一站式論文代寫,英国、美国、澳洲留学生Essay代寫—FreePass代写

Java代寫|算法代寫 - COMP4500/7500 Advanced Algorithms and Data Structures
時間:2020-10-12
COMP4500/7500 Advanced Algorithms and Data Structures School of Information Technology and Electrical Engineering The University of Queensland, Semester 2, 2020 Assignment 2 Due at 4:00pm, Monday the 26th October 2020. This assignment is worth 20% of your final grade. This assignment is to be attempted individually. Please read this entire handout before attempting any of the questions. Submission. Answers to each of the questions Q1(b) and Q1(d) should be clearly labelled and included in a pdf file called a2.pdf. You need to submit (i) your written answers in a2.pdf, as well as (ii) your source code files Recursive.java and Dynamic.java electronically using Blackboard according to the exact instructions on the Blackboard website: http://learn.uq.edu.au/ You can submit your assignment multiple times before the assignment deadline but only the last submission will be saved by the system and marked. Only submit the files listed above. You are responsible for ensuring that you have submitted the files that you intended to submit in the way that we have requested them. You will be marked on the files that you submitted and not on those that you intended to submit. Only files that are submitted according to the instructions on Blackboard will be marked. Submitted work should be neat, legible and simple to understand – you may be penalised for work that is untidy or difficult to read and comprehend. For the programming part, you will be penalised for submitting files that are not compatible with the assignment requirements. In particular, code that is submitted with compilation errors, or is not compatible with the supplied testing framework will receive 0 marks. Late submission. Unless you have been approved to submit an assignment after the due date: late assignments will lose 10% of the marks allocated to the assignment immediately, and a further 10% of the marks allocated to the assignment for each additional calendar day late. Assignments more than 5 calendar days late will not be accepted. If there are medical or exceptional circumstances that will affect your ability to complete an assignment by the due date, then you can apply for an extension as per Section 5.3 of the electronic course profile (ECP). Requests must be made at least 48 hours prior to the submission deadline. Assignment extensions longer than 7 calendar days will not be granted. School Policy on Student Misconduct. You are required to read and understand the School Statement on Misconduct, available at the School’s website at: http://www.itee.uq.edu.au/itee-student-misconduct-including-plagiarism This is an individual assignment. If you are found guilty of misconduct (plagiarism or collusion) then penalties will be applied. COMP4500/7500 Assignment 2 (October 7, 2020) 2 Question 1 (100 marks total) You have been appointed as a site engineer in a mining company. One of your duties is operating a heavy duty hydraulic slurry pump on site for k consecutive hours. Regular maintenance of the slurry pump helps it function optimally and reliably. It will extend the pump’s wear life, helps avoid unplanned downtime, and lowers the cost. There are several services that can be performed for this pump: ? Minor Service (takes 1 hour). Includes changing oil, inspecting the bearings, looking for leakages and examining the stuffing box. ? Regular Service (takes 2 hours). Includes - in addition to the minor service activities - cleaning and checking belts, as well as changing filters. ? Full Service (takes 4 hours). Involves a full inspection of the entire pump. During any service, the pump cannot be used to transport any liquid. The maximum amount (volume) of slurry (liquid) the pump can transport per hour depends on the slurry density. For simplicity, we assume density is constant. The transported volume depends on the last service applied to the pump, and the number of hours since that service. When the last service was a full service, the maximum volume of liquid that can be transported per hour i hours after that activity has concluded (for 0 ≤ i) is described by an array fullServiceCapacity of n non-negative integers (indexed from 0). The hour after the full service activity has finished, the maximum volume per hour of the pump is fullServiceCapacity[0]; the i’th hour after the full service activity is finished (i indices start from zero), the volume per hour of the pump is fullServiceCapacity[i], where 0 ≤ i < n. n hours after a full service is finished, the pump cannot be operated until another service is performed. Similar arrays exist for regular and minor services, called regularServiceCapacity (length m) and minorServiceCapacity (length p) respectively. These describe the maximum volume per hour of the pump i hours after either of these services are completed. Again, if we are m hours after a regular service or p hours after a minor service, then the pump cannot be operated until another service is performed. You have also been given a schedule, represented by an array hourlyVolume of k non-negative integers (indexed from 0). This array gives the volume of liquid that is scheduled to be transported for each of the k hours that you are in charge of the pump. The volume of liquid scheduled to be transported in the first hour (hour 0) is