L03_Basic Feasible Solutions of LPPs are corner point solutions

L03_Basic Feasible Solutions of LPPs are corner point solutions

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Unformatted text preview: Algebraic Solution of LPPs - Simplex Method To solve an LPP algebraically, we first put it in the standard form. This means all decision variables are nonnegative and all constraints (other than the nonnegativity restrictions) are equations with nonnegative RHS. Converting inequalities into equations Consider the LPP Maximize Subject to z = 2 x1 + 3x2 x1 +3 x2 ≤6 3 x1 + 2 x2 ≤6 x1 , x2 ≥0 We make the ≤ inequalities into equations by adding to each inequality a “slack” variable (which is nonnegative). Thus the given LPP can be written in the equivalent form Maximize Subject to z = 2 x1 +3 x2 x1 + 3 x2 + s1 3 x1 + 2 x2 x1 , x2 , s1 , s2 ≥ 0 =6 + s2 = 6 s1 , s2 are slack variables. Thus we seem to have complicated the problem by introducing two more variables; but then we shall see that this is easier to solve. This is one of the “beauties” in mathematical problem solving. The ≥ inequalities are made into equations by subtracting from each such inequality a “surplus” (non-negative) variable. Thus the LPP Maximize Subject to z = 2 x1 + 3x2 x1 +3 x2 ≤6 3 x1 + 2 x2 ≥ 2 x1 , x2 ≥0 is equivalent to the LPP Maximize z = 2 x1 + 3 x2 Subject to x1 +3 x2 + s1 3 x1 +2 x2 x1 , x2 , s1 , s2 ≥0 =6 −s 2 = 2 s1 is a slack variable; s2 is a surplus variable. If in a constraint, the RHS constant is negative, we make it positive by multiplying the constraint by -1. Thus the LPP Maximize Subject to z = 2 x1 + 3 x2 x1 +3 x2 ≤6 3 x1 +2 x2 ≥−2 x1 , x2 ≥0 is equivalent to the LPP Maximize z = 2 x1 + 3 x2 Subject to x1 + 3 x2 ≤6 − x1 −2 x2 ≤ 2 3 x1 , x2 ≥0 Its standard form is the LPP Maximize z = 2 x1 + 3 x2 Subject to x1 +3 x2 + s1 −3 x1 −2 x2 x1 , x2 , s1 , s2 ≥0 =6 + s2 = 2 s1 , s2 are slack variables. Dealing with unrestricted variables If, in an LPP, a decision variable xi is unrestricted (in sign) i.e. it can take positive as well as negative values, then we can, by xi = xi+ − xi− where xi+, xi− writing are (defined below and are) nonnegative, make the LPP into an equivalent LPP where all the decision variables are ≥ 0. Note: | xi | + xi x= ; 2 + i | xi | − xi x= 2 − i xi = x if xi ≥ 0 and − x otherwise + i − i Thus the LPP Maximize Subject to z = x1 + 3x2 x1 + x2 ≤ 2 − x1 + x2 ≤ 4 x1 unrestricted, x2 ≥ 0 is equivalent to the LPP Maximize z = x − x + 3 x2 Subject to + 1 − 1 x − x + x2 + s1 − x + x + x2 + 1 − 1 + 1 − 1 + 1 − 1 =2 + s2 = 4 x , x , x2 , s1 , s2 ≥0 Basic variables, Basic feasible Solutions Consider an LPP (in standard form) with m constraints and n decision variables. We assume m ≤ n. We choose n –m variables and set them equal to zero. Thus we will be left with a system of m equations in m variables. If this m× m square system has a unique solution, this solution is called a basic solution. If further if it is feasible, it is called a Basic Feasible Solution (BFS). The n-m variables set to zero are called nonbasic and the m variables which we are solving for are known as basic variables. Thus a basic solution is of the form x = (x1, x2, …, xn) where n-m “components” are zero and