MAT016: Optimization

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1 MAT016: Optimization M.El Ghami URL: melghami/ March 29, 2011

2 Outline for today The Simplex method in matrix notation Managing a production facility The linear programming problem (LP) or (LO) Examples The Simplex method Matrix notation The primal Simplex method Examples The dual simplex method 1

3 Managing a production facility A company produces three products. The per-unit profit, labor usage, and pollution produced per unit are given in the following table Profit Labor Usage Pollution Product 1 6 NOK 4 hours lb Product 2 4 NOK 3 hours lb Product 3 3 NOK 2 hours lb At most labor hours can be used to produce the three products, and government regulations require that the company produce at most 2lb of pollution. 2

4 Linear programming formulation Let x i =units produced of product i. maximize z = 6x 1 + 4x 2 + 3x 3 subject to 4x 1 + 3x 2 + 2x x x x 3 2 x 1, x 2, x

5 Application of LO Business Logistics Economics Engineering 4

6 The linear programming problem Decision variables (variables) are written as: x j, j = 1, 2,..., n. Objective function: maximize or minimize linear function of decision variables, ζ = c 1 x 1 + c 2 x c n x n. Equivalent: maximize ζ or minimize ζ. Constraints: equality or inequality associated with some linear combination of the decision variables a 1 x 1 + a 2 x a n x n {, =, } b. 5

7 Convert of the constraints Convert of inequality to an other inequality: a 1 x 1 + a 2 x a n x n b, can be written a 1 x 1 a 2 x 2... a n x n b. Convert of inequality to equality: a 1 x 1 + a 2 x a n x n b, can be written a 1 x 1 + a 2 x a n x n + w = b, where w 0, which we call slack variable. 6

8 Convert of the constraints (count) Convert of equality to inequalities: a 1 x 1 + a 2 x a n x n = b, can be written in form of two inequality constraints a 1 x 1 + a 2 x a n x n b, a 1 x 1 + a 2 x a n x n b. 7

9 Formulation of LP Standard form of LP. maximize c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2... a m1 x 1 + a m2 x a mn x n b m x 1, x 2,..., x n 0. n: number of variables. m: number of constraints. (x 1, x 2,..., x n ) is feasible if it satisfies all constraints. It is optimal if it attains the desired maximum. 8

10 Example of infeasible problem maximize 5x 1 + 4x 2 subject to x 1 + x 2 2 2x 1 2x 2 9 x 1, x 2 0. Note that 2x 1 2x 2 9 x 1 + x 2 4.5, which contradict with x 1 + x 2 2. This problem is infeasible. 9

11 Example of unbounded problem maximize x 1 4x 2 subject to 2x 1 + x 2 1 x 1 2x 2 2 x 1, x 2 0. Let x 2 = 0 and x 1 is arbitrarily large. As long as x 1 is greater than 2 the solution will be feasible, and as it gets large the objective function does too. This problem is called unbounded. Definition 1 A problem is unbounded if it has feasible solutions with arbitrarily large objective values. 10

12 Exercises Exercise 1 Jim and Jane have a small furniture workshop in their garage, where they assemble from purchased parts and finish the furniture in preparation for sale. they are presently limiting their production to table and chairs. Each chair requires 4 h to assemble and 2 h to finish, but each table require 2 h to assemble and 4 h to finish. Jim works only on the assembly operation. Jane works only on the finishing operation, and each puts in no more than an 8-h day. It is known that each chair can be sold for 3000 NOK and each table for 2000 NOK. If Jim and Jane wants to maximize the total income. What is a mathematical model for this problem? 11

13 Next How to solve a linear programming? Simplex method. 12

14 Simplex method: Example 1 How the Simplex method works? Example 1: maximize 3x 1 + 2x 2 subject to 2x 1 + 4x 2 8 (I) 4x 1 + 2x 2 8 x 1, x

15 Simplex method: Example (cont) Introduce the slack variables: maximize 3x 1 + 2x 2 = ζ subject to 2x 1 + 4x 2 + w 1 = 8 (II) 4x 1 + 2x 2 + w 2 = 8 x 1, x 2, w 1, w 2, 0. Equivalent to maximize 3x 1 + 2x 2 = ζ subject to 8 2x 1 4x 2 = w 1 (II) 8 4x 1 2x 2 = w 2 x 1, x 2, w 1, w 2, 0. 14

