Task 1: Design the Logical LAN T LAN Topology Given an IP address and mask of 13

Question

Task 1: Design the Logical LAN T LAN Topology Given an IP address and mask of 133 12/2 Subnet Number of Hosts Points 15 Points) Subnet B Next Available 54 The o subnet is used No subnet calculators may be used Al work must be shown on the reverse of this Final Subnet A Points Points Student Specification Number of bits in the subnet umber of bits in the subnet IP mask (binary) New IP mask (decimal) Maximum number of usable subnets (including the O subnet) Number of usable hosts per subnet IP Subnet First IP Host address Last IP Host address Subnet B StudentPoints Specification Input (8 Points) Number of bits in the subnet IP mask (binary New IP mask (decimal) Maximum number of usable subnets (including the Oth subnet) Number of usable hosts per subnet IP Subnet First IP Host address Last IP Host address … Host computers will use the first IP address in the subnet The network router will use the LAST network host address Page 2 of 5 e Copyright© 1992-2007 Cisco Systems, Inc. All rights reserved. This document is Cisco Pubic Information

Explanation / Answer

192.168.12.0/24 which is 11000000.10101000.00001100.00000000

Now 2 subnets need to be created of 37hosts and 64 hosts respectively.

A subnet can only contain number of hosts which are powers of 2, so a subnet cannot have exactly 37 hosts. Consider the next power of 2 for this subnet, which is 64.

Now the problem reduces to create 2 subnets of 64 hosts each.

From the above CIDR notation(192.168.12.0/24), it can be seen that the entire last octet(last 8 bits) is available for use, this is because the "n" in the CIDR notation here is 24, so the rest bits (32-24=8) are available for use.

To create 2 subnets of 64 adresses, we need only 1 bit to be part of subnet id

Now, 64=01000000, the left most bit can be used as subnet ID.

So, the subnet mask is: 255.255.255.128 or 11111111.11111111.11111111.10000000 or /25 in CIDR notation

Starting address of:

first subnet: 192.68.12.0

second subnet: 192.68.12.128

Two subnets in CIDR notation: 192.68.12.0/25 and 192.68.12.128/25

The first and lsat ip address in any network are used as network id and limited broadcast address respectively. So, number of usable hosts is always (total possible hosts-2)

So,Number of usable hosts per subnet=(64-2)=62

For subnet 1:

First usable host address: 192.68.12.1 (first ip, i.e 192.68.12.0, used as network id)

Last usable host address: 192.68.12.126 (last ip, i.e 192.68.12.127, used as limited broadcast address)

For subnet 2:

First usable host address: 192.68.12.129 (first ip, i.e 192.68.12.128, used as network id)

Last usable host address: 192.68.12.254 (last ip, i.e 192.68.12.255, used as limited broadcast address)

I hope I have answered all the questions.

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