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COIT20261 Term 1 Standard Take-Home Examination 2020 Network Routing and Switching — COIT20261 Examination Conditions · This take-home exam should take you up to 3 hours to complete. However, a 24-hour timeframe will be allowed in which you will need to download the exam from the unit’s Moodle website, complete it and upload it back. If you have other exams within or overlapping this period, you will be responsible for managing your time accordingly. · The exam will be made available on 18/06/2020 at 15:00 (3pm) AEST through the Moodle unit website. The unit coordinator will be available for any exam-related queries for one hour after the release of the exam. The unit coordinator can be reached via the Q&A Forum in Moodle, or by emailing
[email protected]. · Your typed answer script in Microsoft Word file format must be uploaded by 19/06/2020 at 15:00 (3pm) AEST. The submission link may become unavailable after this time. Any submission after the deadline may not be accepted. Please ensure you submit the correct file. Reversions to draft to allow resubmission will not be allowed. · All submissions will be processed via Turnitin®. · You are required to do your own work, maintaining academic integrity with all honesty. If the marker suspects a breach of academic integrity, you may be requested to attend a real-time interview via Zoom to confirm your understanding of your submission. Inability to explain or justify given solutions may result in allegations of academic misconduct. · The marks for your assessed answer script will be made available on Certification of Grades day (10 July). Like regular exams, the assessed answer script will not be returned to you, unless you apply to view it as part of the first step of the review of grade process, which is an informal consultation. · If you are unable to take or complete the exam on the specified date due to illness or some other exceptional circumstances, you need to apply for a Deferred Assessment via the Assessment Extension Request Link on the Moodle unit page and supply the necessary documentation. Be aware that Deferred Assessment applications will be refused if there are not sufficient grounds and/or supporting documentation. Note also that a Deferred Assessment is NOT a normal extension. Should your application be accepted, you will be required to take a different assessment at a specified time up to several weeks after the Term 1 grades are certified. A grade of DA (deferred assessment) will be recorded at Certification Date. · For further details please refer to the Assessment Policy and Procedure (Higher Education Coursework), which can be found through this link (https://www.cqu.edu.au/policy). · For any technology-related issues, please contact the Technology and Services Assistance Centre (TaSAC) on the details provided through this link (https://www.cqu.edu.au/about-us/structure/directorates/information-and-technology). Instructions to Students 1. Type all answers clearly in the spaces provided in the downloaded document. You can expand the provided spaces by typing within them. 2. You must upload and submit the same document you downloaded. Answers provided in any other document or form will not be accepted and will not be marked. 3. You may use your own rough paper; however, these are not to be submitted or uploaded with your exam document. 4. Do not copy/paste tables or diagrams from other sources, and do not insert photos or images of any kind to answer any question. Instead just use typed words of your own to answer all questions or use tables already provided in the exam. 5. This examination contains three parts, Parts A, B and C. 6. Answer all questions in all three parts, Parts A, B and C. 7. The total marks available in the examination is 60. 8. You may use a calculator including hand-held calculators with programming and scientific functions, or an electronic calculator provided with your operating system (e.g. Windows or MacOS). 9. This exam is Open Book. You may use the textbook or other printed materials as references only. However, you must correctly cite and reference materials you use. Your answers will be marked on clarity, logic, relevance, and fully addressing all parts of each Question. You must use your own words and must not copy text from any source without attribution. Quotations alone are not acceptable as an answer. 