Discrete (and Continuous) Optimization WI4 131

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1 Discrete (and Continuous) Optimization WI4 131 Kees Roos Technische Universiteit Delft Faculteit Electrotechniek, Wiskunde en Informatica Afdeling Informatie, Systemen en Algoritmiek URL: roos November December, A.D. 2004

2 Course Schedule 1. Formulations (18 pages) 2. Optimality, Relaation, and Bounds (10 pages) 3. Well-solved Problems (13 pages) 4. Matching and Assigments (10 pages) 5. Dynamic Programming (11 pages) 6. Compleity and Problem Reduction (8 pages) 7. Branch and Bound (17 pages) 8. Cutting Plane Algorithms (21 pages) 9. Strong Valid Inequalities (22 pages) 10. Lagrangian Duality (14 pages) 11. Column Generation Algorithms (16 pages) 12. Heuristic Algorithms (15 pages) 13. From Theory to Solutions (20 pages) Optimization Group 1

3 Capter 12 Heuristic Algorithms Optimization Group 2

4 Introduction Many practical problems are NP-hard. In these case one may choose (or even be forced) to use a heuristic or approimation algorithm: a smart method that hopefully finds a good feasible solution quickly. In designing a heuristic, various questions arise: Should one accept any feasible solution, or should one ask a posteriori how far it is from optimal? Can one guarantee a priori that the heuristic produces a solution within ǫ (or α%) from optimal? Can one guarantee a priori that for the class of problems considered the heuristic on the average produces a solution within ǫ (or α%) from optimal? Optimization Group 3

5 Greedy and Local Search Revisited We suppose that the problem can be written as a COP in the form: min S N Consider for eample the 0-1 knapsack problem: z = ma n j=1 {c(s) : v(s) k}. c j j : n j=1 Here is the indicator vector of the set S. Thus, defining a j j b, {0,1} n. c(s) = j S c j, v(s) = j S a j, k = b the 0-1 knapsack problem gets the above form. Also the uncapacitated facility location problem min c ij ij + f j y j : j N i M j N fits in this model if we take c(s) = i M n ij = 1, ij M y j, ij [0,1], y j {0,1} j=1 i M min c ij + f j, v(s) = S, k = 1. j S j S Optimization Group 4

6 Greedy Heuristic min S N {c(s) : v(s) k}. Step 1: Set S 0 = and t = 1. Step 2: Set j t = argmin t c ( S t 1 {j t } ) c ( S t 1) v(s t 1 {j t }) v(s t 1 ). Step 3: If S t 1 is feasible, and the cost does not decrease when passing to S t = S t 1 {j t }, stop with S G = S t 1. Step 4: Otherwise, set S t = S t 1 {j t }. If S t is feasible and the cost function is nondecreasing or t = n: stop with S G = S t. Step 5: If t = n, no feasible solution has been found. Stop. Step 6: Set t t + 1, and return to Step 2. Optimization Group 5

7 Eample: Uncapacitated Facility Location Consider the UFL m = 6 clients and n = 4 depots, and costs as shown below: Greedy solution: (c ij ) = Step 1: Set S 0 = and t = 1. S 0 is infeasible. Step 2: We compute j 1 = argmin t c({j t }) v({j t }). and (f j ) = (21, 16, 11, 24) One has c({1}) = ( ) + 21 = 63, c({2}) = 66, c({3}) = 31, c({4}) = 64. So j 1 = 3, S 1 = {3} and c(s 1 ) = 31. S 1 is feasible. Step 3: Since S 1 is feasible, we check if the cost decreases when passing to S 2 = S 1 {j} for some j / S 1. One has c({3,1}) c({3}) = 18, c({3,2}) c({3}) = 11, and c({3,4}) c({3}) = 11. So the cost is nondecreasing: hence we stop with S G = S 1 = {3}. In many cases the heuristic must be adapted to the problem structure. We give an eample for STSP. Optimization Group 6

