Linear Programming Module- A Conceptual Framework

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Linear Programming Module- A Conceptual Framework

  1. 1. Linear Programming Module- A Conceptual Framework Prepared By P.K. ViswanathanIntroduction:An economist faces the problem of making an optimum allocation of resources amongcompeting projects. A business-planning manager has to decide how many units of eachproduct be produced to maximize profit subject to constraints on production capacity anddemand for these products. We need a model that can help us to find the best solution inthe context of constraints that are operating on the problem. Linear programmingprovides solutions to such problems. The word programming implies planning.Learning Objectives:After reading this module, you will be able to: Know what is Linear Programming (LP)  Formulate Linear Programming Problems  Use Graphical Method to Solve LP  Appreciate Shadow Prices and Interpret  Appreciate Computer Solutions for Large LP problems Contents: 1. What is Linear Programming? 2. Steps in formulating LP Problems 3. Solution-Graphical Method 4. Computer Solution 5. A Comprehensive Case on LP 6. Module Summary 7. Review Questions1. What is Linear Programming?Linear Programming (LP) is concerned with the best way of allocating scarce resourcesamong competing ends. The word optimization is used in the context of eithermaximizing or minimizing an objective function. Organizations do have problems suchas minimizing cost of production, or maximizing the profit. The resources that areavailable in limited quantities are called constraints and the optimization will have to takeplace subject to the constraints the organization has. 1
  2. 2. "Programming" here means planning and has nothing to do with computer programming.The word linear" is very important. Linear implies that the objective function andconstraints are all linear namely the variables are raised to the power 1 only. Theconstraints could be of three types: 1) less than or equal to 2) equal to and 3) greater thanor equal to. Symbolically, they are represented as respectively. , ,To put it succinctly, linear programming is a very widely used OR model that maximizesor minimizes an objective function subject to a number of limiting factors calledconstraints. All relationships among the variables are linear. That is objective function,and constraints are all linear. Another requirement of LP is that all variables must be non-negative (0 or positive). You cannot have a negative solution in linear programmingalthough mathematically you may get negative solutions for the decision variables.Some Examples where LP is applied  An advertising company wants to maximize exposure of a client’s product and has a choice of advertising in different media that include TV, Radio, and magazines. These media charge different advertising costs. The ad company has to achieve its objective of maximizing exposure subject to the constraints of ad budget, minimum and maximum number of ads in the various media. LP can help the company solve this problem.  furniture manufacturer producing a number of items would like to maximize his A profit. He has certain commitments to his customers in the form of minimum and maximum quantities to be supplied for each item. Also in his plant, the available production time in each of the 3 departments is limited. LP can help him find an optimum product mix that can maximize his total profit.  Applications of linear programming are plenty. Product mix problem, media planning, transportation problem, blending problem, scheduling of nurses and doctors to the patients in a hospital, and capital investment are among the many where linear programming can be applied.2. Steps in formulating LP Problems  Identify the decision variables. Decision variables are those for which you are trying to find the solution. Identifying the decision variables is important in solving an LP problem  Formulate the objective function in terms of the decision variables If the organization’s goal is to maximize profit, the objective function will be a maximization case. Likewise, in a blending problem, if the aim is to minimize cost, then the objective function will be a minimization case. At any rate, you must be able to express the objective function in terms of the decision variables. 2
