Chapter two
Download as PPTX, PDF0 likes2,392 views
Linear programming is a technique for choosing the optimal alternative from a set of feasible options to maximize or minimize an objective function subject to constraints. It involves decision variables, an objective function expressed as a linear combination of the variables, and constraints on the variables. The optimal solution can be found graphically or using the simplex method. Graphically, the feasible region is identified and the point optimizing the objective function chosen. Binding constraints affect the optimal solution, while non-binding and redundant constraints do not.
Chapter two
- 1. Linear Programming 5/4/2018 4:12 PM 1
- 2. Linear Programming Linear Programming Method is a technique for choosing the best alternative from a set of feasible alternatives, in situations in which the objective function as well as the constraints can be expressed as linear mathematical functions. 5/4/2018 4:12 PM 2
- 3. Areas of Application i. Product Mix ii. Blending iii. Distribution Problems iv. Purchasing v. Portfolio Selection vi. Advertising Media Mix vii. Production and Inventory Scheduling 5/4/2018 4:12 PM 3
- 4. Total Man-hours Available Description Man-hours per unit Desk Bookcase 640 Assembly 8 4 540 Finishing 4 6 100 Inspection 1 1 Profit per unit $30 $20 5/4/2018 4:12 PM 4 A small company in office furnishings field produces both desks and bookcases. There are three stages of production –assembly, finishing and inspection. A desk requires 8 man-hours, and bookcase 4 man-hours of assembly time. The finishing time needed is 4 man-hours for a desk and 6 man- hours for a bookcase. Finally, each product requires one man-hour per unit of inspection time. For a working week, this company has available 640 man-hours in assembly, 540 man-hours in finishing, and 100 man-hours in inspection. The accountants have determined that, within the range of its production possibilities, each desk can be sold with a profit of $30, and each bookcase for a profit of $20. Objective: Maximum possible Profit!!!
- 5. 5/4/2018 4:12 PM 5
- 6. Structure of LP model It consists of: 1. Decision variables: The courses of action, various activities (alternatives) to achieve optimal values of the objective function. They are denoted as X1, X2, X3, ...., Xn. 2. Objective function: It is expressed in terms of the decision variables to optimize the criteria of optimality; it is a measure of performance and denoted by Z. Z=C1X1 + C2X2 + .... + CnXn. Where C1, C2, ...,Cn are parameters which represent contributions of X1, X2, .., Xn to measure performance. 5/4/2018 4:12 PM 6
- 7. Cont... 3. constraints/limitations: Resources are of limited supply, thus the mathematical relationship to express this limitation is inequality of the decision variables. i.e. X1 + X2 <= b1. 5/4/2018 4:12 PM 7
- 8. General model Max./Min. Z= C1X1 + C2X2 + ... + CnXn Objective function Subject to constraints: a11X1 + a12X2 + ...+ a1nXn (<=,=,>=) b1 a21X1 + a22X2 + ...+ a2nXn (<=,=,>=) b2 X1, X2, Xn>=O non-negativity constraint 5/4/2018 4:12 PM 8
- 9. Definitions 1. Decision Variables: Quantities whose values can be chosen by a decision maker 2. Objective Function: A function of the decision variables whose value is to be made as large (or, in some problems, as small) as possible 3. Constraints: Restrictions on the values which can be taken by the decision variables 4. Non-negativity Constraints: Requirements that the decision variables not take negative values 5. Feasible solution: Any set of values of the decision variables which satisfies all the constraints 6. Optimal Solution: Any feasible solution such that the value of the objective function is as high(or, in minimization problems, as low) as possible 7. Linear Program: A setup in which the objective function and the constraints involve only linear functions of the decision variable 5/4/2018 4:12 PM 9
- 10. Some assumptions of LPP 1. Proportionality: proportionality must exist in the objective function and the constraint inequalities. i.e. If unit contributes $10 to profit, total contribution must be equal to 10X1. 2. Additivity: both in the objective function and constraint inequalities, the total of all activities is given by the sum total of each activity conducted separately. (total profit is sum of individual profits). 3. Continuity: all decision variables are continuous. i.e. Fractional values are possible. 5/4/2018 4:12 PM 10
- 11. Cont... 4. Certainty: coefficients of objective function, coefficients of constraint inequalities, and constraints (resources) are known with certainty. 5. Finite choices: LP model assumes that a limited number of choices are available to a decision maker and decision variables do not assume negative values. Only non-negative levels of activity are feasible. 5/4/2018 4:12 PM 11
