...Subnetting Exercise Create 7 subnets from the following Class C IP address: 194.27.56.0 subnet mask: 255.255.255.0 Answer the following questions and upload your answers to blackboard. See the next page for an example of how to perform the exercise. 1. How many subnets? 2. How many hosts per subnet? 3. What is the new subnet mask? 4. What is the block size? 5. Create the subnet table. 6. What are the useable subnet address ranges? Example Create 2 subnets from the following Class C IP address: IP Address: 192.168.1.0 Subnet mask: 255.255.255.0 How many subnets? Steal enough bits from the fourth octet to acquire the desired number of subnet. 2n where n equals 2 - 11000000 22 = 4 subnets (Note useable subnets are 22 - 2 = 2 usable subnets), so stealing 2 bits is enough. How many hosts per subnet? 2y – 2 = number of hosts per subnet. Where y is the number of unmasked bits, or the zeros bits in the fourth octet. - 11000000 26 -2 = 62 hosts per subnet What is the new subnet mask? 11111111.11111111.11111111.11000000 - 255.255.255.192 What is the block size? 256 – subnet mask = block size 256 - 192 = 64 What are the blocks? Start at zero and add the block size to the number to get the next number. Stop when you reach the block size. 0, 64, 128, 192 Create the subnet table. Subnet 0 64 128 192 First Host 1 65 129 193 Last Host 62 126 190 194 Broadcast 63 127 191 255 What are the useable subnet address...
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...Subnetting 1 2 3 By Adam Chee W.S Ever get stressed out because you know that there would be subnetting question(s) in the next exam you are taking and that these questions easily take up 10 to 20 minutes of your precious exam time? What if there is more than one question? The process of converting the subnet to binary and decimal can drive the unfamiliar insane, not to mention the waste of precious time and brain power which can be utilized for other areas of exam preparation. Let's take a look at a shortcut method that will cut down the time needed to answer these questions without the need for a calculator. Subnet Basics: This article assumes that you know how to perform subnetting in the traditional method but it is important to stress that there are only 3 classes of usable IP addresses which are Class Range Subnet mask Host bit Subnet Class A (127 is reserved for loopback) 255.0.0.0 24 8 Class B 128 - 191 255.255.0.0 16 16 Class C 192 - 223 255.255.255.0 8 24 1 - 126 You must understand and remember this table well in order to master the shortcut. Note: You must borrow at least 2 bits and must leave at least 2 bits The 'Subnet Table' 1 2 3 4 5 6 7 8 Bit Value 128 64 32 16 8 4 2 1 Subnet Mask 128 192 224 240 248 252 254 255 0 2 6 14 30 62 126 254 Bits Borrowed (N) Number of Subnets ((2^N)-2) ...
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...Unit 7 Assignment Given an IP Address of 172.16.10.1 use the guide to get 1000 hosts on each of your 50 networks: 1. What Class is this IP address? _B_ 2. How many bits would you use for networking? _6_ 3. What is the CIDR or Slash value: _/22__ 4. What subnet mask would you generate? __255.255.252.0_____ 5. What is the first subnetwork range created? ________172.16.8.0_____________________ 6. What is the last subnetwork range created? ______172.16.255.0_________________________ Given an IP Address of 172.16.14.1 use the guide to get 500 hosts on each of your 100 networks: 7. What Class is this IP address? _B__ 8. How many bits would you use for networking? _7_ 9. What is the CIDR or Slash value: __/23___ 10. What Subnet Mask would you generate? ___255.255.254.0_____ 11. What is the first Subnetwork range created? ______172.16.14.0____________________ 12. What is the last Subnetwork range created? _____172.16.254.0______________________ Given an IP Address of 172.16.15.1 use the guide to get 200 hosts on each of your 200 networks: 13. What Class is this IP address? _B_ 14. How many bits would you use for networking? __8__ 15. What is the CIDR or Slash value: __/24___ 16. What subnet mask would you generate? ___255.255.255.0____ 17. What is the first subnetwork range created? ____172.16.15.0___________________ 18. What is the last subnetwork range created?...
