Introduction to Linear Programming and Graphical Method

Transcription

1 Course: COMMERCE (CBCS) Subject: Business Mathematics Lesson: Introduction to Linear Programming and Graphical Method Authors Name: Dr. Gurmeet Kaur, Associate Professor, Daulat Ram College,University of Delhi University of Delhi 1

2 Lesson: Introduction to Linear Programming and Graphical Method Table of Contents: 1. Learning Outcomes 2. Introduction to Linear Programming 2.1 Concept of Linear Programming 2.2 Assumptions of Linear Programming Model 2.3 Steps for Solving a Linear Programming Problem 3. Formulation of Linear Programming Problem 3.1 Steps in the Formulation of LPP 3.2 Standard Form of LPP 3.3 General Form of LPP 4. Graphical Method for Solving LPP 5. Exceptional Cases in the Graphical Solution of LPP 5.1 Multiple Optimal Solutions 5.2 Unbounded Solution 5.3 Infeasible Solution 5.4 Redundancy Summary Glossary Exercises Suggested Readings 1. Learning Outcomes: After you have read this chapter, you should be able to: understand the concept of linear programming, formulate a linear programming problem, appreciate the steps for solving such problems, use graphical method to solve linear programming problems, identify exceptional cases in linear programming problems. University of Delhi 2

3 2. Introduction to Linear Programming: 2.1 Concept of Linear Programming Linear means a proportionate relationship of two or more variables in the model. Linear Programming is a mathematical technique used for optimum allocation of limited or scarce resources by choosing the best alternative from a set of feasible solutions in a situation in which objective function as well as constraints can be expressed as linear mathematical relationships. In a linear programming model, the objective function is set up by a simple addition of the contribution by each of activities. Similarly the inequalities, which represent the constraints, are formed as the aggregate of each activity conducted separately. Value Addition 1: Did you know? History of Linear Programming George Dantzig is the founder of Linear Programming. Visit the link to know detailed history of linear programming. Source: Assumptions of Linear Programming Model a. Proportionality One of the basic assumptions of linear programming models is existence of proportionality in both the objective function and in the inequalities. It indicates that there is a constant rate of change in a variable. Thus in a linear objective function the contribution from each decision variable in the objective function is proportional to the value of the decision variable and the contribution for any variable is independent of the other decision variables. In economics terms, proportionality implies constant returns to scale, without reaping any economies of scale. b. Additivity The assumption of additivity refers to the ability of one variable representing an activity being added to another variable representing another activity. For instance, the total amount of a resources consumed will be equal to the sum of the resource values used by individual activities. Thus, each activity is independent of any other activity in a situation. Implication of Proportionality and Additivity Proportionality and additivity taken together entail that every constraint in a linear programming problem must be either a linear inequality or linear equation. c. Continuity (or Divisibility) Another assumption of LPP is that the decision variables should be divisible and continuous. This assumption ensures that the decision variables can acquire fractional values. For example, in production problems the outputs may turn out to be fractional values such as 25 3 units of a product per day. 4 University of Delhi 3

4 d. Certainty It is assumed that the objective function coefficients, the coefficients of the inequality (or equality) and the constraint values of a linear programming model are known with certainty, thus making linear programming deterministic rather than probabilistic in character. e. Finite choices A decision maker has a finite number of choices in the case of a linear programming problem and the decision variables cannot have negative values. 2.3 Steps for Solving a Linear Programming Problem 1. Formulate the Problem: (i) Identify the decision variables whose values have to be determined. (ii) Formulate the objective function in the form of a linear function of decision variables. (iii) State all the limitations on the resources as linear equations or in equations in terms of the identified decision variables. (iv) Add non-negativity restrictions. 2. Find Basic Solution: It is the set of all values of the decision variables that satisfy the set of constraints. It can be found using the Graphical method or the Simplex method. 3. Identify the Feasible Solution: Feasible solution values are identified out of the basic solution values that satisfy the non-negativity restrictions. 4. Ascertain the Optimal Solution: The feasible solution that optimizes the objective function is identified as the optimal solution. A Linear Programming Problem may have a unique optimal solution, or as an exception it may have multiple optimal solutions or unbounded solution or no solution at all. Figure 1: Steps for Solving the Linear Programming Problem Formulation of LPP Finding the solution by using any of the two methods: Graphical Method Simplex Method Identifying the Feasible Solution Ascertaining the Optimal Solution University of Delhi 4

5 3. Formulation of Linear Programming Problem: Formulation of linear programming problem refers to the conversion of a given real world situation into a mathematical format consisting of equations and inequalities that represent the objective function and the constraints. 3.1 Steps in the Formulation of LPP: a. Identification of the Decision Variables: Decision variables are the quantities that have to be calculated. They are generally represented as x 1, x 2, x 3,... or x, y, z,.. Example 1: Formulation of a Linear Programming Problem Production Problem: A manufacturer produces two products A and B. The resource requirement for manufacturing one unit of product A is 6.0 kg of raw material and 4 labour hours. Similarly, for manufacturing one unit of product B 4.5 kg of raw material and 6 labour hours are required. The manufacturer has 1000 kg of raw material and 500 labour hours available. One unit of product A gives a profit of 650 and that product B 400. Formulate the given situation as a Linear Programming Problem to determine the number of units of the products to be manufactured to maximize profits. Identification of the decision variables The given production problem we have to determine the number of units of the products that should be manufactured. Therefore the decision variables would be the number of units of products A and B to be produced by the manufacturer and they would be represented as x and y respectively. b. Formulate Objective Function: Objective Function The objective function specifies the contribution of each of the decision variable to the value to be optimized (maximised or minimised) in the linear programming problem. General Form of Objective Function of LPP Maximize or Minimize Z = where n i=0 c i x i c i = coefficient of contribution of the i th variable to the decision variable x = i th decision variable c i or the coefficients of the decision variables in an objective function indicate the contribution to the value of the objective function of one unit of the corresponding variable. For example, if the objective is to maximize the sales revenue of a company, and x i is the i th product offered for sale, then c i is the sales revenue generated by product i. University of Delhi 5

