Classless Inter-Domain Routing....can someone please help clarify

leeland

Diamond Member
Dec 12, 2000
3,659
0
76
thanks for stopping in,
i am going over Classless Inter-Domain Routing in my networking class and one of the questions we have been going over is the following

Using Classless Inter-Domain Routing, your organization has been assigned a /15 network adress of 195.24.0.0 How many /24 subnetworks can your organization have ?


so do you take the /15 and subtract it from 32 bits ? so 32 - 15 = 17

the part i am lost on is the / 24

our notes suck and the teacher is not very helpful...so if anyone could put this process in noob terms i would greatly appreciate it !!!!


thanks a million

leeland



ps here are the basics of my homework that i have so far




Write your answers in the space provided. Each question is worth 2 points.

What Class network is 145.32.59.24?

Class B

How do you know this?

The first octet of this address (145) is the network portion and falls within the (Class B range 128 -191) range. To double check, convert this portion of the address to binary the first octet is 10 which falls in the Class B network.


Your organization has been assigned a Class B network with the address 132.45.0.0. What is the subnet mask needed if the organization needs to have 10 subnets?

2^4 = 16

255.255.240.0


Your organization has been assigned a Class C network with the address 200.35.1.0. What is the subnet mask needed if the organization wants a maximum of 20 hosts on each subnet?

25.255.255.248

Using Classless Interdomain Routing (CIDR) your organization has been assigned a /15 network address of 195.24.0.0. How many /24 sub networks can your organization have?

This is the one i have a problem with

Using CIDR, your organization wants to have 7 sub networks with a maximum of 15 hosts on each network. What slash (/) network number should be assigned to your organization such that the use of address space is efficiently allocated?

/25
 

cmetz

Platinum Member
Nov 13, 2001
2,296
0
0
2^(24-15) = 2^9 = 512.

Another way to look at it that might be more obvious:

(2^(32-15)) / (2^(32-24)) = (2^17) / (2^8) = 2^(17-8) = 2^9 = 512.
 

jjyiz28

Platinum Member
Jan 11, 2003
2,901
0
0
Your organization has been assigned a Class C network with the address 200.35.1.0. What is the subnet mask needed if the organization wants a maximum of 20 hosts on each subnet?

25.255.255.248

i assume you meant "255.255.255.248"

but shouldn't it be 255.255.255.224???

if its .248, you only have 2^3 = 8 hosts per subnet.