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...NTC 248 Week 5 Individual Learning Team Assignment Click Link Below To Buy: http://hwaid.com/shop/module-2-case-supply-and-demand-assignment/ Create a 1-page network diagram showing the components and devices needed to build a simple office network based on the following scenario: • 50 employees, half Sales office and half Business office users • 2 subnets - Separate LANs within the office 30 hosts per subnet (25 employees, 1 printer and 2 servers, and 2 more available addresses. • 2 network printers, one per subnet • 1 connection to the WAN/Internet • One core router with 2 Ethernet interfaces (one for each existing subnet) and one Serial for WAN • Each subnet using one switch for all hosts • Plan for growth for up to 6 subnets - which do not need to be in the diagram Identify the different collision and broadcast domains on your network diagram. In your diagram, draw a green circle around each collision domain and a red circle around each broadcast domain. Develop a subnet scheme using this address: You have been assigned an IP address and subnet mask: 192.168.111.0 /24. Plan to support two subnets, and allow for growth to as many as 6 subnets in the future. Subnet 1 (not subnet zero) will be used to provide a range of IP addresses for the Sales office users connected to a switch. The switch will connect to an Ethernet interface on the core router. Subnet 2 will be used to provide a range of IP addresses connected to the other Ethernet interface...
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...computer can be accessed by people on the network Public folder sharing – When it’s on, people on the network including homegroup members, can access files in the public folder Media streaming – When on, people and devices can access pictures, music, and videos on this computer File sharing connections – Windows 7 uses 128-bit encryption to help protect file sharing connections. Some devices don’t support 128-bit encryption and must use 40- or 56-bit encryption. Lab03_worksheet Table 3-2 TCP/IP setting – Enabled IPv4 address – 10.26.108.143 Subnet mask – 255.255.252.0 Default gateway – 10.26.110.1 DNS servers – 10.26.110.11, 192.168.110.13, 192.168.110.10 Exercise 3.5 IP Address | Subnet mask | Network address | Host address | CIDR | 192.168.1.1 | 255.255.255.0 | 192.168.1.0 | .1 | /24 | 10.1.1.1 | 255.0.0.0 | 10.1.1.1 | .1.1.1 | /8 | 172.16.1.1 | 255.255.0.0 | 172.16.1.1 | .1.1 | /16 | 192.168.1.35 | 255.255.255.224 | 192.168.1.35 | .3 | /24 | 10.10.1.5 | 255.255.0.0 | 10.10.1.5 | .1.5 | /8 | 172.18.5.6 | 255.255.255.0 | 172.18.5.6 | .6 | /16 | 11.25.5.1 | 255.255.192.0 | 11.25.5.1 | .1.1 | /8 | 74.12.3.1 | 255.0.0.0 | 74.0.0.0 | .12.3.1 | /8 | 195.167.5.4 | 255.255.224.0 | 195.167.0.0 | .5.4 | /24 | 172.19.5.4 | 255.0.0.0 | 172.0.0.0 | 19.5.4 | /16 | 192.168.7.2 | 255.255.0.0 | 192.168.0.0 | .7.2 | /24 | 195.167.3.63 | 255.224.0.0 | 195.160. | .7.3.63 | /24 | 172.18.32.1 | 255.255.240.0 | 172.18.32 | .1 | /16 | 15.4.2.1 | 255.255.255.0...
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...Chapter 5: Mini Case 2 - Central University Central University Suppose you are the network manager for Central University, a medium-size university with 13,000 students. The university has 10 separate colleges (e.g., business, arts, journalism), 3 of which are relatively large (300 faculty and staff members, 2,000 students, and 3 buildings) and 7 of which are relatively small (200 faculty and staff, 1,000 students, and 1 building). In addition, there are another 2,000 staff members who work in various administration departments (e.g., library, maintenance, and finance) spread over another 10 buildings. There are 4 residence halls that house a total of 2,000 students. Suppose the university has the 128.100.xxx.xxx address range on the Internet. How would you assign the IP addresses to the various subnets? How would you control the process by which IP addresses are assigned to individual computers? You will have to make some assumptions to answer both questions, so be sure to state your assumptions. As a network manager, I would assign the IP addresses to various subnets by associating a subnet base address with each physical network. Then I would sequentially assign hosts particular IP addresses within the subnet. Determining host addresses is really quite simple, once we know the subnet address. You should be able to substitute the numbers for the host ID bits in the subnet address. Then convert the address to decimal form. By controlling which IP addresses assigned to...
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...#2 Subnet 18.54.0.0/16 into 10 subnets If you take 255-10 (255 being a full octave and ten being how many you need) you are left with 245. 245 in binary is displayed as 11110101. Since you can’t use this as a subnet mask you have to use the first 0 as your cut off. This would be 11110000 or 240. This will provide you with a range of 16 hosts or 0-15. After subnetting the new subnet mask would be 255.255.255.240 or 11111111.11111111.11111111.11110000 this would allow for the 10 subnets (16 to be exact) this can be confirmed with the Octal set up for Binary: 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Since the first address is the subnet’s network address then the first address network would be: 18.54.0.0 And since the last address is the subnet’s broadcast address then the first broadcast address would be: 18.54.0.15 The range of available IP addresses would be: 18.54.0.1 through 18.54.0.14 With this information I can now set up all the ranges for 16 separate subnets with the next being: 18.54.0.16 Network, 18.54.0.31 Broadcast, 18.54.0.17 through 18.54.0.30 Available IP address 18.54.0.32 Network, 18.54.0.47 Broadcast, 18.54.0.33 through 18.54.0.46 Available IP address 18.54.0.48 Network, 18.54.0.63 Broadcast, 18.54.0.49 through 18.54.0.62 Available IP...
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...What to do if a customer needs a static IP loaded onto gateway 1. Check in billing to see if the rate code is there - In ACSR “CCO Static IP 1” (or 5, 13) will be used as the code - In DST select their IP Gateway by going to “Change box data” (11), note the rate codes (UH 1, @5 1, e.g.) then hit F5 to open up the window where you can check rates, hit 5 and enter and scroll through to verify there is a static IP on the account; if none found—transfer to MACD to get Static IP on the account; if found… proceed to Step 2 2. Look up the Business Name in IPControl - login to IPControl and click the Management Tab at the top [pic] - then go to the right where it says “Search” and type the business name … the field will start responding with names of businesses matching your results… make sure to pay attention to where the business is located to verify you have the correct one. If you cannot find the business name in here, transfer to T2 because they probably need to have the static IP created - when you’re clicking on the business, click “Details” to verify the correct business address against the billing address to make sure you have the correct business (sometimes businesses have numerous IPs at different locations—it’s a good idea to check to verify you’re using the correct one) [pic] - click the IP address that is bold and underlined (in this case 74.94.247.84) …...
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...02/8/2015 Pick a network address from the list below. Everybody must pick a unique network address and number of subnet. Please reply to the Subnetting Discussion Topic under Week 3 Discussions with your selection. Don’t select a network address that somebody else already selected. For each of them, I need the following information: * The new subnet mask after the subnetting (10%) * The following information for the four subnets identified: * Subnet’s network address (3%) * Subnet’s broadcast address (3%) * Subnet’s range of available IP addresses (4%) * The calculations on how you get to the answers (50%). This is very important. If you don't provide the calculations or the way you get the answer, you will lose 50%. Your goal is to subnet them with as little subnet as possible but still meeting the requirement. In other word, maximize the number of hosts that is available for each subnet. 26. 134.84.0.0/16 subnetted to 97 subnets and provide information for subnets #1, #8, #65, and #97 134.84.0.0/16 10000110 01010100 00000000 00000000 Network address 11111111 11111111 00000000 00000000 Subnet Mask The formula to find out how many bits to borrow 2 to the power off 7 = 128 need to borrow 7 bits. 7 bits = 0000001 New 10000110 01010100 00000000 00000000 Network Address 134.84.0.0/16 11111111 11111111 11111110 00000000 Subnet Mask 255.255.254.0 Subnet Mask #1 Network Address 10000110...
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...Unit 6 Subnetting Lab 6 Answer the following questions listed. Using the Class B Subnetting Guide, answer the following: Given an IP address of 172.16.8.1 use the guide to get 30 hosts on each of your 2000 networks: 1. What Class is this IP address? B 2. How many bits would you borrow or take? 11 3. What subnet mask would you generate? 255.255.255.224 4. What is the first subnetwork range created? 172.16.0.1 to 172.16.0.30 5. What is the last subnetwork range created? 172.16.255.225 to 172.16.255.254 Given an IP address of 172.16.4.1 use the guide to get 62 hosts on each of your 1000 networks: 6. What Class is this IP address? B 7. How many bits would you borrow or take? 10 8. What subnet mask would you generate? 255.255.255.192 9. What is the first subnetwork range created? 172.16.0.1 to 172.16.0.62 10. What is the last subnetwork range created? 172.16.255.193 to 172.16.255.254 Given an IP address of 172.16.5.1 use the guide to get 100 hosts on each of your 500 networks: 11. What Class is this IP address? B 12. How many bits would you borrow or take? 9 13. What subnet mask would you generate? 255.255.255.128 14. What is the first subnetwork range created? 172.16.0.1 to 172.16.0.126 15. What is the last subnetwork range created? 172.16.255.129 to...