8 Distance matri. Eample: Symmetric TSP Let us start at node 1. The nearest neighbor of node 1 is 3, yielding P 1 = 1 3. The nearest neighbor of the node set {1,3} is 6, with distance 6 to 3, yielding the three possible paths shown in the table. We use the cheapest insertion: P 2 = The nearest neighbor of the node set {1,3,6} is 4, with distance 6 to node 6. Possible insertions are shown in the table. Now node 2 is nearest to P 3, and the possible insertions are as indicated. We use : P 4 = , which has length 20. Finally, the last node (node 5) must be connected to this path so as to obtain a tour. This can be done in 5 ways: Greedy Solution: The pure greedy heuristic has been applied to this instance in Chapter 2. Here we use a nearest neighborhood insertion heuristic: Starting from an arbitrary node we subsequently build paths P t containing that node by inserting the node nearest to this path in the current path, until we obtain a tour (this happens after n 1 insertions). node node insertions length new path P P P P P possible insertions of node 5 increment = = = = = 47 The shortest tour results when inserting 5 between the nodes 2 and 1 in P 4 (and adding the arc {4,2}): , and the length of this tour is = 52. Optimization Group 7

9 Other Variants of Greedy Insertion for Symmetric TSP The choice of the node to insert in the current path can be made in many different ways: nearest node (as done in the eample); random node; farthest node. When solving the same STSP instance with cheapest insertion of the farthest node we obtain (starting at node 1 again): P 1 = 1 5 (length 12); P 2 = (length 20); P 3 = (length 28); P 4 = (length 27); tour (length 49). The tour length is shorter than when inserting the nearest node! Suprising, but this often happens. Can you understand why? Optimization Group 8

10 Local Search Heuristic Local search can be more conveniently discussed when the problem has the form min S N {c(s) : g(s) = 0}, where g(s) 0 represents a measure for infeasibility of S. E.g., the constraint v(s) k can be represented by using g(s) = (k v(s)) +. For a local search heuristic we need: a solution S N; a local neighborhood Q(S) for each solution S N; a goal function f(s), which can be either c(s) when S is feasible, and infinite otherwise, or a composite function of the form c(s) + αg(s) (α 0). Local search Heuristic: Choose an intial solution S. Search for a solution S Q(S) that minimizes f(s ). If f(s ) = f(s), stop. Then S H = S is locally optimal. Otherwise, set S = S and repeat. Optimization Group 9

11 Local Search Heuristic (cont.) Local search Heuristic: Choose an intial solution S. Search for a solution S Q(S) that minimizes f(s ). If f(s ) = f(s), stop. Then S H = S is locally optimal. Otherwise, set S = S and repeat. Q(S 1 ) Q(S 2 ) Q(S k ) Appropriate choice of the neighborhoods depend on the problem structure. A simple choice is just to add or remove one element to or from S. This neighborhood has O(n) elements. Finding the best solution in the neighborhood can thus be done in O(n) time. If all feasible sets have the same size, a useful neighborhood is obtained by replacing one element of S by an element not in S. This requires O(n 2 ) time. In the case of the STSP this leads to the well known 2-echange heuristic. Optimization Group 10

12 2-Echange Heuristic for STSP With the greedy insertion heuristic we found the tour of length 49. A 2-echange removes two (nonadjacent, why?) edges. The two resulting pieces are connected by two other edges (this can be done in only one way, why?!). If this yields a better tour, we accept it and repeat, until we find a so-called 2-optimal tour. We have a 6-city problem. So a tour has 6 edges. There are = 9 possible 2-echanges. For each 2-echange we give the two new edges and the increase of the tour length Current tour. Length no. 2 deleted arcs 2 new arcs increment 1 {1,3}, {2,5} {1,2}, {3,5} = 15 2 {1,3}, {5,6} {1,5}, {3,6} = 3 3 {1,3}, {6,4} {1,6}, {3,4} 40 5 = 35 4 {3,2}, {5,6} {3,5}, {2,6} = 24 5 {3,2}, {6,4} {3,6}, {2,4} = 15 6 {3,2}, {4,1} {3,4}, {2,1} = 23 7 {2,5}, {6,4} {2,6}, {5,4} = 43 8 {2,5}, {4,1} {2,4}, {5,1} = 13 9 {5,6}, {4,1} {5,4}, {6,1} = New tour. Length 46. The second echange gives an improvement: It deceases the length of the tour to z L = 46. The tour is One may verify that this tour is 2-optimal. Optimization Group

13 Improved Local Search Heuristics How do we escape from a local minimum, and thus potentially do better than a local search heuristic? This is the question addressed by two heuristics that we first discuss briefly: Tabu search Simulated annealing After this we deal with Genetic algorithms Rather than working with individual solutions, genetic algorithms work with a finite population (set of solutions) S 1,..., S k, and the population evolves (changes somewhat randomly) from one generation (iteration) to the net. Finally we discuss two special purpose heuristics for the STSP problem and some heuristics for MIPs. Optimization Group 12