6 In the Example 1, the objective is to maximize profits; the profit is given as 650 per unit of product A and 400 per unit of product B. Thus the objective can be stated as Maximize Z = 650x + 400y which is the total profit generated by both the products. c. Identify and Formulate the Constraints: The constraints denote the limited or scarce resources that have to be allocated. They are identified and expressed as inequalities or equations written in the terms of the identified decision variables. In the Example 1, the manufacturer has 1000 kg of raw material available and each unit of product A and B require 6 kg and 4.5 kg of raw material respectively. Thus the constraint can be expressed as: 6.0x 4.5y 1000 (Stock of Raw Material available) Similarly, the manufacturer has 500 labour hours available and each unit of product A and B require 4 and 6 labour hours respectively. Thus the constraint can be expressed as: d. Add Non-Negativity Restriction: 4x 6y 500 (Labour Hours available) According to the assumption of finite choices, the decision variables cannot have negative values. Therefore the non-negativity restriction is added to ensure that the decision variables cannot assume negative values. In the solution to Example 1, we shall express the formulated LPP as: Let x and y be the number units of products A and B to be produced respectively. Maximize Z = 650x + 400y Subject to the constraints: 6.0x 4.5 y x 6y 500 x, y 0 Value Addition 2: Quiz Fundamentals of Linear Programming Visit the web page f4tf.html to know test your knowledge of linear programming. Source: University of Delhi 6

7 Figure 2: Steps in the Formulation of Linear Programming Problem Identify the decision variables Formulate the objective function Identify and formulate the constraints Add non-negativity restriction 3.2 Standard Linear Programming Problem: A standard linear programming problem has following characteristics: i. It is a maximisation problem. ii. It has less than equal to i.e., type constraints. iii. It has a non-negativity restriction. Standard Form of LPP Maximize Z= c 1 x 1 + c 2 x 2 + c 3 x c n x n Subject to constraints: a 11 x 1 + a 12 x 2 + a 13 x a 1n x n b 1 a 21 x 1 + a 22 x 2 + a 23 x a 2n x n b 2 a 31 x 1 + a 32 x 2 + a 33 x a 3n x n b 3 a m1 x 1 + a m2 x 2 + a m3 x a mn x n b m x 1, x 2, x 3,, x n 0 Any non-standard problem, i.e., a minimisation LPP or with constraints or with variables having unrestricted sign can be converted into the standard form. Value Addition 3: Video Tutorial on Linear Programming Click on the link below to watch a tutorial on linear programming. Source: University of Delhi 7

8 3.3 General Form of Linear Programming Problem: In general form, a linear programming problem can be expressed as: Optimize Z=c 1 x 1 + c 2 x 2 + c 3 x c n x n Subject to constraints (Structural or resource constraints): a 11 x 1 + a 12 x 2 + a 13 x a 1n x n, =, b 1 a 21 x 1 + a 22 x 2 + a 23 x a 2n x n, =, b 2 a 31 x 1 + a 32 x 2 + a 33 x a 3n x n, =, b 3 a m1 x 1 + a m2 x 2 + a m3 x a mn x n, =, b m x 1, x 2, x 3,, x n 0 Where, x j (x 1, x 2, x 3,, x n ) = Decision or Choice Variables, c j (c 1, c 2, c 3,, c n ) = Cost or Profit Coefficients, a ij (i =1 to n, j=1 to m) = Exchange Coefficients, and b i (b 1, b 2, b 3,, b m ) = Resource Values. Illustration 1: Investment Problem Mr. Anil is planning to invest up to 2,50,000 in various securities. He has finalised two alternatives: Security A with 6.5% p.a. and PPF yielding 8.5% p.a. As per the income tax regulation, he can invest at most of 1,50,000 in PPF and he has decided to invest at least 50,000 in Security A. He also wants the amount invested in Security A to be at least equal to the amount invested in PPF. Formulate the given situation as Linear Programming Problem to maximize the return from the investment. Solution: i. Identification of the decision variables In the given problem we have is to determine the amount to be invested in the two investment alternatives. Let x 1 and x 2 be the amount invested in Security A and PPF respectively. The total interest income from the investment can be stated as: 0. 65x x 2 ii. Formulate objective function: Since the objective is to maximize the return from the investment, therefore the objective function is given by aximize Z = 0. 65x x 2, where Z refers to total return from the investment. iii. Identify and formulate the constraints: 1. The total amount available for investment is given as up to 2,50,000. Stated mathematically: x 1 + x The amount to be invested in Security A should be at least 50,000. Mathematically, it can be expressed as: x At most 1,50,000 can be invested in PPF, i.e. x Amount that can be invested in Security A should be at least equal to the amount invested in PPF. Put mathematically, x 1 x 2 x 2 x 1 0. University of Delhi 8

9 iv. Add non-negativity restriction: Since the amount invested is spending and not borrowing, therefore the value of x 1 and x 2 should be positive, i.e. x 1, x 2 0 Formulation: The above mathematical expressions can be organised to formally express the LPP as: Maximize Z = 0.65x x 2 Subject to the constraints: x 1 + x x x x 2 x 1 0 x 1, x 2 0 Illustration 2: Personnel Management Problem The nurses in a hospital work six consecutive days and have one day off and their five days of work can start on any day of the week. Based on the past experience the hospital has calculated the following minimum number of nurses required on a particular week day: Mon Tue Wed Thurs Fri Sat Sun Formulate the above as a linear programming problem in order to minimise the number of nurses hired by the hospital. Solution: i. Identification of the decision variables In the given problem we have is to determine the number of nurses that should be employed for each of the seven days. Let x 1, x 2, x 3, x 4, x 5, x 6 and x 7 be the number of nurses that report for work on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday respectively. ii. Formulate objective function: The objective is to minimize the total number of nurses hired by the hospital. Thus, the objective function can be stated as: Minimize Z = x 1 + x 2 + x 3 + x 4 + x 5 +x 6 + x 7 iii. Identify and formulate the constraints: There are seven constraints related to the number of nurses required during a particular day of the week. The nurses work for six consecutive days and they have one day off. Therefore the nurses who start the week on Monday would work for 6 days i.e. till Saturday. Similarly those who start their week on Tuesday would work for 6 days i.e. till Sunday and so on. Accordingly, the seven constraints for the seven days of the week can be stated as follows: University of Delhi 9

10 1. Monday: x 1 is the number of nurses who start their 6-days work schedule on Monday. Besides this, those who start their week on Wednesday, Thursday, Friday, Saturday and Sunday will also be working on Monday. Also, the minimum number of nurses required on Monday is specified as 22. Therefore the constraint can be mathematically expressed as: 2. Tuesday: x 1 + x 3 + x 4 + x 5 +x 6 + x x 2 is the number of nurses who start their 6-days work schedule on Tuesday. Besides this, those who start their week on Thursday, Friday, Saturday, Sunday and Monday will also be working on Tuesday. The minimum number of nurses required on Tuesday is 30. Therefore the constraint is: 3. Wednesday: x 2 + x 4 + x 5 +x 6 + x 7 + x x 3 is the number of nurses who start their 6-days work schedule on Wednesday. Besides this, those who start their week on Friday, Saturday, Sunday, Monday and Tuesday will also be working on Wednesday. The minimum number of nurses required on Wednesday is 45. Therefore the constraint is: 4. Thursday: x 3 +x 5 + x 6 + x 7 + x 1 + x x 4 is the number of nurses who start their 6-days work schedule on Thursday. Besides this, those who start their week on Saturday, Sunday, Monday, Tuesday and Wednesday will also be working on Thursday. The minimum number of nurses required on Thursday is 40. Therefore the constraint is: 5. Friday: x 4 +x 6 + x 7 + x 1 + x 2 +x 3 40 x 5 is the number of nurses starting their 6-days work schedule on Friday. Besides this, those who start their week on Sunday, Monday, Tuesday, Wednesday and Thursday will also be working on Friday. The minimum number of nurses required on Friday is 45. Therefore the constraint is: 6. Saturday: x 5 + x 7 + x 1 + x 2 +x 3 + x 4 45 x 6 is the number of nurses starting their 6-days work schedule on Saturday. Besides this, those who start their week on Monday, Tuesday, Wednesday, Thursday and Friday will also be working on Saturday. The minimum number of nurses required on Saturday is 30. Therefore the constraint is: 7. Sunday: x 6 + x 1 + x 2 +x 3 + x 4 + x 5 30 x 7 is the number of nurses starting their 6-days work schedule on Sunday. Besides this, those who start their week on Tuesday, Wednesday, Thursday, Friday and Saturday will also be working on Sunday. The minimum number of nurses required on Sunday is 20. Therefore the constraint is: iv. Add non-negativity restriction: x 7 + x 2 + x 3 + x 4 + x 5 + x 6 20 Since the number of nurses has to be positive, the non-negativity restriction is: x 1, x 2, x 3, x 4, x 5, x 6, x 7 0 University of Delhi 10

11 Formulation: Upon organising the objective function and the constraints, we may express the LPP as under: Illustration 3: Minimize Z = x 1 + x 2 + x 3 + x 4 + x 5 +x 6 + x 7 Subject to the constraints: x 1 + x 3 + x 4 + x 5 +x 6 + x 7 22 x 2 + x 4 + x 5 +x 6 + x 7 + x 1 30 x 3 +x 5 + x 6 + x 7 + x 1 + x 2 45 x 4 +x 6 + x 7 + x 1 + x 2 +x 3 40 x 5 + x 7 + x 1 + x 2 +x 3 + x 4 45 x 6 + x 1 + x 2 +x 3 + x 4 + x 5 30 x 7 + x 2 + x 3 + x 4 + x 5 + x 6 20 x 1, x 2, x 3, x 4, x 5 x 6, x 7 0 A firm produces three varieties of bags: Clutches, Wallets and Sling Bags with selling price per unit 500, 900, and 1500 respectively. The cost of the faux leather required for Clutches, Wallets and Sling Bags is 150, 200 and 400 respectively. The capacity (bags per hour) of moulding, sewing and packing each of the bags is as follows: Capacity per hour Machine Clutch Wallet Sling Bag Moulding Sewing Packing The firm possesses only one of each type of machine. The cost per hour to run each of the three machines is 500 for Moulding, 800 for Sewing and 400 for packing. Formulate the above as a linear programming problem to determine the number of bags of each type to be produced in order to maximise profit of the firm. Solution: i. Identification of the decision variables In the given problem we have is to determine the number of bags of each type to be produced. Let x1, x2and x 3 be the number of Clutches, Wallets and Sling Bags respectively, to be manufactured in order to maximize the profit ii. Formulate objective function: The objective is to maximise the profit earned from the bags, therefore it has to be computed (Profit=Selling Price minus Cost per bag) of each variety as follows: University of Delhi 11