  3. 3.  Formulate the constraints in terms of the decision variables. Constraints represent limitations on resources. They will also have to be written in terms of the decision variables. Note: As the solution for the decision variables should not turn out to be negative, you have to specify that they are zero or positive. This is termed as non-negativity constraint. In any LP problem, it is generally understood that all the decision variables are non- negative.Example: One of the divisions of a small-scale unit manufactures two products. Both theproducts require two raw materials and the consumption per unit production is givenbelow: Table -Consumption in kg per unit production Product 1 Product 2 Maximum Availability of Raw Materials (kg)Raw Material1 2 3 18Raw Material2 1 1 8Further, it is known from market research that product 1 cannot be sold more than 4 units.The profit per unit of product 1 is $ 15, and product 2 is $ 21. You will have to help thissmall-scale unit decide how many units of product 1 and product 2 to produce so as tomaximize the total profit.FormulationIdentify the decision variables:Let x1 be the number of units of product 1 to be producedLet x2 be the number of units of product 2 to be producedFormulate the objective function in terms of the decision variables:The profit per unit of product is $. 15 and product 2 is $.21. If you produce x1 units ofproduct 1 and x2 units of product 2, then the total profit =15x1 + 21x2 . This has to bemaximized. In linear programming terminology, we write the objective function asZ = 15x1 + 21x2 . Z is to be maximized.Formulate the constraints in terms of the decision variables:The maximum quantity of product 1 that can be sold is 4 units. Sox1  (Sales constraint) 4 3
  4. 4. The maximum availability of raw material 1 is 18 kg. To produce x1 units of product1 andx2 units of product 2, you need 2x1 + 3x2 kg. So,2x1 + 3x2 18 (raw material 1 availability) (see table of consumption above)Similarly we can work out the raw material availability constraint for raw material 2. Andit isx1 + x2 8 (raw material 2 availability)Complete formulation for the example:Maximize Z = 15x1 + 21x2Subject to:x1 42x1 + 3x2 18x1 + x2 8And x1, x2 0(non-negativity)Inline Question: Linear programming is one of the computer programming languages.True or False.Answer: The statement is false. The word programming means planning and linear meansthe relationships among the variables are raised to the power 1. This has nothing to do withcomputer programming language.Assumptions in Linear Programming: 1) Proportionality: Because we assume linear relationship among variables in LP, proportionality is inevitable. Suppose you need 5 kg of input to produce one unit. Then to produce two units, you need 10 kg of input. This is not necessarily true. 2) Additivity: When we formulate the constraints, we add resource requirements of the decision variables. We ignore the interaction term in the expression. Interaction measures the effect of any combination of decision variables. 3) Divisibility: Divisibility does not guarantee integer solutions. In a particular problem, fractional solution does not make sense. For example, in a product mix problem, you may get a solution that says produce 10.3 units of model 1 and 14.4 4
  5. 5. units of model 2. It does not make sense. Rounding to the whole number sometimes may vitiate the solutions. Of course you have a technique called integer linear programming that can help you solve problems where you are looking for integer solutions.4) Certainty: In linear programming, you assume all variables are known without any doubt. Thus, you assume a deterministic or certainty scenario. But in reality, you may have to assess carefully the magnitude of uncertainty.3. Solution-Graphical MethodGraphical method of solving a linear programming model is possible only for twovariables. To get a feel for linear programming, graphical method is extremely useful.Let us take the same example used for formulating LP model. Example Problem:Graphical SolutionConstraint Lines and Feasibility RegionDrawn in a combined manner x2 8 x1 + x2= 8 7 6 A(0,6) X 1=4 5 4 B (4, 3 1/3) 3 Feasibility Region 2 2 x1 + 3x2 =18 1 D(0,0) C(4,0) x1 1 2 3 4 5 6 7 7 8 8 9 9 10 10 5
