CCNA question dont understand

Mohalg

I have 3 routers and teh questions was Which addressing scheme would
satisfy the needs of this network yet waste the fewest addresses?

network 1 R1 --50 hosts
network 2 R2---20 hosts
network 3 R3----10 hosts

i dont understand the answer though .ie why N2 is 192.168.10.64/ and why Serial link start with 192.168.10.112/

answer .
C. Network 1: 192.168.10.0/26Network 2: 192.168.10.64/27Network 3:
192.168.10.96/28Serial link 1: 192.168.10.112/30Serial link 2: 192.168.10.116/30

vinbuck

Not sure about the bottom half of your question as it's a bit unclear, but here would be the best subnets for those networks

network 1 R1 --50 hosts /26 - 62 Hosts is the closest subnet
network 2 R2---20 hosts /27 - 30 Hosts is the cosest subnet
network 3 R3----10 hosts /28 - 14 Hosts is the closest subnet

Subnetting is like the Price is Right in reverse......you have to get the lowest number OVER (or equal to) what you need for hosts.

Mohalg

i do understand how to get 62 Hosts i,30 and 14. now why Network 2: 192.168.10.64/ why i snot Network 2: 192.168.10.0/ and th eserial link why strat by 112 Serial link 1: 192.168.10.112

halaakajan

It is VLSM. Because as stated by vinbuck


network 1 R1 --50 hosts /26 - 62 Hosts is the closest subnet
network 2 R2---20 hosts /27 - 30 Hosts is the cosest subnet
network 3 R3----10 hosts /28 - 14 Hosts is the closest subnet


192.168.10.0 - 63 would be network 1 w/ subnet mask of /26
192.168.10.64-95 would be network 2 w/ subnet mask of /27
192.168.10.96-111 would be network 3 w/ subnet mask of /28

The serial link is connecting two networks and it is a Point to Point Connection. We need only 2 IPs for that so it is /30

Legacy User

well its a little difficult to decipher what you have typed Mohalg. but network 2 is not 192.168.10.0 because the question is asking what is the best ip addressing scheme without WASTING ip addresses and if you would use 192.168.10.0 you will have 254 host addresses for only 20 users. That is a huge waste of ip addresses.

It is dealing with vlsm I'm pretty sure it was a question on the test and each network would basically start off right where the previous network range left off.

Legacy User

just chart it out for class C 256|128|64|32|16|8|4 --subnets available in class c. easy way to figure out hosts is subtract 2 from the subnet number for broadcast and net id
128 |64 |32|16|8 |4|2 ---last octet in /24

and if you understand how to subnet you should have the same scheme as halaakajan typed:

192.168.10.0 - 63 would be network 1 w/ subnet mask of /26
192.168.10.64-95 would be network 2 w/ subnet mask of /27
192.168.10.96-111 would be network 3 w/ subnet mask of /28

basically once you used 192.168.10.0 - 63 the 0 -63 range for 50 hosts it is no longer available for use so you move on to the next number which is 64 where 192.168.10.64-95 would be network 2 w/ subnet mask of /27 and figure out the next range you need for the specific amount of hosts and continue down the line

Mohalg

We need only 2 IPs for that so it is /30 so 192.168.10. 100-103
192.168.10.104-107
192.168.10. 108-111
192.168.10. 112-115
192.168.10. 116-119
so why he choose 112 and 116 instead of 104 and 108 for serial link /30

Legacy User

well following the original scheme 104-107 and 108-111 fall within the 192.168.10.96-111 ip range so those ip address are already used.

192.168.10.0 - 63 would be network 1 w/ subnet mask of /26
192.168.10.64-95 would be network 2 w/ subnet mask of /27
192.168.10.96-111 would be network 3 w/ subnet mask of /28

Mohalg

Many thanks now i got it in my head . I do appreciated your help /

Legacy User

anytime