DECISION MAKING
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Let x1 = Number of units of M1 produced x2 = Number of units of M2 produced ii) Write the constraints: 4x1 + 2x2 ≤ 80 (Grinding constraint) 2x1 + 5x2 ≤ 180 (Polishing constraint) x1, x2 ≥ 0 iii) Write the objective function: Maximize Z = 3x1 + 4x2 iv) Solve the LP problem graphically or by simplex method to find the optimal solution.
DECISION MAKING
- 1. “Decision Making, Introduction to quantitative tools, Linear programming and Graphic solution to problems” Linear Programming AG ECON- 508 Presented to Dr. Hulas Pathak Speaker (Professor Deptt. Of Agricultural Jyotsana Jogi Economics) M.Sc. (Ag.) Previous year Deptt. Of Agri.Economics
- 2. Decision Making Decision making can be defined as the process of making choices among the possible alternatives. The skills considered important to effective decision making are based on a normative model of decision making, which prescribes how decisions should be made.
- 3. Decision making concepts Objectives must first be established. Objectives must be classified and placed in order of importance Alternative actions must be developed. The alternative must be evaluated against all the objectives. The alternative that is able to achieve all the objectives is the tentative decision. The tentative decision is evaluated for more possible consequences.
- 4. The decisive actions are taken, and additional actions are taken to prevent any adverse consequences from becoming problems and starting both systems (problem analysis and decision making) all over again. There are steps that are generally followed that result in a decision model that can be used to determine an optimal production plan. In a situation featuring conflict, role-playing is helpful for predicting decisions to be made by involved parties.
- 5. Types of decision making Sequential decisions:- Decisions in which the outcome of the decisions influences other decisions are known as sequential decisions. For example, a company trying to decide whether or not to market a new product might first decide to test the acceptance of the product using a consumer panel. Conscious or unconscious decisions:- These decisions are taken under consciousness or unconsciousness. Managerial decisions:- Decisions concerning the operation of the firm, such as the choice of the farm size, firm growth rate, and employee compensation are known as Managerial Decisions.
- 6. Farm Management decisions:- Farm management implies decision making process. Several decisions need to be made by the farmer as a manager in the organizational management decisions. These are some types of Farm Management decisions. 1. Organizational Management decisions:- The Organizational management decisions are further sub-divided into operational management decisions and strategic management decisions. a. operational Management Decisions:- Those decisions which involve less investment and are made frequently, are called operational management decisions. The affect of these decisions is short lived. These decisions can be reversed without incurring a cost or less cost. Decisions like; what to produce? How to produce? How much to produce? are some of important operational management decisions.
- 7. b. Strategic Management Decisions:- These decisions involves heavy investment and are made less frequently. The effect of these decisions is long lasting. These decisions can not be altered. These decisions are also known as basic decisions. Example:- Size of farm, construction of farm building. 2. Administrative Management Decisions:- Beside organizational management decisions, the farmer also make some administrative decisions like financing the farm business, supervision, accounting and adjusting his farm business according to government policies. 3.Marketing Management Decisions:- Marketing decisions are most important under changing environment of Agriculture. These decisions include buying and selling.
- 8. Introduction to quantitative tools The area of quantitative methods for decision making is based on the scientific method for investigating and helping to take decisions about complex problems in modern organizations. Quantitative methods for decision making are also known as operations research. Objectives:- To describe the quantitative analysis approach. To understand the application of quantitative analysis in a real situation. To describe the use of modelling in quantitative analysis. Use computers and spreadsheets models to perform quantitative analysis. Discuss possible problems in using quantitative analysis. To perform a break even point
- 9. Essential Features for Quantitative Tools : Logical sequence and smooth flow of questions. Format conducive to efficient, clear and complete recording of responses. Length and time required to complete survey is manageable/acceptable. Sound sampling procedures that will generate representative samples of clients and comparison groups, thereby producing reliable data for making generalizations about clients.
- 10. Introduction to Linear Programming:- Linear Programming (LP) is a mathematical procedure for determining optimal allocation of scarce resources. LP is a procedure that has found practical application in almost all facets of business, from advertising to production planning. Transportation, distribution, and aggregate production planning problems are the most typical objects of LP analysis. In the petroleum industry, for example a data processing manager at a large oil company recently estimated that from 5 to 10 percent of the firm's computer time was devoted to the processing of LP and LP-like models. LINEAR PROGRAMMING
- 11. Definition of Lp Linear Programming (LP) is a mathematical optimization technique. By optimization technique, it refers to a method which attempts to maximize or minimize some objective, for example, maximize profits or minimize costs. Linear programming is a subset of a larger area of mathematical optimization procedures called mathematical programming, which is concerned with making an optimal set of decisions. In any LP problem, certain decisions need to be made. These decisions are represented by decision variables which are used in the formulation of the LP model.
- 12. Components of Linear Programming Objective function:- The objective function is a mathematical representation of the overall goal stated in terms of the decision variables. The firm’s objective and its limitations must be expressed as mathematical equations or inequalities, and these must be linear equations and inequalities. Constraints:- Constraints are also in terms of the decision variables, and represent conditions which must be satisfied in determining the values of the decision variables. Most constraints in a linear programming problem are expressed as inequalities. They set upper or lower limits, they do not express exact equalities; thus permit many possibilities. Resources must be in limited supply. For example, a furniture plant has a limited number of machine-hours available; consequently, the more hours it schedules for furniture’s, the fewer furniture’s it can make. There must be alternative courses of action, one of which will achieve the objective.