10. When you are ready to submit the exam for marking, remember to click the submit button, then your submission will be final. The exam begins on the next page. PART A20 MARKS/ TEN SHORT ANSWER QUESTIONS Answer all 10 Questions from this part. Type your answers in the space provided. The space will expand as you type. Each Question is worth 2 marks (10 x 2 = 20 marks). Question 1 2 Marks/ A block of addresses has the mask /26. If the organisation creates 4 equal-sized subnets, how many addresses (including the special addresses) are available in each subnet? Show your calculations. Answer: There are total of 64 addresses available in each subnet Calculations: /26 = 255.255.255.192 Binary – 11111111.11111111.11111111.11000000 Subnet bit – 2 Host bit – 6 26 = (20+21+22+23+24+25+26) = 0+2+4+8+16+32 = 62 total number of valid hosts Question 2 2 Marks/ A valid classful address was given as 190.53.13.3. What is the default mask for this address? Answer: 255.255.255.248 Question 3 2 Marks/ Given the classless address 215.150.68.112/25, what is the direct broadcast address? Show your calculations (essential). Answer: Broadcast add – 215.150.68.127 Calculations: /25 = 255.255.255.128 Binary – 11111111.11111111.11111111.10000000 Subnet bit – 1 Host bit – 7 27 = = (20+21+22+23+24+25+26+27) = 0+2+4+8+16+32+64 = 126 total number of valid hosts Question 4 2 Marks/ In an IPv4 Header, the value of the Header Length field was decimal 7. What is the total length of the Header in bytes? Explain your answer or show your calculations. Answer: In IPv4 header length is 4 bytes. We can see the the IPv4 header (32 bits) is containing 4bits of version, 4-bits of header length, 8-bits of service and 16-bits d header length. Question 5 2 Marks/ A packet was detected in Wireshark as having a Source port address of 45,800, and a Destination port address of 13. Is this packet travelling from client to server, or server to client? Briefly explain your answer (essential). Answer: The packet is travelling from client to server. Where server port is 13 nad client port is 45800. They are sync daytime service that is used to sync daytime. Question 6 2 Marks/ Consider the mask 255.255.254.0. Is this the mask of a classful address, a classless address, an IPv6 address, or a limited broadcast? Give reasons for your choice as well as reasons why the others are not correct. Answer: It is a classless address that is subneted with /23 CIDR Question 7 2 Marks/ In your own words, when browsers need to locate a website, briefly explain what information they need from DNS servers. Answer: When a browser locates a website, it sends request to the DNS server to resolve its FQDN into corresponding IP address. So that it is searched over public IP network Question 8 2 Marks/ In the context of routing and forwarding, what does “indirect delivery” mean? Explain in your own words. Answer: In the indirect delivery the send packet go from host to host or router to router and travel until it didn’t reached its destination host or final destination. Question 92 Marks/ Classful addressing was referred to as “two-level addressing”. Briefly explain the two levels referred to. Answer: A classful address is having two part of its address. One is host-id and second is net-id. The net-id represent the belonging network while the host-id represent the end host. Question 10 2 Marks/ Enabling Options in IPv4 changes the header length but enabling Options in IPv6 does not change the header length. Briefly explain the reason for this difference. Answer: This is because IPv4 and IPv6 both have different header format and length, size End of Part A Scroll down for Part B PART B25 MARKS/ FIVE PRACTICAL & THEORY PROBLEMS Answer all 5 Questions from this part. Type your answers in the spaces provided. The space will expand as you type. Show your working. Question 15 Marks/ The ABEL Company has requested a block of IP addresses from an ISP. The block that was allocated to ABEL included the address 191.56.125.34/25. Calculate the number of addresses in this block, the network address, the direct broadcast address, and the range of addresses that would be available for use by hosts. The number of addresses available in this block: (1 mark/ ) Total number of addresses - 128 The network address: (1 mark/ ) Network address – 191.56.125.0 The direct broadcast address: (1 mark/ ) Broadcast address – 191.56.125.127 The range of addresses that can be used by hosts: (1 mark/ ) From:__191.56.125.1__ to: _191.56.125.126___ Calculations: (1 mark/ ) /25 = 255.255.255.128 Binary – 11111111.11111111.11111111.10000000 Net bit – 25 Host bit – 7 27 = = (20+21+22+23+24+25+26+27) = 0+2+4+8+16+32+64 = 126 total number of valid hosts So there will be 128 total number of addresses in which very first address is nett address and last one is broadcast address. The remaining addresses are used as host address, as provided details Question 25 Marks/ A company has been granted a block of classless addresses which starts at 198.87.24.0/23. Create the following 4 subnets by calculating the subnet address including the prefix for each of the following subnets. Create a table to organise your answer. Show your working in the space provided. a) 1 subnet with 128 addresses b) 1 subnet with 64 addresses c) 2 subnets with 32 addresses each Answer: 198.87.24.0/23 subnet has total number of 512 host in which 510 are valid hosts, each subnet. a) 1 subnet (128 addresses) – 198.87.24.1 – 198.87.24.128 b) 2 subnet (64 addresses) – 198.87.26.1 – 198.87.26.64 c) 3 subnet (32 addresses) – 198.87.28.1 – 198.87.28.32 Calculations: