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# Formulation Lpp

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### Formulation Lpp

1. 1. LINEAR PROGRAMMING PROBLEM
2. 2. Linear : form meant a mathematical expression of the type a 1 x 1 + a 2 x 2 + ……….+ a n x n where a 1 , a 2 , ….., a n are constants, and x 1 , x 2 , ………, x n are variables. Programming : refers to the process of determining a particular program or plan of action . Linear Programming Problem(LPP) : Technique for optimizing(maximizing/minimizing) a linear function of variables called the ‘OBJECTIVE FUNCTION’ subject to a set of linear equations and/or inequalities called the ‘CONSTRAINTS’ or ‘RESTRICTIONS’.
3. 3. FORMULATION OF LP PROBLEMS
4. 4. LP Model Formulation <ul><li>Objective function </li></ul><ul><ul><li>a linear relationship reflecting the objective of an operation </li></ul></ul><ul><ul><li>most frequent objective is to maximize profit or to minimize cost . </li></ul></ul><ul><li>Decision variables </li></ul><ul><ul><li>an unknown quantity representing a decision that needs to be made. It is the quantity the model needs to determine </li></ul></ul><ul><li>Constraint </li></ul><ul><ul><li>a linear relationship representing a restriction on decision making </li></ul></ul>
5. 5. Steps in Formulating the LP Problems 1. Define the objective. (min or max) 2. Define the decision variables. 3. Write the mathematical function for the objective. 4. Write the constraints. 5. Constraints can be in < , =, or > form.
6. 6. Example Two products : Chairs and Tables Decision : How many of each to make this month? Objective : Maximize profit
7. 7. Data <ul><li>Other Limitations : </li></ul><ul><ul><li>Make no more than 450 chairs </li></ul></ul><ul><ul><li>Make at least 100 tables </li></ul></ul>1000 1 hr 2 hrs Painting 2400 4 hrs 3 hrs carpentry \$5 \$7 Profit Contribution Hours Available Chairs (per chair) Tables (per table)
8. 8. Solution Decision Variables : T = Num. of tables to make C = Num. of chairs to make Objective Function : Maximize Profit Maximize \$7 T + \$5 C
9. 9. Constraints <ul><li>Have 2400 hours of carpentry time available </li></ul><ul><li>3 T + 4 C < 2400 (hours) </li></ul><ul><li>Have 1000 hours of painting time available </li></ul><ul><li>2 T + 1 C < 1000 (hours) </li></ul>
10. 10. <ul><li>More Constraints : </li></ul><ul><li>Make no more than 450 chairs </li></ul><ul><li> C < 450 </li></ul><ul><li>Make at least 100 tables </li></ul><ul><li> T > 100 </li></ul><ul><li>Non negativity : </li></ul><ul><li>Cannot make a negative number of chairs or tables </li></ul><ul><li>T > 0 </li></ul><ul><li>C > 0 </li></ul>
11. 11. Model Max 7 T + 5 C Subject to the constraints : 3 T + 4 C < 2400 2 T + 1 C < 1000 C < 450 T > 100 T , C > 0
12. 12. General Formulation of LPP Max/min z = c 1 x 1 + c 2 x 2 + ... + c n x n subject to: a 11 x 1 + a 12 x 2 + ... + a 1n x n (≤, =, ≥) b 1 a 21 x 1 + a 22 x 2 + ... + a 2n x n (≤, =, ≥) b 2 : a m1 x 1 + a m2 x 2 + ... + a mn x n (≤, =, ≥) b m x 1 ≥ 0, x 2 ≥ 0,…….x j ≥ 0,……., x n ≥ 0. x j = decision variables b i = constraint levels c j = objective function coefficients a ij = constraint coefficients
13. 13. Example Cycle Trends is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be made from aluminum and steel alloys. The anticipated unit profits are \$10 for the Deluxe and \$15 for the Professional. The number of pounds of each alloy needed per frame is summarized on the next slide. A supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the steel alloy weekly. How many Deluxe and Professional frames should Cycle Trends produce each week?
14. 14. Example Pounds of each alloy needed per frame Aluminum Alloy Steel Alloy Deluxe 2 3 Professional 4 2
15. 15. Solution <ul><li>Define the objective </li></ul><ul><ul><li>Maximize total weekly profit </li></ul></ul><ul><li>Define the decision variables </li></ul><ul><ul><li>x 1 = number of Deluxe frames produced weekly </li></ul></ul><ul><ul><li>x 2 = number of Professional frames produced weekly </li></ul></ul>
16. 16. Solution <ul><ul><li>Max Z = 10x 1 + 15x 2 </li></ul></ul><ul><ul><li>Subject To </li></ul></ul><ul><ul><ul><li>2x 1 + 4x 2 < 100 </li></ul></ul></ul><ul><ul><ul><li>3x 1 + 2x 2 < 80 </li></ul></ul></ul><ul><ul><ul><li>x 1 , x 2 > 0 </li></ul></ul></ul>
17. 17. Example A firm manufactures 3 products A, B and C. The profits are Rs.3, Rs.2, and Rs.4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product. Machine G and H have 2000 and 2500 machine-minutes respectively. The firm must manufacture 100 A’s, 200 B’s and 50 C’s, but not more than 150 A’s. Set up an LP problem to maximize profit. 4 2 2 H 5 3 4 G Machines C B A Product
18. 18. Solution <ul><li>Define the objective </li></ul><ul><ul><li>Maximize profit </li></ul></ul><ul><li>Define the decision variables </li></ul><ul><ul><li>x 1 = number of products of type A </li></ul></ul><ul><ul><li>x 2 = number of products of type B </li></ul></ul><ul><ul><li>x 3 = number of products of type C </li></ul></ul>
19. 19. Solution Max Z = 3x 1 + 2x 2 + 4x 3 Subject To 4x 1 + 3x 2 + 5x 3 ≤ 2000 2x 1 + 2x 2 + 4x 3 ≤ 2500 100 ≤ x 1 ≤ 150 x 2 ≥ 200 x 3 ≥ 50 x 1 , x 2 , x 3 ≥ 0
20. 20. Example The Sureset Concrete Company produces concrete. Two ingredients in concrete are sand (costs \$6 per ton) and gravel (costs \$8 per ton). Sand and gravel together must make up exactly 75% of the weight of the concrete. Also, no more than 40% of the concrete can be sand and at least 30% of the concrete be gravel. Each day 2000 tons of concrete are produced. To minimize costs, how many tons of gravel and sand should be purchased each day?
21. 21. Solution <ul><li>Define the objective </li></ul><ul><ul><li>Minimize daily costs </li></ul></ul><ul><li>Define the decision variables </li></ul><ul><ul><li>x 1 = tons of sand purchased </li></ul></ul><ul><ul><li>x 2 = tons of gravel purchased </li></ul></ul>
22. 22. Cont… <ul><ul><li>Min Z = 6x 1 + 8x 2 </li></ul></ul><ul><ul><li>Subject To </li></ul></ul><ul><ul><ul><li>x 1 + x 2 = 1500 </li></ul></ul></ul><ul><ul><ul><li>x 1 < 800 </li></ul></ul></ul><ul><ul><ul><li>x 2 > 600 </li></ul></ul></ul><ul><ul><ul><li>x 1 , x 2 > 0 </li></ul></ul></ul>