hourlyVolume[0], the volume scheduled for the second hour (hour 1) is hourlyVolume[1] etc. If the amount of liquid needing to be transported in an hour (given by hourlyVolume) is greater than the amount that can be transported in that hour (zero during any service, otherwise determined by fullServiceCapacity, regularServiceCapacity or minorServiceCapacity), then the remaining liquid must be discarded. Hence the cost/loss to your company is the amount of liquid that could not be transported. A service can be scheduled to take place at the start of any hour you are in charge of the pump. The pump will be out of action for the duration of the service (e.g 2 hours for a regular service) and cannot transport any liquid during this time. Once the service is complete, the pump can begin transporting liquid again. Only one service can be in progress at any given time, but otherwise there is no limit to the number of services that can be performed, or the order in which they can be done. For the k hours that you are in charge, it is up to you to determine when these services should occur. Given arrays fullServiceCapacity, regularServiceCapacity, minorServiceCapacity and hourlyVolume, and the knowledge that the last service performed on the pump was a full service that concluded the hour before you were put in charge of the pump, your task is to find the least loss that can be incurred by your company over the k hours that you operate the slurry pump. COMP4500/7500 Assignment 2 (October 7, 2020) 3 Example As an example, consider the following scenario: hourlyVolume = [50, 40, 90, 10, 5, 100, 40, 20, 50] fullServiceCapacity = [100, 90, 80, 70, 60, 50, 40, 30, 20, 10] regularServiceCapacity = [70, 50, 40, 30, 20, 10] minorServiceCapacity = [50, 40, 20, 10] where you are in charge for k = 9 hours, and a full service took place the hour before you were put in charge of the pump. For each of the k hours (i.e hours 0, 1, 2 · · · kk1) you are in charge of the pump, you can either choose to perform one of the three possible services, or no service at all. Your choices can have an impact on the loss incurred by the company. For example, ? If you choose to perform no services at all: Hourly Capacity = [100, 90, 80, 70, 60, 50, 40, 30, 20] Total loss = 0 + 0 + 10 + 0 + 0 + 50 + 0 + 0 + 30 = 90 ? If you choose to perform a full service starting at hour 3 (i.e. the fourth hour): Hourly Capacity = [100, 90, 80, 0, 0, 0, 0, 100, 90] Total loss = 0 + 0 + 10 + 10 + 5 + 100 + 40 + 0 + 0 = 165 ? If you choose to perform a regular service starting at hour 3 (i.e the fourth hour) and a minor service starting at hour 7 (i.e the eighth hour) Hourly Capacity = [100, 90, 80, 0, 0, 70, 50, 0, 50] Total loss = 0 + 0 + 10 + 10 + 5 + 30 + 0 + 20 + 0 = 75 Where ‘hourly capacity’ refers to the maximum amount of liquid that could be transported in a given hour. This is zero during any service activity and follows the relevant array after a service is finished. There are many other possible service schedules. For this example, the least loss that can be incurred by the company for the k = 9 hours that you operate the pump is 75. a. (20 marks) Implement the public static method optimalLossRecursive from the Recursive class in the assignment2 package that is available in the zip file that accompanies this handout, to provide a recursive algorithm to determine the least cost that can be incurred by the company over the k hours (i.e. hour 0 to hour k k 1) that you operate the slurry pump. To implement that method, you will need to provide an implementation for the private static method optimalLossRecursive from the same class. The recursive solution does NOT need to find a schedule of services that produces the least loss – it just needs to determine the least loss. Efficiency is not at all a concern for this part, so focus on an elegant solution. b. (20 marks) It is expected that your recursive algorithm will not be polynomial-time in the worst case. For the case where the number of hours that you are responsible for the slurry pump is k, give an asymptotic lower bound on the worst-case time complexity of your recursive algorithm in terms of parameter k. Make your bound as tight as possible. Give an argument explaining why the time-complexity is exponential in terms of a (lower bound) recurrence derived from your algorithm. [Make your answer as concise as possible – it should be no more than half a page using minimum 11pt font. Longer answers will not be marked.] COMP4500/7500 Assignment 2 (October 7, 2020) 4 c. (30 marks) Develop a bottom-up dynamic programming solution to the problem (not memoi?sed) by implementing the public static method optimalLossDynamic in the Dynamic class from the assignment2 package that accompanies this handout. Your dynamic programming solution should run in polynomial time (in terms of k, the number of hours you are responsible for the slurry pump). This dynamic solution does NOT need to find a schedule of activities that produces the least loss – it just needs to determine the least loss. d. (10 marks) Provide an asymptotic upper bound on the worst-case time complexity of your dynamic programming solution for part (c) in terms of the parameter k, the number of hours you are responsible for the slurry pump. Make your bounds as tight as possible and justify your solution. You should assume for this analysis that hourlyVolume is the shortest array that you are provided. [Make your answer as concise as possible – it should be no more than half a page using minimum 11pt font. Longer answers will not be marked.] e. (20 marks) Extend your bottom-up dynamic programming solution from part (c) to calculate an optimal schedule of activities (that produces the least loss) by implementing the public static method optimalServicesDynamic in the Dynamic class from the assignment2 package. Like method optimalLossDynamic, your implementation of this method should run in polynomial time (in terms of k). It should be a bottom-up dynamic programming (not memoised) solution. Practicalities Do not change the class name of the Recursive or Dynamic classes or the package to which those files belong. You many not change the signatures of the methods that you have to implement in any way or alter their specifications. (That means that you cannot change the method name, parameter types, return types or exceptions thrown by the those methods.) Do not modify any of the other classes or interfaces or enumerated types defined in package assignment2. You are encouraged to use Java 8 SE API classes, but no third party libraries should be used. (It is not necessary, and makes marking hard.) Don’t write any code that is operating-system specific (e.g. by hard?coding in newline characters etc.), since we will batch test your code on a Unix machine. Your source file should be written using ASCII characters only. You may not write and submit any additional classes. Your solution to Q1(a) should be self-contained in the Recursive class. Similarly your solution to parts Q1(c) and Q1(e) should be self-contained in the Dynamic class. Both of these classes will be tested in isolation and should not depend upon each other. The zip file for the assignment also some junit4 test classes to help you get started with testing your code. The Junit4 test classes as provided in the package assignment2.test are not intended to be an exhaustive test for your code. Part of your task will be to expand on these tests to ensure that your code behaves as required. Your programming implementations will be tested by executing our own set of junit test cases. Code that is submitted with compilation errors, or is not compatible with the supplied testing framework will receive 0 marks. A Java 8 compiler will be used to compile and test the code. The Recursive class will be tested in isolation from the Dynamic class. Implementations that do not satisfy the assignment requirements will receive 0 marks even if they pass some of the test cases (e.g. if the solution given to Q1(c) is not a bottom-up dynamic programming solution, then it will receive 0 marks.) COMP4500/7500 Assignment 2 (October 7, 2020) 5 You may lose marks for poorly structured, poorly documented or hard to comprehend code, or code that is not compatible with the assignment requirements. Line length should be less than or equal to 80 characters so that it can be printed – please use spaces to indent your code instead of tabs to ensure compatibility with different machines. Don’t leave print statements in your submitted code. Evaluation Criteria Question 1 ? Question 1 (a) (20 marks) Given that your implementation satisfies the requirements of the question, your implementation will be evaluated for correctness by executing our own set of junit test cases. 20 : All of our tests pass 16 : at least 80% of our tests pass 12 : at least 60% of our tests pass 8 : at least 40% of our tests pass 4 : at least 20% of our tests pass 0 : less than 20% of our test pass or work with little or no academic merit Note: Code that is submitted with compilation errors, or is not compatible with the supplied testing framework will receive 0 marks. A Java 8 compiler will be used to compile and test the code. Implementations that do not satisfy the assignment requirements will receive 0 marks even if they pass some of the test cases. The Recursive class will be tested in isolation from the Dynamic class. ? Question 1 (b) (20 marks) For this part of the question, the analysis should be no more than 1/2 of a page using minimum 11pt font. Longer solutions will receive 0 marks. Also, if a plausible, neat, legible and simple to understand solution to Q1(a) has not been given, this question will receive 0 marks. Otherwise the following marking criteria applies. 20 : A correct asymptotic lower bound on the worst-case time complexity the recursive algorithm from Q1(a) is given in terms of parameter k. The lower bound, which should be exponential in k, should be as tight as reasonably possible for the algorithm at hand. The time-complexity given should be clearly justified by giving and solving a correct (lower bound) recurrence derived from your algorithm. Any assumptions made in the analysis are reasonable and clearly stated. Asymptotic notation should be used correctly and the asymptotic time complexity given has been simplified to remove lower order terms and unnecessary constant factors. 