- 13. Decision variables:- Decision variables are the unknown quantities in the linear programming formulations. Non- negativity conditions- non-negativity conditions are special constraints which require all variables to be either zero or positive.
- 14. Uses of linear programming in different fields 1. Industrial application a. Product mix problem b. Blending problem c. Production scheduling problem d. Assembly line balancing e. Make- or- buy problem, 2. Management application a. Media selection problem b. Port folio selection problems c. Profit planning problems d. Transportation problems,
- 15. 3. Miscellaneous application a. Diet problems b. Agriculture problem c. Flight scheduling problems d. Facilities location problems.
- 16. Assumptions of Linear Programming 1.Linearity:- The objective function and constraints are all linear functions; that is, every term must be of the first degree. Linearity implies the next two assumptions. It implies the products of two variables such as X1X2, powers of variables such as x2 , combination of variables such as- ax1 + 2.5x2 =5000. 2.Proportionality:- For the entire range of the feasible output, the rate of substitution between the variables is constant. It means that profit per unit is directly proportional to number of units sold. 3.Additivity:- All operations of the problem must be additive with respect to resource usage, returns, and cost. This implies independence among the variables. For example if a company use T1 hours on machine A to make product 1, and T2 hours to make product 2 then the time on machine A denoted to product 1 and 2 is additive i.e. T1+T2.
- 17. 4.Divisibility:- Non-integer solutions are permissible. It implies the care of products and resources to be used. 5.Certainty:- All coefficients of the LP model are assumed to be known with certainty. Remember, LP is a deterministic model. 6.Multiplicativity:- If we take one hour on a single item on a given machine it take ten hours to make ten parts. The total profit by selling given number of units of a product is the unit profit multiplied by the number of units sold. Profit of 1 unit = 10 Therefore the Profit of 10 unit will be = 10 x 10 = 100.
- 18. Formulation of problems Formulate the linear problems. Plot the constraints on a graph. Identify the feasible solution region. Plot the two objective function lines. Determine the direction of improvement. Determine find the most attractive corner. Determine the co-ordinates of the MAC. Find the value of the objective.
- 19. A manufacturer produces two types of models M1 and M2.Each model of the type M1requires 4 hours of grinding and 2 hours of polishing; where as each model of M2requires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinder sand 3 polishers. Each grinder works for 40 hours a week and each polisher works 60hours a week. Profit on M1 model is Rs.3.00 and on model M2 is Rs.4.00.Whatever produced in a week is sold in the market. How should the manufacturer allocate his production capacity to the two types of models, so that he makes maximum profit in a week?
- 20. i) Identify and define the decision variable of the problem Let X1 and X2 be the number of units of M1 and M2 model. ii) Define the objective function Since the profits on both the models are given, the objective function is to maximize the profit. Max Z = 3X1 + 4X2 iii) State the constraints to which the objective function should be optimized (i.e.Maximization or Minimization) There are two constraints one for grinding and the other for polishing. The grinding constraint is given by 4X1 + 2X2 < 80
- 21. No of hours available on grinding machine per week is 40 hrs. There are two grinders. Hence the total grinding hour available is 40 X 2 = 80 hours. The polishing constraint is given by 2X1 + 5X2 < 180 No of hours available on polishing machine per week is 60 hrs. There are three grinders. Hence the total grinding hour available is 60 X 3 = 180 hours
- 22. Finally we have, Max Z = 3X1 + 4X2 Subject to constraints, 4X1 + 2X2 < 80 2X1 + 5X2 < 180 X1, X2 > 0
- 23. Example - Solve the following LPP by graphical method Minimize Z = 20X1 + 40X2 Subject to constraints 36X1 + 6X2 ≥ 108 3X1 + 12X2 ≥ 36 20X1 + 10X2 ≥ 100 X1 X2 ≥ 0
- 24. Solution: The first constraint 36X1 + 6X2 ≥ 108 can be represented as follows. We set 36X1 + 6X2 = 108 When X1 = 0 in the above constraint, we get 36 x 0 + 6X2 = 108 X2 = 108/6 = 18 Similarly when X2 = 0 in the above constraint, we get, 36X1 + 6 x 0 = 108 X1 = 108/36 = 3 The second constraint3X1 + 12X2 ≥ 36 can be represented as follows, We set 3X1 + 12X2 = 36 When X1 = 0 in the above constraint, we get, 3 x 0 + 12X2 = 36 X2 = 36/12 = 3
- 25. Similarly when X2 = 0 in the above constraint, we get, 3X1 + 12 x 0 = 36 X1 = 36/3 = 12 The third constraint20X1 + 10X2 ≥ 100 can be represented as follows, We set 20X1 + 10X2 = 100 When X1 = 0 in the above constraint, we get, 20 x 0 + 10X2 = 100 X2 = 100/10 = 10 Similarly when X2 = 0 in the above constraint, we get, 20X1 + 10 x 0 = 100 X1 = 100/20 = 5
- 27. Point X1 X2 Z = 20X1 + 40X2 0 0 0 0 A 0 18 Z = 20 x 0 + 40 x 18 = 720 B 2 6 Z = 20 x2 + 40 x 6 = 280 C 4 2 Z = 20 x 4 + 40 x 2 = 160* Minimum D 12 0 Z = 20 x 12 + 40 x 0 = 240 The Minimum cost is at point C When X1 = 4 and X2 = 2 Z = 160