14 Tabu Search In order to escape from local minima one has to accept so now and then a solution with a worse value than the incumbent. Since the (old and better) incumbent may belong to the neighborhood of the new incumbent it may happen that cycling occurs, i.e., that the algorithm returns to the same solution every two or three steps: S 0 S 1 S 0 S To avoid cycling, certain solutions or moves are forbidden or tabu. Comparing the new solution with all previous incumbents would require much memory space and may be very time consuming. Instead, a tabu list of recent solutions, or solution modifications, is kept. A basic version of the algorithm is Step 1: Initialize a tabu list. Step 2: Get an initial solution S. Step 3: While the stopping criteria is not satisfied: 3.1: Choose a subset Q (S) Q(S) of non-tabu solutions. 3.2: Let S = argmin { f(t) : T Q (S) }. 3.3: Replace S by S. Step 4: On termination, the best solution found is the heuristic solution. Optimization Group 13

15 Tabu Search (cont.) Step 1: Initialize a tabu list. Step 2: Get an initial solution S. Step 3: While the stopping criteria is not satisfied: 3.1: Choose a subset Q (S) Q(S) of non-tabu solutions. 3.2: Let S = argmin {f(t) : T Q (S)}. 3.3: Replace S by S. Step 4: On termination, the best solution found is the heuristic solution. The parameters specific to tabu search are (i) The choice of Q (S). If Q(S) is small, one takes the whole neighborhood. Otherwise, Q (S) can be a fied number of neighbors of S, chosen randomly or by some heuristic rule. (ii) The tabu list consists of a small number t of most recent solutions (or modifications). If t = 1 or t = 2, it is not surprising that cycling is still common. The magic value t = 7 is often a good choice. (iii) The stopping rule is often just a fied number of iterations, or a certain number of iterations without any improvement of the goal value of the best solution found. Optimization Group 14

16 Tabu Search (cont.) If the neighborhood consists of single element switches: Q(S) = {T N : T \ S = 1}, S N, then the tabu list might be a list of the last t elements {i 1,..., i t } added to the incumbent and a list of the last t elements {j 1,..., j t } removed from the incumbent. A neighbor T is then tabu if T = S \ {i q } or if T = S {j q } for some q = 1,..., t. So, in forming the new incumbent one is not allowed to modify the current incumbent S by adding some i q or by removing some j q. When implementing tabu search, the performance can be improved by using common sense. E.g., there is no justification to make a solution tabu if it is the best solution found by other researchers. In other words, tabu search can be viewed as a search strategy that tries to take advantage of the history of the search and the problem structure intelligently. Optimization Group 15

17 Simulated Annealing Simulation is less direct, and less intelligent. The basic idea is to choose a neighbor randomly. The neighbor replaces the incumbent with probability 1 if it is better, and with some (changing) probability (0,1) if it has a worse value. The probability for accepting a worse solution is taken proportional to the difference in goal values. So, if the number of iterations is large enough, one can escape from every local minimum. On the other hand, to guarantee convergence of the algorithm, the probability of accepting worse solutions decreases over time. A more formal description is as follows: Step 1: Get an initial solution S. Step 2: Get an initial temperature T, and a cooling ratio r (0 < r < 1). Step 3: While not yet frozen, do the following: 3.1: Perform the following loop L times: 3.1.1: Pick a random neighborhood S of S : Let = f(s ) f(s) : If 0, replace S by S : If > 0, replace S by S with probability e T. 3.2: Set T rt. (Reduce the temperature.) Step 4: Return the best solution found, this is the heuristic solution. Optimization Group 16

18 Simulated Annealing (cont.) Step 1: Get an initial solution S. Step 2: Get an initial temperature T, and a cooling ratio r (0 < r < 1). Step 3: While not yet frozen, do the following: 3.1: Perform the following loop L times: 3.1.1: Pick a random neighborhood S of S : Let = f(s ) f(s) : If 0, replace S by S : If > 0, replace S by S with probability e T. 3.2: Set T rt. (Reduce the temperature.) Step 4: Return the best solution found, this is the heuristic solution. Note that the probability of accepting worse solutions decreases in time, because the temperature T decreases in time. Also, the probability decreases if increases, i.e. if the quality of the solution becomes worse. Just as for other local search heuristics, one has to define an initial solution, a neighborhood for each solution and the value f(s) of a solution. The parameters specific for SA are (i) The initial temperature T. (ii) The cooling ration r. (iii) The loop length L. (iv) The definition of frozen, i.e., a stopping rule. Optimization Group 17

