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- CHAPTER TWO LINEAR PROGRAMMING MODELS(LPM) 1
- Chapter objectives • After studying this lesson, you should be able to: – Define the basic concepts in LP; – Formulate Linear Programming Problem – Identify the characteristics of linear programming problem – Make a graphical analysis of the linear programming problem – Solve the problem graphically – Understand the basics of simplex method – Explain the simplex calculations – Describe various solutions of Simplex Method – Understand Big-M method – Realize the special cases of LP model; – Perform sensitivity analysis of an LP model – Understand the Dual Linear programming Problem – Formulate and solve the Dual Problem 2
- 2.1. INTRODUCTION • In 1947, George Danzig developed the use of algebra for determining solutions to problems that involved the optimal allocation of scarce resources. • In spite of numerous potential applications in business, response to this new technique was low due to substantial computational burden, which is now removed with subsequent advances in computer technology and related software during the last three decades. 3
- Cont’d • The term linear implies that all the mathematical relations used in the problem are linear or straight-line relations. • The term programming refers to the method of determining a particular program or plan of action, i.e., the use of algorithms that is a well defined sequence of steps that will lead to an optimal solution. • Taken as a whole, the term linear programming refers to a family of mathematical techniques for determining the optimum allocation of resources and obtaining a particular objective when there are alternative uses of the limited or constrained resources. 4
- 2.2. CHARACTERISTICS OF LPM • Linear programming models are mathematical representations of LP problems. • Linear programming models have certain characteristics in common. • Knowledge of these characteristics enables us to recognize problems that are amenable to a solution using LP models, and to be able to correctly formulate an LP model. • These characteristics can be grouped as components and assumptions. • The components relate to the structure of a model, where as the assumptions reveal the conditions under which the model is valid. 5
- 2.1.1. COMPONENTS OF LP MODELS • There are four major components of LP models including: Objective function, decision variables, constraints and parameters. 1. Objective and Objective Function-The objective in problem solving is the criterion by which all decisions are evaluated. • Because we are dealing with optimization, the objective will be either maximization or minimization. • Hence, every LP problem will be either maximization or a minimization problem. • An LP model consists of a mathematical statement of the objective called the objective function. 6
- Cont’d 2. Decision variables- They represent unknown quantities to be solved for.. For example, how much of each product should be produced in order to obtain the greatest profit? 3. Constraints- However, the ability of a decision maker to select values of the decision variables in an LP problem is subject to certain restrictions or limits coming from a variety of sources. – The restrictions may reflect availabilities of resources (e.g., raw materials, labor time, etc.), legal or contractual requirements (e.g., product standards, work standards, etc.), technological requirements (e.g., necessary compressive strength or tensile strength) or they may reflect other limits based on forecasts, customer orders, company policies, and so on. 7
- Cont’d In LP model, the restrictions are referred to as constraints. Only solutions that satisfy all constraints in a model are acceptable and are referred to as feasible solutions. The optimal solution will be the one that provides the best value for the objective function. • Constraints can be arranged into three groups: – System constraints – involve more than one decision variable, – Individual constraints – involve only one variable, and – Non-negativity constraints – specify that no variable will be allowed to take on a negative value. The non- negativity constraints typically apply in an LP model, whether they are explicitly stated or not. 8
- Cont’d 4. Parameters- The objective function and the constraints consist of symbols that represent the decision variables (e.g., X1, X2, etc.) and numerical values called parameters. • The parameters are fixed values that specify the impact that one unit of each decision variable will have on the objective and on any constraint it pertains to as well as the numerical value of each constraint. • The following simple example illustrates the components of LP models: 9
- 10
- 2.1.2 ASSUMPTIONS OF LP MODELS • A linear programming model is valid under the following assumptions: • 1. Linearity (proportionality) • The linearity requirement is that each decision variable has a linear impact on the objective function and in each constraint in which it appears. • In terms of a mathematical model, a function or equation is linear when the variables included are all to the power 1 (not squared, cubed, square root, etc.) and no products (e.g., x1x2) appear. • On the other hand, the amount of each resource used (supplied) and its contribution to the profit (or cost) in the objective function must be proportional to the value of each decision variable. • For example, if production of one unit requires 5 hours of a particular resource, then making 3 units of that product requires 15 hours (3x5) of that resource. 11
