Homework 4 Solutions CSE 101 Summer 2017

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1 Homework 4 Solutions CSE 101 Summer Scheduling 1. LPT Scheduling (a) Find the Upper Bound for makespan of LPT Scheduling for P C max. (b) Find a tight worst-case example for the makespan achieved by LPT Scheduling in comparison to the optimum makespan for m = 5 processors. (c) Prove that if optimal makespan of LPT schedule, C max < 3.p k, then each processor has at most two jobs. (d) Prove that if each processor has at most two jobs, then LPT is always optimal. (a) Refer to Lemma 7 in Extra Reading (b) To achieve a tight worst-case for the makespan by LPT Scheduling with 5 processors, we choose n = 2(5) + 1 = 11 jobs as follows: 2 jobs of size 2(5) 1 = 9, 2 jobs of size 2(5) 2 = 8, 2 jobs of size 2(5) 3 = 7, 2 jobs of size 2(5) 4 = 6(= 5 + 1), 3 jobs of size 5. The Gantt-Chart for this list {9, 9, 8, 8, 7, 7, 6, 6, 5, 5, 5} using LPT Scheduling in comparison to the optimum makespan is: 1

2 ( m )C max = ( 19 T 15 )15 = 19 = CLP max. (c) Refer to Corollary 3 in Extra Reading (d) Refer to Corollary 4 in Extra Reading 2. Single Processor Scheduling (a) The following n = 8 jobs with given processing times have to be scheduled on a single processor with the objective of minimizing the maximum lateness L max. d j is the maximum delay of each process. A B C D E F G H p j d j Calculate the values of L max and U j (i) If you schedule the jobs in the given order - A,B,..,H (ii) If you schedule the jobs using EDD algorithm (iii) If you schedule the jobs using the Moore s algorithm (b) Prove that there is no better solution than the one obtained using Earliest Due Date (EDD) algorithm, to the problem of minimizing maximum lateness. (c) Prove that Moore s algorithm provides the optimal solution for 1 U j. (a) (i) A B C D E F G H p j d j C j L j Just scheduled in that order would give: L max = 43 (job H) Uj = 6 (jobs C, D, E, F, G, H) 2

3 (ii) D F E A G C H B p j d j C j L j Scheduling using EDD algorithm would give: L max = 40 (job B) Uj = 7 (jobs A, B, C, E, F, G, H) (iii) The jobs are initially scheduled in EDD order. Scheduling using Moore s algorithm would give: L max = 40 (job B) Uj = 5 (jobs A, C, F, G, H) (b) Proof (by contradiction). Assume that, for a given set of jobs, schedule S is optimal even though the jobs are not scheduled according to EDD. Then there must be at least two jobs, say j1 and j2 having delays d j1 > d j2, and j1 being scheduled before j2, as shown: 3

4 On swapping jobs j1 and j2, we get In this schedule S, L j2 = C j2 d j2 < C j2 d j2 (since C j2 > C j1) L j1 = C j1 d j1 = C j2 d j1 < C j2 d j2 (since d j1 > d j2 ) Then, for every other job j, (j j1,j2) This means, L j = C j d j < C j d j (since they are not rescheduled in S ) max{l j1, L j2, max{l j}} max{l j1, L j2, max{l j }} = L max S L max S But this is a contradiction to the assumption that S is optimal. It follows that EDD is optimal for 1 L max (c) Proof (by induction). Step 1: If there exists a schedule such that there are no late jobs, then there won t be any late job in the schedule obtained by following the EDD algorithm. [Refer 2(b] Step 2: Assume that there exists a schedule S with U j = k. We need to prove that there can t be another schedule S such that U j = k + 1. This is easy to see since the total processing time of on-time jobs is as small as possible, given that the job with the largest processing time is removed always. 2 Graphs 3. You are given an undirected unweighted graph G = (V, E), V = n, E = m. Determine, whether this graph contains a circuit or not. (a) We will use a DFS algorithm to solve this problem. Let s color each node to two colors: red and blue. If the node hasn t been visited yet by DFS, then the color of the node is red. Otherwise, the color of the node is blue. If at some step DFS tries to visit the node with the blue color, then there exists a circuit. 4

