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Quantitative methods for business Chapter 7: Linear Programming Models: Graphical Methods
Learning Objectives After completing this chapter, students will be able to: • Basic concepts and assumptions • Graphically solve any LP problem that has only two variables by both the corner point and isoprofit line methods • Understand special issues in LP such as infeasibility, unboundedness, redundancy, and alternative optimal solutions • Understand the role of sensitivity analysis
Machinery, labor, money, time, warehouse space, raw materials Introduction MANGEMENT DECISION • A widely used mathematical modeling techniquedesigned to help managers in planning and decision making relative to resource allocation. • Applied over the past 50 years. LINEAR PROGRAMMING
Steps in Formulating LP Problems • 1. Completely understand the managerial problem being faced Developing a mathematical modelto represent the managerial problem • 2. Identify the objective and constraints • 3. Define the decision variables • 4. Use the decision variables to write mathematical expressions for the objective function and the constraints
Requirements of LP Problems and LP Basic Assumptions Requirements • Decision variables - mathematical symbols representing levels of activity of a firm. • Objective function - a linear mathematical relationship describing an objective of the firm, in terms of decision variables - this function is to be maximized or minimized. • Constraints – requirements or restrictions placed on the firm by the operating environment, stated in linear relationships of the decision variables. • Parameters - numerical coefficients and constants used in the objective function and constraints.
Requirements of LP Problems and LP Basic Assumptions Assumptions • Proportionality - a change in a variable results in a proportionate change in that variable's contribution to the value of the function. • Non-negativity • Divisibility (Continuity) - the decision variables can be divided into non-integer values, taking on fractional values. • Additivity - the function value is the sum of the contributions of each term. • Certainty; that is, that the coefficients are known and constant.
Example - Flair Furniture Company The Flair Furniture Company produces inexpensive tables and chairs. Processes are similar in that both require a certain amount of hours of carpentry work and in the painting and varnishing department • Each table takes 4 hours of carpentry and 2 hours of painting and varnishing • Each chair requires 3 of carpentry and 1 hour of painting and varnishing • There are 240 hours of carpentry time available and 100 hours of painting and varnishing per week. • Each table yields a profit of $70 and each chair a profit of $50 What is the best combination of chairs and tables per week to maximize profit?
Example: Flair Furniture Company • 1. Completely understand the managerial problem being faced • The objective is to Maximize profit • The constraints are • The hours of carpentry time used cannot exceed 240 hours per week • The hours of painting and varnishing time used cannot exceed 100 hours per week • The decision variables representing the actual decisions we will make are T = number of tables to be produced per week C = number of chairs to be produced per week • 2. Identify the objective and constraints • 3. Define the decision variables • 4. Use the decision variables to write mathematical expressions for the objective function and the constraints
subject to 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) T, C≥ 0 (nonnegativity constraint) Example: Flair Furniture Company • The company wants to determine the best combination of tables and chairs to produce to reach the maximum profit • The complete problem stated mathematically Maximize profit = $70T + $50C
subject to 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) T, C≥ 0 (nonnegativity constraint) Example: Flair Furniture Company • The complete problem stated mathematically Maximize profit = $70T + $50C
C 100 – – 80 – – 60 – – 40 – – 20 – – – Number of Chairs subject to T, C≥ 0 (nonnegativity constraint) 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Graphical Representation of a Constraint This Axis Represents the Constraint T≥ 0 First constraint: T, C≥ 0 This Axis Represents the Constraint C≥ 0
C 100 – – 80 – – 60 – – 40 – – 20 – – – Number of Chairs (30, 40) (70, 40) subject to (30, 20) T, C≥ 0 (nonnegativity constraint) 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Graphical Representation of a Constraint • Any point on or below the constraint plot will not violate the restriction (feasible region) • Any point above the plot will violate the restriction (infeasible region) Second constraint: 4T + 3C <= 240 Graph of carpentry constraint equation