- 2. Divisibility (Continuity) • The divisibility requirement pertains to potential values of decision variables. • It is assumed that non-integer values are acceptable. However, if the problem concerns, for example, the optimal number of houses to construct, 3.5 do not appear to be acceptable. • Instead, that type of problem would seem to require strictly integer solutions. In such cases, integer-programming methods should be used. It should be noted, however, that some obvious integer type situations could be handled under the assumption of divisibility. • For instance, suppose 3.5 to be the optimal number of television sets to produce per hour, which is unacceptable, but it would result in 7 sets per two hours, which would then be acceptable. 12
- 3. Certainty • -This requirement involves two aspects of LP models. • One aspect relates to the model parameters, i.e., the numerical values. It is assumed that these values are known and constant. • In practice, production times and other parameters may not be truly constant. Therefore, the model builder must make an assessment as to the degree to which the certainty requirement is met. • Large departures almost surely will have a significant effect on the model. • The other aspect is the assumption that all relevant constraints have been identified and represented in the model. 13
- 5. Non-negativity • -It assumes that negative values of variables are unrealistic and, therefore, will not be considered in any potential solutions. • Only positive values and zero will be allowed and the non-negativity assumption is inherent in LP models. • For example X=-12, can’t be a solution as it doesn’t give any meaning as negative number of units can’t be produced 14
- 2.3. SOLVING LP MODELS • Linear programming model can be solved by either of: • Graphical method or • Simplex or algebraic method 15
- 2.3.1.GRAPHICAL METHODS • Graphical linear programming is a relatively straightforward for determining the optimal solution to certain linear programming problems involving only two decision variables. • Although graphic method is limited as a solution approach, it is very useful in the presentation of LP, in that it gives a “picture” of how a solution is derived thus a better understanding of the solution. • Moreover, graphical methods provide a visual portrayal of many important concepts. • In this method, the two decision variables are considered as ordered pairs (X1, X2), which represent a point in a plane, i.e, X1 is represented on X-axis and X2 on Y-axis. • Graphical method has the following advantages: – It is simple – It is easy to understand, and – It saves time. 16
- STEPS IN GRAPHICAL SOLUTION METHODS Step 1. Formulate the LPM of the problem Steps in formulating LP models: a)Identify the decision variables. b)Determine the objective function. c)Identify the constraints. d)Build and validate the model. Step 2. Plot each of the constraints on the graph Step 3. Identify the feasible region -The feasible region is the solution space that satisfies all the constraints simultaneously. Step 4. Locate the optimal solution- Graphic approach to find a solution for LP model consisting of only two decision variables: a) The extreme point enumeration method b) The objective (Iso-profit or cost) function line approach 17
- Formulating the LPM of the Problem • Example1 • ABC Company is to start a production of two spare parts. The production manager would like to produce so many parts to earn per unit profit of Birr 60 from the sale of type I and Birr 50 from the sale of type II parts. However, the total available time to assemble the items, to inspect them and the space for their storage is limited to 100 hours, 22 hours and 39 square feet respectively. Four hours of assembly time, two hours of inspection time and three square feet of storage space is required to produce each unit of type I spare part. On the other hand 10 hours of assembly time, 1 hour of inspection time and 3 square feet of storage space is needed to produce each unit of type II spare parts. How many of each item should the Company produce and sale to maximize its profit? • Required: Formulate the LPM of the Problem 18
- Example 2 • An individual investor has Birr 70,000 to divide among several investments. The alternative investments are municipal bonds with an 8.5% return, certificates of deposits with a 10% return, Treasury bill with a 6.5% return, and income bonds with a 13% return. The amount of time until maturity is the same for each alternative. However, each investment alternative has a different perceived risk to the investor; thus it is advisable to diversify. The investor wants to know how much to invest in each alternative in order to maximize the return. The following guidelines have been established for diversifying the investment and lessening the risk perceived by the investor. – No more than 20% of the total investment should be in an income bonds. – The amount invested in certificates of deposit should not exceed the amount invested in other three alternatives. – At least 30% of the investment should be in treasury bills and certificates of deposits. – The ratio of the amount invested in municipal bonds to the amount invested in treasury bills should not exceed one to three. – The investor wants to invest the entire Birr 70,000. • Required: Formulate a LP model for the problem. 19