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...Device | Interface | IP Address | Subnet Mask | Default Gateway | Building1 | G0/0 | 172.31.103.1 | 255.255.255.240 | N/A | | G0/1 | 172.31.103.33 | | N/A | | S0/0/0 | 172.31.103.97 | | N/A | Building2 | G0/0 | 172.31.103.65 | 255.255.255.240 | N/A | | G0/1 | 172.31.103.81 | 255.255.255.240 | N/A | | S0/0/0 | 172.31.103.98 | 255.255.255.252 | N/A | ASW-1 | VLAN 1 | 172.31.103.2 | 225.255.255.224 | | ASW-2 | VLAN 1 | 172.31.103.34 | | | ASW-3 | VLAN 1 | 172.31.103.66 | 255.255.255.240 | | ASW-4 | VLAN 1 | 172.31.103.95 | | | Host-A | NIC | 172.31.103.30 | | 172.31.103.1 | Host-B | NIC | 172.31.103.62 | | 172.31.103.33 | Host-C | NIC | 172.31.103.78 | | 172.31.103.65 | Host-D | NIC | 172.31.103.94 | | 172.31.103.94 | 172.31.103.0/24 172.31.103.0 255.255.255.224 172.31.103.32 172.31.103.64 172.31.103.96 172.31.103.128 Subnet Number Subnet Address First Usable Host Address Last Usable Host Address Broadcast Address 0 192.168.100.0 192.168.100.1 192.168.100.30 192.168.100.31 1 192.168.100.32 192.168.100.33 192.168.100.62 192.168.100.63 2 192.168.100.64 192.168.100.65 192.168.100.94 192.168.100.95 3 192.168.100.96 192.168.100.97 192.168.100.126 192.168.100.127 4 192.168.100.128 192.168.100.129 192.168.100.158 192.168.100.159 5 6 7 8 9...
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...Associate Level Material Appendix E TCP/IP Network Planning Table Refer to appendix E1. Identify the problems with the TCP/IP network and complete the table. |Problem |Explanation of Proposed Solution | |Group A and D have incorrect subnet addresses. |All subnet addresses should be corrected to: 255.255.0.0 | | |This would be corrected by changing the subnet addresses from computer A to match computers B and C.| |Group H, I, J, and K have incorrect IP addresses. |D also needs to be changed to the correct subnet to match the other computers. | | | | |Group B has an incorrect gateway address. |All of the IP addresses for Groups: H, I, J, K should start with the same numbers: 135.137.0. 0. | | |Changing this will now make it so the gateway address has to be corrected. Computer 1, Router 2 IP | | ...
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...Chapter 2 Solutions Minimum frame size is the same as the maximum round-trip delay on the LAN. 1. Maximum round-trip delay = (2 x max. I-way propagation delay) + repeater delay = (500 x 5 x 2)/200) + 25 (seconds = 50 (seconds At 10 Mbps, generated bytes = 50 Mbps x 10 (sec = 500 bits ~ 64 bytes, which is the minimum frame size. 2. Maximum round-trip delay is the time it takes a bit to traverse between two farthest end stations. Add to this two traversals of repeater, 1 (s each way. Max. R-T delay = (400 /2 x 108) + 2 = 4( sec. Min. number of bits to detect collision = 4 x 10-6 x 109 = 4000 bits. Minimum bytes is the next higher 2n bytes is 512 bytes. 3. Choice 1: Switched Ethernet - Replace regular hub to switched hub. This will increase the maximum capacity to about 6 times. No modifications need to workstations. Easy to install. Switch the hub and plug the cables into the new hub. Choice 2: Full duplex - Convert NICs on the 12 workstations and replace the hub to full duplex operation. This requires hardware and configuration changes to the hub and workstations. Will double the capacity. However, this is a dead-end approach. Choice 3: Convert the network to 100Bast-T Fast Ethernet. Need to replace the NICs in the workstations and replace hub for 100BaseT. Increases capacity by ten times. The speed at each workstation increases ten times. Requires 12 NICs for the workstation and a new hub. Choice 4: Split the workstation...
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...Autoclaved Aerated Concrete Abstract This report investigates the Autoclaved Aerated Concrete (AAC). AAC, or known as Autoclave Cellular Concrete (ACC) is an innovative concrete material that used for all types of structures. In this paper, its history (development), manufacturing process, some typical attributes and advantages, properties and limitations would be interpreted; in order to give an idea why it is so popular to be building material nowadays. Introduction AAC is a precast building single material made with all-natural raw materials, and it is also a lightweight product to provide structure, insulation, and fire and termite resistance. AAC is available in many configurations include blocks, wall panels, floor slabs and roof panels, and lintels. It is a very popular construction material due to its features of sustainable, economical, efficient and relative environment- friendly. History In 1914, the Swedes discovered a mixture of cement, lime, water and sand that expands by adding aluminum powder. This kind of material was then developing, till mid-1920s, Dr. Axel Eriksson, an architect working with Professor Henrik Kreüger at the Royal Institute of Technology was invented AAC. It went into Swedish production in 1929 in a factory in Hällabrottet, and became quite popular. In the 1940s the trade mark Ytong was introduced, but often referred to as “blue concrete”. This version of Ytong was produced from alum slate, which due to its combustible carbon content was...
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