16 Example (cont) Simplex method is an iterative process in which we start with a solution x 1, x 2, w 1, w 2 and then look for new solution x 1, x 2, w 1, w 2 which is better in the following sense 3x 1 + 2x 2 3 x x 2 we continue until we arrive at the solution that can t improve. Initial feasible solution (starting point) of this example is x 1 = 0, x 2 = 0, w 1 = 8, w 2 = 8, the objective value is ζ = 0. 15

17 Example (cont) This solution can be improved? If yes. How? The coefficients of x 1 and x 2 are positive (Coefficient of x 1 is large than coefficient of x 2 ). Increase the value of x 1 and then the value of ζ increase. Make sure that all slack variables stay nonnegative. Since x 2 = 0. Then we have to make sure that { 8 x 1 2 = 4, 8 } 4 = 2 = x 1 2. New solution is and ζ = 6. (x 1 = 2, x 2 = 0, w 1 = 4, w 2 = 0). 16

18 Example (cont) Next step Rewrite the equations of our system in such a way that x 1, w 1, and ζ. are expressed as functions of w 2, and x 2. By using the second constraint we have x 1 = x w 2. Performing the appropriate row operations, we can eliminate x 1 from other equations. Then we have x w 2 = ζ x w 2 = x 1 (III) 4 3x w 2 = w 1. 17

19 Example (cont) w 2, and x 2 are called independent variables and x 1, and w 1 are called dependent variables. Increase x 2, the value of ζ increase also (coefficient of x 2 is positive). Make sure that x 1, and w 1 remain nonnegative. { x 2 4, 4 }, 3 then x New solution is ( x 1 = 4 3, x 2 = 4 ) 3, w 1 = 0, w 2 = 0. and ζ =

20 Example (cont) We will write our equations: ζ, x 1, and x 2 as functions of w 1, and w 2. Solving the last equation in (III) for x 2, we get x 2 = w w 2. Performing the appropriate row operations, we eliminate x 2 from other equations. We have w w 2 = ζ w w 2 = x 1 (IV) w w 2 = x 2. Note that all coefficients of the variables presented in the objective function (w 1 and w 2 ) are negative. This means that we can t improve the objective function. 19

21 Example (cont) Definition 2 The systems of equations (II), (III), and (IV ) are called dictionaries. The dependent variables are called basic variables and B denote the set of indices corresponding to the basic variables. The independent variables are called nonbasic variables and N denote the set of indices corresponding to the nonbasic variables. The solutions we have obtained by setting the nonbasic variables to zero are called basic feasible solutions. 20

22 Simplex method We consider a standard (LP). maximize subject to nj=1 c j x j nj=1 a ij x j b i, i = 1, 2,..., m x j 0, j = 1, 2,..., n. 21

23 Simplex method (cont) Introduce slack variables and a name for the objective function. maximize subject to nj=1 c j x j = ζ nj=1 a ij x j + w i = b i, i = 1, 2,..., m x j 0, j = 1, 2,..., n. w i 0, i = 1, 2,..., m. For simplicity, we denote (x 1,..., x n, w 1,..., w m ) = ( x 1,..., x n, x n+1,..., x n+m ). 22

24 Simplex method (cont) Our problem become maximize nj=1 c j x j = ζ subject to b i n j=1 a ij x j = x n+i, i = 1, 2,..., m We have N = {1, 2,..., n} and B = {n + 1, n + 2,..., n + m}. But this will change later (just after one iteration). maximize j N c j x j + ζ = ζ subject to b i j N ā ij x j = x n+i, i B. We put bars over the coefficients to indicate that they change as the algorithm progresses. After one iteration: One variable goes from basic to nonbasic (leaving variable). One goes from nonbasic to basic (entering variable). 23

25 Simplex method (cont) Entering variable: x k such that k N + = { j N : c j > 0 }. If N + = then the current solution is optimal. If N + contains more than one element, we will pick an index k having the largest coefficient c k. Leaving variable: chosen to preserve nonnegativity of the current basic variables. If x k is the entering variable, its value will be increased from zero to a positive value. The basic variables will change: We must ensure that x i = b i ā ik x k, i B. x k = x i 0 b i ā ik x k 0, i B. min i B:ā ik >0 b i ā ik x k = Thus Leaving variable: pick l from Pivot: Step from one dictionary to the next. ( max i B ) 1 ā ik. b i { i B : āik b i is maximal }. 24

