Use an IP calculator, such as the one available at http:lljodies.delipcalc to co

Question

Use an IP calculator, such as the one available at http:lljodies.delipcalc to complete the following exercises. Note: There are many online IP calculators. Some good, some bad. YMMV The one at the link above has been around a long time and has the functionality needed to complete the problems included below. If you use a different one, you are on your own- make sure you understand how it works. Give all IP addresses in dotted decimal. 1. Given IP address 137.156.48.0 and netmask 255.255.0.0 a) Write the network address using CIDR notation. b) What is the maximum number of hosts that could be in this network? 2. Given IP address 94.31.208.14 and netmask 255.255.255.192 a) Write the network address using CIDR notation. b) What is the maximum number of hosts that could be in this network? Given a host with IP addresses 140.50.6.210/21 a) Is a host with IP address 148.58.8.18/21 part of the same network? 3. b) Is a host with IP address 140.50.8.47/21 part of the same network? 4. Divide the 199.15.64.0/19 address block into four subnets a) List the network addresses for each of the four subnets. b) How many hosts can be in each subnet? c) If you add up the total number of hosts for all the subnets it is less than the total hosts possible in the original network. Why is this the case? d) List the range of host IP addresses for the first subnet (lowest IP addresses). 5. Consider the address block 284.85.32.8/23. The goal is to divide the available address range into the maximum number of subnets while having enough addresses in each subnet to accommodate the indicated number of hosts For each of the following 0 Indicate the netmask to accomplish this and (ii) indicate how many subnets are possible a) At least 25 hosts per subnet b) At least 50 hosts per subnet c) At least 100 hosts per subnet

Explanation / Answer

4. To divide the 199.15.64.0/19 address block into 4 subnets, it require two additional bits(2^2) from Host bits.

Therefore in total 21 bits are required for network mask, to identify these four subnets. Thus number of bits available for hosts are 32-21=11.

With 11 bits , 2^11 hosts are available in each subnet.

a. Network addresses of each subnet is:

First Subnet: 199.15.64.0/21 --- 11000111.00001111.010 00 000.00000000

Second Subnet: 199.15.72.0/21 --- 11000111.00001111.010 01 000.00000000

Third Subnet : 199.15.80.0.21 --- 11000111.00001111.010 10 000.00000000

Fourth Subnet: 199.15.99.0.21 --- 11000111.00001111.010 11 000.00000000

b. Number of hosts in each subnet is :2048-2 = 2046

With 11 bits for host , 2^11 hosts are possible in each subnet. But each subnet should have one reserved for broadcast addresses (with host part all ones(1's)and one for network addresses (with host part all zeros(0's). Thus total available host = 2^11-2 = 2046.

c.With 4 subnets each with 2046 hosts, in total there are 4*2046 = 8184 hosts available.

Whereas the total number of hosts possible in the original network without subnetting is:

with /19 as network mask, 32-19=13 bits available for host. Therefore 2^13 = 8192 are the total possible number of hosts. With one for network address and one for broadcast address reserved for the network, total available number of hosts = 8192-2=8190.

Thus it is greater than 8184. With subnetting there are in total 4 broadcast addresses and 4 network addresses one for each subnet reserved. Thus 8 addresses are reserved when it is subnetted. Whereas only 2 addresses are reserved for original network. Hence total number of available host are greater in original network than the total number of available host in subnetted network.

d. Range of host IP address for first subnet:

First host address : 199.15.64.1 ---- 11000111.00001111.010 00 000.00000001

Last host address : 199.15.71.254 ---- 11000111.00001111.010 00 111.11111110

5. For address 204.85.32.0/23

a. For atleast 25 hosts per subnet

It requires 5 bits for host part(2^5 = 32). As 2^4 = 16 is less than 25, the next number 5 is taken and checked to satisfy the condition of atleast 25(25 < 32), therefore 5 bits are taken for host part.

With /23 as mask, the remaining 9 (32-23) bits are divided into subnet part and host part. Host part requires 5 bits.

Therefore subnet part will have 9-5=4 bits.

23 : 4 : 5

N/w : subnet : host

i. Netmask to accomplish is should be original network mask + subnet mask (23+4) = 27 (/27- network mask) bits all set to ones and remaining 5 bits set to zeros.

Netmask: 255.255.255.244 ---- 11111111.11111111.11111111.11100000 (/27)

ii. Number of subnets possible is 2^4 =16 subnets

b. For atleast 50 hosts per subnet

It requires 6 bits for host part(2^6 = 64). As 2^5 = 32 is less than 50, the next number 6 is taken and checked to satisfy the condition of atleast 50(50 < 64), therefore 6 bits are taken for host part.

With /23 as mask, the remaining 9 (32-23) bits are divided into subnet part and host part. Host part requires 6 bits.

Therefore subnet part will have 9-6=3 bits.

23 : 3 : 6

N/w : subnet : host

i. Netmask to accomplish is should be original network mask + subnet mask (23+3) = 26 (/26- network mask) bits all set to ones and remaining 6 bits set to zeros.

Netmask: 255.255.255.192 ---- 11111111.11111111.11111111.11000000 (/26)

ii. Number of subnets possible is 2^3 =8 subnets

c. For atleast 100 hosts per subnet

It requires 7 bits for host part(2^7 = 128). As 2^6 = 64 is less than 100, the next number 7 is taken and checked to satisfy the condition of atleast 100(100 < 128), therefore 7 bits are taken for host part.

With /23 as mask, the remaining 9 (32-23) bits are divided into subnet part and host part. Host part requires 7 bits.

Therefore subnet part will have 9-7=2 bits.

23 : 2 : 7

N/w : subnet : host

i. Netmask to accomplish is should be original network mask + subnet mask (23+2) = 25 (/25- network mask) bits all set to ones and remaining 7 bits set to zeros.

Netmask: 255.255.255.128 ---- 11111111.11111111.11111111.10000000 (/25)

ii. Number of subnets possible is 2^2 =4 subnets

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