C 100 – – 80 – – 60 – – 40 – – 20 – – – Number of Chairs subject to T, C≥ 0 (nonnegativity constraint) 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Graphical Representation of a Constraint (T = 0, C = 100) Graph of painting and varnishing constraint equation Third constraint: 2T + 1C <= 100 (T = 50, C = 0)
C 100 – – 80 – – 60 – – 40 – – 20 – – – Number of Chairs subject to T, C≥ 0 (nonnegativity constraint) 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Graphical Representation of a Constraint • Feasible solution region for Flair Furniture Painting/Varnishing Constraint Which points lie on the feasible region? (a) T = 30, C=20 (b) T= 70, C=40 (c) T = 50, C=5 Carpentry Constraint Feasible Region
subject to T, C≥ 0 (nonnegativity constraint) 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) Graphical Representation of a Constraint • A feasible solution does not violate any of the constraints: For the point (30, 20) • Aninfeasible solution violates at least one of the constraints: For the point (70, 40)
subject to T, C≥ 0 (nonnegativity constraint) 4T+ 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) Graphical Representation of a Constraint • For the point (50, 5)
C 100 – – 80 – – 60 – – 40 – – 20 – – – Number of Chairs | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Isoprofit Line Solution Method • Isoprofit line at $2,100 $2,100 = $70T + $50C (0, 42) (30, 0)
C 100 – – 80 – – 60 – – 40 – – 20 – – – Number of Chairs | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Isoprofit Line Solution Method • Four isoprofit lines $3,500 = $70T + $50C $2,800 = $70T + $50C $2,100 = $70T + $50C $4,200 = $70T + $50C
C 100 – – 80 – – 60 – – 40 – – 20 – – – Number of Chairs | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Isoprofit Line Solution Method • Optimal solution to the Flair Furniture problem Maximum Profit Line Optimal Solution Point (T = 30, C = 40) $4,200 = $70T + $50C
C 100 – – 80 – – 60 – – 40 – – 20 – – – 1 2 Number of Chairs 4 3 | | | | | | | | | | | | 0 20 40 60 80 100 T Number of Tables Corner Point Solution Method • Four corner points of the feasible region
Corner Point Solution Method To find the coordinates for Point accurately we have to solve for the intersection of the two constraint lines • Using the simultaneous equations method, we multiply the painting equation by –2 and add it to the carpentry equation 4T + 3C = 240 (carpentry line) – 4T – 2C = –200 (painting line) C = 40 • Substituting 40 for C in either of the original equations allows us to determine the value of T 4T + (3)(40) = 240 (carpentry line) 4T + 120 = 240 T = 30
Point : (T = 0, C = 0) Profit = $70(0) + $50(0) = $0 Point : (T = 0, C = 80) Profit = $70(0) + $50(80) = $4,000 Point : (T = 50, C = 0) Profit = $70(50) + $50(0) = $3,500 Point : (T = 30, C = 40) Profit = $70(30) + $50(40) = $4,100 1 2 • Because Point returns the highest profit, this is the optimal solution 3 3 3 4 Corner Point Solution Method
Using QM for Windows • Computer screen for input of data
Using QM for Windows • Computer screen for input of data
Using QM for Windows • Computer screen for output of solution
Slack and surplus The term slackis used for the amount of a resource that is not used. Slack = (Amount of resource available) – (Amount of resource used) Example: If the company decided to produce 20 tables and 25 chairs instead of the optimal solution, the amount of carpentry time used (4T + 3C) would be: 4(20) + 3(25) = 155 Slack time in carpentry = 240 – 155 = 85 The term surplus is used with greater-than-or-equal-to constraints to indicate the amount of by which the right-hand-side of a constraint is exceeded. Surplus = (Actual amount) – (Minimum amount) Example: If there is a constraint requiring the total numbers of chairs and tables combined to be at least 42 units. For the optimal solution (30,40): the surplus would be: 70-42 = 28
Solving Minimization Problems • Minimizing an objective such as cost instead of maximizing a profit function • Minimization problems can be solved graphically by first setting up the feasible solution region and then using • the corner point method • or an isocost line approach
Holiday Meal Turkey Ranch The Holiday Meal Turkey Ranch is considering buying two different brands of turkey feed and blending them to provide a good, low-cost diet for its turkeys. Each feed contains, in varying proportions, some or all of the three nutritional ingredients (A, B, and C) essential for fattening turkeys: at least 90 ounces, 48 ounces, and 1.5 ounces for ingredients A, B, and C respectively. • Each pound of brand 1 purchased, for example, contains 5 ounces of ingredient A, 4 ounces of ingredient B, and 0.5 ounces of ingredient C. • Each pound of brand 2 contains 10 ounces of ingredient A, 3 ounces of ingredient B, but no ingredient C. • The brand 1 feed costs 2 cents a pound, while the brand 2 feed costs 3 cents a pound. What is the lowest-cost combination diet that meets the minimum monthly intake requirement for each nutritional ingredient?