  6. 6. Please see the feasibility region in the diagram bounded by thick lines with corner pointsco-ordinates marked by A(0,6), B(4, 3 1/3), C(4,0) and D(0,0). This is the region thatsatisfies all the three constraints. Co-ordinates for B is obtained by solving the twoequations X1 = 4 and 2X1+3X 2 =18. Do not read the co-ordinates from the graph. You maynot get the accuracy. The word feasibility is important here. It says that every pair of co-ordinate points in this region is a feasible solution to the problem given. But, what is theoptimum solution? See next. Optimal Solution In any linear programming problem the optimum solution when it exists is always one of the corner points in the feasibility region. Select that corner point which gives the maximum Z value. (Note: For minimization case select that corner point that gives the minimum Z value). In our case we need to select the maximum Z value. Corner Point Z Value : Z= 15x1 + 21x2 A(0,6) 126 B(4, 3 1/3) Optimum 130 C(4,0) 60 D(0,0) 0 6
  7. 7. The optimum solution that maximizes the profit is to produce 4 units of product 1 and3 1/3 units of product 2. The maximum profit is $. 130.Inline question: If the example problem discussed is a minimization problem, what is theoptimum solution?Answer: Take that corner point that gives the minimum value of Z. Omit the trivialsolution D(0,0) that says don’t produce any thing. Z is minimum for C(4,0). The value of Z=60. The optimum solution is to produce 4 units of product 1 and 0 units of product 2.Assignment: Solve the following LP problem graphically:Maximize Z = 8X1 + 10X2Subject to: 8X1 + 4X2 240 6X1 +12X2 360 X1, X2 04. Computer Solution:Graphical method just provides some insights into linear programming. It can solveproblems having two variables only. In practice, complex problems involving more thantwo variables are encountered. The efficient way of solving such problems is to use“Solver” in Microsoft Excel. You can get output giving insights into shadow prices andsensitivity analysis apart from the usual optimum solution. The method used for solvinglinear programming problems is called “simplex method” discovered by Dantiz. Doing thismanually in practice is not only time consuming but also error prone. Computer solution issimple and best.The emphasis in this module is on your ability to formulate a linear programming modelfor a given problem and then solve it using “Solver”. The technical terms associated withlinear programming purely from a mathematical perspective will be avoided. For amanager, a problem well formulated is half solved. The simple slogan I would popularize is“Excel in formulating any LP problem and solve it using Excel”.Before we move on to solving some complex LP problems, let us first solve a simpleproblem using solver. Take the example used for graphical method. Follow the step-by-stepprocedure. The formulation of this problem on the spreadsheet is given below: 7
  8. 8. Model Formulation on the spreadsheetExplanation on the Formulation:This is a crucial step in using solver. Column A is used for labeling the linear programmingvocabulary. Decision Variables appear in cell A2. In Column B and C, Product1 andProduct 2 are labeled. Corresponding to Row 2 and columns B and C values 1 and 1 areentered. These are the initial values of the decision variables X1 and X 2 we have used whileformulating the LP model. These initial values are varied by solver to get the optimumvalues of X1 and X2. Please remember, solver cannot accept X1 and X2 in the respectivecells. Please note that you are at liberty to use your own labeling and Excel is flexible inthis regard.In A4 cell Objective Function is written. Against this label, under Column B and C values15 and 21 are entered. These are the unit profits of the two products. F3 is labeled as TotalProfit and F4 gives the profit that is obtained by multiplying the coefficients of theobjective function with the initial values of the decision variables. That is (15)(1)+21(1)=36. In order that Excel understands this, you create a formula in F4(target cell)=B4*B2+C4*C2. Excel will vary the values of B2 and C2 to get the optimum values ofProduct1 and Product2. 8