19 Genetic Algorithms Rather than working with individual solutions, genetic algorithms work with a finite population (set of solutions) S 1,..., S k, and the population evolves (changes somewhat randomly) from one generation (iteration) to the net. An iteration consists of the following steps: (i) Evaluation. The fitness of the individuals is evaluated. (ii) Parent Selection. Certain pairs of solutions (parents) are selected based on their fitness. (iii) Crossover. Each pair of parents combines to produce one or two new solutions (offspring). (iv) Mutation. Some of the offspring solutions are randomly modified. (v) Population Selection. Based on their fitness, a new population is selected replacing some or all of the original population by an identical number of offspring solutions. Each of the five steps is discussed in more detail in the book. Optimization Group 18

20 We consider the integer knapsack problem Worst-case Analysis of (some) Heuristics z = ma Integer Knapsack Heuristic n j=1 c j j : n j=1 a j j b, Z n where a j Z + for all j and b Z +. We suppose that the variables are ordered so that c j a is non-increasing and, moreover, that a j j b for all j. We consider the following simple heuristic: take H 1 = b a1 and H j = 0 for j 2, yielding the value z H = c b 1 a1. Theorem 1 z H 1 2 z. Proof: The solution of the linear relaation is 1 = a b, and 1 j = 0 for j 2. This provides the upper bound z LP = c 1b a z. As a 1 1 b, we have b a1 1. Setting b a = [ ] b 1 a1 + b a1 f we have 0 [ ] b a1 < 1. Hence f z H b = c 1 a 1 = b a1 z LP = b a 1 b a1 b a1 + [ b a1 ]f z LP b a1, b a1 + b a1 z LP = 1 2 zlp 1 2 z. Optimization Group 19

21 Eucledian STSP We say that an STSP in graph G = (V, E) is Eucledian if for any three edges e, f and g forming a triangle, one has the so-called triangle inequality: c e c f + c g. Proposition 1 Let G be a complete graph whose edge lengths satisfy the triangle inequality, and let G contain a subgraph H = (V, Ē) which is Eulerian. Then G contains a Hamilton cycle of length at most c(h) = e Ē c e. Proof: Note that any Eulerian circuit in H passes eactly once through every edge in Ē, and hence has length e Ē c e. Any such circuit C passes through all the nodes in V, probably more than once through some or all of them. Using C we construct a Hamilton circuit as follows. We choose a node v 1 V, and starting at that node we walk along the edges of C (in one of the two possible directions). The first new node that we encounter after leaving v 1 is called v 2. After leaving v 2, the first node different from v 1 and v 2 is called v 3, and so on. Proceeding in this way we get a sequence v 1, v 2..., v n containing all the nodes in V. We claim that the tour T : v 1 v 2... v n v 1 has length at most e Ē c e For this two observations are necessary. First, G contains the edges { } v i, v i+1 for i = 1 to i = n 1 and also the edge {v n, v 1 }, since G is a complete graph. So T is a tour indeed. Second, the nodes on T partition the Euler circuit C in n subsequent edge-disjoint pieces. Each edge on the tour T is a shortcut for the corresponding piece, due to the triangle inequality. Thus the length of the tour is at most equal to the length of C. Optimization Group 20

22 Geometric proof We consider the piece of the Eulerian tour between the nodes v i and v i+1 of the Hamilton tour. It will be convenient to call the nodes on the corresponding piece of the Eulerain tour 1, 2,..., k. So 1 v i and k v i+1. The figure depicts the situation for k = 6. The red edges are part of the Eucledian tour Using the triangle inequality we show that the shortcut 1 6 is shorter than the path : c 16 c 15 + c 56 c 14 + c 45 + c 56 c 13 + c 34 + c 45 + c 56 c 12 + c 23 + c 34 + c 45 + c 56. Optimization Group 21