5 (b) If the DFS algorithm is trying to enter a node v with the blue color from node u, this means that we have visited v before. If we take the path from v to u (which was constructed by a DFS tree) and add to this path edge (u, v), which exists as we are trying to visit node v from u, then we will get a circuit. If there exists a circuit in the graph, then, at some point, DFS will visit one of its nodes. Let that node be v. As node v is part of a circuit, it has at least two edges (v, a) and (v, b), such that there is a path from a to b that doesn t go through v. This path will be processed sooner or later by DFS and we will get to node b from a. Then, we will try to visit v, which is going to have the blue color. (c) The complexity of the solution equals the complexity of DFS, i.e. O(n + m) 4. There are n cities on a plane with coordinates (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ). All cities are distinct and located at different points on the plane. The city 1 has access to electricity. You need to connect cities with wires, such that all cities will have access to electricity. A city has access to electricity if it is city 1 or if it is connected with a wire to another city, that has access to electricity. The length of the wire that connects the two cities equals the euclidean distance between these cities. Your task is to connect the cities with wires, such that the total length of all wires is minimized. (a) Let matrix d[i][j] be equal to the euclidean distance between cities i and j, i.e. d[i][j] = (xi x j ) 2 + (y i y j ) 2. As a result, we have a full graph G = (V, E) with nodes as cities and the weight on edges as euclidean distances between the cities. In order to find the answer, we will find the minimum spanning tree of this graph. This tree will be the answer to our problem. (b) Let G = (V, E ) be a graph that connects all nodes in G and the sum of weights in this graph is as small as possible. Let s find the minimum spanning tree of G. The sum of weights in the minimum spanning tree of G will be less than or equal to the sum of weights of graph G, thus it will be optimal. But as G is a subgraph of G, then the total sum of weights of the spanning tree of G is less than or equal to the sum of all weights in the spanning tree of G. Thus, it is also optimal. (c) The time complexity of this solution equals the time complexity of finding the minimum spanning tree in a full graph. If we apply Prim s algorithm to this purpose, then the time complexity is O(n 2 ). 5. You have a chess board n m. There is a knight on cell (x s, y s ). The knight can make moves from cell (x, y) to cells (x + 1, y + 2), (x 1, y + 2), (x + 1, y 2), (x 1, y 2), (x + 2, y 1), (x 2, y 1), (x + 2, y + 1) and (x 2, y + 1) (all cells must be on the board). Find the minimum number of steps the knight needs to make to get from cell (x s, y s ) to cell (x f, y f ). (a) If we assume that each cell of the board is a node, then we can connect a node that corresponds to cell (x, y) with nodes that correspond to cells (x + 1, y + 2), (x 1, y + 2), (x + 1, y 2), (x 1, y 2), (x + 2, y 1), (x 2, y 1), (x + 2, y + 1) and (x 2, y + 1) (all cells must be on the board) with undirected edges. As a result, we will have an undirected unweighted graph. We then launch BFS from the node which corresponds to cell (x s, y s ) and find the distance to the node which corresponds to cell (x f, y f ). 5

6 (b) Let s assume that we have an optimal sequence of steps from cell (x s, y s ) to (x f, y f ). As we can see, this sequence will correspond to some path in the constructed graph from the node that corresponds to cell (x s, y s ) to the node that corresponds to cell (x f, y f ). But as BFS always finds the minimum distance from the node from which it starts to all other nodes, then the path found in BFS will also be optimal. (c) The time complexity equals the time complexity of BFS. There are nm nodes in the constructed graph and less than 4nm edges. The overall time complexity is O(nm). 6. You are given an undirected connected weighted graph G = (V, E), V = n, E = m. The weights of all edges are positive integer numbers. Your task is to delete edges in this graph so that the resulting graph will be connected and the product of all weights of all edges in this graph will be as small as possible. (a) Let s construct another graph G = (V, E ), such that V = V, E = E and w(e ) = log(w(e)) for all e E and e E. Then, we will find the minimum spanning tree in G with Prim s algorithm. We will take the same tree G (the same nodes and edges). This tree in G will be the graph that we are looking for. (b) When we are looking for the minimum spanning tree in G, we are minimizing the following function log(w(e 1)) + log(w(e 2)) log(w(e k )) = log(w(e 1)w(e 2)...w(e k )), where k is the number of edges in the resulting graph. But minimizing log(x) for x > 0 means minimizing x. Thus, we are minimizing function w(e 1)w(e 2)...w(e k ), when we search for the minimum spanning tree. (c) The time complexity equals the number of operations required to get a new graph plus the time complexity of Prim s algorithm, i.e. O(n + m) + O(n 2 ) = O(n 2 ). 7. There is a set of n courses. Each course has its prerequisites (from the same set). So course i is assigned a set of courses that are required to be completed before course i starts. You are given a set of n courses, and each course is provided with a list of prerequisites. Find a schedule (a list) of courses, if such exists, such that all n courses are in this list and they all can be completed. If there is no such a schedule, just output 1. For example, let s say we have 5 courses and a list (A, D, C, B, E). We go from left to right and try to complete a current course. If all prerequisites for the current course are completed, then the current course can also be completed. (a) Let s construct a graph G = (V, E), V = n, E = m, such that each node represents some course. There is a directed edge from node u to node v if the course for node u is a prerequisite for the course for node v. Now, we have a directed graph. Let s find topological sorting of this graph. If there is such, then the list will be courses given in a topological sorted order of their corresponding nodes. 6