- Example 3: • A chemical corporation produces a chemical mixture for the customer in 1000- pound batches. The mixture contains three ingredients- Zinc, mercury and potassium. The mixture must conform to formula specifications (i.e., a recipe) supplied by a customer. The company wants to know the amount of each ingredient to put in the mixture that will meet all the requirements of the mix and minimize total cost. • The customer has supplied the following formula specifications for each batch of mixture. 1. The mixture must contain at least 200 lb of mercury 2. The mixture must contain at least 300 lb of zinc 3. The mixture must contain at least 100 lb of potassium • The cost per pound of mixture is Birr4, of zinc Birr 8 and of potassium Birr 9. • Required: Formulate LPM for the problem 20
- 1.MAXIMIZATION CASE • Example 1 of the previous example 1 • After formulating the LPM of the Problem, the next steps will be: • Plotting each constraints on the graph, then • Identifying the feasible solution and finally • Locating optimal solution 21
- Cont’d • 22
- Figure 2-1 Feasible Region Based on a Plot of the First Constraint (assembly time) and the Non-negativity Constraint 23
- Figure 2–2 A Completed Graph of the Server Problem Showing the Assembly and Inspection Constraints and the Feasible Solution Space 24
- Figure 2–3 Completed Graph of the Server Problem Showing All of the Constraints and the Feasible Solution Space 25
- Finding the Optimal Solution • The extreme point approach – Involves finding the coordinates of each corner point that borders the feasible solution space and then determining which corner point provides the best value of the objective function. – The extreme point theorem – If a problem has an optimal solution at least one optimal solution will occur at a corner point of the feasible solution space. 26
- The Extreme Point Approach 1. Graph the problem and identify the feasible solution space. 2. Determine the values of the decision variables at each corner point of the feasible solution space. 3. Substitute the values of the decision variables at each corner point into the objective function to obtain its value at each corner point. 4. After all corner points have been evaluated in a similar fashion, select the one with the highest value of the objective function (for a maximization problem) or lowest value (for a minimization problem) as the optimal solution. 27
- Figure 2–5 Graph of Server Problem with Extreme Points of the Feasible Solution Space Indicated 28
- Table 2–3 Extreme Point Solutions for the Server Problem 29
- SLACK VERSUS SURPLUS • Slack is the amount of a scarce resource that is unused by a given solution. Slack can potentially exist in a < constraint. – Slack variables are considered in the objective function by using a coefficient of zero for each of them. When all the constraints are written as equalities after adding a slack variable to each of them, the linear program is said to be in standard form. • Surplus on the other hand is the amount by which the optimal solution causes a > constraint to exceed the required minimum amount. It can be determined in the same way that slack can, i.e., substitute the optimal values of the decision variables into the left side of the constraint and solve. – The difference between the resulting value and the original right hand side amount is the amount of surplus. – Surplus should also be accounted for in the objective function by using coefficients of zero likewise. 30
- Table 3–5 Computing the Amount of Slack for the Optimal Solution to the Server Problem 31
- 2.MINIMIZATION CASE • Minimize Z with inequalities of constraints in > form • Example: • Suppose that a machine shop has two different types of machines; machine 1 and machine 2, which can be used to make a single product .These machines vary in the amount of product produced per hr., in the amount of labor used and in the cost of operation. • Assume that at least a certain amount of product must be produced and that we would like to utilize at least the regular labor force. How much should we utilize each machine in order to utilize total costs and still meets the requirement? 32
- 33
- 34
- Cont’d 35
- Figure 2–8 A Comparison of Maximization and Minimization Problems 36
- 3. MIXED CONSTRAINT CASE • It is only in some cases that the maximization problems consist of constraints connected to the RHS value with only <. By the same token, constraints of minimization problems are not always connected to their RHS with>. This is to mean that both maximization and minimization problems may consist of constraints connected to RHS with a mix of algebraic signs ( <, >, =). • Example Minimize Z=1500x+2400y Subjected to: 4x+Y>24 …I 2x+3y>42 …II X+4y>36 …III X<14 …IV y<14 …V x, y>0 37
- 38
- 39
- Exercise 1. Formulate the LP ABC Furniture manufacturer produces two products: Beds and Chairs. Each unit of Bed requires 3 hrs in molding unit, 4hrs in painting unit, and 1 hr in finishing. On the other hand, each unit of Chair requires 3 hrs in molding unit, 2 hrs in the paint shop and 2 hours in finishing. Each week, there are 210 hrs available in molding, 200hrs in painting, and 120 hrs in finishing unit. The demand for Beds cannot exceed 40 units per week. Each unit of Bed contributes Birr 20 to profit, while each unit of chair contributes Birr 30. 2. Maximize Z 3x + 4y S.T. 5x + 4y ≤ 200 3x + 5y ≤ 150 5x + 4y ≤ 100 8x + 4y ≤ 80 X,y ≥ 0 40 3. Minimize Z: 10x + 4y S.T. 3x + 6y ≥ 250 15x + y ≥ 150 2x + 9y ≥ 200 X,y ≥ 0
- 2.3.2.Some Special Issues a) No Feasible/Infeasible Solutions – Occurs in problems where to satisfy one of the constraints, another constraint must be violated. 41 In the graph, there is no common point in the shaded area. All constraints cannot be satisfied simultaneously and there is no feasible solution to the problem.
- b) Unbounded Problems – Exists when the value of the objective function can be increased without limit. 42 Note Here that the two corners of the region are A(0,3) and .B(2,1). The value of Max Z(A)=6 and Max Z(B)=8. But there exist number of points in the shaded region for which the value of the objective function is more than 8. For example, the point (10, 12) lies in the region and the function value at this point is 70 which is more than 8. Remark: An unbounded solution does not mean that there is no solution to the given LPP, but implies that there exits an infinite number of solutions.
- c)Redundant Constraints – A constraint that does not form a unique boundary of the feasible solution space; its removal would not alter the feasible solution space. – If a constraint when plotted on a graph doesn’t form part of the boundary making the feasible region of the problem that constraint is said to be redundant. 43 Note: The packaging hour’s constraint does not form part of the boundary making the feasible region. Thus, this constraint is of no consequence and is therefore, redundant. The inclusion or exclusion of a redundant constraint does not affect the optimal solution of the problem.
- d) Multiple Optimal Solutions –Problems in which different combinations of values of the decision variables yield the same optimal value. 44 Both C and D are optimal solutions. Any point on the line segment CD will also lead to the same optimal solution. Multiple optimal solutions provide more choices for management to reach their objectives.
- 2.3.2.THE SIMPLEX APPROACH • The graphical approach is useful in solving linear programming models having only two variables. • When the model has more than two variables, the appropriate approach is the Simplex procedure. • It begins with a feasible solution which is not optimal, and the solution is improved through continuous algebraic manipulations (iterations) until the optimal solution is determined. 45
- Cont’d • The graphical method to solving LPPs provides fundamental concepts for fully understanding the LP process. However, the graphical method can handle problems involving only two decision variables (say X1 and X2). • Simplex Method which is an efficient approach to solve applied problems containing numerous constraints and involving many variables that cannot be solved by the graphical method. • The simplex method is an ITERATIVE or “step by step” method or repetitive algebraic approach that moves automatically from one basic feasible solution to another basic feasible solution improving the situation each time until the optimal solution is reached at. 46
- STEPS IN SIMPLEX APPROACH • Step1. Formulate the linear programming model of the real world problem • Step2. Express the mathematical model of L.P. problem in the standard form • Step 3. Set up the initial simplex tableau • Step 4.Check if the current solution is optimum or not. • Step 5. Further iterate towards an optimum solution. • Step 6. Repeat Step3-5 until optimal solution is reached 47
- A.MAXIMIZATION CASE Let us take the example of a spare part production problem – Step 1 Formulate LPP Model – Step 2 Standardize the problem i.e. Convert constraint inequality into equality form by introducing a variable called Slack variable. • Slack Variables: –A slack variable(s) is added to the left hand side of a < constraint to covert the constraint inequality in to equality. 48
- Cont’d – The value of the slack variable shows unused resource. – Slack variables represent unused resource or idle capacity. – Thus, they don’t produce any product and their contribution to profit is zero. – A slack variable emerges when the LPP is a maximization problem. – Slack variables are added to the objective function with zero coefficients. – Let that s1, s2, and s3 are unused labor, machine and marketing hrs. respectively. 49
- 50 Example: Taking the microcomputer problem its standard form is as follows: Solve the problem using the simplex approach Zmax = 60X1 + 50X2 4X1 + 10X2 100 2X1 + X2 22 3X1 + 3X2 39 X1, X2 0 The standard form is:
- Standardized Form Zmax = 60X1 + 50X2 + 0S1 + S2 + 0S3 4X1 + 10X2 + S1 = 100 2X1 + X2 + S2 = 22 3X1 + 3X2 + S3 = 39 X1, X2, S1, S2, S3 0 51
- Cont’d – To represent the data, the simplex method uses a table called the simplex table or the simplex matrix. – In constructing the initial simplex tableau, the search for of the optimal solution begins at the origin. Indicating that nothing can be produced; – Thus, first assumption, No production implies that x1 =0 and x2=0 52
- Cont’d 4X1 + 10X2 + S1 +0 s2+ 0 s3= 100 4(0) +0 + s1 +0 s2+ 0 s3= 100 s1= 100 – Unused assembly time. 2x1+1x2 +0 s1 + s2+ 0 s3= 22 2(0) +1(0) + 0s1 + s2+ 0 s3= 22 s2= 22 – Unused inspection time. 3x1+3x2+ 0s1 +0s2+ s3= 39 3(0 )+3(0)+ 0s1 +0 s2+ s3= 39 s3= 39 – Unused storage space. 53 Therefore, Max.Z=60x1 +50x2 + 0 s1 +0 s2+ 0 s3 =60(0) +50(0) + 0(100) +0(22) + 0(39) = 0
- Step 3. Construct the initial tableau The initial tableau always represents the “Do Nothing” strategy, so that the decision variables are initially non- basic. • List the variables across the top of the table and write the objective function coefficient of each variable jut above it. • There should be one row in the body of the table for each constraint. • List the slack variables in the basis column, one per raw. • In the Cj column, enter the objective function coefficient of zero for each slack variable. (Cj - coefficient of variable in the objective function) • Compute values for row Zj • Computer values for Cj – Zj. 54
- 55
- Interpretation • Cj = the coefficient of the variables in the objective function in the standard form. • Cb = the coefficient of basic variables in the obj. fun. • X1, X2, ..Xn = decision variables . • S1, s2, Sn = basic variables (slack). • a11, a12,… coefficient of DV in the constraint set. • 1, 0, 1 are coefficient of basic variables in the constraints set. • RHSV = the right hand side value of the constraints. 56 56
- Step 3 cont’d Basic variable Cj 60 X1 50 X2 0 S1 0 S2 0 S3 RHSV Ratio S1 0 4 10 1 0 0 100 100/4 = 25 S2 0 2* 1 0 1 0 22 22/2 = 11 S3 0 3 3 0 0 1 39 39/3 = 13 Zj 0 0 0 0 0 0 Cj- Zj 60 50 0 0 0 0 57
- Evaluate, the initial Tableau, does it optimal or not? 58 Based on the above explanation it is not optimal: Current Solution is x1= X2 =0, and S1= 100, S2= 22 and S3 =39 Decision: S2 departing and X1 is entering variable
- Step 4. • Choose the “incoming” or “entering” variables • Note: The entering variable is the variable that has the most positive value in the Cj - Zj row also called as indicator row. Or the entering variable is the variable that has the highest contribution to profit per unit. a) X1 in our case is the entering variable b) The column associated with the entering variable is called key or pivot column ( X1 column in our case ) 59
- Step 5 • Choose the “leaving “or “outgoing” variable • In this step, we determine the variable that will leave the solution for X1 (or entering variable) Note: – The row with the minimum or lowest positive (non-negative) replacement ratio shows the variable to leave the solution. Replacement Ratio (RR) = Solution Quantity (Q) Corresponding values in pivot column Note: RR>0 – The variable leaving the solution is called leaving variable or outgoing variable. – The row associated with the leaving variable is called key or pivot row (s3 column in our case) – The element that lies at the intersection of the pivot column and pivot row is called pivot element(No 2 in our case) 60
- Step 6: Develop subsequent tableaus • Repeat step 3-5 till optimum basic feasible solution is obtained. Or: repeat step 3-5 till no positive value occurs in the Cj - Zj row. • Note: o Divide each element of the pivot row by the pivot element to find new values in the key or pivot row. o Perform row operations to make all other entries for the pivot column equal to zero. Raw operation= old Raw Value – Key column Value (New raw value 61
- Cont’d 62 Basic variable Cj 60 X1 50 X2 0 S1 0 S2 0 S3 RHS V Ratio S1 0 0 8 1 -2 0 56 56/8 = 7 X1 60 1 1/2 0 1/2 0 11 11/. 5 = 22 S3 0 0 3/2 0 - 3/2 1 6 6/1.5 = 4 Zj 60 30 0 30 0 660 Cj-Zj 0 20 0 -30 0 0
- Optimal table 63 Basic variable Cj 60 X1 50 X2 0 S1 0 S2 0 S3 RHS V Ratio S1 0 0 0 1 6 -16/3 24 X1 60 1 0 0 1 -1/3 9 X2 50 0 1 0 -1 2/3 4 Zj 60 50 0 10 40/3 740 Cj-Zj 0 0 0 -10 -40/3
- Cont’d • Since all the Cj - Zj < 0 optimal solution is reached at X1 = 9 X2 = 4 S1 = 24 hrs Z = Birr 740 • “A simplex solution is a maximization problem is optimal if the Cj – Zj row consists entirely of zeros and negative numbers (i.e., there are no positive values in the bottom row).” 64