23 The Tree Heuristic for Eucledian STSP Step 1: Find a minimum-length spanning tree T, with edge set E T and length z T = e E T c e. Step 2: Double each edge in T to form a connected Eulerian graph. Step 3: Using the previous proposition, taking any Eulerian tour, construct a Hamilton circuit of length z H. Proposition 2 z H 2 z. Proof: Since every tour consists of a spanning tree plus an edge, we have the lower bound z T z. The length of the Eulerian tour is 2z T, by construction. The construction of the Hamilton circuit guarantees that z H 2z T. Thus we have z H 2 z T z z H. Optimization Group 22

24 Eample of the Tree Heuristic for Eucledian STSP Below the distances are given between the capitals of 11 of the 12 provinces in the Netherlands Amsterdam Arnhem Assen Den Haag Groningen Den Bosch Leeuwarden Maastricht Middelburg Utrecht Zwolle We want to find a minimum length Hamilton circuit along these cities. Optimization Group 23

25 Eample of the Tree Heuristic for Eucledian STSP We first find a minimal weight spanning tree. The minimal weight tree is shown in the graph. It has weight = 674. By doubling each edge in the tree the graph becomes Eulerian An Eulerian tour is (e.g.) , which has length From this we obtain the Hamilton circuit , with length = N.B. When applying 2-echanges to the above tour the length 9 6 reduces to The farthest neighbor insertion heuristic, when started at node 10, yields a tour of length When applying 2-echanges this length reduces to 990. This is optimal: For the nearest neighborhood insertion heuristic, 8 starting at node 8, the length is 1045, which can be reduced to 1032 by 2-echanges. Optimization Group 24

26 The Tree/Matching Heuristic for Eucledian STSP Step 1: Find a minimum-length spanning tree T, with edge set E T and length z T = e E T c e. Step 2: Let U be the set of odd nodes in (V, E T ). Find a perfect matching M of minimum length z M in the subgraph G = (U, E ) of G induced by the subset U, so E contains all edges in G with both end points in U. By construction, (V, E T M) is an Eulerian graph. Step 3: From any Eulerian tour in (V, E T M), construct a Hamilton circuit of length z C. Proposition 3 (Christofides, 1976) z C 3 2 z. Proof: As above, z T z. Let the nodes be ordered such that n 1 is { an optimal } tour (of length z). Let j 1 <... < j 2k be the nodes of U, and let e i = ji, j i+1, with jn+1 = j 1. Due to the triangle inequality, one has 2k i=1 e i z. The sets M 1 = {e i : i odd} and M 2 = {e i : i even} are matchings in G = (U, E ). Hence z M z M1 and z M z M2. Consequently, since z M1 + z M2 = 2k i=1 e i z, 2z M z. Hence either z M1 1 2 z, or z M z. Suppose z M z. Then (V, E T M 1 ) is Eulerian, and has weight 3 2z. As we have seen before, we can then construct a Hamiltonian circuit of length z C 3 2 z. Optimization Group 25

27 Geometric proof We are given an optimal Hamilton circuit C : n 1 of length z in the graph G = (V, E) and the set U of nodes having odd degree in a minimum weight spanning tree T of G. The nodes in U partition C in pieces, we connect these nodes by the edges that are shortcuts of these pieces, as indicated in the figure below. These edges form an even length circuit, because the number of nodes in U is even. We alternately assign the edges to the sets M 1 (red) and M 2 (green). Then M 1 and M 2 are matchings in the subgraph of G induced by U. u 2 u 1 u 3 u 6 u 5 u 4 Obviously, due to the triangle inequality, z M1 + z M2 z, Hence either z M1 1 2z, or z M2 1 2z. By adding the shortest matching to T we get an Eulerian graph whose Eulerian circuits have length 1 2 z. Optimization Group 26

28 Eample of the Tree/Matching Heuristic for Eucledian STSP We use the minimal weight spanning tree that we found earlier. The odd degree nodes in the tree form the set {7,8,9,10}. The indiced subgraph on these nodes is depicted below: The mimimal weight matching consists of the arcs {8, 9} and {7,10}. Adding these arcs to the tree the graph becomes Eulerian. An Eulerian tour is (e.g.) From this we obtain the Hamilton circuit , with length = This route is 2-optimal. Note that we know that a minimum length tour can not be shorter than = Optimization Group 27

