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New Member

Need solution and explanation

Good Day All Cisco community,

I am new to the CCNA certificate, and I just finished studying the course, now I am in practising phase, I am trying to follow and practise on all Cisco laps which comes along with packet tracer, on lap 6.8.1. I am facing a problem of understanding the question, I cannot figure out which IP should should be assigned to each device, although I have a good command of subnetting but I didn't figure out the IPs.

Can you please help me?

1 ACCEPTED SOLUTION

Accepted Solutions
Bronze

Hi,1st subnet: 192.168.23.0

Hi,
1st subnet: 192.168.23.0/26 (192.168.23.0 - 192.168.23.63, 255.255.255.192)
2nd subnet: 192.168.23.64/27 (192.168.23.64 - 192.168.23.95, 255.255.255.224)
3rd subnet: 192.168.23.96/28 (192.168.23.96 - 192.168.23.111, 255.255.255.240)
4th subnet: 192.168.23.112/29 (192.168.23.112 - 192.168.23.119, 255.255.255.248)
5th subnet: 192.168.23.120/30 (192.168.23.120 - 192.168.23.123, 255.255.255.252)
6th subnet: 192.168.23.124/30 (192.168.23.124 - 192.168.23.127, 255.255.255.252)
7th subnet: 192.168.23.128/30 (192.168.23.128 - 192.168.23.131, 255.255.255.252)
For example (2nd subnet): 192.168.23.64 is network address (unusable) and 192.168.23.95 is broadcast (unusable). All other addresses in range are valid for hosts.


Regarding point-to-point links (5th - 7th). For CCNA level they want above values.
But on CCNP level you can learn to use /31 masks to save IP addresses or even ip unnumbered. But this is only for information.

Server - 192.168.23.109
R1-ISP Fa0/0 - 192.168.23.110
R1-ISP S0/0/0 - 192.168.23.122
R2-Central S0/0/0 - 192.168.23.121
R2-Central Fa0/0 - 192.168.23.62
Hosts 1A, 1B - 192.168.23.1, 192.168.23.2

2 REPLIES
Bronze

Hi,1st subnet: 192.168.23.0

Hi,
1st subnet: 192.168.23.0/26 (192.168.23.0 - 192.168.23.63, 255.255.255.192)
2nd subnet: 192.168.23.64/27 (192.168.23.64 - 192.168.23.95, 255.255.255.224)
3rd subnet: 192.168.23.96/28 (192.168.23.96 - 192.168.23.111, 255.255.255.240)
4th subnet: 192.168.23.112/29 (192.168.23.112 - 192.168.23.119, 255.255.255.248)
5th subnet: 192.168.23.120/30 (192.168.23.120 - 192.168.23.123, 255.255.255.252)
6th subnet: 192.168.23.124/30 (192.168.23.124 - 192.168.23.127, 255.255.255.252)
7th subnet: 192.168.23.128/30 (192.168.23.128 - 192.168.23.131, 255.255.255.252)
For example (2nd subnet): 192.168.23.64 is network address (unusable) and 192.168.23.95 is broadcast (unusable). All other addresses in range are valid for hosts.


Regarding point-to-point links (5th - 7th). For CCNA level they want above values.
But on CCNP level you can learn to use /31 masks to save IP addresses or even ip unnumbered. But this is only for information.

Server - 192.168.23.109
R1-ISP Fa0/0 - 192.168.23.110
R1-ISP S0/0/0 - 192.168.23.122
R2-Central S0/0/0 - 192.168.23.121
R2-Central Fa0/0 - 192.168.23.62
Hosts 1A, 1B - 192.168.23.1, 192.168.23.2

New Member

Hi Milos, 

Hi Milos

Yes I got your answer and now I understand how these IPs should be calculated, Thanks so much for your kind help. I will try to solve another example for confirmation.

Thanks again & appreciate :)

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