  9. 9. Then we label in Column A in Row 6 Constraints. Under this we further label theindividual constraints namely Sales for Product1, Raw material1 availability, and Rawmaterial2 availability. In Column B and C we enter the coefficients of the decisionvariables corresponding to these constraints. The value 1 is entered under column Bcorresponding to Sales for Product1 and value 0 is entered under column C for Product2. InColumn D corresponding to Sales for product1, the value 1 appears. This is calculated bythe formula =B7*B2+C7*C2 and this is 1. In Column E corresponding to Sales forProduct1, we label the symbol < = meaning less than or equal to constraint. This labeling isdone for our own understanding and Excel ignores this. In Column F corresponding toSales for Product1, we have entered the value 4. Now if you look at the entire Row 4, forthe constraint Sales for Product1, you are saying B7*B2+C7*C2 < = 4. This is same assaying X1 < =4. Likewise the constraints on raw material 1 and raw material 2 are workedout. For better clarity, spreadsheet showing model equations are given next:Spreadsheet Giving Model Equations 9
  10. 10. Solution: Now you are ready to invoke solver. Click Tools, and click Solver. You getIn “Set Target Cell” highlight F4 that contains the objective function. In “Equal To” clickMax because you want to maximize the objective function. “By Changing Cells” isSolver’s way of saying that the initial values of the decision variables are varied untiloptimal solution is reached. You highlight the cells B2:C2. You now getSolution continues: 10
  11. 11. You now come to “Subject to the Constraints”. Solver accepts all the three types ofconstraints (< =, =, > =). When you develop a model, you better arrange the constraintsaccording to types in the spreadsheet itself. That is all < = are done first; then > = and then=. This will save a lot of time. You can enter a set of constraints together in Solver byhighlighting the appropriate cells. Here all the constraints are < = and are arranged in thatorder. You click “Add” and you getIn “Cell Reference” you highlight D7:D9. Click < =. Then highlight cells F7:F9 forconstraint values. You have now entered all constraints. You get nowClick OK and you getComplete Formulation of LP Model for the Example 11
  12. 12. The last two things that remain to be done by you is to tell Solver that you are solving alinear programming model and the decision variables are non-negative. Click Options.Then click Assume Linear Model and Assume Non-Negative. After this, you click Solveand you getNow click OK. You get the solution shown next.Model Giving Optimal SolutionAs you can see the optimum solution is to produce 4 units of Product1 and 3.33 units ofProduct2. The maximum total profit is $. 130. Constraints 1 and 2 are fully exhaustedmeaning that the resources are fully utilized. In the LP parlance, the word slack is used.Here slack is 0. Slack represents the unused quantity of a resource. For constraint3, there is 12
  13. 13. a slack of 0.67 kg that was unutilized. This cannot be helped as things stand. Any attemptwe make to use more of this will not give a profit better than $.130.Sensitivity AnalysisFinding the solution to a linear programming model is only a first step. A manager wouldlike to know how sensitive the solution is to changes in inputs and assumptions. Let usinterpret the sensitivity analysis output given by solver for the LP problem just solved.When you click solve, you actually getIf you now press OK, you get the optimum solution. Before clicking OK, highlightSensitivity and then click OK. You get 13
  14. 14. Shadow Price:  The shadow price gives the incremental increase in profit when the right hand side value of a constraint is increased by one unit, and the decrease in profit when the right hand side value of a constraint is decreased by one unit. Shadow prices play a key role in evaluating the worth of increasing the resources and also new product evaluation. Let us interpret the shadow prices for our example.  Shadow price corresponding to constraint 1 namely Sales for Product1 is seen to be =1. This means that the objective function (total profit) will increase or decrease by $1 if the sale quantity for product1 is increased or decreased by $1. From a marketing angle every additional unit sold on product 1 will increase the profit by $1.  Shadow price corresponding to constraint 2 is 7. This means that an increase or decrease of 1 kg in availability of raw material 2 will produce an increase or decrease of $7 in the objective function. So, the moral of the story is that every additional kg of raw material 2 made available will increase the profit by $7.  Shadow price for raw material 2 is 0. It does not make any impact on the objective function.  Thus shadow prices for Sale of Product 1, Raw Material 1, and Raw Material 2 are $1, $7, and $0 respectively.  