29 MIP-based Heuristics Dive-and-Fi Aim: find quickly a feasible solution, which gives a tight lower bound in the search tree. Suppose we have a mied 0-1 problem in variables i R and y j {0,1}. Given a solution (, y ) of the linear relaation, let F = { j : y j / {0,1} }. Initialization Take the solution (, y ) of the linear relaation at some node in the search tree. Basic Iteration As long as F, do the{ following: [ Let i = argmin j F min y j,1 yj ]} (find the fractional variable closest to integer). If yi < 0.5, fi y i = 0 (if close to 0, fi to 0). If yi 0.5, fi y i = 1 (if close to 1, fi to 1). Solve the resulting LO problem. If the LO problem is infeasible, stop (the heuristic has failed). Otherwise, let (, y ) be the new linear solution. Termination If F =, (, y ) is a feasible mied integer solution. Optimization Group 28

30 Two other MIP-based Heuristics We discuss two other heuristics for mied integer problems that are hard to solve. For simplicity we describe the heuristic for an IP of the following form. { z = ma c 1T 1 + c 2T 2 : A A 2 2 = b, 1 Z n 1 +, 2 Z n } 2 +. It is supposed that the variables 1 j for j N 1 are more important than the variables 2 j for j N 2, where N i = n i for i = 1,2. The idea is to solve two (or more) easier LO problems or MIPs. The first one allows us to fi or limit the range of the more important 1 variables, whereas the second allows us choose good values for the 1 variables. Optimization Group 29

31 z = ma Rela-and-Fi { } c 1T 1 + c 2T 2 : A A 2 2 = b, 1 Z n 1 +, 2 Z n 2 +. Rela: Solve the relaation { z = ma c 1T 1 + c 2T 2 : A A 2 2 = b, 1 Z n 1 +, 2 R n 2 + Let ( 1, 2) be a solution of this problem. Fi: Fi the (important) 1 variables to their value in 1 and solve the restriction z = ma { c 1T 1 + c 2T 2 : A 2 2 = b A 1 1, 1 = 1, 2 Z n 2 + If this problem is infeasible, the heuristic fails. Otherwise, let ( 1, 2) be a solution of this problem. Heuristic: Use as heuristic solution H = ( 1, 2), whose value satisfies z = c T H z z. } }.. Optimization Group 30

32 z = ma Cut-and-Fi { } c 1T 1 + c 2T 2 : A A 2 2 = b, 1 Z n 1 +, 2 Z n 2 + We assume that a strong cutting plane algorithm is available, which generates strong cuts C C 2 2 γ, so after adding these cuts, at least some of the integer variables take values close to integer or to their optimal values. Cut: Using a strong cutting plane algorithm find a solution of the tight linear relaation { } z = ma c 1T 1 + c 2T 2 : A A 2 2 = b, C C 2 2 γ, 1 R n 1 +, 2 R n 2 +. Let ( 1, 2) be a solution of this problem. N.B. The cuts are generated in the course of the algorithm solving the linear relaation of the given problem. Fi or Bound : Choose ǫ. For j N 1, set l j = 1 j + ǫ and u j = 1 j ǫ. Solve the restriction z = ma c 1T 1 + c 2T 2 : A A 2 2 = b, C C 2 2 γ, l 1 u, 1 Z n 1 +, 2 Z n 2 +. If this problem is infeasible, the heuristic fails: one may try again with ǫ decreased. Otherwise, let ( 1, 2) be a solution of this problem. Heuristic: Use as heuristic solution H = ( 1, 2), whose value satisfies z = c T H z z. Observe that if ǫ is small and positive, 1 variables taking values within ǫ of integer in the linear relaation are fied in the restricted problem, while others are forced to either the value 1 j or the value 1 j. On the other hand, if ǫ is negative, all the 1 variables can still take two values in the restricted problem.. Optimization Group 31

33 More Courses on Optmization Code Name Docent WI3 031 Niet-Lineaire Optimalisering C. Roos WI4 051TU Introduction to OR H. van Maaren WI4 060 Optimization and Engineering C. Roos WI4 062TU Transportation, Routing and Scheduling Problems C. Roos WI4 063TU Network Models and Algorithms J.B.M. Melissen WI4 064 Discrete Optimization C. Roos WI4 087TU Optimization, Models and Algorithms H. van Maaren WI4 131 Discrete and Continuous Optimization G.J. Olsder/C. Roos IN4 082 Local (Heuristic) Search Methods H. van Maaren IN4 077 Computational Logic and Satisfiability H. van Maaren/C. Witteveen IN4 081 Randomized algorithms H. van Maaren Optimization Group 32