Holiday Meal Turkey Ranch • Holiday Meal Turkey Ranch data
Let X1 = number of pounds of brand 1 feed purchased X2 = number of pounds of brand 2 feed purchased Holiday Meal Turkey Ranch Minimize cost (in cents) = 2X1 + 3X2 subject to: 5X1 + 10X2≥90 ounces (ingredient constraint A) 4X1 + 3X2≥ 48 ounces (ingredient constraint B) 0.5X1≥ 1.5 ounces (ingredient constraint C) X1≥ 0 (nonnegativity constraint) X2≥ 0 (nonnegativity constraint)
– 20 – 15 – 10 – 5 – 0 – X2 Pounds of Brand 2 a b c | | | | | | 5 10 15 20 25 X1 Pounds of Brand 1 Holiday Meal Turkey Ranch • Using the corner point method • First we construct the feasible solution region • The optimal solution will lie at on of the corners as it would in a maximization problem • Point ais the intersection of ingredient constraints C and B 4X1 + 3X2 = 48 X1 = 3 • Substituting 3 in the first equation, we find X2 = 12 • Solving for point bwith basic algebra we find X1 = 8.4 and X2 = 4.8 • Solving for point cwe find X1 = 18 and X2 = 0 Ingredient C Constraint Feasible Region Ingredient B Constraint Ingredient A Constraint
Holiday Meal Turkey Ranch • Substituting these value back into the objective function we find Cost = 2X1 + 3X2 Cost at point a = 2(3) + 3(12) = 42 Cost at point b = 2(8.4) + 3(4.8) = 31.2 Cost at point c = 2(18) + 3(0) = 36 • The lowest cost solution is to purchase 8.4 pounds of brand 1 feed and 4.8 pounds of brand 2 feed for a total cost of 31.2 cents per turkey
– 20 – 15 – 10 – 5 – 0 – X2 Pounds of Brand 2 | | | | | | 5 10 15 20 25 X1 Pounds of Brand 1 Holiday Meal Turkey Ranch • Using the isocost approach • Choosing an initial cost of 54 cents, it is clear improvement is possible Feasible Region 54¢ = 2X1 + 3X2 Isocost Line Direction of Decreasing Cost 31.2¢ = 2X1 + 3X2 (X1 = 8.4, X2 = 4.8)
Holiday Meal Turkey Ranch • QM for Windows can also be used to solve the Holiday Meal Turkey Ranch problem
Holiday Meal Turkey Ranch • Setting up Solver to solve the Holiday Meal Turkey Ranch problem
Holiday Meal Turkey Ranch • Solution to the Holiday Meal Turkey Ranch problem using Solver
X2 8 – – 6 – – 4 – – 2 – – 0 – Region Satisfying Third Constraint | | | | | | | | | | 2 4 6 8 X1 Four Special Cases in LP • No feasible solution (1/4) • Exists when there is no solution to the problem that satisfies all the constraint equations X1 + 2X2 <= 6 2X1 + X2 <= 8 X1 >=7 X1, X2, >=0 Region Satisfying First Two Constraints
X2 15 – 10 – 5 – 0 – | | | | | 5 10 15 X1 Four Special Cases in LP • Unboundedness (2/4) • In a maximization problem, one or more solution variables, and the profit, can be made infinitely large without violating any constraints • In a graphical solution, the feasible region will be open ended • This usually means the problem has been formulated improperly Maximize profit = $3X1 + $5X2 X1 >= 5 X2 <=10 X1 + 2X2 >=10 X1, X2 >= 0 X1≥ 5 X2≤ 10 Feasible Region X1 + 2X2≥ 15
X2 30 – 25 – 20 – 15 – 10 – 5 – 0 – | | | | | | 5 10 15 20 25 30 X1 Four Special Cases in LP • Redundancy (3/4) • A redundant constraint is one that does not affect the feasible solution region • Eliminating redundant constraints simplifies the model. 2X1 + X2≤ 30 Maximize profit = $1X1 + $2X2 X1 + X2 <= 20 2X1 + X2 <=30 X1 <= 25 X1, X2 >= 0 Redundant Constraint X1≤ 25 X1 + X2≤ 20 Feasible Region
X2 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 – 0 – A B | | | | | | | | 1 2 3 4 5 6 7 8 X1 Four Special Cases in LP • Alternate Optimal Solutions (4/4) • Occasionally two or more optimal solutions may exist • Allow management great flexibility in deciding which combination to select as the profit is the same at each alternate solution. Maximize profit = $3X1 + $2X2 6X1 + 4X2 <= 24 X1 <= 3 X1, X2 >= 0 Optimal Solution Consists of All Combinations of X1 and X2 Along the AB Segment Isoprofit Line for $8 Isoprofit Line for $12 Overlays Line Segment AB Feasible Region
Sensitivity Analysis • Sensitivity analysis is used to determine effects on the optimal solution within specified ranges for • The objective function coefficients, • Constraint coefficients, • Right hand side (RHS) values. • Sensitivity analysis provides answers to certain what-if questions. • What if the profit on product 1 increases by 10%? • What if less money available in the advertising budget constraint? • What if new technology will allow a product to be wired in one-third the time it used to take?
X2 (receivers) 60 – – 40 – – 20 – 10 – 0 – Optimal Solution at Point a X1 = 0 CD Players X2 = 20 Receivers Profits = $2,400 | | | | | | 10 20 30 40 50 60 X1 (CD players) Sensitivity Analysis Maximize profit = $50X1 + $120X2 Subject to 2X1 + 4X2≤ 80 (hours of electrician’s time available) 3X1 + 1X2 ≤ 60 (hours of audio technician’s time available) X1, X2 ≥ 0 • The High Note Sound Company graphical solution a = (0, 20) b = (16, 12) Isoprofit Line: $2,400 = 50X1 + 120X2 c = (20, 0)
40 – 30 – 20 – 10 – 0 – X2 | | | | | | 10 20 30 40 50 60 X1 Sensitivity Analysis 1/3 Changes in the Objective Function Coefficient • What if a technical breakthrough help raise the profit per receiver (X2) from $120 to $150? • What if the coefficient for X2 is only $80? Profit per receiver (X2) increased from $120 to $150,the solution at point a (0, 20) is still optimal because the lines still pass through point a. The new profit = 0($50) + 20($150) = $3000 Profit Line for 50X1 + 80X2 (Passes through Point b) Profit Line for 50X1 + 120X2 (Passes through Point a) Optimal point b a Profit Line for 50X1 + 150X2 (Passes through Point a) c If X2’s profit coefficient is now only $80, the slope of the profit line changes enough to cause a new corner point b (16,12) to become optimal. The profit = 16($50) + 12($80) = $1760
X2 X2 X2 60 – 40 – 20 – – 60 – 40 – 20 – – 60 – 40 – 20 – – 2 X1 + 1X2≤ 60 3X1 + 1X2≤ 60 3X1 + 1X2≤ 60 Stereo Receivers Optimal Solution Still Optimal Optimal Solution a a d f b 16 g 2X1 + 5 X2≤ 80 2X1 + 4X2≤ 80 2X1 + 4X2≤ 80 c e c | | | 0 20 40 | | | 0 20 40 | | | 0 20 40 | 30 X1 X1 X1 CD Players 2/3- Changes in the Technological Coefficients • Reflect changes in the state of technology If fewer or more resources are needed to produce a product such as a CD player or receiver, coefficients in the constraint equations will change. • These changes will have no effect on the objective function, but it can produce a significant change in the shape of the feasible region may cause a change in the optimal solution (a) Original Problem (b) Change in Circled Coefficient (c) Change in Circled Coefficient Fewer hrs of technicians for a CD More hrs of electricians for a receiver
3/3 Changes in Resources or Right-Hand-Side Values • The right-hand-side values of the constraints often represent resources available to the firm • Reflect how much should be paid for additional resources and how much more of a resource would be useful. • If the right-hand side of a constraint is changed, the feasible region will change (unless the constraint is redundant) Often the optimal solution will change • The amount of change in the objective function value that results from a unit change in one of the resources available is called the dual price or dual value • The dual price is relevant only within limits.