The allowable increase and allowable decrease shown in the output indicates the range for which the shadow prices will hold true.Inline Question: For the example discussed, if the small-unit’s R&D department hascreated a new product called Product 3. It is quite profitable, $25 per unit. It requires 4kgof raw material 1 and 1 kg of raw material 2. Should Product 3 be produced?Solution: The opportunity cost of producing one unit of Product 3 is:(Shadow price of raw material 1)(Consumption of raw material 1 per unit production)+(Shadow price of raw material 2)(Consumption of raw material 2 per unit production)=$7 4+0 1=$28This represents the opportunity cost that the company will have to forego by not producingProduct 1 and Product 2. The profit per unit of Product 3 is $25. Since this is less than$28(opportunity cost of not producing Product 1 and Product 2), Product 3 should not beproduced. 14
  15. 15. 5. Comprehensive Case on LPProgrammers Recruitment PlanningA software company conducts a short applied certificate course in computerprogramming for trainee programmers. Trained programmers are used as instructors inthe certificate course in the ratio of one for every ten trainee programmers recruited. Thecertificate course lasts for one month. From past experience, it has been found that out often trainee programmers hired, only six complete the certificate course successfully (theunsuccessful trainee programmers are sent out).Trained programmers are also needed for companys computer assignments and thecompanys requirements for the next three months are as follows: Month 1 200 Month 2 300 Month 3 350In addition, the company requires 400 trained programmers in the beginning of Month 4.There are 230 trained programmers available as of now. Salary Details are: Each Trainee Programmer Rs. 4000 Each Trained Programmer Rs. 7000 (Doing company’s computer assignments or as instructor) Each Trained Programmer who remains idle Rs. 5000 (Union agreement prevents firing Trained Programmers)Decide and set up the hiring and training schedule of programmers, which will meet thecompanys requirements at minimum costProblem Formulation for the Case:•Identify the decision variablesLet X1 =Trained programmers functioning as instructors in Month 1Let X2 =Trained programmers who remain idle in Month 1Let X3= Trained programmers functioning as instructors in Month 2Let X4= Trained programmers who remain idle in Month 2Let X5= Trained programmers functioning as instructors in Month 3Let X6 =Trained programmers who remain idle in Month 3 15
  16. 16. •Formulate the objective functionMinimize Z = 4000(10X1+10X3+10X5) +7000(X1+X3+X5)+5000(X2+X4+X6)Z = Total Variable Cost. The salary to be paid to programmers in Month 1, Month 2, andMonth 3 are fixed cost that cannot be reduced. Hence, in the objective function, it is notincluded. Simplifying Z = 47000X1+5000X2+47000X3+5000X4+47000X5+5000X6Note: Because trained programmers are used as instructors in the course in the ratio ofone for every 10 trainee programmers recruited, the following are true: 10X1 trainee programmers are recruited in Month 1 10X3 trainee programmers are recruited in Month 2 10X5 trainee programmers are recruited in Month 3•Formulate the constraints 230 = 200+X1+X2 (Month 1 constraint) 230+6X1 =300+X3+X4 (Month 2 constraint) 230+6X1+6X3 =350+X5+X6 (Month 3 constraint) 230+6X1+6X3+6X5 = 400 (Month 4 constraint)Note: Out of 10X1 trainee programmers recruited, 6X1 are successful in the course inMonth 1 and are available for company’s computer assignments or functioning asinstructors in Month 2. Similarly those who are successful in completing the course inMonth 2, and Month 3 are available in the subsequent month for company’s computerassignments or functioning as instructorsUpon simplification, the constraints are:Month 1 X1+X2 = 30Month 2 6X1-X3-X4 = 70Month 3 6X1+6X3-X5-X6 = 120Month 4 6X1+6X3+6X5 = 170 X1,X2,X3,X4,X5,X6 are 0 (Non-negativity)Complete LP Formulation for the Example:Minimize Z=47000X1+5000X2+47000X3+5000X4+47000X5+5000X6Subject to constraintsX1+X2 = 30(Month 1)6X1-X3-X4 = 70(Month 2)6X1+6X3-X5-X6 = 120(Month 3)6X1+6X3+6X5 = 170(Month 4)X1, X2, X3, X4, X5, X6 are 0 (Non-negativity) 16
  17. 17. Inline Question: Obtain optimum solution for the case problem above using solver. Youshow the initial model formulation on the spreadsheet and then the model solution.Solution: Follow the step-by step procedure discussed in this module. It is just anextension of the two-decision variables case to six-decision variables case. Every stepwill be the same. Spreadsheet Displaying Model Formulation 17
  18. 18. Spreadsheet Displaying Optimum SolutionAs you can see that the minimum cost is Rs 1416531. The optimum values of thedecision variables are:X1 =Trained programmers functioning as instructors in Month 1 = 13X2 =Trained programmers who remain idle in Month 1 = 17X3= Trained programmers functioning as instructors in Month 2 = 8X4= Trained programmers who remain idle in Month 2 = 0X5= Trained programmers functioning as instructors in Month 3 = 7X6 =Trained programmers who remain idle in Month 3 = 0We have rounded the values of the decision variables to integers as fractional values ofprogrammers are not possible. Please note that all constraints are fully satisfied when youtake the fractional values. 18
  19. 19. 6. Module SummaryThis module has introduced you to a very useful resource optimization tool called linearprogramming. Specifically this module focused on:  Meaning, definition, and examples of linear programming  Steps in formulating linear programming model  Explaining the LP model formulation steps through an example  Problem formulation is key to solution  Computer solution using Solver of Microsoft Excel  Explaining computer solution through an example  Sensitivity analysis as an important tool for mangers to see the changes in solution as a result of changes in inputs and assumptions.  The role of shadow price in evaluating the worth of increasing or decreasing resource quantities as well as the worth of a new product  comprehensive case discussion on LP model formulation and computer solution A7. Review Questions (Test Items)Discussion TopicAnalyze, criticize, and explain the following statement:“ Linear programming is an oversimplification of real life problems. It assumes only oneobjective function when organizations are facing multiple objectives”. 1) What is linear programming? Answer: Linear programming is a technique in management science that maximizes or minimizes an objective function subject to a number of limiting factors called constraints. 2) What are the assumptions of a linear programming model? 19
  20. 20. Answer: The model assumes proportionality, additivity, divisibility, and deterministic scenario.3) What are the steps in a linear programming model formulation? Answer: 1) Identify the decision variables 2) Formulate the objective function in terms of the decision variables 3) Formulate the constraints in terms the decision variables 4) Superimpose the non-negativity requirements on decision variables (0 or positive)4) What is sensitivity analysis in LP? Answer: Sensitivity analysis means how the linear programming model solution reacts to changes in input parameters such as costs, prices, and resources. It attempts to answer how sensitive the solution is to changes in inputs and assumptions.5) What is shadow price? Answer: Shadow price gives the incremental increase in profit when the right hand side value of a constraint is increased by one unit, and the decrease in profit when the right hand side value of a constraint is decreased by one unit. Shadow prices play a key role in evaluating the worth of increasing the resources and also new product evaluation.6) Mini Case:Mr. Joseph, investment manager of Wizard Finance has $120000 to invest in new ITstocks. Data on four chosen stocks are given below:Stock Dividend Growth % Price per Share1 1.5 30 352 0.75 25 453 2.0 28 404 0.5 35 55Mr. Joseph has been asked to maximize growth in $ value excluding dividends of thisportfolio of shares in the coming year. It is Wizard’s policy to get at least $2500 asannual dividend from this portfolio of shares. Wizard also wants to play safe bystipulating the following conditions. Amount invested in stock 4 cannot be more than55% of amount invested in stock 1. Further the company requires that at least 15% oftotal investment be in stock 2.Formulate and solve this case as a linear programming problem and obtain theoptimum solution-using Solver. 20
