Feasible Region D demand constraint G z = 100 carpentry constraint NON-CONVEX H F z = 180 z = 60 E A C 10 20 40 50 60 80 X1 31
Are these sets convex or non-convex? A E B A B A B C D (a) (b) (c) (d) 32
Convex: a and b Non-convex: c and d A E B A B A B C D (a) (b) (c) (d) 33
What are the extreme points of the convex sets below? A E B A B A B C D (a) (b) (c) (d) 34
In figure (a), each point on the circumference of the circle is an extreme point of the circle. In figure (b), A, B, C, and D are extreme points of S. Point E is not an extreme point since E is not an end point of the line segment AB. A E B A B A B C D (a) (b) (c) (d) 35
It can be shown that: The feasible region for any LP will be a convex set. The feasible region for any LP has only a finite number of extreme points. Any LP that has an optimal solution has an extreme point that is optimal. 36
100 80 60 40 20 X2 B finishing constraint Feasible Region D For the Giapetto problem, the optimal solution (Point G) is an extreme point of the feasible region. G z = 100 demand constraint carpentry constraint H F z = 180 z = 60 E A C 10 20 40 50 60 80 X1 37
Example 2: Dorian Auto Dorian Auto manufactures luxury cars and trucks. The company believes that its most likely customers are high-income women and men. To reach these groups, Dorian Auto has started an ambitious TV advertising campaign and will purchase 1-minute commercial spots on two type of programs: comedy shows and football games. 38
Each comedy commercial is seen by 7 million highincome women and 2 million high-income men and costs $50,000. Each football game is seen by 2 million high-income women and 12 million high-income men and costs $100,000. Dorian Auto would like for commercials to be seen by at least 28 million high-income women and 24 million highincome men. Use LP to determine how Dorian Auto can meet its advertising requirements at minimum cost. 39
Example 2: Solution Dorian must decide how many comedy and football ads should be purchased, so the decision variables are x 1 = number of 1-minute comedy ads x 2 = number of 1-minute football ads Dorian wants to minimize total advertising cost. Dorian s objective functions is min z = 50 x 1 + 100x 2 Dorian faces the following the constraints Commercials must reach at least 28 million high-income women. (7x 1 + 2x 2 28) Commercials must reach at least 24 million high-income men. (2x 1 + 12x 2 24) The sign restrictions are necessary, so x 1, x 2 0. 40
Solution continued X2 14 B 12 High-income women constraint To solve this LP graphically begin by graphing the feasible region. 10 8 6 4 2 D z = 320 A E Feasible Region (unbounded) z = 600 High-income men constraint 2 4 6 8 10 12 14 C X1 41
Solution continued Like the Giapetto LP, The Dorian LP has a convex feasible region. The feasible region for the Dorian problem, however, contains points for which the value of at least one variable can assume arbitrarily large values. Such a feasible region is called an unbounded feasible region. 42
Solution continued Since Dorian wants to minimize total advertising costs, the optimal solution to the problem is the point in the feasible region with the smallest z value. An isocost line with the smallest z value passes through point E and is the optimal solution at x 1 = 3.6 and x 2 = 1.4. Both the high-income women and high-income men constraints are satisfied, both constraints are binding. 43
Does the Dorian problem meet the four assumptions of linear programming? The Proportionality Assumption is violated because at a certain point advertising will yield diminishing returns. Even though the Additivity Assumption was used in writing: (Total viewers) = (Comedy viewer ads) + (Football ad viewers) many of the same people might view both ads, double-counting of such people would occur thereby violating the assumption. The Divisibility Assumption is violated if only 1-minute commercials are available. Dorian is unable to purchase 3.6 comedy and 1.4 football commercials. The Certainty assumption is also violated because there is no way to know with certainty how many viewers are added by each type of commercial. Despite these violations, analysts are using similar models to help companies determine their optimal media mix. Why? 44
3.3 Special Cases The Giapetto and Dorian LPs each had a unique optimal solution. Some types of LPs do not have unique solutions. Some LPs have an infinite number of solutions (alternative or multiple optimal solutions). Some LPs have no feasible solutions (infeasible LPs). Some LPs are unbounded: There are points in the feasible region with arbitrarily large (in a max problem) z-values. 45
60 50 40 30 20 10 Some LPs have multiple optima. Consider the following formulation: X2 B D Feasible Region max z = 3x 1 + 2x 2 s.t. 1 40 x 1 1 60 x 2 1 E 1 50 x 1 1 50 x 2 1 z = 100 z = 120 x 1 x 2 0 z = 60 Any point (solution) falling on line segment AE will yield an optimal solution of z =120. F 10 20 30 40 A C 50 X1 46