26 Unboundedness Let x k be the leaving variable, and let l { i B : āik b i is maximal If ālk 0 i B, āik 0. Then our problem is unbounded. (the b l b i entering variable can be increased indefinitely to produce large objective value). We consider the following dictionary: }. ζ = 5 + x 3 x 1 x 2 = 5 + 2x 3 3x 1 x 4 = 7 4x 1 x 5 = x 1. The entering variable is x 3 and the ratios are: 2 5, 0 7, 0 0. This problem is unbounded. Because all these ratios are nonpositive. 25

27 Geometry: Example 1 We consider the following linear problem maximize 3x 1 + 2x 2 subject to 4x 1 + 2x 2 8 2x 1 + 4x 2 8 x 1, x 2 0. See Blackboard 26

28 We consider a LP in standard form Matrix notation maximize subject to nj=1 c j x j nj=1 a ij x j b i, i = 1, 2,..., m x j 0, j = 1, 2,..., n. Introduce slack variables x n+i = b i Problem in matrix form n j=1 a ij x j, i = 1, 2,..., m. maximize c T x subject to Ax = b x 0, 27

29 Matrix notation (cont) Where A = a 11 a a 1n 1 a 21 a a 2n a m1 a m2... a mn 1 b = b 1. b m, c = c 1. c n 0. 0, and x = x 1. x n x n+1. x n+m 28

30 Matrix notation (cont) In Component notation, the ith component of Ax can be broken up into a basic part and a nonbasic part: n+m j=1 We can writ the Matrix A a ij x j = A = j B [ a ij x j + B N ]. j N a ij x j. Separations of x and c into basic and nonbasic parts, then Constraints: Objective function: x = x B x N c = c B c N. Ax = b Bx B + Nx N = b. ζ = c T x c T B x B + c T N x N. Matrix B is m m and invertible! Why? 29

31 Matrix notation (cont) x B and ζ in terms of x N x B = B 1 b B 1 Nx N ζ = c T B x B + c T N x N = c T B B 1 b ( (B 1 N ) T cb c N ) T xn. Dictionary ζ = c T B B 1 b ( (B 1 N ) T cb c N ) T xn x B = B 1 b B 1 Nx N. 30

32 Dual variables (z 1,..., z n, y 1,..., y m ) ( ) z 1,..., z n, z n+1,..., z n+m The dual dictionary corresponding to the primal dictionary is ζ = c T B B 1 b ( B 1 b ) T zb z N = ( B 1 N ) T cb c N + ( B 1 N ) T zb. Dual solution associated to this dictionary is: z B = 0 z N = ( B 1 N ) T cb c N. Solution: ζ = c T B B 1 b 31

33 Dictionary Primal dictionary: ζ = ζ z N T x N x B = x B B 1 Nx N. Dual dictionary ζ = ζ z N T x N z N = z N + B 1 Nz B. 32

34 Example We consider a LP. maximize 5x 1 + 4x 2 + 3x 3 subject to 2x 1 + 3x 2 + x 3 5 4x 1 + x 2 + 2x 3 11 (I) 3x 1 + 4x 2 + 2x 3 8 x 1, x 2, x 3 0. First we introduce the slack variables. Our problem is equivalent to maximize 5x 1 + 4x 2 + 3x 3 = ζ subject to 2x 1 + 3x 2 + x 3 + x 4 = 5 4x 1 + x 2 + 2x 3 + x 5 = 11 (II) 3x 1 + 4x 2 + 2x 3 + x 6 = 8 x 1, x 2, x 3, x 4, x 5, x

35 Matrix notation for this example Constraints: x 1 x 2 x 3 x 4 x 5 x 6 5 = Objective function: [ ] x 1 x 2 x 3 x 4 x 5. x 6 34

36 Example (cont) Basic variables: x 1, x 5, x 6. Nonbasic variables: x 2, x 3, x 4. Ax = = = 2x 1 + 3x 2 + x 3 + x 4 4x 1 + x 2 +2x 3 +x 5 3x 1 + 4x 2 + 2x 3 +x 6 2x 1 +3x 2 + x 3 + x 4 4x 1 + x 5 +x 2 +2x 3 3x 1 + x 6 +4x 2 + 2x = Bx B + Nx N. x 1 x 5 + x x 2 x 3 x 4 35

37 Example (cont) B = = B 1 = B 1 b = B 1 N = =

38 Example (cont) ( B 1 N ) T cb c N = = c T B B 1 b = [ ] = 12.5 Associated Primal Solution: x N = 0 x B = B 1 b =

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