Introduction to Discrete Optimization

Transcription

1 Prof. Friedrich Eisenbrand Martin Nieeier Due Date: March 9 9 Discussions: March 9 Introduction to Discrete Optiization Spring 9 s Exercise Consider a school district with I neighborhoods J schools and G grades at each school. Each school j has a capacity of C j g for grade g. In each neighborhood i the student population of grade g is S i g. Finally the distance of school j fro neighborhood i is d i j. Forulate a linear prograing proble whose objective is to assign all students to schools while iniizing the total distance traveled by all students. You ay ignore the fact that nubers of students ust be integer. Let x i j g be the aount of students fro neighborhood i of grade g travelling to schoolj. Then for an assignent of students to schools the total distance travelled by all students is given as d i j x i j g. i I j J g G For a feasible assignent every student of every neighborhood and grade ust be assigned to a school this gives the constraint x i j g = S i g i I g G. j J The nuber of students each school can take of the respective grades is bounded by C j g thus x i j g C j g j J g G i I ust hold. Finally there can be no negative nubers of assignents: x. This gives the following linear progra: in subject to i I j J g G d i j x i j g j J x i j g = S i g i I x i j g C j g x i I g G j J g G Exercise Consider the vectors x = x = x = x = x =.

2 Let A = {x... x }. Find two disjoint subsets A A A such that Hint: Recall the proof of Radon s lea conva conva. Since we have points in R Radon s lea states that the subsets A and A exist. To copute the we review the proof of Radon s lea. We construct the set { A x = x x x x The vectors of A are linearly dependent and we can copute a nontrivial linear cobination of the all zero vector. xi = λ i. i= As shown in the proof if the define sets P := {i : λ i } and N := {i : λ i < } then the sets A := {x i : i P} and A := {x i : i N } have the desired property. To copute a nontrivial linear cobination of the all zero vector using points fro A we solve the following linear progra: λ + λ + λ + λ + λ = λ + λ + λ + λ = λ + λ + λ + λ + λ = λ + λ + λ + λ + λ = Using standard ethods e.g. gaussian eliination one can copute the solution set as: S = {a } a a a : a R. } We take the solution S which gives the nontrivial linear cobination x = x + x x x + As shown in the proof of Radon s lea the sets A := {x x x } and A := {x x } have the required property i.e conva conva. As a certificate the proof gives that is contained in both conv A and conv A. v = x + x = x + x =..

3 Exercise Consider the vectors The vector x = x = x = x = v = x + x + x + x + x = x = is a conic cobination of the x i. Write v as a conic cobination using only three vectors of the x i. Hint: Recall the proof of Carathéodory s theore Let X = {x... x n }. Observe that v conex. Since x... x R Carathéodory s theore states that we can write v as a conic cobination using at ost three vectors of X. How to copute this conic cobination? Recall the proof of Carathéodory s theore. The nuber of vectors in the conic cobination v = i= λ i x i can be reduced by one with the following ethod: Copute a nontrivial linear cobination of the all zero vector i.e. copute µ...µ R not all of the zero such that i= µ i x i = holds. Thus v = i= λ i ɛµ i x i for each ɛ >. As described in the proof one can find an ɛ such that λ i ɛ µ i for each i =... and λ i ɛ µ i = for at least one i. Thus we get a new conic cobination of v using one vector less than before. We now applying the idea to the exercise. We first copute a nontrivial linear cobination of the all zero vector by solving the following syste of linear equations: µ + µ + µ + µ + µ = µ + µ + µ + µ = µ + µ + µ + µ + µ =. Using standard ethods e.g. gaussian eliination one can copute the solution set as: S = { a ba + b ab : ab R} We take the solution 6 S which gives a nontrivial linear cobination i.e. = x + 6x + x + x. What is the axial ɛ such that v = + ɛx + x + 6ɛx + ɛx + ɛx is a conic cobination? Each coefficient has to be nonnegative thus observe that. ɛ = is the axiu. We get the new conic cobination v = 8 x + x + x + x + 8 x.

4 Observe that since the coefficient of x is zero we can reove it fro the conic cobination. We need to reove one ore vector to get a conic cobination using only three vectors. Again we copute a nontrivial linear cobination of the all zero vector using the reaining vectors x x x x : We copute the solution set which is µ + µ + µ + µ = µ + µ + µ + µ = µ + µ + µ + µ =. S = { a aa : a R} We take the solution S which gives a nontrivial linear cobination i.e. = x x + x. What is the axial ɛ such that is a conic cobination? It is given by ɛ = 8. The new conic cobination is v = 8 + ɛx + x + + ɛx + 8 ɛx v = x + x x x = 6 x + x + 6 x + x. Since the coefficient of x is zero we can reove it and obtain the desired convex cobination of v using only three vectors. Exercise Show that a basic solution can be associated to two different bases i.e. give an exaple of a solution x to a linear progra in{c T x : Ax = b x } such that there are two bases A B and A B with A B x B = b A B x B = b and x i = i {j =...n : j B B }. Consider the linear progra in n i= x i Ax = b x Where A := and b = Let B = {} and B = {}. Set x := T. Observe that x has all desired properties..

5 Exercise Recall the naive algorith given in the lecture to solve a linear progra by generating all basic solutions. Consider linear progras of the for in{c T x : Ax = b x } where A Q n b Q and c Q n. Assue that you have a coputer that for every subset J {...n} can check whether A J is a basis copute x = A J b check whether x and copute c T x in sec. If n = what is the largest such that this coputer can calculate an optial solution of the linear progra using the naive algorith in. one inute. one day. one year 6 days The tie needed is given by the nuber of sets we have to test: The naive algorith considers each subset of {...n} of cardinality. For each such set sec is needed. Since there n are = such sets the naive algorith needs a running tie of sec. 8 One inute has 6 sec. = 86 and = 876. Thus = 9 is the 9 largest that can be processed in less than one inute. 8 One day has 86 sec. = 66 and = 7. Thus = is the largest that can be processed in less than one day. 6 8 One year has 6 sec. = 97 and = Thus = 8 is the largest that can be processed in less than one year.