15 : A correct asymptotic lower bound on the worst-case time complexity the recursive algorithm from Q1(a) is given in terms of parameter k. The lower bound should be exponential in k. The time-complexity given should be mostly clearly justified by giving and solving a correct (lower bound) recurrence derived from your algorithm. Any assumptions made in the analysis are mostly reasonable and clearly stated. COMP4500/7500 Assignment 2 (October 7, 2020) 6 10 : A reasonable attempt has been made to give a tight asymptotic lower bound on the worst-case time complexity of the recursive algorithm from Q1(a) in terms of parameter k, and to justify it with respect to a recurrence derived from the algorithm, however the analysis or justification may contain minor mistakes or omissions or lack clarity. 5 : An attempt has been made to give an asymptotic lower bound on the worst-case time complexity of the recursive algorithm from Q1(a) in terms of parameter k, and justify it, however it contains either a major mistake or many mistakes, gives an unreasonably loose lower bound, or is not clearly justified by giving and solving a correct (lower bound) recurrence derived from your algorithm. 0 : Work with little or no academic merit. ? Question 1 (c) (30 marks) Given that your implementation satisfies the requirements of the question (i.e. it is a bottom-up dynamic programming (not memoised) solution that runs in polynomial time in terms of k), your implementation will be evaluated for correctness and efficiency by executing our own set of junit test cases. 30 : All of our tests pass 24 : at least 80% of our tests pass 18 : at least 60% of our tests pass 12 : at least 40% of our tests pass 6 : at least 20% of our tests pass 0 : less than 20% of our test pass or work with little or no academic merit Note: Code that is submitted with compilation errors, or is not compatible with the supplied testing framework will receive 0 marks. A Java 8 compiler will be used to compile and test the code. Implementations that do not satisfy the assignment requirements will receive 0 marks even if they pass some of the test cases. The Dynamic class will be tested in isolation from the Recursive class. ? Question 1 (d) (10 marks) For this part of the question, the analysis should be no more than 1/2 of a page using minimum 11pt font. Longer solutions will receive 0 marks. Also, if a plausible, neat, legible and simple to understand solution to Q1(c) has not been given, this question will receive 0 marks. Otherwise the following marking criteria applies. 10 : A correct asymptotic upper bound on the worst-case time complexity of the algorithm from Q1(c) is given in terms of parameter k. The upper bound, which should be polynomial in k, should be as tight as reasonably possible for the algorithm at hand. The time-complexity given should be clearly justified with respect to the algorithm. Any assumptions made in the analysis are reasonable and clearly stated. Asymptotic notation should be used correctly and the asymptotic time complexity given has been simplified to remove lower order terms and unnecessary constant factors. 7 : A correct asymptotic upper bound on the worst-case time complexity the algorithm from Q1(c) is given in terms of parameter k. The upper bound should be polynomial in k. The time-complexity given should be mostly clearly justified with respect to the algorithm. Any assumptions made in the analysis are mostly reasonable and clearly stated. COMP4500/7500 Assignment 2 (October 7, 2020) 7 5 : A reasonable attempt has been made to give a tight asymptotic upper bound on the worst-case time complexity of the algorithm from Q1(c) in terms of parameter k, and to justify it, however the analysis or justification may contain minor mistakes or omissions or lack clarity. 3 : An attempt has been made to give an asymptotic upper bound on the worst-case time complexity of the algorithm from Q1(c) in terms of parameter k, and justify it, however it contains either a major mistake or many mistakes, gives an unreasonably loose lower bound, or is not clearly justified. 0 : Work with little or no academic merit. ? Question 1 (e) (20 marks) Given that your implementation satisfies the requirements of the question (i.e. it is a bottom-up dynamic programming (not memoised) solution that runs in polynomial time in terms of k), your implementation will be evaluated for correctness and efficiency by executing our own set of junit test cases. 20 : All of our tests pass 16 : at least 80% of our tests pass 12 : at least 60% of our tests pass 8 : at least 40% of our tests pass 4 : at least 20% of our tests pass 0 : less than 20% of our test pass or work with little or no academic merit Note: Code that is submitted with compilation errors, or is not compatible with the supplied testing framework will receive 0 marks. A Java 8 compiler will be used to compile and test the code. Implementations that do not satisfy the assignment requirements will receive 0 marks even if they pass some of the test cases. The Dynamic class will be tested in isolation from the Recursive class. Change Log ? 7/10/2020 - Slight modification in the wording of the paragraph describing the fullServiceCapacity array to make it more clear how the i indices work. This change was made to make it more explicit that the indices of the array start from zero.

在線客服

售前咨詢
售后咨詢
微信號
Essay_Cheery
微信
专业essay代写|留学生论文,作业,网课,考试|代做功課服務-PROESSAY HKG 专业留学Essay|Assignment代写|毕业论文代写-rushmyessay,绝对靠谱负责 代写essay,代写assignment,「立减5%」网课代修-Australiaway 代写essay,代写assignment,代写PAPER,留学生论文代写网 毕业论文代写,代写paper,北美CS代写-编程代码,代写金融-第一代写网 作业代写:CS代写|代写论文|统计,数学,物理代写-天天论文网 提供高质量的essay代写,Paper代写,留学作业代写-天才代写 全优代写 - 北美Essay代写,Report代写,留学生论文代写作业代写 北美顶级代写|加拿大美国论文作业代写服务-最靠谱价格低-CoursePass 论文代写等留学生作业代做服务,北美网课代修领导者AssignmentBack