  21. 21. Solution:Identify the decision variables Let X1 = $ amount invested in Stock 1 Let X2 = $ amount invested in Stock 2 Let X3 = $ amount invested in Stock 3 Let X4 = $ amount invested in Stock 4Formulate the objective functionFrom the wording of the problem, it is clear that Wizard would like to maximize thegrowth of this portfolio of shares.Maximize Z = 0.30X1+0.25X2+0.28X3+0.35X4 Formulate the constraintsStock Dividend Growth % Price per Share1 1.5 30 352 0.75 25 453 2.0 28 404 0.5 35 55 The amount set aside for investment is $120000. Hence the total amount invested in the four shares has to be less than or equal to $120000. X1+X2+X3+X4 < = 120000(Investment amount constraint) The company specifies an annual dividend of at least $2500 for this portfolio of shares. This is a bit tricky. Since we know dividends per share as well as the price of each share. The annual dividend from this portfolio is =   1.5  0.75    2.0   0.5    X1    X2    X3   . This must be greater than or equal to X4   35 45    40   55 $2500. Hence this constraint is   1.5  0.75    2.0   0.5    X1    X2    X3   > =2500(Dividend constraint). Upon X4   35 45   40   55 simplification this constraint becomes 0.43X1+0.017X2+0.050X3+0.009X4 > =2500(Dividend constraint) 21
  22. 22. Amount invested in Stock 4 cannot be more than 55% of amount invested inStock 1. This constraint can be written asX4 < = 0.55X1. Simplifying this is written asX4-0.55X1< =0(Safety constraint)The last constraint is at least 15% of total amount invested must be in Stock2.This can be written asX2 > = 0.15(X1+X2+X3+X4). Simplifying this expression, this constraint becomes–0.15X1+0.85X2-0.15X3-0.15X4 > = 0(Minimum amount in Stock 2 constraint)Finally X1, X2, X3, X4 > = 0(Non-negativity)Complete Model Structure (arranged according to the type of constraints) Maximize Z = 0.30X1+0.25X2+0.28X3+0.35X4 Subject to X1+X2+X3+X4 < = 120000(Investment amount constraint) X4-0.55X1< =0(Safety constraint) 0.43X1+0.017X2+0.050X3+0.009X4 > =2500(Dividend constraint) –0.15X1+0.85X2-0.15X3-0.15X4 > = 0(Minimum amount in Stock 2 constraint) X1, X2, X3, X4 > = 0(Non-negativity)Let us now get the solution from Excel using Solver. 22
  23. 23. Spreadsheet Displaying Model formulation Optimum solution is given next. 23
  24. 24. Spreadsheet Displaying Optimum SolutionThe optimum solution is to invest $65806 in Stock 1, $18000 in Stock 2, $0 inStock 3, and $36194 in Stock 4. The maximum growth = $36909. 68Glossary:Additivity: This is one of the assumptions of linear programming in which theresources for the decision variables are added for getting the total resource.Certainty: In linear programming, we assume all variables are known withoutany doubt. That is, we assume a deterministic or certainty scenario.Constraints: Restrictions in the form of resource limitations or requirement in alinear programming problem.Decision variables: Variables those are unknown for which solution is sought.Divisibility: One of the assumptions of linear programming by whichthe decision variables will have fractional solutions. This may not be meaningfulin situations where we need integer solutions. 24
  25. 25. Linear: This is one of the requirements of linear programming, which stipulatesthat the objective function and constraints are all linear namely the variables areraised to the power 1 only.Linear Programming (LP): It is a very widely used OR model that maximizesor minimizes an objective function subject to a number constraints. LinearProgramming is concerned with the best way of allocating scarce resourcesamong competing ends.Nonnegativity Constraints: This is a requirement in linear programming thatsays that the decision variables cannot have negative solutions. That is, they canhave only positive or zero values.Objective Function: It is the goal of the organization written mathematically interms of the decision variables. This is either maximized or minimized.Optimization: This word is used in the context of either maximizing orminimizing an objective function.Optimum Solution: The best values of the decision variables for a given linearprogramming problem.Programming: Programming in the context of LP means planning and hasnothing to do with computer programming.Proportionality: This is another assumption in linear programming that says thatthe value of the resources utilized are in direct proportion to the values of thedecision variables. This assumption is a sequel to the assumption of linearity ofobjective function and constraints in LP.Shadow Price: This gives the incremental increase in profit when the right handside value of a constraint is increased by one unit, and the decrease in profit whenthe right hand side value of a constraint is decreased by one unit. Shadow pricesplay a key role in evaluating the worth of increasing the resources and also newproduct evaluation. 25

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