It is possible for an LP s feasible region to be empty, resulting in an infeasible LP. Because the optimal solution to an LP is the best point in the feasible region, an infeasible LP has no optimal solution. 47
60 50 40 30 20 10 The following formulation is infeasible: X2 No Feasible Region x 1 30 x1 >= 0 max z = 3x 1 + 2x 2 s.t. 1 40 x 1 1 50 x 1 1 60 x 2 1 1 50 x 2 1 xx2 2 30 >=0 x 1 30 x 2 20 30 x 1 x 2 0 No feasible region exists 10 20 30 40 50 X1 48
Some LPs are unbounded. For a max problem, an unbounded LP occurs if it is possible to find points in the feasible region with arbitrarily large z-values. This corresponds to arbitrarily large profits or revenue. For a minimization problem, an LP is unbounded if there are points in the feasible region producing arbitrarily small z-values. Consider the formulation shown to the right and the graph on the next slide. max z = 2x 1 x 2 s.t. x 1 x 2 1 2x 1 +x 2 6 x 1, x 2 0 49
The constraints are satisfied by all points bounded by the x 2 axis and on or above AB and CD. The isoprofit lines for z = 4 and z = 6 are shown. Any isoprofit line drawn will intersect the feasible region because the isoprofit line is steeper than the line x 1 x 2 = 1. Thus there are points in the feasible region which will produce arbitrarily large z- values (unbounded LP). X2 6 5 4 3 2 1 D Feasible Region z = 4 B z = 6 A C 1 2 3 4 5 6 X1 50
Hence, every LP with two variables must fall into one of the following four cases. The LP has a unique optimal solution. The LP has alternative or multiple optimal solutions: Two or more extreme points are optimal, and the LP will have an infinite number of optimal solutions. The LP is infeasible: The feasible region contains no points. The LP unbounded: There are points in the feasible region with arbitrarily large z-values (max problem) or arbitrarily small z-values (min problem). 51
Another Example 52
The Data Gathered 53
Definition of the Problem Determine what the production rates should be for the two products in order to maximize their total profit, subject to the restrictions imposed by the limited production capacities available in the three plants. 54
Data for the Problem 55
Formulation as a Linear Programming Decision Variables: Model x 1 = number of batches of product 1 produced per week x 2 = number of batches of product 2 produced per week Objective Function: Maximize Z = total profit per week (in thousands) from producing these two products 56
LP Formulation 57
Graphical Solution Since the LP has only two decision variables (two dimensions), the problem can be solved graphically. Let s identify the values of x 1 and x 2 that are permitted by the constraints. Constraint 1: x 1 4 58
Constraint 1: The Feasible Region x 1 4 Constraint 2: 2x 2 12 Constraint 3: 3x 1 + 2x 2 = 18 Constraints 4 and 5: x 1 0, x 2 0 59
Maximizing the value of Z Slope of the objective function line = -3/5 60
Optimal Solution 61
The Optimal solution is x 1 = 2, x 2 = 6 with Z = 36. This solution indicates that the Wyndor Glass Co. should produce products 1 and 2 at the rate of 2 batches per week and and 6 batches per week, respectively, with a resulting total profit of $36,000 per week. 62
3.4. A Diet Problem (Stigler, 1939) My diet requires that all the food I get come from one of the four basic food groups. At present, the following four foods are available for consumption: brownies, chocolate ice cream, cola and pineapple cheesecake. Each brownie costs 50, each scoop of ice cream costs 20, each bottle of cola costs 30, and each piece of pineapple cheesecake costs 80. Each day, I must ingest at least 500 calories, 6 oz of chocolate, 10 oz of sugar, and 8 oz of fat. The nutritional content per unit of each food is given. Formulate a linear programming model that can be used to satisfy my daily nutritional requirements at minimum cost. 63
64
Solution As always, begin by determining the decisions that must be made by the decision maker: how much of each type of food should be eaten daily. Thus we define the decision variables: x 1 = number of brownies eaten daily x 2 = number of scoops of chocolate ice cream eaten daily x 3 = bottles of cola drunk daily x 4 = pieces of pineapple cheesecake eaten daily 65
Solution continued My objective is to minimize the cost of my diet. The total cost of any diet may be determined from the following relation: (total cost of diet) = (cost of brownies) + (cost of ice cream) +(cost of cola) + (cost of cheesecake). 66
67
68
69
Solution continued Can you find the optimal solution graphically? A version of the diet problem with a more realistic list of foods and nutritional requirements was one of the first LPs to be solved by computer. 70