12 Particulars Clutches ( ) Wallets ( ) Sling Bags ( ) A. Selling Price B. Cost: (i) Faux Leather (ii) Moulding: = Cost per hr. of run X Machine time in hrs. allocated per bag = = = 20 = Cost per hr. of run (iii) Sewing: (ii) Packing: = Cost per hr. of run = Cost per hr. of run 1 hr. Capacity in no. of bags 1 Capacity 1 Capacity Thus, the objective function can be stated as: iii. Identify and formulate the constraints: Maximize Z = 312 x x x 3 Here, the capacity of the three machines given in hours has to be translated into the structural constraints. The machines can be used for any of the three bags. The time of a machine allocated per unit of a bag can be calculated as 1 hour Capacity expressed in no. of bags produced per hour = = = = = = 20 Cost (i)+(ii)+(iii) Profit (A-B) which gives the capacity of the machine utilised per unit of bag. The proportion of time of a machine that can be allocated among the three bags cannot exceed 1, therefore the constraints associated with the three machines may be specified as: x x x 3 1 (Moulding constraint) x x x 3 1 (Sewing constraint) 3. iv. Add non-negativity restriction: 1 50 x x x 3 1(Packing constraint) Since the production cannot be negative, the value of x 1, x 2 and x 3 should be positive, i.e. x 1, x 2, x 3 0. University of Delhi 12

13 Formulation: The above mathematical expressions can be organised to formally express the LPP as: Maximize Z = 312 x x x 3 Subject to the constraints: 4. Graphical Method for Solving LPP: 1 50 x x x x x x x x x 3 1 x 1, x 2, x 3 0 A linear programming problem can be solved by either by graphical or by simplex method. Although graphical method is easier than simplex method but it can be used only to solve two variable problems. The graphical method involves the following steps: 1. Plotting the Constraints: The graph of each of the constraints is plotted by treating the inequalities as equations. These graphs depict the availability of the resources or the structural constraints. 2. Identify the Feasible Region: The feasible region consists of the area corresponding to the feasible solutions represented by each constraint. It is identified by to putting the coordinates of origin i.e. (0, 0) in a constraint and examining whether it satisfies the inequality representing the constraint. If it does, then the corresponding side of the constraint line which contains the origin is the feasible area corresponding to that particular constraint. 3. Identify the Common Feasible Region: The overlapping area on the graph of feasible region that is common to all the constraints is the common feasible region. It contains the feasible solution and it may acquire the shape of a quadrilateral, triangle, polygon, line segment, single point, an empty space or an unbounded region. 4. Find the Optimal Solution: According to the Extreme Point Theorem: An optimum solution to a Linear Programming Problem, if any, exists at one of the extreme points of the feasible region/solution space. The optimal solution which maximizes or minimizes the objective function can be obtained either by corner point method or iso-profit/iso-cost method. The optimal solution is generally unique i.e. one solution that optimizes the objective function. The technique of graphical method for solving a Linear Programming Problem can be applied to a maximisation problem, a minimisation problem and certain exceptional cases such as multiple optimal, infeasibility, unbounded solution and redundancy. (i) Corner Point Method: This method uses the following steps to find the optimal solution: a. Locate the corner points or the extreme points or the vertices of the common University of Delhi 13

14 feasible region. b. Evaluate the objective function at each corner point. c. Select the maximum or minimum value for a maximisation or minimization problem respectively. This is the optimal solution for the given LPP. (ii) Iso-Profit/Iso-Cost Method: In this method the following steps help us to find the optimal solution: a. Select any two convenient values of the objective function which lie within the common feasible region. b. Plot the objective function lines based on the selected values to determine the direction of improvement of the objective function. c. In a maximization problem, the improvement in the objective function is in the direction of greater value and in a minimisation problem it is in the direction of lesser value. d. Draw a parallel line to the objective function in the direction of its improvement so that it passes through the farthest corner point. This gives us the optimal solution. Both the methods stated above, give us the same optimal solution. We will follow the corner point method for solving LPP graphically. Figure 3: Steps involved in solving a LPP by using Graphical Method Plot the constraints as linear equations Identify feasible region corresponding to each of the constraint Identify the Common Feasible Region Find the Optimal Solution by either the Corner Point Method or the Iso-Cost/Iso-Profit Method Value Addition 4: Video Fundamentals of Graphical Method Visit the web page Fundamentals-of-Operations-Research/3# to see a video lecture on the Graphical Method for solving a linear programming problem. University of Delhi 14

15 Example 2: Solving a Standard LPP by Graphical Method Solve the following problem graphically: Maximize Z = 5x 1 + 7x 2 Subject to the constraints: 6x 1 + 9x x 1 + 4x x x x 1, x 2 0 Solution: Step1: Rewrite the constraints as equations as shown below: 6x 1 + 9x 2 = x 1 + 4x 2 = x x 2 = Step2: Plot the graph of the equations: 6x 1 + 9x 2 = x x x 1 + 4x 2 = x x x x 2 = x x Step3: Determine the feasible set of solutions: i. Identify the feasible region corresponding to each of the plotted line by checking whether a point (generally (0, 0)) on either side of the line satisfies the constraint inequality. For instance, substitute coordinates (0, 0) in the constraint 1. We obtain which is true. It implies that the left side area of the line (containing the origin point (0, 0)) is the feasible area corresponding to the line represented by 1. Follow this procedure for each of the constraint and obtain the feasible area corresponding to each of them. ii. Identify the common feasible region which satisfies all the constraints simultaneously. University of Delhi 15

16 Common Feasible Region Step4: Obtain the corner points: (i) A (at the intersection of y axis (x 1 =0) and line represented by equation 1) x 2 = 540 x 2 = 60 (ii) B (at the intersection of lines represented by equations 1 and 3) Solving the two: 6x 1 + 9x 2 = x x 2 = Multiplying 3 by 2/5 6x 1 + 4x 2 = 280 and subtracting from 1 we get, x 2 = = 52 & x 1 = = 12 (iii) C (at the intersection of lines represented by equations 2 and 3) Solving the two: 12x 1 + 4x 2 = x 1 + x 2 = x x 2 = x 1 + 2x 2 = 140 Subtracting 2 from 3 x 2 = 20 & x 1 = = (iii) D (at the intersection of lines represented by equations 2 and x-axis i.e., x 2 =0) 12x 1 + 4(0) = x 1 = 40 3 University of Delhi 16

17 Answer: Step5: Locate the optimal solution: Evaluate the Objective Function at the corner points as shown in the table below: Corner points Coordinates Z=5x 1 + 7x 2 A (0,60) B (12,52) C (40,33.33) D (40,0) Since the maximum value of Z occurs at point B, therefore Z=424 is the optimal solution at x 1 = 12 and x 2 = 52. Example 3: Solving a Maximization LPP with Mixed Constraints using Graphical Method Solution: Solve the following problem graphically: Maximize Z = 15x x 2 Subject to the constraints: x 1 + 4x x 1 + 3x x 1 + x x 1, x 2 0 Step1: Rewrite the constraints as equations as shown below: x 1 + 4x 2 = x 1 + 3x 2 = x 1 + x 2 = 78 3 Step2: Plot the graph of the equations: x 1 + 4x 2 = x x x 1 + 3x 2 = x x x 1 + x 2 = 78 3 x x Step3: Determine the feasible set of solutions: i. Identify the feasible region corresponding to each of the plotted line by checking whether a point (generally (0, 0)) on either side of the line satisfies the constraint inequality. For instance, substitute coordinates (0, 0) in the constraint 1. We obtain University of Delhi 17

18 0 100 which is false. It implies that the right side area of the line (not containing the origin point (0, 0)) is the feasible area corresponding to the line represented by 1. Follow this procedure for each of the constraint and obtain the feasible area corresponding to each of them. ii. Identify the common feasible region which satisfies all the constraints simultaneously. Common Feasible Region Step4: Obtain the corner points: (i) A (at the intersection of lines represented by equations 1 and 3) Solving the two: x 1 + 4x 2 = x 1 + x 2 = 78 3 x 1 = x 2 = (ii) B (at the intersection of lines represented by equations 2and 3) Solving the two: 2x 1 + 3x 2 = x 1 + x 2 = 78 3 x 1 = 14 x 2 = 36 (iii) C (at the intersection of lines represented by equations 1 and 2) Solving the two: x 1 + 4x 2 = x 1 + 3x 2 = University of Delhi 18

19 Step5: Locate the optimal solution: x 1 = 60 x 2 = 10 Evaluate the Objective Function at the corner points as shown in the table below: Answer: Corner points Coordinates Z=15x x 2 A (19.28,20.18) B (12,42) C (60,10) Since the maximum value of Z occurs at point B, therefore Z=1230 is the optimal solution at x 1 = 12 and x 2 = 42. Example 4: Solving Maximization Linear Programming Problem where Feasible Region is a line segment using Graphical Method Solution: Solve the following problem graphically: Maximize Z = 3x 1 + 5x 2 Subject to the constraints: 5x 1 + 3x 2 = x 1 + x x 1 + 3x x 1, x 2 0 Step1: Rewrite the constraints as equations as shown below: 5x 1 + 3x 2 = x 1 + x 2 = 18 2 Step2: Plot the graph of the equations: 5x 1 + 3x 2 = 30 1 x x x 1 + x 2 = 18 2 x x x 1 + 3x 2 = 15 3 x x x 1 + 3x 2 = 15 3 Step3: Determine the feasible set of solutions: i. Identify the feasible region corresponding to each of the plotted line. For the constraint 1 which is an equation, only the points lying on the line segment will University of Delhi 19

20 be feasible. For the other two constraints, check whether the origin (0, 0) satisfies the constraint inequality. For the constraint 2, 0 18 this is true. It implies that the left side area of the line (containing the origin point (0, 0)) is the feasible area corresponding to the line represented by 2. For the constraint which is false. It implies that the right side area of the line (not containing the origin point (0, 0)) is the feasible area corresponding to the line represented by 3. ii. Here, the common feasible region which satisfies all the constraints simultaneously is the line segment AB. Common Feasible Region Step4: Obtain the corner points: (i) A (at the intersection of lines represented by equations 1 and 3) Solving the two: 5x 1 + 3x 2 = 30 1 x 1 + 3x 2 = 15 3 x 1 = 3.75 x 2 = 3.75 (ii) B (at the intersection of lines represented by equations 1 and x-axis) Solving the two: 5x 1 + 3(0) = 30 1 x 1 = 6 x 2 = 0 University of Delhi 20

21 Step5: Locate the optimal solution: Evaluate the Objective Function at the corner points as shown in the table below: Answer: Corner points Coordinates Z=3x 1 + 5x 2 A (3.75,3.75) 30 B (6,0) 18 Since the maximum value of Z occurs at point A, therefore Z=30 is the optimal solution at x 1 = 3.75 and x 2 = Example 5: Solving a Maximization LPP with x or y Constraints using Graphical Method Solution: Solve the following problem graphically: Maximize Z = 11x x 2 Subject to the constraints: 2x 1 + 3x x x x 1 x 2 4 x 1, x 2 0 Step1: Rewrite the constraints as equations as shown below: 2x 1 + 3x 2 = x 1 = 60 2 x 2 = x 1 = x 2 4 Step2: Plot the graph of the equations: 2x 1 + 3x 2 = x x x 1 = 60 2 x x x 2 = 30 3 x x x 1 = x 2 4 x x University of Delhi 21

22 Step3: Determine the feasible set of solutions: i. The feasible region corresponding to the constraints can be identified in the same manner as explained in the previous examples. However, for the constraint 4 i.e., 5x 1 x 2 we cannot take origin (0, 0) to determine the feasible region because the constraint itself passes through it. Therefore we will take a point on one of the sides of the line and decide feasible area. Say, take (100, 0), substituting it in 4 gives which is true. Therefore the area containing (100, 0) is feasible region with respect to constraint 4. ii. Identify the common feasible region which satisfies all the constraints simultaneously. Step4: Obtain the corner points: (i) A (0, 0) (ii) B (at the intersection of lines represented by equations 3 and 4) Solving the two: x 2 = x 1 = x 2 4 x 1 = 6 x 2 = 30 (iii) C (at the intersection of lines represented by equations 1 and 3) Solving the two: 2x 1 + 3x 2 = University of Delhi 22

