# Discrete (and Continuous) Optimization Solutions of Exercises 1 WI4 131

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1 Discrete (and Continuous) Optimization Solutions of Exercises 1 WI4 131 Kees Roos Technische Universiteit Delft Faculteit Informatietechnologie en Systemen Afdeling Informatie, Systemen en Algoritmiek URL: roos November December, A.D. 2002

2 Course Schedule 1. Formulations (18 pages) 2. Optimality, Relaxation, and Bounds (10 pages) 3. Well-solved Problems (13 pages) 4. Matching and Assigments (10 pages) 5. Dynamic Programming (11 pages) 6. Complexity 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 Exercise Suppose that you are interested in choosing a set of investments {1,..., 7} using 0-1 variables x i, i = 1,..., 7. Model the following constraints. (i) You cannot invest in all of them. (ii) You must choose at least one of them. (iii) Investment 1 cannot be chosen if investment 3 is chosen. (iv) Investment 4 can be chosen only if investment 2 is also chosen. (v) You must choose either both investments 1 and 5 or neither. (vi) You must choose either at least one of the investments 1,2,3, or at least two from 2,4,5,6. Solution: (i) x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 6. (ii) x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 1. (iii) x 1 + x 3 1. (iv) x 2 x 4 (N.B. x 4 = 1 x 2 = 1). (v) We must have either x 1 + x 5 2 or x 1 + x 5 0. We introduce for each alternative a 0-1 variable, y 1 and y 2, respectively. The model is (with M = 2): x 1 + x 5 2y 1 x 1 + x 5 2(1 y 2 ) y 1 + y 2 = 1 y 1, y 2 {0,1} x 1 + x 5 2y 1 x 1 + x 5 2y 1 y 1 {0,1} (vi) We must have either x 1 + x 2 + x 3 1 or x 2 + x 4 + x 5 + x 6 2. Using 0-1 variables y 1 and y 2, one for each alternative, the model becomes (with M = 2): x 1 + x 2 + x 3 y 1 y 1 + y 2 = 1 x 1 + x 4 + x 5 + x 6 2y 2 y 1, y 2 {0,1} Optimization Group 2

4 Exercise Formulate the following as mixed integer problems: (i) u = min {x 1, x 2 }, assuming that 0 x i C for j = 1,2. (ii) v = x 1 x 2, with 0 x i C for j = 1,2. (iii) the set X \ {x }, where X = {x Z n : Ax b} and x X. Solution: (i): u = max {t : t x 1, t x 2, 0 x 1 C, 0 x 2 C}. (ii): u = min {t : t x 1 x 2 t, 0 x 1 C, 0 x 2 C}. (iii): For each i define X i = {x Z n : x i x i 1, Ax b}, X + i = {x Z n : x i x i + 1, Ax b}. With M a large enough number, and using a binary vanriable y i, the union of these sets has the linear formulation Since X i X + i = {x Z n : Ax b, 1 + x i x i My i, 1 + x i x i M(1 y i ), y i {0,1}}. X \ {x } = n i=1 ( X i X + i this set can be linearly represented by using binary variables z i such that z z n = 1 and x X i X + i if and only if z i = 1. ), Optimization Group 3

5 Exercise 1.9.3: Modelling disjunctions (i) Extend the formulation of discrete alternatives of Section 1.5 to the union of two polyhedra P k = { x R n : A k x b k, 0 x u } for k = 1,2, where max k max i { a k i x b k i : 0 x u } M. (ii) Show that the extended formulation for P 1 P 2 is x = z 1 + z 2, y 1 + y 2 = 1 A k z k b k y k, 0 z k uy k, k = 1,2 z k R n, y k {0,1}, k = 1,2. Solution: (i) Using the method described on page 11 we introduce binary variables y 1 and y 2. Then a suitable model is as follows: A k x b k M(1 y k ), k = 1,2 y 1 + y 2 = 1, y k {0,1}, k = 1,2 0 x u. (ii) The formulation in (ii) is also valid. Because we have either y 1 = 1 and y 2 = 0 or y 1 = 0 and y 2 = 1. In the first case we have z 2 = 0, whence x = z 1. Then the constraint A k z k b k y k gives A 1 x b 1 and 0 0, which is equivalent to x P 1. Similarly, the second case occurs if and only if x P 2. Since either y 1 = 1 or y 2 = 1, the given constraints are satisfied if and only if x P 1 P 2. Optimization Group 4

6 Exercise Show that P i, i = 1,2,3 are formulations for the same set X = {0,1} 4 P i, i = 1,2,3, where P 1 = { x R 4 : 97x x x x } P 2 = { x R 4 : 2x 1 + x 2 + x 3 + x 4 3 } P 3 = { x R 4 : x 1 + x 2 + x 3 2, x 1 + x 2 + x 4 2, x 1 + x 3 + x 4 2 }. Solution: In a brute force way we determine for each of the 16 vectors x {0,1} 4 whether or not one has x P i. The table below shows that the answer is the same for each x and for each i. x 1 x 2 x 3 x 4 P 1 P 2 P 3 (1) P 3 (2) P 3 (3) Optimization Group 5

