UNIT 3 An introduction to Linear Programming
An LP problem An engineering factory makes 4 products (PROD1 to PROD4) on the following machines: 4 grinders, 2 drills and 1 planer. Each product yields a certain contribution to profit (defined as selling price in €/unit minus cost of raw materials). These quantities together with the unit production times (hours) required on each process are given in the table below. PROD 1 PROD 2 PROD 3 PROD 4 Contribution to profit 10 6 8 4 Grinding 0.5 0.7 - - Drilling 0.1 0.2 - 0.3 Planing - - 0.01 - If there are 8 working hours in a day, what should the factory produce (daily) in order to maximize the total profit?
An LP model
Canonical form of a Linear Programming (LP) problem + + If it is an LP maximization Max c x ... c x 1 1 n n + + ≤ problem, all the constraints s . t . a x ... a x b must be ≤ 11 1 1 n n 1 + + ≤ a x ... a x b 21 1 2 2 n n ... If it is a minimization + + ≤ a x ... a x b 1 1 problem, the constraints m mn n m ≥ x , x ,..., x 0 must be ≥ 1 2 n All the variables must be non-negative
Canonical form of a Linear Programming (LP) problem t Max c x A : Technical matrix ≤ . . s t A x b c : Coefficients vector ≥ b : RHS vector x 0
Standard form of an LP problem t Max c x = s . t . A x b All the constraints must be = ≥ x 0 An LP problem can be transformed into its standard form adding slack variables.
Characteristics of LP problems S is always a convex set (but it may be bounded or unbounded). An LP problem can be feasible and not have an optimal solution (it can be unbounded) or be bounded and not have an optimal solution (infeasible). All the optima (if any) are global optima (because the Local-Global theorem can always be applied). A solution is optimal if and only if it is a K-T point. Optimal solutions are always boundary (border) points (never interior points), and at least one of them will be a vertex (corner point) of S. If two solutions are optimal, all the solutions in the segment that joins them are optimal too.
Basic feasible solutions (BFS) Definition: A solution is basic if: x A = x b • . • n-m of its elements are zero. • The submatrix of A associated with basic variables (basic matrix) is regular (its determinant is not zero). • All the variables satisfy the non-negativity constraints Non-basic variables The remaining variables, which can take non-zero values A basic feasible solution is called degenerate if any of its basic variables takes value 0.
Example + + Max 4 x 5 y Max 4 x 5 y + ≤ + + = s . t . 2 x y 8 . . 2 8 s t x y s ≤ + = y 5 y t 5 ≥ ≥ x , y 0 x , y , s , t 0 x y s t 2 1 1 0 = Technical matrix A 0 1 0 1 m=2 equations n=4 variables
Example = Is a BFS? x ( 0 , 0 , 8 , 5 ) 0 2 1 1 0 0 8 = = • A x 0 1 0 1 8 5 5 • 2 basic variables: s,t (2 non-basic variables: x,y) 1 0 = = ≠ • Basic matrix B B 1 0 0 1 ≥ x , y , s , t 0 Basic feasible solution •
Obtaining BFSs Given a set of basic variables, if |B| is not 0, the value of these basic variables can be obtained (if |B|=0, there is no basic feasible solution associated with this set of basic variables).
Obtaining BFSs: example + Max 4 x 5 y + + = Basic variables: y,t s . t . 2 x y s 8 + = y t 5 ≥ x , y , s , t 0 1 0 = = ≠ B B 1 0 1 1 = 8 y 1 0 y 8 = = − = y 8 , t 3 + = 1 1 t 5 y t 5 (0,8,0,-3) is not a basic feasible solution, since t<0.
Obtaining BFSs: example + Max 4 x 5 y + + = Basic variables: x,s s . t . 2 x y s 8 + = y t 5 ≥ x , y , s , t 0 2 1 = = B B 0 0 0 There is no basic feasible solution with the basic variables x,s
Obtaining BFSs: example + Max 4 x 5 y + + = Basic variables: x,y s . t . 2 x y s 8 + = y t 5 ≥ x , y , s , t 0 2 1 = = B B 2 0 1 + = 3 2 1 x 8 2 x y 8 = = = y 5 , x = 0 1 5 y 5 y 2 (5,1.5,0,0) is a BFS
Fundamental theorem of Linear Programming If an LP problem has at least one feasible solution, then it will also have at least one BFS. If an LP problem has optimal solution(s), then at least one optimal solution will be a BFS.
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