23 x 2 = 30 3 x 1 = 30 x 2 = 30 (iv) D (at the intersection of lines represented by equations 1 and 2) Solving the two: 2x 1 + 3x 2 = x 1 = 60 2 x 1 = 60 x 2 = 10 (v) E (at the intersection of lines represented by equation 2and x-axis) Step5: Locate the optimal solution x 1 = 60 2 x 2 = 0 Evaluate the Objective Function at the corner points as shown in the table below: Answer: Corner points Coordinates Z=11x x 2 A (0,0) 0 B (6,30) 456 C (30,30) 720 D (60,10) 790 E (60,0) 660 Since the maximum value of Z occurs at point D, therefore Z=790 is the optimal solution at x 1 = 60 and x 2 = 10. Example 6: Solving a Minimization LPP using Graphical Method Solve the following problem graphically: Minimize Z = 35x x 2 Subject to the constraints: 8x 1 + 2x x 1 + 2x x 1 + 3x x 1, x 2 0 Solution: Step1: Convert the constraints into equations 8x 1 + 2x 2 = x 1 + 2x 2 = x 1 + 3x 2 = University of Delhi 23

24 Step2: Plot the equations by taking two points (Put (i) x 1 =0 and calculate x 2, (ii) x 2 =0 and calculate x 1 ) and joining the same 8x 1 + 2x 2 = x x x 1 + 2x 2 = x x x 1 + 3x 2 = x x Step3: Find the feasible solution Mark the feasible region for each line by checking whether a point (generally (0, 0)) on either side of the line satisfies the inequality. Then shade the common feasible region satisfying all the constraints. University of Delhi 24

25 Step4: Determine the corner points (i) A (at the intersection of lines represented by equations 1 and 2) Solving the two: 8x 1 + 2x 2 = x 1 + 2x 2 = x 1 = 20 x 2 = 70 (ii) B (at the intersection of lines represented by equation 1 and y-axis) 8(0) + 2x 2 = x 1 = 0 x 2 = 250 (iii) C (at the intersection of lines represented by equation 3 and y axis) 5(0) + 3x 2 = x 1 = 0 x 2 = 300 (iv) D (at the intersection of lines represented by equations 2 and 3) Solving the two: x 1 + 2x 2 = x 1 + 3x 2 = Step5: Locate the optimal solution x 1 = x 2 = Evaluate the Objective Function at the corner points as shown in the table below: Answer: Corner points Coordinates Z=35x x 2 A (20,170) B (0,250) C (0,300) D (102.86,128.57) Since the minimum value of Z occurs at point A, therefore Z=10900 is the optimal solution at x 1 = 20 and x 2 = Exceptional Cases in the Graphical Solution of LPP: So far we have observed in the solution of a Linear Programming Problem that there is a unique optimal solution that optimizes the objective function. We have already illustrated the application of the technique of graphical method to a standard LPP, maximisation problem with mixed constraints as well as minimisation problem. Besides, there are certain exceptional cases such as multiple optimal, infeasibility, unbounded solution and redundancy. University of Delhi 25

26 5.1 Multiple Optimal Solutions: It is a special case where the objective function is parallel to one of the binding constraints and it yields the same optimal value for more than one feasible solution. Thus we get multiple optimal solutions to the LPP. Illustration 4: Solve the following problem graphically: Maximize Z = 16x 1 + 4x 2 Subject to the constraints: 8x 1 + 2x x 1 + 2x x 1 + 3x x 1, x 2 0 Solution: Step1: Rewrite the constraints as equations as shown below: 8x 1 + 2x 2 = 80 1 x 1 + 2x 2 = x 1 + 3x 2 = Step2: Plot the graph of the equations 8x 1 + 2x 2 = 80 1 x x x 1 + 2x 2 = 72 2 x x x 1 + 3x 2 = x x Step3: Determine the feasible set of solutions i. The feasible region corresponding to the constraints can be identified in the same manner as explained in the previous examples. ii. Identify the common feasible region which satisfies all the constraints simultaneously. University of Delhi 26

27 Step4: Determine the corner points (i) A (at the intersection of lines represented by equations 1 and 2) Solving the two: 8x 1 + 2x x 1 + 2x x 1 = 1.6 x 2 = 33.6 (ii) B (at the intersection of lines represented by equation 3 and y axis) 4(0) + 3x 2 = 120 x 1 = 0 x 2 = 40 (iii) C (at the intersection of lines represented by equation 3 and x axis) 4x 1 + 3(0) = x 1 = 30 x 2 = 0 (iv) D (at the intersection of lines represented by equation 2 and x axis) 3x 1 + 2(0) 72 2 x 1 = 24 x 2 = 0 University of Delhi 27

28 Step5: Locate the optimal solution Evaluate the Objective Function at the corner points as shown in the table below: Answer: Corner points Coordinates Z=16x 1 + 4x 2 A (1.6,33.6) 160 B (0,40) 160 C (30,0) 480 D (24,0) 384 Since the minimum value of Z occurs at point A as well as point B, therefore Z=160 is the optimal solution at x 1 = 1.6, x 2 = 33.6 and x 1 = 0, x 2 = 40. This is a special case of multiple optimal solution. All the points lying on the line segment AB will give same optimal solution of 160. Let us take a point E(12,28) on the segment AB and evaluate the objective function Z. Z = = 160 Similarly, all the points lying on segment AB will give us Z= 160 which is the optimal value for the objective function. 5.2 Unbounded Solution: If the value of the decision variables can be increased indefinitely in the solution space without violating any of the constraints then it is the case of an unbounded solution. Here, an infinite number of feasible solution points exist in the unbounded feasible region. The value of objective function will increase (in maximisation case) and decrease (in minimization case) indefinitely solution space. Such a situation occurs when the LPP is not constructed properly or one or more non-redundant constraints have not been accounted for. However, in the case of a minimization problem the optimal solution may be found at one of the corner points even if the solution space is unbounded. Illustration 5: Maximization Problem with an Unbounded Solution Solution: Solve the following problem graphically: Maximize Z = 7x 1 + 9x 2 Subject to the constraints: 5x 1 + 6x x 1 + 2x x x 1, x 2 0 Step1: Rewrite the constraints as equations as shown below: 5x 1 + 6x 2 = x 1 + 2x 2 = x 2 = 80 3 University of Delhi 28