- 12. Formulation of LPP: Max. case Example 2.1: A firm is engaged in producing two products A & B. Each unit of product A requires 2 kg of raw materials and 4 hours of labour for processing, while each unit of B requires 3 kg of raw materials and 3 hours of labour of the same type. Every week, the firm has an availability of 60 kg of raw materials and 96 labour hours. One unit of product A sold yields $40 and B gives $35 as profit. Formulate the LPP to determine how many units of each product should be produced per week so that the firm earns max. Profit. Assume that there is no marketing constraint so that all that is produced can be sold. 5/4/2018 4:12 PM 12
- 13. Formulation cont... Solution: Max. Z= 40X1 +35X2 profit S.T. 2X1 + 3X2<=60 raw materials constraint 4X1 + 3X2<=96 labour hours constraint X1,X2>=0 non-negativity constraint 5/4/2018 4:12 PM 13
- 14. Cont... Min. case Example 2.2: Agricultural research institute suggested to a farmer to spread out at least 4800 kg of a special phosphate fertilizer and not less than 7200 kg of a special nitrogen fertilizer to raise productivity of crops in field. There are two sources for obtaining these mixture A & B. Both of these are available in bags weighting 100kg each and they cost $40 and $24 respectively. Mixture A contains phosphate and nitrogen equivalent of 20kg and 80kg respectively, while mixture B contains these ingredients equivalent of 50kg each. Formulate this as LPP and determine how many bags of each type should the farmer buy in order to obtain the required fertilizer at min. Cost. 5/4/2018 4:12 PM 14
- 15. formulation Solution: Min. Z = 40X1 + 24X2 cost S.T. 20X1 + 50X2>=4800 phosphate requirement 80X1 + 50X2>=7200 nitrogen requirement X1,X2>=0 non-negativity constr. 5/4/2018 4:12 PM 15
- 16. Solution of LPP LPP can be solved either by graphical method or algebraic (simplex) method. graphical method can be used only when two variables are involved. 5/4/2018 4:12 PM 16
- 17. Graphical method To use graphical method for solving LPP, the following steps are required: 1. Identify the problem. The decision variables, the objective function and the constrain restrictions. 2. Draw a graph that includes all the constraints/restrictions and identify the feasible region. 3. Obtain the point on the feasible region that optimizes the objective function---optimal solution. 4. Interpret the result. 5/4/2018 4:12 PM 17
- 18. Max. case Consider example 2.1 again. For this problem, decision variables are X1 and X2 for products A and B. Thus: max. Z= 40X1 + 35X2 S.T. 2X1 + 3X2<=60 4X1 + 3X2<=96 X1,X2>=0 5/4/2018 4:12 PM 18
- 19. 5/4/2018 4:12 PM 19
- 20. Cont... N.B: the region of acceptable values of the decision variables in relation to the given constraints ( and non- negativity restriction) is called feasible region. Interpretation: the optimal solution is to produce 18 units of product A and 8 units of product B every week to get the profit of $1000. No other product mix under the given conditions could yield more profit than this. 5/4/2018 4:12 PM 20
- 21. Cont... alternately Extreme point method is possible Point X1 X2 Z O 0 0 0 P 0 20 700 Q 18 8 1000 maximum (optimum) R 24 0 960 5/4/2018 4:12 PM 21
- 22. Minimization Case Example 2.2 reconsidered Min. Z = 40X1 + 24X2 total cost S.t 20X1 + 50X2 >= 4800 phosphate requirement 80X1 + 50X2 >= 7200 nitrogen requirement X1 , X2 >= 0 5/4/2018 4:12 PM 22
- 23. 5/4/2018 4:12 PM 23
- 24. Cont... Decision variables X1 & X2 represent , respectively, the number of bags of mixture A and of mixture B to be bought. Interpretation: the total cost is minimum at the optimal point and the optimal solution is to buy 144 bags of mixture B only and none of mixture A. This would entail a total cost of $3456. 5/4/2018 4:12 PM 24
- 25. Binding and non-binding constraints Constraints is termed as binding if the left hand side of the right hand side of it are equal when optimal values of decision variables are substituted into the constraint it is termed non-binding otherwise. e.g 2X1 + 3X2 2 x 18 + 3 x 8 = 60 = RHS binding 4X1 + 3X2 4 x 18 + 3 x 8 = 96 = RHS binding 20X1 + 50X2 20x0 + 50x144 =7200 ≠4800 RHS non binding 80X1 + 50X2 80x0 + 50x144=7200=RHS binding 5/4/2018 4:12 PM 25
- 26. Redundant Constraint If and when a constraint, when plotted, doesn`t form part of the boundary making the feasible region of the problem, it is said to be redundant. The inclusion or exclusion of a redundant constraint obviously doesn`t affect the optimal solution to the problem. 5/4/2018 4:12 PM 26
- 27. Cont.. Example: Continuing with example 2.1, suppose packaging is required. Every unit of product A requires 4 hours while every unit of B needs 3.5 hours for packaging. Suppose that in packaging department, 105 hours are available every week. What product mix would maximize the profit? Soln.: max. Z=40X1 + 35X2 S.T. 2X1 + 3X2<=60 raw materials constraint 4X1 + 3X2<=96 labour hours constraint 4X1 + 3.5X2<=105 packaging hours constr. X1,X2>=0 non-negativity constraint 5/4/2018 4:12 PM 27
- 28. 5/4/2018 4:12 PM 28
- 29. Cont... Interpretation: The inclusion of the packaging hours constraint does not provide a side of the polygon representing the feasible region. Thus this constraint is of no consequence and hence redundant. optimal solution of the problem remains the same as that of example 2.1. 5/4/2018 4:12 PM 29