CHAPTER 3 LINEAR PROGRAMMING: SIMPLEX METHOD

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1 CHAPTER 3 LINEAR PROGRAMMING: SIMPLEX METHOD Linear programming is optimization problem where the objective function is linear and all equality and inequality constraints are linear. This problem was first defined mathematically in the thirties in the field of economics. It became important tool of allocating resources during World War II. Linear programming problems can be easily solved geometrically for the case of two variables and few constraints. However, finding a solution becomes increasingly difficult to solve as dimensionality of the problem increase. For example, consider the problem of airline planning allocation a limited number of planes to serve a fixed schedule of travel between multiple cities while minimizing fuel cost. As of 1979 (before the introduction of personal computers), 25% of all computational time worldwide was dedicated to solving linear programming problems. A breakthrough was achieved by George Dantzig who proposed the simplex method based on his work for the US Airforce. His solution as we will see is both simple and elegant. The simplex method remained the standard for solving linear programming problem until Narendra Karamarkar of Bell Labs developed an alternative that is 50 times faster. George Dantzig ( ), 5-3/5-3.html Narendra Karamarkar (1957- ), n-mathematicians

2 Standard Form of the Linear Programming Problem There are two possible forms to express the linear programming problem: Scalar Form: Minimize, n Subject to, f(x) = c i x i i=1 n a 1i x i = b 1 i=1 n a mi x i = b m x i 0 i = 1,2,, n where x are the decision variables. c, b, and a are constant coefficients. i=1 Matrix Form: Minimize, f(x) = c T x Subject to, ax = b x 0 In this case, a will be m x n matrix. Comments: The problem is formulated as minimization. If you want to maximize, simply minimize the negative of the objective function. Other clarifications of the problem will be presented later. 2

3 Geometry of Linear Programming Problems As mentioned earlier, the two variable problem can be solved geometrically. In this case, the feasible area is a polygon in the upper right corner of the x 1-x 2 plane. The objective function will be a family of parallel straight line forming a plan in the case of two variables and a hyperplane when the number of variables is more than two. The extremum will be at one if the vertices of the polygon. A proof will be presented later but for now, inspection can show this. x 2 Various values of the objective function x 1 3

4 Special Case #1: Sometime the problem is ill-conditioned and is unbounded as can be seen in the figure below. An example may be maximizing a function while including no relevant upper bound constraints on the variables. x 2 x 1 Special Case #2: The objective function has the same slope as one of the constraints. If the minimum is related to this constraint, there is no unique solution. x 2 x 1 4

5 Example 3.1: Consider this quality control problem items has to be inspected in 8-hours day. The factory has two types of inspectors: Grade 1 and Grade 2 inspectors. Grade 1 inspectors can inspect 25 pieces per hour. Their accuracy is 98%. Their hourly wage is $4.00 per hour. Grade 2 inspectors can inspect 15 pieces per hour. Their accuracy is 95%. Their hourly wage is $3.00 per hour. An error cost the factory $2 per item. The factory has access to 8 Grade 1 inspectors and 10 Grade 2 inspectors. Find a way to minimize the total cost of inspection. 5

6 Formulation: Variables: x 1: Number of Grade 1 inspectors x 2: Number of Grade 2 inspectors Constraints: Lower limit: x 1 0 x 2 0 Upper limit: x 1 8 x 2 10 Number of items inspected constrained: 8(25)x 1 + 8(15)x x 1 + 3x 2 45 Objective function: Minimize, C = 8(4x 1 + 3x 2 ) + 8(2)((25)(1 0.98)x 1 + (15)(1 0.95)x 2 ) C = 40x x 2 6

7 Solution: 5x 1 +3x 2 >=45 x 2 x 1 <=8 x 2 <=10 x 1 The polygon is a triangle with three vertices as two of the constraints were rendered obsolete. x 1 0 x 2 0 Possible solutions are: Vertex Value of the Objective Function (C) (3, 10) 3(40) + 10(36) = = 480 (8, 10) 8(40) + 10(36) = = 680 (8, 5/3) 8(40) + (5/3)(36) = = 380 Notes: Lower limits of constraints are made redundant since they lead to trivial solution: no inspectors. Since we are minimizing, it is clear that (8, 5/3) is the optimal solution. However, it may be difficult to find 5/3 of Grade 2 inspectors! (why?). Therefore, we will switch to the nearest feasible integer solution, which is (8, 2). In this case the objective function value is, 8(40) + (2)(36) = = 392 This will lead to a case of integer programming, which we are not covering within this course. 7

8 Example 3.2: Consider this oil refinery problem. The refinery deals with two kinds of crude oil: Crude A and Crude B. Crude A costs $30 per barrel. There are 20,000 barrels available. Crude B costs $36 per barrel. There are 30,000 barrels available. Both crudes can be refined to produce gasoline and lube oil. The yields of both crudes are listed in the table below. Yield/Barrel (A) Yield/Barrel (B) Sale Price ($) Market Capacity (Barrels) Gasoline ,000 Lube Oil ,000 Maximize the profit of the refinery. 8

9 Formulation: Variables: x 1: Number of Crude A barrels x 2: Number of Crude B barrels Constraints: Lower limit: x 1 0 x 2 0 Upper limit: x x Market capacity constraints: 0.6x x x x Objective function: Maximize, P = 50(0.6x x 2 ) + 120(0.4x x 2 ) 30x 1 36x 2 P = 48x x 2 9

10 Solution: 0.4x x 2 >= x 2 x 1 <= x x 2 >= x 2 <=30000 x 1 The polygon is a triangle with three vertices as two constraints intersect at the same point. Additionally, two of the constraints were rendered obsolete. x 1 0 x 2 0 Possible solutions are: Vertex Value of the Objective Function (P) (10000, 30000) 10000(48) (28) = 1,320,000 (20000, 30000) 20000(48) (28) = 1,800,000 (20000, 10000) 20000(48) (28) = 1,240,000 Since we are maximizing, it is clear that (20000, 30000) is the optimal solution. Notes: Lower limits of constraints are made redundant since we look for profit. Market capacity of gasoline constraint is made redundant by the market capacity of lube oil constraint. This problem is unrealistic as refining cost and cost of storage excess are not included. Homework: 3.7, 3.9 Formulate these problems and solve graphically: 3.52, 3.94,