WS 14 (Subnet 2)

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Pennsylvania State University *

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220

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Industrial Engineering

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Dec 6, 2023

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IST 220: Networking and Telecommunications Work Sheet 14 (Subnet 2 and CIDR) Instructions: - Use your class notes and text book to complete this work sheet. - You can discuss questions with other students and instructor. - Submit your answers (Word Document) to Canvas using the drop box before the deadline Student Name: Open Ended: 1. If an organization wants to divide a network into 5 subnets with the subnet mask of 24 bits, how many bits should be reserved for host ID ? 24 (total bits) - 3 (subnet bits) = 21 bits for host IDs. 2. Refer to previous question, how many maximum numbers of subnets an organization can create. 2^ (3-1) = 4 3. Refer to previous question (Question # 1), what should be the subnet mask for each new subnet? 24 + 3 = 27 bits 4. What is the purpose of subnet mask? The subnet mask is essential in IP addressing because it facilitates network organization, efficient use of IP address space, and data routing between multiple subnets. 5. What is the difference between subnet mask and network prefix? Page 1 of 3
Mask of the subnet A subnet mask is a 32-bit dotted-decimal value (for example, 255.255.255.0). It is most encountered in the context of IPv4 addressing. The subnet mask is used to determine the network and host components of an IP address . Network Prefix: A network prefix is frequently coupled with Classless Inter-Domain Routing (CIDR) notation in IPv4 or used directly in IPv6 addressing. In CIDR notation, the network prefix is stated as an IP address followed by a forward slash and the prefix length (e.g., 192.168.1.0/24). Problem Solving – Show all steps (do not just write the answer) 6. An organization would like to divide its network so that there are five separate subnets for each department. They will need 23 hosts in each subnet. The organization main IP address is 192.168.162. 0/24. Complete each of the following: NOTE: If you create more than five subnets, list the extra ones too. 24 (original bits) + 5 = 29 bits. 2^(32 - subnet mask) = 2^3 = 8 Subnet Network address Host addresses Broadcast address Subnet mask: 255.255.255. ( or 24 in CIDR ) First subnet 192.168.162. 192.168.162. - 192.168.162. 192.168.162. Second subnet 192.168.162. 192.168.162. - 192.168.162. 192.168.162. Third subnet 192.168.162. 192.168.162. - 192.168.162. 192.168.162. Fourth subnet 192.168.162. 192.168.162. - 192.168.162. 192.168.162. Fifth subnet 192.168.162. 192.168.162. - 192.168.162. 192.168.162. Sixth subnet ? 192.168.162.40/29 192.168.162.41 - 192.168.162.46 : 192.168.162.47 ? ? ? 7. An organization would like to divide its network (219.7.9.0/24) so that there are 32 separate subnets. They will need 6 hosts in each subnet. Complete the following table: NOTE: Because there are so many subnets, don't write them all out (unless you just want to). If you can do the first ten and know what the last one is, you get the idea. Subnet Network address Host addresses Broadcast address Subnet mask: 255.255.255 ( or 24 in CIDR ) Page 2 of 3
First subnet 219.7.9./28___ 219.7.9. 1___- 219.7.9. _14__ 219.7.9. 15___ Second subnet 219.7.9. /28 219.7.9. _17__ - 219.7.9.30___ 219.7.9. 31___ Third subnet 219.7.9. /28 ___ ___ 219.7.9. 33__ - 219.7.9.46 ___ 219.7.9.47 ___ Fourth subnet 219.7.9. /28___ 219.7.9.49 ___ - 219.7.9.62 ___ 219.7.9.63 ___ Fifth subnet 219.7.9. /28___ 219.7.9.65 ___ - 219.7.9.78 ___ 219.7.9.79 ___ Sixth subnet 219.7.9./28 219.7.9. 81 - 219.7.9.94 219.7.9.95 Seventh subnet 219.7.9./28 219.7.9.97 - 219.7.9.110 219.7.9. 111 Eighth subnet 219.7.9./28 219.7.9.113 - 219.7.9. 126 219.7.9. 127 Ninth subnet 219.7.9./28 219.7.9. 129- 219.7.9. 142 219.7.9. 143 Tenth subnet 219.7.9./28 219.7.9. 145- 219.7.9.158 219.7.9. 159 . . Thirty-second subnet 219.7.9./28 219.7.9. 481- 219.7.9. 494 219.7.9. 495 Page 3 of 3
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