7 Exercise John Dupont is attending a summer school where he must take four courses per day. Each course lasts an hour, but because of the large number of students, each course is repeated several times per day by different teachers. Section i of course k denoted (i, k) meets at the hour t ik, where courses start on the hour between 10 a.m. and 7 p.m. John s preferences for when he takes courses are influenced by the reputation of the teacher, and also the time of the day. Let p ik be his preference for section (i, k). Unfortunately, due to conflicts, John cannot always choose the sections he prefers. (i) Formulate an integer problem to choose a feasible course that maximizes the sum of John s preferences. (ii) Modify the formulation, so that John never has more than two consecutive hours of classes without a break. (iii) Modify the formulation, so that John chooses a schedule in which he starts his day as late as possible. N.B. What is the meaning of the word section? It is stated that each course is repeated several times per day. I suppose that the pairs (i, k) and (j, k) refer to the same course k, given at distinct hours, starting at t ik and t jk respectively. Since the starting times are on the hour between 10 a.m. and 7 p.m. we may assume that i runs from i = 1 (t 1k = 10 a.m.) to i = 9 (t 9k = 6 p.m.). Moreover there are four courses, so k runs from 1 to 4. We assume furthermore that each course k is given at every hour. Otherwise, we should introduce for each course k a set H k of hours when course k is given. Optimization Group 6

8 Exercise (cont.) Solution: Use binary variables x ik, with x ik = 1 if John chooses section (i, k) and x ik = 0 otherwise. The constraints for a feasible schedule are 9 x ik = 1, k = 1,..., 4, i=1 The answer to (i) is to maximize 9 i=1 k=1 4 x ik 1, i = 1,..., 9. k=1 4 x ik p ik over these constraints. In (ii) we are asked to avoid that courses are scheduled at three or more consecutive hours. This can be modelled with the additional constraint 4 ( ) 4 2 xik + x i+1,k + x i+2,k = x i+j,k 2, i = 1,..., 7. k=1 k=1 j=0 In (iii) the first course should start as late as possible. This can be modelled by changing the objective function to 9 9 x ik (p ik + i) or to ix ik p ik. i=1 If John gives more priority to the lateness condition than to his previous preferences he might replace the constraint 4 k=1 x ik 1 by 4 k=1 x ik y i for i = 1,..., 9, where the y i are binary variables, and use the objective function 9 max iy i. Note that an optimal solution will automatically satisfy 9 i=1 y i = 4. i=1 i=1 Optimization Group 7

9 Exercise Prove that the set of feasible solutions to the formulation of the travelling salesman problem is precisely the set of incidence vectors of tours. Solution: The assignment constraints guarantee that x is the incidence vector of a collection of subtours. We only need to show that the subtour elimination constraints eliminate these subtours indeed. These constraints are: or, alternatively, i S j / S i S j / S x ij 1, S N, S. x ij S 1, S N, 2 S n 1. Suppose that the solution x contains a subtour. Take for S the set of vertices on this subtour. Then there is no edge {i, j} with i S and j / S such that x ij = 1. Hence i S j / S x ij = 0, which gives a contradiction with the first set of subtour elimination constraints. On the other hand, the subtour containing the nodes in S yields precisely S edges {i, j} with i S and j S such that x ij = 1, wherefore i S j / S x ij = S, which gives a contradiction with the second set of subtour elimination constraints. Optimization Group 8

10 Exercise The QED Company must draw up a production program for the next nine weeks. Jobs last several weeks and once started must be carried out without interruption. During each week a certain number of skilled workers are required to work full-time on the job. Thus if job i lasts p i weeks, l i,j workers are required in week j (after the start of this job) for j = 1,..., p i. The total number of workers available in week t is L t. Typical job data ( ) i, p i, l i1,..., l ipi is shown below. Job Length Week 1 Week 2 Week 3 Week (i) Formulate the problem of finding a feasible schedule as an IP. (ii) Formulate when the objective is to minimize the maximum number of workers used during any of the nine weeks. (iii) Job 1 must start at least two weeks before job 3. Formulate. (iv) Job 4 must start not later than one week after job 5. Formulate. (v) Jobs 1 and 2 both need the same machine, and cannot be carried out simultaneously. Formulate. Optimization Group 9

11 We introduce the following variables: Exercise 1.9.7: Solution I: set of all jobs {1,2,3,4,5}; T: set of all weeks {1,..., 9}; Y i : set {t : t + p i 1 T } of possible starting weeks for job i; x it : 1 if job i starts in week t, 0 otherwise; w t : number of workers in week t T. (i) To ensure that every job is handled we require for each i: t Y i x it = 1. The number of required workers in week t can be computed from: w t = l it x is. i I s Y i :s t<s+p i This number should not exceed the number of available workers: w t L t for each t T. Finally, the variables x it are binary: x it {0,1} for all i I and all t T. Optimization Group 10

12 Exercise 1.9.7: Solution (cont.) (ii) This can be achieved by adding constraints ω t w for each t T, and taking min w as objective. (iii) Job 1 will start at least two weeks before job 3 if we require t 2 s=1 x 1s x 3t for t 3 and x 1t = x 2t = 0. (iv) Job 4 will start not later than one week after job 5 if we require t+1 x 4s x 5t for t 8. s=1 (v) We need to avoid that jobs 1 and 2 are carried out simultaneously. We define y it = 1 if job i is carried out in week t and y it = 0 otherwise. Then we have p i t y it = x i,t+1 j = x i,s. Hence, using the convention that j=1 s=t+1 p i x is = 0 if s 0, we achieve the desired result by adding the constraints t s=t+1 p 1 x 1s + t s=t+1 p 2 x 2s 1, t T. Optimization Group 11

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