- B.MINIMIZATION PROBLEMS • There are two methods to solve minimization LP problems: 1. Direct method/Big M-method/ –Using artificial variables 2. Conversion method –Minimization by maximizing the dual • Surplus Variable (-s): A variable inserted in a greater than or equal to constraint to create equality. It represents the amount of resource usage above the minimum required usage. Surplus variable is subtracted from a > constraint in the process of converting the constraint to standard form. Neither the slack nor the surplus is negative value. They must be positive or zero. 65
- Cont’d • Example: • 2x1+x2 < 40 ==>is a constraint inequality x1= 12 and x2= 11==> 2x1+x2+s = 40 ==>2(12)+11+s= 40 ==> s=5 unused resource • 5x1+3x2 < 45 x1= 12 and x2= 11==> 5x1+3x2+s= 45 ==>5(12)+3(11)+s = 45 ==> s=0 unused resource (No idle resource) • 5x1+2x2 >20 x1= 4.5 and x2= 2==> 5x1+2x2- s = 20 ==>5(4.5)+2(2)-s = 20 ==> s=6 unused resource • 2x1+x2 >40 x1= 0 and x2= 0(No production)==> 5x1+2x2- s = 20 ==>5(4.5)+2(2)-s= 20 ==> s=-6 (This is mathematically unaccepted) 66
- Artificial variable • Thus, in order to avoid the mathematical contradiction, we have to add artificial variable (A) • Artificial variable is a variable that has no meaning in a physical sense but acts as a tool to create an initial feasible LP solution. • Note: Type of constraint To put into standard form < --------------------------------------------- Add a slack variable = ---------------------------------------------Add an artificial variable > ---------------------- Subtract a surplus variable and add artificial variable 67
- Big M-Method • The Big-M Method is a method which is used in removing artificial variables from the basis. • In this method; we assign coefficients to artificial variables, undesirable from the objective function point of view. • If objective function Z is to be minimized, then a very large positive price (called penalty) is assigned to each artificial variable. • Similarly, if Z is to be maximized, then a very large negative price (also called penalty) is assigned to each of these variables. 68
- Characteristics of Big-M Method: A. High penalty cost (or profit) is assumed as M B. M is assigned to artificial variable A in the objective function Z. C. Big-M method can be applied to minimization as well as maximization problems with the following distinctions: – Minimization problems -Assign +M as coefficient of artificial variable A in the objective function Z – Maximization problems: -Here –M is assigned as coefficient of artificial variable A in the objective function Z D. Coefficient of S (slack/surplus) takes zero values in the objective function Z E. For minimization problem, the incoming variable corresponds to the highest negative value of Cj-Zj. F. Solution is optimal when there is no negative value of Cj-Zj.(For minimization case) 69
- Example: • 1. Minimize Z=25x1 +30x2 Subject to: 20x1+15x2 > 100 2x1+ 3x2 > 15 x1, x2 > 0 70
- Solution Step 1: Standardize the problem Minimize Z=25x1 +30x2 +0s1+0s2 +MA1+MA2 Subject to: 20x1+15x2- s1+A1 = 100 2x1+ 3x2 –s2+A2 = 15 x1, x2 , s1, s2 ,A1 ,A2 > 0 Step 2: Initial simplex tableau The initial basic feasible solution is obtained by setting x1= x2= s1= s2=0 No production, x1= x2= s1=0==>20(0) +15(0) - 0+A1 = 100 ==> A1 = 100 x1= x2= s2=0==>0(0)+3(0) - 0+A2 =15==> A2 = 15 71
- Cont’d Initial simplex tableau 72 Note: Once an artificial variable has left the basis, it has served its purpose and can therefore be removed from the simplex tableau. An artificial variable is never considered for re-
- Cont’d 2nd Simplex Tableau 73
- Cont’d 3rd Simplex Tableau 74
- Cont’d Cj - Zj > 0==>Optimal solution is reached X1=5/2 X2=10/3 and MinZ=162.5 Note: As long as an “A” variable is available in the solution variable column, the solution is infeasible. 75
- C.MAXIMIZATION WITH MIXED CONSTRAINTS • The simplex method requires that all the constraints be in standard form. • Constraints that are ≤ can be put in to standard form by adding a slack variable in the constraint. • Constraints with ≥or = sign are handled a bit differently. 76
- Cont’d • To change equality constraints to standard form add artificial variables • To convert this inequality to standard form subtract surplus variable first and add artificial variable 77
- Solve using simplex method Maximize Z= 6X + 8Y Subject to: Y ≤4 X+ Y =9 6X+2Y ≥ 24 X, Y ≥ 0 78
- In standard form Max! 6X + 8Y + 0S1 + 0S3 – MA2 – MA3 Subject to: Y+ S1 = 4 X+Y + A2 = 9 6X+ 2Y –S3 + A3 = 24 All variables ≥ 0 79
- Initial table BV CBV X Y S1 S3 A2 A3 6 8 0 0 -M -M Quantity S1 0 A2 -M A3 -M 0 1 1 0 0 0 1 1 0 0 1 0 6 2 0 -1 0 1 4 9 24 Z C-Z -7M -3M 0 M -M -M 6+ 7M 8+3M 0 -M 0 0 -33M 80
- Second table BV CBV X Y S1 S3 A2 A3 6 8 0 0 -M -M Quantity S1 0 A2 -M X 6 0 1 1 0 0 0 0 2/3 0 1/6 1 -1/6 1 1/3 0 -1/6 0 1/6 4 5 4 Z C-Z 6 2-2M/3 0 -1M/6 -1 -M 1+M/6 0 6+2M/3 0 1+ M/6 0 -1-m/6 24 - 5M 81
- Third table BV CBV X Y S1 S3 A2 6 8 0 0 -M Quantity Y 0 A2 -M X 6 0 1 1 0 0 0 0 -2/3 1/6 1 1 0 -1/3 -1/6 0 4 7/3 8/3 Z C-Z 6 8 6 + 2M/3 -M/6 -1 -M 0 0 -6-2M/3 1+ M/6 0 48 -7M/3 82
- Fourth table BV CBV X Y S1 S3 6 8 0 0 Quantity Y 8 S3 0 X 6 0 1 1 0 0 0 -4 1 1 0 -1 0 4 14 5 Z C-Z 6 8 2 0 0 0 -2 0 62 83
- Special case in simplex method a) Unbounded solution – Zmax 84 Cj 6 9 0 0 Basic X1 X2 S1 S2 Q X2 5 -1 1 2 0 30 S2 0 -2 0 -1 1 10 Zj -9 9 18 0 270 Cj - Zj 15 0 -18 0 -
- b) Multiple optimal solution • Zmax 85 Cj 2 4 0 0 Basic X1 X2 S1 S2 Q X2 4 1/2 1 1/2 0 5/2 S2 0 1/2 0 -1/2 1 3/2 Zj 2 4 3/2 0 10 Cj - Zj 0 0 -3/2 0 -
- c) Infeasible solution 86 Cj 5 8 0 0 M Basic X1 X2 S1 S2 A2 Q X1 5 1 0 1 3 0 200 X2 8 0 1 -1 -2 0 100 A2 M 0 0 0 -1 1 20 Zj 5 8 -3 -1-M M 1800+20M Cj - Zj 0 0 3 1+M 0 -
- d) Degeneracy 87 Cj 2 3 0 0 0 Basic X1 X2 S1 S2 S3 Q 0 S1 2 5 1 0 0 20 0 S2 1 1 0 1 0 5 0 S3 1 2 0 0 1 8 Zj 0 0 0 0 0 0 Cj - Zj 2 3 0 0 0 -
- 2.4. POST OPTIMALITY ANALYSIS • Carried out after the optimal solution is found • Is begins with the final simplex tableau • There are two scenarios in this case –Duality and –Sensitivity analysis 88
- 2.4.1. Sensitivity Analysis • Sensitivity analysis carries the LP analysis beyond the determination of the optimal solution and begins with the final simplex tableau. • Its purpose is to explore how changes in any of the parameters of a problem, such as the coefficients of the constraints, coefficients of the objective function, or the right hand side values; would affect the solution. • For this, instead of resolving the entire problem as a new problem with new parameters, we may consider the original optimal solution as an initial solution for the purpose of knowing the ranges, both lower and upper, within which a parameter may assume a value. 89
- a) A change in the RHS of a constraints • Simultaneous changes are not contemplated • Change in RHS or Q of one constraint is considered at a time • The range of feasibility is the range over which the RHS value of a constraint can be changed and still have the same shadow prices. • Starts with Considering shadow price • Shadow price: is a marginal value; it indicates the impact that a one unit change in the value of the constraint would have on the value of the objective function. • Shadow prices are the values in the Zj-row of slack columns 90
- Determining Range of Feasibility: How? Feasibility Ratio (FR) = quantity ÷ respective slack variable coefficient: in the final optimal solution tableau The determine the range of feasibility according to the following rule The effect of a RHS change on the objective function depends on whether the change leads to relaxation or tightening of the constraint ≤ ≥ Allowable RHS increase The negative FR closest to zero Smallest positive FR Allowable RHS decrease Smallest positive FR The negative FR closest to zero 91
- Coun’t… • Relaxation of a constraint has favorable impact on the objective function. • However, tightening has unfavorable impact on the objective function 92
- Example The LPM of the micro computer problem is: Max Z: 60X1 + 50X2 Subject: Assembly time: 4X1 + 10X2 ≤100 Inspection time: 2X1 + X2≤22 Storage space: 3X1 + 3X2 ≤39 X1 and X2 ≥ 0 If the amount of assembly time was increased by one hour, there would be no effect on profit; if the inspection time was increased by one hour, the profit would have increased by $10, and if storage space was increased by one cubic foot, profit would increase by $40/3. The reverse holds true. one –unit decrease in inspection time would decrease profit by $10 since its shadow price is $ 10. 93
- Consider the previous optimal solution and compute 1. Compute the range of feasibility for the constraints in the above illustration. 2. The manager in the above problem is contemplating one of two possible changes in the level of the storage constraint. One change would be an increase of 3 cubic feet in its level and another change will be an increase of 8 cubic feet. Determine the revised optimal solution for each possible change. 94
- Example 3. Suppose the manager in the above problem is contemplating a decrease in the storage space due to an emergency situation. There are two possibilities being considered, a decrease of 6 cubic feet and a decrease of 9 cubic feet; what will be the impact on the optimal solution? 95
- Determine the FR and range of feasibility for each CONS. 1. Recall the original value of the resources Original value constraints S1 S2 S3 100 S1 1 6 -16/3 22 S2 0 -1 -1/3 39 S3 0 -1 2/3 2. Ratio = Q/respective slack values S1= 24/1= 24 S2= 24/6= 4 S3= 24/-16/3= -4.5 9/0=undefined 9/-1= -9 9/-1/3= -27 4/0= undefined 4/-1= -4 4/2/3= 6 NB: If there is no suitable negative/positive FR, then the allowable increase/decrease is ∞ (undefined) 96
- 3. Find the range of feasibility Constraints Original value Lower limit Upper limit Range of feasibility S1 100 100-24= 76 100+∞ 76-∞ S2 22 22-4= 18 22+4= 26 18-26 S3 39 39-6 = 33 39+4.5= 43.5 33-43.5 Therefore: Constraint one (assembly line): 100-24 up to 100+∞= (76-∞) Constraint two (inspection time): 22-4 up to 22-4= (18-26) Constraint three (storage space): 39-6 up to 39+4.5= (33-43.5) 97
- b) A change of coefficient of objective function • Two cases Case 1: Range of insignificance : The range over which the non basic variables objective function coefficient can change without making these variables entering in the solution Case 2. Range of optimality The range over which the objective function coefficient of basic variables can change without changing the optimal values i.e. without changing basic and non basic variables but change the optimal function value. 98
- Range of optimality • Range of optimality is calculated for only original decision variables because slack/surplus variables and artificial variables do not have any contribution for the objective function. 99
- Cont’d Determine the range of optimality for the following final simplex tableaue Min Z = 10X1 + 3X2 Subject to: 2X1 + X2 ≥80 X1 + 4X2 ≥ 200 X1 and X2 ≥ 0 100
- Range of Insignificance (RoI) –The range over which a non-basic variable’s objective function coefficient can change without causing that variable to enter in to the solution mix is called its range of insignificance –In a maximization problem if a variable is not currently in solution, its objective function coefficient would have to increase by an amount that exceeds the Cj –Zj value in order for it to enter as a basic variable in the optimal solution. Any lesser value of its objective function coefficient would keep it out of solution –In a minimization problem if a variable is not in solution, its objective function coefficient would have to decrease by an amount that exceeds the Cj –Zj value for that variable in the final tableau in order for it to enter as a basic variable in the optimal solution. 101
- The final tableau of the above problem is: Basic variable 10 X1 3 X2 0 S1 0 S2 r.h.s. S2---0 0 0 -4 1 120 X2---3 2 1 -1 0 80 Zj 6 3 -3 0 160 Cj- Zj 4 0 3 0 240 102
- Cont’d • X1 is non-basic variable in this optimal table, so, determine RoI for X1 Since the problem is minimization the objective function coefficient of x1 must decrease by an amount that exceeds 4 (i.e the Cj-Zj row coefficient of x1 in the optimal table.) • Therefore the range of insignificance for X1 is: • Its current OF coefficient-4 up to its current of coefficient + ∞ • 10 – 4 up to 10+ ∞: 6 - ∞ • Interpretation: The cost per unit of Product A must be less than 6 before the company starts manufacturing product A 103
- 2.4.2.Duality • Every LPP has another LPP associated with it, which is called its dual. • The given problem is called the primal. • The dual of a dual is the primal 104 Primal Dual Objective is minimization Objective is maximization and vice versa > type constraints < type constraints No of columns No of rows No of rows No of columns No of decision variables No of constraints No of constraints No of decision variables Coefficient of Object function RHS value RHS value Coefficient of Object function
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