29 Step2: Plot the graph of the equations 5x 1 + 6x 2 = x 1 + 2x 2 = x 2 = 80 3 x x x x x x Step3: Determine the feasible set of solutions i. The feasible region corresponding to the constraints can be identified in the same manner as explained in the previous examples. ii. Identify the common feasible region which satisfies all the constraints simultaneously. Step4: Determine the corner points (i) A (at the intersection of lines represented by equations 1 and 3) Solving the two: University of Delhi 29

30 5x 1 + 6x 2 = x 2 = 80 3 x 1 = 36 x 2 = 20 (ii) B (at the intersection of lines represented by equations 1 and 2) 5x 1 + 6x 2 = x 1 + 2x 2 = x 1 = 15 x 2 = 37.5 (iii) C (at the intersection of lines represented by equation 2 and y axis) 3x 1 + 2x 2 = (0) + 2x 2 = Step5: Locate the optimal solution x 1 = 0 x 2 = 60 Evaluate the Objective Function at the corner points as shown in the table below: Answer: Corner points Coordinates Z=7x 1 + 9x 2 A (36,20) 432 B (15,37.5) C (0,60) 540 The maximum value of Z occurs at the corner point C however since the solution space is unbounded, Z=540 may not be the optimal solution. Let us take a point D(50,60) in the unbounded solution space and evaluate the objective function Z. Z = = 890 It yields a higher value for the objective function than the value that occurs at the corner points. Similarly, all the points lying in the unbounded solution space will give a value of the optimal solution which is the more than the optimal value for the objective function. This is a special case of unbounded solution. Thus, the maximum value of the objective function is in the solution space. Therefore, the given LPP does not have an optimal solution and it is unbounded. Illustration 6: Minimization Problem with an Unbounded Solution Space Solve the following problem graphically: Minimize C = 2x 1 + 3x 2 Subject to the constraints: 6x 1 + 5x x 1 + 3x x 1 + 5x x 1, x 2 0 University of Delhi 30

31 Solution: Step1: Rewrite the constraints as equations as shown below: 6x 1 + 5x 2 = x 1 + 3x 2 = x 1 + 5x 2 = Step2: Plot the graph of the equations 6x 1 + 5x 2 = x 1 + 3x 2 = x 1 + 5x 2 = x x x x x x Step3: Determine the feasible set of solutions i. The feasible region corresponding to the constraints can be identified in the same manner as explained in the previous examples. ii. Identify the common feasible region which satisfies all the constraints simultaneously. University of Delhi 31

32 Step4: Determine the corner points (i) A (at the intersection of lines represented by equation 3 and x axis) 2x 1 + 5(0) = x 1 = 50 x 2 = 0 (ii) B (at the intersection of lines represented by equations 1 and 3) 6x 1 + 5x 2 = x 1 + 5x 2 = x 1 = 20 x 2 = 12 (iii) C (at the intersection of lines represented by equations 1 and 2) 6x 1 + 5x 2 = x 1 + 3x 2 = x 1 = 8.57 x 2 = 25.7 (iv) D (at the intersection of lines represented by equation 2 and y axis) 5(0) + 3x 2 = x 1 = 0 x 2 = 40 Step5: Locate the optimal solution Evaluate the Objective Function at the corner points as shown in the table below: Answer: Corner points Coordinates C = 2x 1 + 3x 2 A (50,0) 100 B (20,12) 76 C (8.57,25.7) D (0,40) 120 Since the minimum value of the objective function occurs at the corner point B therefore C=76 is the apparently the optimal solution at x 1 = 20, x 2 = 12. However, the graph shows that this is a special case of unbounded solution. In the given problem, the feasible region is unbounded therefore, there exist an infinite number of points in the unbounded area that yield feasible solutions. Let us take a point E(30,40) in the unbounded solution space and evaluate the objective function. C = = 180 It gives a higher value for the objective function than the value that occurs at the corner points. Similarly, all the points lying in the unbounded solution space will yield a value of the optimal solution which is the more than the optimal value for the objective function. Thus, point B gives optimal solution with Minimum C=76 at x 1 = 20, x 2 = 12. Thus, in case the LPP is a minimization problem and the value of the objective University of Delhi 32

33 function increases in the unbounded solution space, the optimal solution exists at one of the corner points. 5.3 Infeasible Solution: A given LPP is said to be infeasible if all the given constraints cannot be simultaneously satisfied. Infeasibility may occur in case of incorrect specification of the problem situation or formulation where the constraints conflict each other. Infeasibility cannot occur if all the constraints are type. Graphically, the common feasible region cannot be identified and therefore the feasible solution does not exist. Illustration 7: Solve the following problem graphically: Minimize C = 78x x 2 Subject to the constraints: 5x 1 + 2x x x x 1, x 2 0 Solution: Step1: Rewrite the constraints as equations as shown below: 5x 1 + 2x 2 = x 2 = x 1 = Step2: Plot the graph of the equations 5x 1 + 2x 2 = x 2 = x x x x x x x Step3: Determine the feasible set of solutions i. The feasible region corresponding to the constraints can be identified in the same manner as explained in the previous examples. ii. In the present case, there is no common feasible region satisfying all the constraints simultaneously. Therefore the problem is infeasible or it has no feasible solution. University of Delhi 33