the remaining m components form the unique solution of the square system (formed by the m constraint equations). n Note that we may have a maximum of m basic solutions. Consider the LPP: Maximize Subject to z = 2 x1 + 3 x2 x1 +3 x2 ≤6 3 x1 + 2 x2 ≤6 x1 , x2 ≥0 This is equivalent to the LPP (in standard form) Maximize Subject to C z = 2 x1 + 3 x2 x1 + 3 x2 + s1 3x1 + 2 x2 x1 , x2 , s1 , s2 ≥ 0 =6 + s2 = 6 s1 , s2 are slack variables. Nonbasic Basic Basic (zero) variables solution variables (x ,x , s , s 1 2 1 2 ) Assoc Feasible? Object-iated ive value, corner z point O B E D A C Yes Yes No No Yes Yes 0 6 4 48/7 Optimal ( x1 , x 2 ) ( s1 , s 2 ) ( x1 , s1 ) ( x2 , s 2 ) (0,0,6,6) (0,2,0,2) (0,3,-3,0) (6,0,0, -12) (2,0,4,0) ( x1 ,s 2 ) ( x2 ,s1 ) ( x 2 , s1 ) ( x1 ,s 2 ) ( x2 ,s 2 ) ( x1 ,s1 ) ( s1 ,s 2 ) ( x1 , x 2 ) 6 12 (, ,0,0) 77 Graphical solution of the above LPP x2 Thus every Basic Feasible Solution corresponds to a corner(=vertex) of the set SF of all feasible solutions. Optimal point E (0,3) B (0,2) O (0,0) (6/7, 12/7) C SF A (2,0) D (6,0) x1 Question 6 (Problem set 3.2A – Page 79) Consider the LPP: Maximize Subject to z = x1 + 3 x2 x1 +x2 ≤ 2 − 1 +x2 ≤ 4 x x1 unrestricted , x2 ≥0 This is equivalent to the LPP(in standard form) Maximize Subject to + 1 z = x − x + 3 x2 − 1 + 1 − 1 x − x + x2 + s1 − x + x + x2 + 1 − 1 + 1 − 1 =2 + s2 = 4 x , x , x2 , s1 , s2 ≥0 Nonbasic Basic Basic Assoc- Feasible? Objective (zero) variables solution iated value, z variables ( x1+ , x1− , x2 , s1 , s2 ) corner point ( x , x , x2 ) ( s1 , s2 ) + 1 − 1 (0,0,0,2,4) (0,0,2,0,2) (0,0,4,-2,0) (0,-2,0,0,6) O B E D C Yes Yes No No Yes Yes 0 6 -4 8 ( x , x , s1 ) ( x2 , s2 ) ( x1+ , x1− , s2 ) ( x2 , s1 ) + 1 − 1 ( x , x2 , s1 ) ( x , s2 ) + 1 + 1 − 1 − ( x , x2 , s2 ) ( x1 , s1 ) (0,4,0,6,0) ( x , s1 , s2 ) ( x , x2 ) + 1 − 1 (0,1,3,0,0) Optimal value Nonbasic Basic Basic (zero) variables solution variables +− Associ- Feasible ated ? corner ( x1 , x1 , x2 , s1 , s2 ) point (2,0,0,0,6) (-4,0,0,6,0) (-1,0,3,0,0) N0 Solution A Yes No No - Objective value, z ( x , x2 , s1 ) ( x , s2 ) ( x1+ , s1 ) ( x , x2 , s 2 ) − 1 − 1 + 1 2 - ( x , s1 , s2 ) ( x , x2 ) ( x2 , s1 , s2 ) ( x , x ) + 1 − 1 − 1 + 1 Hence note that the number of Basic Solutions can be less than n m z = x1 + 3 x2 Direction of increasing z C z maximum 8 at (-1,3) E B D 0 A Example: Convert the following optimization problem into a LPP: Maximize z = max{| 2 x1 + 3 x2 |, | 3 x1 − 7 x2 |} Subject to x1 +x2 ≤ 2 − 1 +x2 ≤ 4 x x1 , x2 ≥0 Note that here the objective function is NOT linear. Let us put y = max{| 2 x1 + 3 x2 |, | 3 x1 − 7 x2 |} Hence y ≥ | 2 x1 + 3 x2 | and y ≥ | 3 x1 − 7 x2 | Which is equivalent to y ≥ 2 x1 + 3 x2 , y ≥ − (2 x1 + 3 x2 ) y ≥ 3 x1 − 7 x2 , y ≥ − ( 3 x1 − 7 x2 ) Hence the given optimization problem is equivalent to the LPP: Maximize z = y Subject to x1 +x2 ≤ 2 − 1 +x2 ≤ 4 x 2 x1 + 3x2 − y ≤ 0, − 2 x1 − 3x2 − y ≤ 0 3x1 − 7 x2 − y ≤ 0, − 3x1 + 7 x2 − y ≤ 0. x1 , x2 , y ≥ 0 ...
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This note was uploaded on 02/23/2011 for the course MATH 112 taught by Professor Ritadubey during the Spring '11 term at Amity University.

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