7 (b) Let s say v 1, v 2,..., v n is a topological order of nodes in graph G. Then, from the properties of the topological sort we know, that there is no edge from v i to v k, when i > k, i.e. all corresponding courses can be completed. This list also includes all nodes, which makes it a valid schedule. Now, let s say u 1, u 2,..., u n is a list of nodes, which correspond to courses from the answer to this problem. Then, this list will also be the one of the topological orders of graph G. Indeed, as this is a valid schedule, all courses can be completed, i.e. there can only be edges (v i, v k ), such that i < k. (c) The time complexity is to construct a graph and to find its topological order, i.e. O(n + m). 8. You are given a rooted tree G = (V, E), V = n, E = m. You have q online queries (u, v). For each query (u, v) you need to answer whether u is an ancestor of v in the tree G. (a) Let s use DFS to solve this problem. We launch DFS from the root. We will need two additional values for each node v: in[v] and out[v]. in[v] is a time when DFS visits node v. out[v] is a time when DFS leaves node v. The time starts with zero and increases by one each time we enter or quit some node in DFS. Then, if in[a] < in[b] and out[b] < out[a], a is an ancestor of b. (b) If in[a] < in[b], this means that we have visited a sometime before we have visited b. If out[b] < out[a], this means that by the time we have left b, we are still in the subtree, in which a is a root. Now, if a is an ancestor of b, then we will visit a first, then we will visit b, then we will quit b and, finally, we will quit a. This leads to our inequality. (c) We need O(n + m) for DFS and O(q) for all queries. Each query takes O(1) time to answer. The overall complexity will be O(n + m + q). 9. You are given a weighted directed graph G = (V, E), V = n, E = m with positive weights. You are also given a node s in this graph. Find such node v, such that there is a path from v to s and that among all possible such nodes, the shortest path from v to s is as large as possible. (a) Let s construct graph G = (V, E ) such that G is a copy of G, but all edges are directed in an opposite direction. Let s find shortest paths from node s to all other nodes in G. We can do this with Dijkstra s algorithm. Then, among all these paths, we choose the one with the maximum value. The node, to which this path leads is the answer to our problem. (b) Any path from s to v in G is a path from v to s in G and vice versa. This means that the shortest path from v to s in G is the shortest path from s to v in G. This proves the correctness of our solution. (c) The time complexity equals the time complexity of building graph G and using Dijkstra s algorithm on it, i.e. O(n + m) + O(n 2 ) = O(n 2 ) 7

8 3 Linear Programming 10. A company manufactures two products A and B. The company earns $3 for each piece of A and $2 for each piece of B. Raw materials V1 and V2 are necessary to build these products. But there are only 8 parts of V1 and 6 parts of V2 available. To manufacture single piece of product A, 2 parts of V1 and 1 part of V2 is required. To manufacture single piece of product B, 1 part of V1 and 1 part of V2 is required. How many pieces of A and B does the company have to manufacture when the objective function is to maximize the earning? (a) Formulate the mathematical program. (b) Solve the problem graphically. (c) Solve the problem with the Simplex algorithm. (d) Give an interpretation of the dual values: If the company could have access to one more piece of V1, how would the production program and the earning change? If the company could have access to one more piece of V2, how would the production program and the earning change? see lecture slides 8