34 x 1 Answer: The given LPP is infeasible or it has no feasible solution because common feasible region satisfying all the constraints simultaneously does not exist. 5.4 Redundancy Generally the constraints in a Linear Programming Problem bind the feasible region and generate a set of feasible solutions. A redundant constraint is the one which does not bind the feasible solution space, therefore it need not be considered for finding the feasible solution. The feasible solution remains unchanged even if the redundant constraint is removed from the problem. Illustration 8: Solve the following problem graphically: Solution: Minimize C = 3x 1 + 2x 2 Subject to the constraints: 2x 1 + x x 1 + 5x x 1 + x x 1, x 2 0 Step1: Rewrite the constraints as equations as shown below: University of Delhi 34

35 2x 1 + x 2 = x 1 + 5x 2 = x 1 + x 2 = Step2: Plot the graph of the equations: 2x 1 + x 2 = x x x 1 + 5x 2 = 45 2 x x x 1 + x 2 = x x Step3: Determine the feasible set of solutions i. The feasible region corresponding to the constraints can be identified in the same manner as explained in the previous examples. ii.identify the common feasible region which satisfies all the constraints simultaneously. iii. As reflected in the graph, the constraint 2 does not bind the common feasible region therefore it is redundant constraint. The feasible solution will not be affected if this constraint is removed from the LPP. University of Delhi 35

36 Step4: Determine the corner points (i) A (at the intersection of lines represented by equation 3 and x axis) 4x = x 1 = 26 x 2 = 0 (ii) B (at the intersection of lines represented by equations 1 and 3) 2x 1 + x 2 = x 1 + x 2 = x 1 = 2 x 2 = 96 (iii) C (at the intersection of lines represented by equations 1and x axis) 2x = Step5: Locate the optimal solution x 1 = 50 x 2 = 0 Evaluate the Objective Function at the corner points as shown in the table below: Corner points Coordinates C = 3x 1 + 2x 2 A (26,0) 78 B (2,96) 198 C (50,0) 150 Answer: The minimum value of the objective function occurs at the corner point A therefore C=78 is the optimal solution at x 1 = 26, x 2 = 0 and the graph shows that the constraint 2 is redundant. Even if it is deleted from the problem the optimal solution can be reached. This is a special case of redundancy. Summary: Linear Programming is a mathematical technique used for optimum allocation of limited or scarce resources by choosing the best alternative from a set of feasible solutions in a situation in which objective function as well as constraints can be expressed as linear mathematical relationships. Assumptions of Linear Programming Model are proportionality, additivity, continuity (or divisibility), certainty and finite choices. First step to solve a LPP includes formulate the problem by identifying the decision variables, formulate the objective function in the form of a linear function of decision variables, state all the limitations on the resources as linear equations or inequalities in terms of the identified decision variables, and finally add non-negativity restrictions. University of Delhi 36

37 Next, find Basic Solution that satisfies the set of constraints using the Graphical method or the Simplex method. Then, identify the Feasible Solution out of the basic solution values that satisfy the non-negativity restrictions. Next, ascertain the Optimal Solution that optimizes the objective function. A Linear Programming Problem may have a unique optimal solution, or as an exception it may have multiple optimal solutions or unbounded solution or no solution at all. A standard linear programming problem is a maximisation problem with less than equal to i.e., type constraints and a non-negativity restriction. Any non-standard problem, i.e., a minimisation LPP or with constraints or with variables having unrestricted sign can be converted into the standard form. The Common Feasible Region in a feasible solution is the one that is common corresponding to all the constraints. It contains the feasible solution and it may acquire the shape of a quadrilateral, triangle, polygon, line segment, single point, an empty space or an unbounded region. Optimal Solution follows the Extreme Point Theorem which states that the optimum solution to a Linear Programming Problem, if any, exists at one of the extreme point of the feasible region/solution space. The optimal solution which maximizes or minimizes the objective function can be obtained either by corner point method or iso-profit/iso-cost method. The optimal solution is generally unique i.e. one solution that optimizes the objective function. The technique of graphical method for solving a Linear Programming Problem can be applied to a maximisation problem, a minimisation problem and certain exceptional cases such as multiple optimal, infeasibility, unbounded solution and redundancy. Multiple Optimal Solutions is a special case where the objective function is parallel to one of the binding constraints and it yields the same optimal value for more than one feasible solution. Unbounded Solution exists if the value of the decision variables can be increased indefinitely in the solution space without violating any of the constraints then it is the case of an unbounded solution. Here, an infinite number of feasible solution points exist in the unbounded feasible region. The value of objective function will increase (in maximisation case) and decrease (in minimization case) indefinitely solution space. Such a situation occurs when the LPP is not constructed properly or one or more non-redundant constraints have not been accounted for. However, in the case of a minimization problem the optimal solution may be found at one of the corner points even if the solution space is unbounded. In case the LPP is a minimization problem and the value of the objective function increases in the unbounded solution space, the optimal solution exists at one of the corner points. Infeasible Solution is a situation where if all the given constraints in a LPP cannot be simultaneously satisfied. Infeasibility may occur in case of incorrect specification of the problem situation or formulation where the constraints conflict each other. Infeasibility cannot occur if all the constraints are type. Graphically, the common feasible region cannot be identified and therefore the feasible solution does not exist. The given LPP is infeasible or it has no feasible solution because common feasible region satisfying all the constraints simultaneously does not exist. In the case of redundancy one or more of the constraints do not bind the feasible solution space, therefore it need not be considered for finding the feasible solution. The feasible solution remains unchanged even if the redundant constraint is removed from the problem. University of Delhi 37