Your organization has been assigned the 172.16.0.0 network. The organization was
ID: 3793316 • Letter: Y
Question
Your organization has been assigned the 172.16.0.0 network. The organization was a requirement for 500 subnets, with at least 100 hosts per subnet. a. What subnet mask would you use? How many subnets will your subneting allow? How many usable hosts can you have on each subnet? Consider the network 192.168.1.0/27. a. How many additional subnet bits exist beyond the classful boundary? How many subnets can you get from the network? How many usable hosts addresses are there per subnet? Consider the subnetwork that node 192.168.1.172/26 belongs to. a. What is the Network ID for node 192.168.1.172? What is the broadcast address for the network that contain node 192.168.1.172? What is the last subnet ID on the network?Explanation / Answer
6)
a)172.16.0.0 belongs to class B addresses
subnet mask used is 255.255.0.0
b) CIDR (Classless Inter Domain Routing)
CIDR is a slash notation of subnet mask. CIDR tells us number of on bits in a network address.
Class A has default subnet mask 255.0.0.0. In slash notation it would be written as /8, means address has 8 bits on. Example 3.0.0.0/8
Class B has default subnet mask 255.255.0.0. In slash notation it would be written as /16, means address has 16 bits on. Example 172.16.0.0/16
Class C has default subnet mask 255.255.255.0. In slash notation it would be written as /24, means address has 24 bits on. Example 192.168.1.0/24
for above problem that IP belongs to Class B so it will 2^16 subnet masks because host id is 2^16
subnet masks(2^16)=hostID(2^16)
C) Total hosts are the hosts available per subnet. To calculate total hosts use formula 2^H = Total hosts. H is the number of host bits. For address in address 172.16.0.0/16 we have 32-16[Total bits in IP address - Bits consumed by network address] = 16. Total hosts per subnet would be (2^16)-2=65536-2=65534
7)192.168.1.0/27
a)By looking at address we can determined that this address is belong to class C and class C has default subnet
mask 255.255.255.0 [/24 in CIDR].In given address we borrowed 27 - 24 = 3 additional subnet bits exist beyond the classfull boundry
b)we 2^3=8 subnets from the network
c) Total IP address is 32 bits and subnet bits is 27
total usable hosts=32-27=5
(2^5)=32 hosts
8)
a) 192.168.1.172/26 is class C address
network id is 192.168.1 host is 172
b) address 192.168.1.172 11000000.10101000.00000001.10101100
subnet 192.168.1.192 11000000.10101000.00000001.11000000 (two subnet bits extended in host bits)
Borrowed host bits are 2 [26-24].
total subnets are 2^2 = 4. subnet mask would be 192.168.1.192
Block size would be 256-192 = 64.
So broad cast address is 192.168.1.255
c)last subnet id is 255.255.255.255
thus add 64 to 192
192+64=255
IP Class Default Subnet Network bits Host bits Total hosts Valid hosts A 255.0.0.0 8 24 2^24 (2^24)-2 B 255.255.0.0 16 16 2^16 (2^16)-2 C 255.255.255.0 24 8 2^8 (2^8)-2Related Questions
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